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Mechanical free energy devices => RomeroUK pulse motor Muller generator => Topic started by: tinman on December 09, 2015, 01:59:10 PM
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Today i made a some what confusing discovery that could have large ramifications.
Yes-another look at the humble pulse motor.
For all those that think you can measure some pulsed systems accurately-think again,as it aint just cut and dried.
In this thread,i will show you that using a CVR and scope will not always lead to accurate measurements. I mentioned some time back,that using resistors(CVR's) dose not always give accurate result's,and in some system's your scope could be leading you up the garden path.
The simple pulse motor can be a real bitch to measure P/in and P/out accurately-unlike some of the EE guys may tell you.
In my video series to come,i will show you (using an incandescent bulb) how our scope shows us one thing,and our incandescent bulb shows us the opposite.
Simply by placing a cap within the pulse circuit,we can throw accurate power measurements out the window-->even though our CVR remains on the input side. With pulsed systems,we can increase the current flowing through an incandescent bulb,raise the voltage across that incandescent bulb,but have the bulb dissipate less power-as in the way of heat and light. This is the opposite as to what should be seen across a resistor,as an incandescent bulb is just that-->a resistor. The only way to increase the heat and light output of an incandescent bulb is to increase the current flowing through it,and this in turn will show a higher voltage across it-as the resistance rises as the heat (and light) increases.
This is very important to know,and i will show you that it is wrong. This also means that what you may see on your scope across your CVR may not be actually what is really happening-->the incandescent bulb shows this to be true.
Or is there really some truth to the claim of radiant energy,or some form of cold electricity within the humble pulse motor?.
!! Dose ohm's law always apply with pulsed system's?.
I will post the first video here as soon as it is uploaded.
Brad.
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Sounds interesting. I am looking forward to watching your video.
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With pulsed systems,we can increase the current flowing through an incandescent bulb,raise the voltage across that incandescent bulb,but have the bulb dissipate less power-as in the way of heat and light. This is the opposite as to what should be seen across a resistor,as an incandescent bulb is just that-->a resistor. The only way to increase the heat and light output of an incandescent bulb is to increase the current flowing through it,and this in turn will show a higher voltage across it-as the resistance rises as the heat (and light) increases.
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Hi Brad,
This is surely going to be a good topic.
Why I put in bold above your statement is that I think an incandescent bulb is a nonlinear resistor, unfortunately and this fact can indeed lead a tinkerer up the garden path, especially when the duty cycle of the pulse motor current also changes. ;)
EDIT I added the voltage-current curve of an incandescent lamp, taken from this video at random:
https://www.youtube.com/watch?v=rNsykkSR3wg (https://www.youtube.com/watch?v=rNsykkSR3wg) Ohm's law is not valid of course.
Gyula
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.... even changing (+) and (-) tips of your multimeter may completely change the scene. (-) is connected with the shield and has higher capacitive load to ground as the (+) tip.
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Hi Brad,
This is surely going to be a good topic.
Why I put in bold above your statement is that I think an incandescent bulb is a nonlinear resistor, unfortunately and this fact can indeed lead a tinkerer up the garden path, especially when the duty cycle of the pulse motor current also changes. ;)
Gyula
Hi Gyula
Both voltage and current will rise together with an incandescent bulb,in that the more current flowing through it,the higher the voltage across it.. This makes them linear. The incandescent bulb is just a resistor,and ohms law should tell us how much power that resistor is disipating when we know the voltage across it,and the current flowing through it--which we do !! apparently !! :o
With an LED,voltage will hit the rated voltage,and then the current will continue to rise,while the voltage remains close to the same. This makes the LED a non linear device.
Brad
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.... even changing (+) and (-) tips of your multimeter may completely change the scene. (-) is connected with the shield and has higher capacitive load to ground as the (+) tip.
In the first video,the pulse motor will be running from a battery,and we will be using the scope to measure the voltage across a CVR,and the voltage across the incandescent bulb.
If the avreage voltage across the CVR rises,then that tells us the average current has risen. If you raise the avarage current flowing through an incandescent bulb,then you also raise the avaerage voltage across that incandescent bulb-which means you should be now dissipating more power from that bulb-->in the way of heat and light. The more power you disipate from an incandescent bulb,the brighter it shine's-the more light it puts out.
Brad.
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Hi Gyula
Both voltage and current will rise together with an incandescent bulb. This makes them linear.
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Just uploaded the current-voltage characteristic of an incandescent lamp in my previous post. You are right that both voltage and current will rise but NOT linearly with each other like with a normal resistor.
LEDs are also nonlinear loads, and I agree with what you wrote on them.
EDIT Yes, when we know the average current via a light bulb and we know the average voltage across it, we can use Ohm's law in that case. 8)
Gyula
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Just uploaded the current-voltage characteristic of an incandescent lamp in my previous post. You are right that both voltage and current will rise but NOT linearly with each other like with a normal resistor.
LEDs are also nonlinear loads, and I agree with what you wrote on them.
Gyula
Gyula
That is a lumens output curve per V/A,not a voltage/current curve.
Anyway,regardless of that,we know ohms law still applies,in that we can safely say that the dissipated power from the bulb should increase if the current flowing through it increases,and the voltage across it increases. This should result in more light output-should it not?,as this is how incandescent bulbs work.
Brad
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OK,here is the first video,and i am off to make the second.
https://www.youtube.com/watch?v=QfqvIjABXZ8
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Dear Brad,
In the meantime I added a sentence to my previous post and the use of Ohm's law is ok for the average current and voltage for an incandescent bulb. 8)
The VI curve I took from the video link was done from the current and voltage measurements, they changed the voltage across the lamp from zero to about 6 V and measured the current with an Ammeter in series with the lamp. Of course the brightness of the lamp changes as the voltage across the lamp changes.
Gyula
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Works out to be the same a joule thief, because of dc pulsing visually more lumens
when you connect the cap the pulse is filtered dc voltage and current is higher but
the lamp show less light because there is no pulsing..light illusion . BUT a circuit diagram
will get me out of the bush..
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AGAIN A LITTLE BIT OF THAT WHITE'S MAN MAGIC STUFF..
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Works out to be the same a joule thief, because of dc pulsing visually more lumens
when you connect the cap the pulse is filtered dc voltage and current is higher but
. BUT a circuit diagram
will get me out of the bush..
First ,we are using an incandescent bulb-not an LED.
the lamp show less light because there is no pulsing..light illusion
yes,your eyes can be fooled with LED's,but incandescent bulbs do not work that way.
What if we use a solar panel to measure light output--can a solar panel be fooled like your eyes can?
So we shall see.
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The resistance R of the bulb will change as it heats up. Raising V will cause I to increase, creating heat which will change R untill it reaches it's stable temperature
My 2p worth
Regards
Mike 8)
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In this thread,i will show you that using a CVR and scope will not always lead to accurate measurements. I mentioned some time back,that using resistors(CVR's) dose not always give accurate result's,and in some system's your scope could be leading you up the garden path.
Yes, an inductive CVR can lead you up the garden path.
Non-inductive CSRs (CVRs) do not lead up the garden path.
OK,here is the first video,and i am off to make the second.
https://www.youtube.com/watch?v=QfqvIjABXZ8 (https://www.youtube.com/watch?v=QfqvIjABXZ8)
In this video you seem to be using a wirewound CVR that is inductive. (http://falstad.com/circuit/circuitjs.html?cct=$+1+0.000005+11.558428452718767+50+5+50%0At+704+464+640+464+0+-1+0+-0.6290204441864482+1000%0At+704+464+768+464+0+-1+2.269347595663698+-0.6290204441863052+100%0Aw+704+464+704+512+0%0Aw+640+480+640+512+0%0Aw+640+512+704+512+0%0Ar+640+448+640+384+0+100%0Ar+768+448+768+384+0+100%0Aw+640+384+768+384+0%0Ar+768+480+768+576+0+150%0Aw+640+512+576+512+0%0Aw+768+480+832+480+0%0As+832+480+832+576+0+1+false%0Ar+768+576+832+576+0+10%0Aw+768+576+768+640+1%0Ag+768+640+768+656+0%0Ax+607+464+628+467+0+16+Q1%0Ax+780+465+801+468+0+16+Q2%0Ai+544+688+544+752+0+0.01%0Ad+400+384+400+512+1+0.805904783%0Aw+400+384+640+384+0%0A174+576+512+512+528+0+1000+0.1436+Resistance%0Ag+544+752+544+768+0%0Aw+544+576+544+528+0%0Ai+464+672+464+752+0+0.01%0A174+496+560+432+560+0+1000+0.8465+Resistance%0Aw+464+672+464+640+0%0Ag+464+752+464+768+0%0Aw+512+512+400+512+0%0Aw+400+512+400+560+0%0Aw+400+560+432+560+0%0Aw+496+560+640+560+0%0Aw+640+560+640+512+0%0Aw+464+672+432+672+0%0Aw+464+752+432+752+0%0Ar+432+752+432+672+0+1000000%0As+464+640+464+576+0+0+false%0Ar+576+752+576+688+0+1000000%0Aw+544+688+576+688+0%0Aw+544+752+576+752+0%0As+544+688+544+576+0+0+false%0AR+640+384+640+304+0+0+40+5+0+0+0.5%0Ao+13+64+0+33+7.62939453125e-55+0.011692013098647223+0+-1%0A)
Also, the light bulb has a coiled filament which makes it inductive - maybe that amount of inductance does not matter at these frequencies but it will at higher ones.
I recommend automotive dome light bulbs with straight filament, which are not inductive and are much better candidates for power measurement via optical coupling to the photovoltaic cell.
P.S.
Could you draw a schematic of your setup? It can be on a sheet of paper or a mouse-scrach in MS-Paint.
It is hard to follow even that amount of wires from the video ...especially at high zoom.
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First ,we are using an incandescent bulb-not an LED.
yes,your eyes can be fooled with LED's,but incandescent bulbs do not work that way.
What if we use a solar panel to measure light output--can a solar panel be fooled like your eyes can?
So we shall see.
Why not just get a Lumen meter - for less than $20 and it will be more accurate. It is said that a light bulb output needs to be nearly double for the human eye to notice a difference. I use a Lumen meter and have found them to be quite telling where I could not even perceive a difference.
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This will be a good topic.
Im making some changes to and electric bike hub motor. Its from a Tidalforce M-750. I have not been able to get much info on the motor architecture till I found this paper this week. Always searching for info on this as the original company dropped out of the motorized vehicle industry some years ago. So I figured if I keep looking, and there is a lot out there, I might find something. Was checking through a forum and a guy posted the pdf below. It is pretty much what I needed to work with.
Im removing the controller board and replacing it with an arduino setup. The pdf gives me a pretty good idea of what I will be aiming for.
The M-750 is limited to 750w and 20mph. Cant pedal faster than 20 with the throttle engaged. But the M-750X is 1kw and 30mph max. From what I have read it is all in the programming. So really, I can get this thing to pull wheelies and do over 30mph, probably for limited amounts of time. Just call above 30 turbo mode.
Im thinking I want to pulse this motor(pwm) but the pdf seems to describe a sine input to the coils. Just went over the pdf 1 time and need to read again. So What IM thinking is the sine is dc biased and sends smooth sine instead of brute pulsing. So I have to decide. With sine there will be no bemf to capture. Pulsed I can. But again, that will all be in the programming. ;)
Its a 7 phase motor. the coils are in pairs and in series, so both are operating at the same time. The motor uses optical sensing for timing and the transistors are very low ohm .008ohm.
Ill be making a topic on the build as I get everything together. But this topic here will come in handy if I go pulsed mode. ;D
Mags
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Here is the second video showing dissipated power by the incandescent bulb using a solar panel to measure output changes,in stead of our eyes.
https://www.youtube.com/watch?v=lElIrYuKGag
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Yes, an inductive CVR can lead you up the garden path.
Non-inductive CSRs (CVRs) do not lead up the garden path.
In this video you seem to be using a wirewound CVR that is inductive. (http://falstad.com/circuit/circuitjs.html?cct=$+1+0.000005+11.558428452718767+50+5+50%0At+704+464+640+464+0+-1+0+-0.6290204441864482+1000%0At+704+464+768+464+0+-1+2.269347595663698+-0.6290204441863052+100%0Aw+704+464+704+512+0%0Aw+640+480+640+512+0%0Aw+640+512+704+512+0%0Ar+640+448+640+384+0+100%0Ar+768+448+768+384+0+100%0Aw+640+384+768+384+0%0Ar+768+480+768+576+0+150%0Aw+640+512+576+512+0%0Aw+768+480+832+480+0%0As+832+480+832+576+0+1+false%0Ar+768+576+832+576+0+10%0Aw+768+576+768+640+1%0Ag+768+640+768+656+0%0Ax+607+464+628+467+0+16+Q1%0Ax+780+465+801+468+0+16+Q2%0Ai+544+688+544+752+0+0.01%0Ad+400+384+400+512+1+0.805904783%0Aw+400+384+640+384+0%0A174+576+512+512+528+0+1000+0.1436+Resistance%0Ag+544+752+544+768+0%0Aw+544+576+544+528+0%0Ai+464+672+464+752+0+0.01%0A174+496+560+432+560+0+1000+0.8465+Resistance%0Aw+464+672+464+640+0%0Ag+464+752+464+768+0%0Aw+512+512+400+512+0%0Aw+400+512+400+560+0%0Aw+400+560+432+560+0%0Aw+496+560+640+560+0%0Aw+640+560+640+512+0%0Aw+464+672+432+672+0%0Aw+464+752+432+752+0%0Ar+432+752+432+672+0+1000000%0As+464+640+464+576+0+0+false%0Ar+576+752+576+688+0+1000000%0Aw+544+688+576+688+0%0Aw+544+752+576+752+0%0As+544+688+544+576+0+0+false%0AR+640+384+640+304+0+0+40+5+0+0+0.5%0Ao+13+64+0+33+7.62939453125e-55+0.011692013098647223+0+-1%0A)
Also, the light bulb has a coiled filament which makes it inductive - maybe that amount of inductance does not matter at these frequencies but it will at higher ones.
I recommend automotive dome light bulbs with straight filament, which are not inductive and are much better candidates for power measurement via optical coupling to the photovoltaic cell.
As can be seen on the scope traces,there is no inductance shown. If inductance was a problem,then channel 1(across the globe) would show it in the form of a reversed voltage polarity-and we do not see that.
It is hard to follow even that amount of wires from the video ...especially at high zoom.
Sorry about what looks like high zoom,but that is as far as my camera zooms out.
P.S.
Could you draw a schematic of your setup? It can be on a sheet of paper or a mouse-scrach in MS-Paint.
Will do tonight when i get home from work.
Yes, an inductive CVR can lead you up the garden path.
Non-inductive CSRs (CVRs) do not lead up the garden path.
We shall see.
Brad.
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It is said that a light bulb output needs to be nearly double for the human eye to notice a difference. I use a Lumen meter and have found them to be quite telling where I could not even perceive a difference.
Why not just get a Lumen meter - for less than $20 and it will be more accurate.
I have a lumen meter,and we will get to that. I can only do so much in a given time.
I need to go through each step,so as others that may not have a lumen meter can follow along with a pulse motor they have.
Brad
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Those that are following this thread, take some time to think about what you see.
It is said and stated by ohms law that the power disipated by any resistor is determined by the current flowing through it, and the value of that resistor. Now think about the incandescent bulb, and what is needed in order for the resistance of that bulb to increase. Then think about what is needed to raise the light output from that globe, and what must increase in order to raise the dissipated power from that bulb.
When you have the answers to the above, ask your self how is it we see the opposite happening on the scope. How dose ohms law explain more power being dissipated across a resistor with less current flowing through it, and less power being dissipated across it with more current flowing through it.
Can we really dissipate more power from a pulsed resistor or resistive load than we could using straight DC.
Brad.
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I think that when you connect the capacitor to the probe side of the bulb, you have contaminated the reading and are no longer reading simply the voltage drop across the bulb.
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I think that when you connect the capacitor to the probe side of the bulb, you have contaminated the reading and are no longer reading simply the voltage drop across the bulb.
:)
And if the bulb was removed TK,and we had just the CVR,would not that contamination you speak of still exist at the probe side of the CVR?,as it is just a resistor like our bulb. We have done nothing more than remove one of the two resistors that are in series,and now the cap is conected to the probe side of the CVR.
What is this contamination?-->how can a voltage be seen across a resistor that dosnt actually represent the true voltage across that resistor?.
I will get deeper into this as we go-->this (what you call) contamination.
But first a few more video's on testing done to find the answer. Next wee will be connecting a DC current source to our bulb,and increasing the voltage until we get the same power output from our solar panel. We can then see what is truth,and what is not. I will replace the current resistor that is across the solar panel with a higher value one,so as we can get a more accurate voltage reading across that resistor. Then we will run the pulse motor again,and obtain our two reading's and values. then we send a DC current through the light bulb,and keep increasing the voltage until we get the same output from the solar panel.
Brad
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OK, I've essentially duplicated your results, using a " no-Bedini Atoll " circuit as illustrated below.
I found that the value of the inductor-resistor combination determines whether the bulb gets brighter, or dimmer, when the capacitor cliplead is connected.
The values shown in my circuit image produce _no change_ in the visible bulb brightness when the clip lead is connected, even though the scope shows over twice the power through the CVR-bulb when the cap is connected than when it is not.
A slightly higher resistance produces less brightness when the cap is connected; a slightly lower resistance produces more brightness when the cap is connected.
With 21.3 ohms, it is easy to see that the bulb is less bright when the cap is connected than when it is not.
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I've made a video showing my results but it will probably be an hour or so before it's ready to view on YouTube.
I don't know if my "contamination" idea is correct or not. It is certainly an interesting effect that TinMan has discovered, and as my testing shows it doesn't depend on a Bedini motor or any pulse motor at all, just a pulsed circuit. I don't know if the inductor is even necessary either.
Ah... where is MarkE when you need him. Sigh. He'd set us straight on this, I'm sure, but alas it is not to be.
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I've made a video showing my results but it will probably be an hour or so before it's ready to view on YouTube.
I don't know if my "contamination" idea is correct or not. It is certainly an interesting effect that TinMan has discovered, and as my testing shows it doesn't depend on a Bedini motor or any pulse motor at all, just a pulsed circuit.
Ah... where is MarkE when you need him. Sigh. He'd set us straight on this, I'm sure, but alas it is not to be.
Thank you TK for taking the time to carry out these test.
As i stated at the start of the thread--Yes,it is with pulsed system's,not just the bedini circuit.
I don't know if the inductor is even necessary either.
During this thread,and after doing some other tests,it will come to the stage where we will find this out--next couple of video i hope i will get to it,but my experimenting time is short during the working week. Feel free to carry on as you will ,and carry out tests as you see fit--oh,and post your results ;). The ramifications of the pending outcome could be high--as far as accurate measurements of pulsed systems go.
Brad
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http://www.youtube.com/watch?v=Xh16cxceN9w
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http://www.youtube.com/watch?v=Xh16cxceN9w
TK
Try an inductor with a core.
I think the results will be better and more evident.
Brad
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Here is the pulse motor circuit from video 1 and 2,showing scope probe placement as well.
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Here is video 3-with circuit attached.
I have switched over to pulsing the inductor with my FG,so as to get a cleaner wave form,and also have raised the frequency to 100Hz.
As my scope and FG share a common ground,i had to change the CVR and globe over to the negative side of the circuit.
Next video,we will remove the inductor,and see it we still get the same effect.
https://www.youtube.com/watch?v=bMHk3m6DsFE
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TK
Try an inductor with a core.
I think the results will be better and more evident.
P.S--forgot to ask--no power measurements mentioned on the video?.
Brad
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IT SEEMS THAT WE ARE LOOKING AT REFLECTIVE POWER..
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Tinman,
Surely you are not serious...
Incandescent bulbs have a very large positive temperature coefficient. As such, they present very non-linear I versus V behavior.
Incandescent bulbs tend to regulate the current flowing thru them. As more voltage is applied to the bulb and it attempts to draw more current, the temperature of the filament increases, which increases the filaments resistance, which reduces the current flowing thru it. This inherent behavior has been used for many years in unique applications such as current limiters and sine wave oscillator AGC circuits.
Incandescent bulb filaments have thermal mass and, therefore, thermal inertia. Because of this thermal inertia, the temperature of the filament, and therefore its resistance, will be different for applied voltages having different frequency, duty cycle, and peak voltage values, even if the average value of those applied voltages are identical.
Incandescent bulbs are very inefficient emitters of visible light. The bulk of their emissions is in the far infrared and their emission bandwidth (spectrum) is highly dependent on the temperature of the filament. The limited spectral response of a typical solar cell is not going to respond to the long wavelength IR which represents the bulk of the bulb's emissions.
Lesson to be learned, always use a low temperature coefficient, non-inductive resistor for your CVR...
PW
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Tinman,
Surely you are not serious...
Incandescent bulbs have a very large positive temperature coefficient. As such, they present very non-linear I versus V behavior.
This inherent behavior has been used for many years in unique applications such as current limiters and sine wave oscillator AGC circuits.
Incandescent bulb filaments have thermal mass and, therefore, thermal inertia. Because of this thermal inertia, the temperature of the filament, and therefore its resistance, will be different for applied voltages having different frequency, duty cycle, and peak voltage values, even if the average value of those applied voltages are identical.
Lesson to be learned, always use a low temperature coefficient, non-inductive resistor for your CVR...
PW
I am very serious PW,and i think you missed the boat here some where.
Did you watch the first two video's ? ,do you know what this thread is all about.
Incandescent bulbs tend to regulate the current flowing thru them. As more voltage is applied to the bulb and it attempts to draw more current, the temperature of the filament increases, which increases the filaments resistance, which reduces the current flowing thru it.
Please post a short video here showing us how you can raise the voltage across an incandescent bulb without the current also rising. Yes,the resistances rises as the element rises in temperature,but in order to gain a temperature rise,the current flowing through that bulb must increase,and as the temperature rises,so will the visible light output.
Incandescent bulbs are very inefficient emitters of visible light. The bulk of their emissions is in the far infrared and their emission bandwidth (spectrum) is highly dependent on the temperature of the filament. The limited spectral response of a typical solar cell is not going to respond to the long wavelength IR which represents the bulk of the bulb's emissions.
This is telling me that you have not watched the first video,where there was no solar cell being used,and the rise and fall of visible light was very apparent. Once again,the only way to increase the visible light output of an incandescent bulb is to increase the current flowing through it-->which will increase the voltage across that bulb.
And then there is the CVR-what of that-->and please do not go on about how those wire wound resistors are no good because they have inductance,as no inductance at all can be seen on the scope from that CVR at these low frequencies. How is it that in the last video,the voltage across the CVR remained the same,indicating that the current flowing through the system was the same,and yet my very reliable amp meter says the current increased by 70mA when the cap was conected.
In this thread,only those that can ! show ! what they say to be true will be taken notice of. So your first task is to show Quote: Incandescent bulbs tend to regulate the current flowing thru them. As more voltage is applied to the bulb and it attempts to draw more current, the temperature of the filament increases, which increases the filaments resistance, which reduces the current flowing thru it.
So i would like you to show us how you can increase the voltage across an incandescent bulb,while maintaining or decreasing the current flowing through it. Show us how you can dissipate more power from an incandescent bulb with less current flowing through it,and dissipate less power with more current flowing through it-->as i have shown.
I have shown the effect in the form of experiment's,and presented those experiments and results by way of video. In this thread,those that choose to argue the point will do so with actual experiments-->(! not text book physics !),and will present there experiment right here on this thread. No credibility will be given here to words without experimental data to back up there claims.
Words are no longer good enough.
P.S
I would like to add this quote from ION
Quote: Even the lowly incandescent bulb can be viewed as a measuring device.
Brad
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I am very serious PW,and i think you missed the boat here some where.
Did you watch the first two video's ? ,do you know what this thread is all about.
Please post a short video here showing us how you can raise the voltage across an incandescent bulb without the current also rising. Yes,the resistances rises as the element rises in temperature,but in order to gain a temperature rise,the current flowing through that bulb must increase,and as the temperature rises,so will the visible light output.
This is telling me that you have not watched the first video,where there was no solar cell being used,and the rise and fall of visible light was very apparent. Once again,the only way to increase the visible light output of an incandescent bulb is to increase the current flowing through it-->which will increase the voltage across that bulb.
And then there is the CVR-what of that-->and please do not go on about how those wire wound resistors are no good because they have inductance,as no inductance at all can be seen on the scope from that CVR at these low frequencies. How is it that in the last video,the voltage across the CVR remained the same,indicating that the current flowing through the system was the same,and yet my very reliable amp meter says the current increased by 70mA when the cap was conected.
In this thread,only those that can ! show ! what they say to be true will be taken notice of. So your first task is to show Quote: Incandescent bulbs tend to regulate the current flowing thru them. As more voltage is applied to the bulb and it attempts to draw more current, the temperature of the filament increases, which increases the filaments resistance, which reduces the current flowing thru it.
So i would like you to show us how you can increase the voltage across an incandescent bulb,while maintaining or decreasing the current flowing through it. Show us how you can dissipate more power from an incandescent bulb with less current flowing through it,and dissipate less power with more current flowing through it-->as i have shown.
I have shown the effect in the form of experiment's,and presented those experiments and results by way of video. In this thread,those that choose to argue the point will do so with actual experiments-->(! not text book physics !),and will present there experiment right here on this thread. No credibility will be given here to words without experimental data to back up there claims.
Words are no longer good enough.
P.S
I would like to add this quote from ION
Quote: Even the lowly incandescent bulb can be viewed as a measuring device.
Brad
@Tinman, you do great work, always!!!
in an Incandescent Bulb, Resistance changes with heat - See: Nonlinear Conduction (http://www.allaboutcircuits.com/textbook/direct-current/chpt-2/nonlinear-conduction/)
Any Resistor that changes in Temperature will suffer changes in resistance.
An Incandescent Bulb is not Linear...
Chris Sykes
hyiq.org
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@Tinman, you do great work, always!!!
in an Incandescent Bulb, Resistance changes with heat - See: Nonlinear Conduction (http://www.allaboutcircuits.com/textbook/direct-current/chpt-2/nonlinear-conduction/)
Any Resistor that changes in Temperature will suffer changes in resistance.
An Incandescent Bulb is not Linear...
Chris Sykes
hyiq.org
Yes,the current/voltage curve is non linear,but the only way to dissipate more power from an incandescent bulb is to increase the amount of current flowing through it,and this increase of current will increase the voltage across that globe.
I show the exact opposite,where i decrease the amount of current flowing through the bulb,decrease the voltage across that bulb,but increase the dissipated power across that bulb.
We are using our CVR to watch this happen,so either the CVR is not telling us the true current flowing through it,or our light bulb has everything back to front--it has to be one of the two,as the current flowing into the system ! MUST! flow through the CVR and bulb first.
Take that inductor away,and what do you think will happen?
Brad
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Tinman,
I watched video two, could not find video one, scanned thru video three. Some of what I heard in the videos and read in this thread just made me shake my head.
Perhaps it is you that is missing the boat. Please reread my post and tell me which of the points I raised you disagree with. I can't believe there is anything I stated in my post that you could possibly disagree with.
Read up on positive and negative temperature coefficient resistors. Look at the data sheets for various resistors, the temperature coefficient is almost always given (usually stated as ppm/C).
Measure the cold resistance of a 100 watt incandescent and calculate its power draw based on that resistance.
PW
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Tinman,
Perhaps you should consider building a very simple Wien bridge sine wave oscillator using an incandescent bulb as the gain control element.
Marvel at how, without even glowing, the varying resistance of the lamp regulates the positive feedback.
Very old school...
PW
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Yes,the current/voltage curve is non linear,but the only way to dissipate more power from an incandescent bulb is to increase the amount of current flowing through it,and this increase of current will increase the voltage across that globe.
I show the exact opposite,where i decrease the amount of current flowing through the bulb,decrease the voltage across that bulb,but increase the dissipated power across that bulb.
We are using our CVR to watch this happen,so either the CVR is not telling us the true current flowing through it,or our light bulb has everything back to front--it has to be one of the two,as the current flowing into the system ! MUST! flow through the CVR and bulb first.
Take that inductor away,and what do you think will happen?
Brad
The Current/Voltage Curve is the Change in the Resistance of the Globe. Because: I = V/R and V = R x I - If R changes, then I and or V must also change because: R = V / I.
Even the CVR will have an amount of Non Linearity to the result.
Current flowing through a Circuit Element is normally proportional to the Applied Voltage across a Circuit Element, not taking into account the Time Domain and the instantaneous measurement in a Non Linear Situation. A Voltage drop Across a Circuit Element, is associated with Heat dissipated from that Circuit Element. Electrical Energy Transformed to Heat. So I am sorry, I don't fully agree.
It is however another good experiment and I may have totally missed something...
Chris Sykes
hyiq.org
P.S: The only reason I know this is because I thought I was onto something and this was what it turned out to be.
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From Wiki:
https://en.wikipedia.org/wiki/Incandescent_light_bulb
Current and resistance
The actual resistance of the filament is temperature dependent. The cold resistance of tungsten-filament lamps is about 1/15 the hot-filament resistance when the lamp is operating. For example, a 100-watt, 120-volt lamp has a resistance of 144 ohms when lit, but the cold resistance is much lower (about 9.5 ohms).[50][93]
PW
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PW/TK,
These familiar "Gotchas" raise their ugly heads from time to time. Last time I involved myself with this one in particular was with Luc on his Recirculating BEMF circuit.
This was back in 2009, and I did up an analysis which explained his observations and showed, as often seems the case, that erroneous assumptions were being made.
Here is that analysis: Luc_flyback01.pdf (http://overunity.com/downloads/sa/view/down/595/)
A lot of time and effort went into this document (as did all my documents uploaded here), and aside from a few too many references to "RMS", it is still relevant, and should I believe shed some light on this scenario as well.
A couple of excerpts in summary:
The Rbulb intensity (or heat) can not be used reliably to indicate the total amount of power used by the entire circuit!
When there are other components in the circuit (such as resistors and coils) and the bulb is in series with them and the power supply, the bulb's intensity is only indicative of the power being dissipated in the bulb itself. It does not indicate how much total power is being taken from the power supply and being used in the whole circuit.
Peace Out.
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Perhaps this is a good opportunity to ask one of the most important questions of all: "How is it that a change in Temperature changes the resistance so much? What is going on to make this happen?"
We know this is Nonlinear Conduction (http://www.allaboutcircuits.com/textbook/direct-current/chpt-2/nonlinear-conduction/) - From this we can see that the "Conductivity" of the Filament has changed. Temperature is the cause. Why?
What else can change the Conductivity of an Element?
Chris Sykes
hyiq.org
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Incandescent bulbs have a very large positive temperature coefficient. As such, they present very non-linear I versus V behavior.
Yes, incandescent bulbs are not linear but that does not mean that they are not monotonic.
Light bulb's brightness depends on the current flowing through it and Tinman is correct that more current always means more brightness....and in his experiment this is all that matters.
I am more concerned how his scopes calculates the average of these pulses, e.g. from the screen data, from memory or from the waveform period...
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Hmmm.... I think some people are still missing some points and being distracted by Red Herrings.
1. The current measured by monitoring the voltage drop across the CVR is also the current flowing through the bulb. The elements are in strict series so the same current is flowing through them both. Right?
2. The brightness of the bulb is an indication of the power being dissipated in the bulb. A dimmer bulb means less power dissipated in the bulb, a brighter bulb means more power dissipated in the bulb. This is true regardless of factors like the temperature coefficient of resistance of the bulb, and the duty cycle of pulsation. Right?
3. The instantaneous power being dissipated in the bulb is Watts=I2R, where I is the current at the instant of measurement and R is the bulb's resistance at that instant. Right? And this is also equivalent to Watts=V2/R, so solving for R we have R=V/I by Ohm's Law. Right?
4. When the capacitor is connected, the current through the bulb is (relatively) constant, so there is no difficulty with the "mean" value of the current. So the power dissipated in the bulb is also constant. The resistance of the bulb can be calculated by R=V/I. We know I from the CVR measurement, but what is V?
Question: Does the voltage drop across the bulb as measured by the scope give us the "V" value for this equation when the voltage is constant, duty cycle 100 percent?
5. When the capacitor is _not_ connected, the current through the bulb is pulsed. So the power being supplied to the bulb is no longer constant. Depending on the thermal lag of the filament, this power is "smeared out" or averaged over some time interval, so the bulb is actually dissipating some power even when the filament is not receiving any current. Hence it can appear to be glowing steadily even though its current supply is pulsed. During the current peaks, as measured by the scope, the power dissipated is related to the _square_ of the current, adjusted somehow for the thermal lag and the temperature coefficient of resistance of the filament. Right?
Question: Is it legitimate to use the simple "mean" value of the current to calculate the average power dissipated by the bulb in this case, since the power supplied during the peaks is related to the square of the current?
Are the Red Herrings starting to jump out of the bucket yet?
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Yes, incandescent bulbs are not linear but that does not mean that they are not monotonic.
Light bulb's brightness depends on the current flowing through it and Tinman is correct that more current always means more brightness....and in his experiment this is all that matters.
I am more concerned how his scopes calculates the average of these pulses, e.g. from the screen data, from memory or from the waveform period...
I believe the scope calculates the measurements from the data displayed on the screen, or, in the case of my Rigol, it can also calculate between cursors. Yesterday I did a rough visual analysis of TinMan's scope traces and I think it is calculating the "mean" value correctly. A better question might be, as I've tried to clarify above, "Is the mean value of the current (and/or voltage drop) the appropriate measurement to use here when estimating the power dissipated in the bulb?"
Looking at the screenshot below, is the "A" area the same as the "B" area? I think it is, pretty close anyway, which tells me that the scope is calculating the mean value correctly. I think.
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In this thread,only those that can ! show ! what they say to be true will be taken notice of. So your first task is to show Quote: Incandescent bulbs tend to regulate the current flowing thru them. As more voltage is applied to the bulb and it attempts to draw more current, the temperature of the filament increases, which increases the filaments resistance, which reduces the current flowing thru it.
So i would like you to show us how you can increase the voltage across an incandescent bulb,while maintaining or decreasing the current flowing through it. Show us how you can dissipate more power from an incandescent bulb with less current flowing through it,and dissipate less power with more current flowing through it-->as i have shown.
I have shown the effect in the form of experiment's,and presented those experiments and results by way of video. In this thread,those that choose to argue the point will do so with actual experiments-->(! not text book physics !),and will present there experiment right here on this thread. No credibility will be given here to words without experimental data to back up there claims.
Words are no longer good enough.
~~~~~
Brad
I agree with this, I want to learn this effect, PW should demonstrate this effect in an experiment as TK did his, otherwise there is a definite credibility gap, as we can say that the effect has been reproduced as TM describes.
after discovering that Pi is not only 'incorrect' but it's also essentially a very inefficient way to calculate (i.e how much human methods seem) i'm sure some people that understand or learned this might be starting to question the reality in which they live?
I'd like to see a simple experiment where:
- an incandescent bulb of this type shines brighter with less or the same current.
note:
(even though bizarrely the last video showed two different currents along the same direct path.)
(however TK experiment reproduced the effect without the CVR did it not? ) (i.e wound R)
ah i'm staring to see the light here the CVR is on the negative, the inductor is the key, of course electrons don't exist, and so called 'electron flow' if this is looked at as a closed magnetic flux system it starts to make more sense.
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Hmmm.... I think some people are still missing some points and being distracted by Red Herrings.
1. The current measured by monitoring the voltage drop across the CVR is also the current flowing through the bulb. The elements are in strict series so the same current is flowing through them both. Right?
2. The brightness of the bulb is an indication of the power being dissipated in the bulb. A dimmer bulb means less power dissipated in the bulb, a brighter bulb means more power dissipated in the bulb. This is true regardless of factors like the temperature coefficient of resistance of the bulb, and the duty cycle of pulsation. Right?
3. The instantaneous power being dissipated in the bulb is Watts=I2R, where I is the current at the instant of measurement and R is the bulb's resistance at that instant. Right? And this is also equivalent to Watts=V2/R, so solving for R we have R=V/I by Ohm's Law. Right?
4. When the capacitor is connected, the current through the bulb is (relatively) constant, so there is no difficulty with the "mean" value of the current. So the power dissipated in the bulb is also constant. The resistance of the bulb can be calculated by R=V/I. We know I from the CVR measurement, but what is V?
Question: Does the voltage drop across the bulb as measured by the scope give us the "V" value for this equation when the voltage is constant, duty cycle 100 percent?
5. When the capacitor is _not_ connected, the current through the bulb is pulsed. So the power being supplied to the bulb is no longer constant. Depending on the thermal lag of the filament, this power is "smeared out" or averaged over some time interval, so the bulb is actually dissipating some power even when the filament is not receiving any current. Hence it can appear to be glowing steadily even though its current supply is pulsed. During the current peaks, as measured by the scope, the power dissipated is related to the _square_ of the current, adjusted somehow for the thermal lag and the temperature coefficient of resistance of the filament. Right?
Question: Is it legitimate to use the simple "mean" value of the current to calculate the average power dissipated by the bulb in this case, since the power supplied during the peaks is related to the square of the current?
Are the Red Herrings starting to jump out of the bucket yet?
A Circuit to study might be a good start:
Tinman, is this Circuit correct? SW1 being the manual Clip lead connection.
Chris Sykes
hyiq.org
Lets figure out what's going on...
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First thing I noticed is that the Wave Forms are completely different, no sign of any Pulsing:
Chris Sykes
hyiq.org
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I think if the Circuit is right, and the Scope Shots are an indication, we can see there is a lot less Voltage on the Yellow Trace especially... The Frequency, Rise/Fall time, there, will change the Lights Luminosity.
The total Applied Voltage across the Globe is very much different.
Chris Sykes
hyiq.org
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Something is wrong with the mean calculation!
At 2 volts per division for the Yellow Trace, its not giving you the right figure.
I agree with what Verpies said:
Yes, incandescent bulbs are not linear but that does not mean that they are not monotonic.
Light bulb's brightness depends on the current flowing through it and Tinman is correct that more current always means more brightness....and in his experiment this is all that matters.
I am more concerned how his scopes calculates the average of these pulses, e.g. from the screen data, from memory or from the waveform period...
Chris Sykes
hyiq.org
JOKE: Try DC Coupling.... Hahahaha
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Compared to:
RMS = Vpeak * 1/sqrt(2) = 8 * 0.707 = 5.6568V
NOTE: The above Calculation is supposed to be only good for Symmetrical Sine Waves. I realise its not totally correct to use it here. This is just to give some idea of the difference.
Correction: Pk to Pk , not Peak:
RMS = Vpeak to peak * 1/sqrt(2) = 8 * 0.35355 = 2.8284V
The Globe is seeing a Difference of 8 Volts at a frequency of 68.4 Hz.
Chris Sykes
hyiq.org
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Verpies & TK
It is good to see some understand as to what it is we are looking at here.
PW
I in no way disagree with what you say about the resistance increasing across the bulb when.the temperature rises on the filament. What I am saying is the the current flowing through that bulb must increase in order for there to be a temperature rise which causes the rise in resistance.
I am showing a reduction of current flowing through that bulb, but an increase in light and heat output, which means the bulb is dissipating more power. Remember, the voltage also drops across the bulb when the current flowing through it drops-but the dissipated power in the form of heat and light increases.-as TK has also shown, and he is using what you would class as an acceptable CVR.
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Verpies & TK
It is good to see some understand as to what it is we are looking at here.
PW
I in no way disagree with what you say about the resistance increasing across the bulb when.the temperature rises on the filament. What I am saying is the the current flowing through that bulb must increase in order for there to be a temperature rise which causes the rise in resistance.
I am showing a reduction of current flowing through that bulb, but an increase in light and heat output, which means the bulb is dissipating more power. Remember, the voltage also drops across the bulb when the current flowing through it drops-but the dissipated power in the form of heat and light increases.-as TK has also shown, and he is using what you would class as an acceptable CVR.
Yep, I must admit, took me a bit to get it... Little bit of study required.
Chris Sykes
hyiq.org
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Tinman - Can you scope C2 in run mode and post pic please?
Chris Sykes
hyiq.org
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On connection of C2 via the Clip Lead (SW1 in the Circuit) this adds a large Capacitive reactance to the Circuit. Its hard to get any data from the screenshots of the scope in the video because the Cap Smooth's out the Pulsing very well!!!
But, this should introduce a Current Lead into the Circuit. Hard to see without more info. This could also be part of the cause of the Higher current seen.
All be it very small as we see mostly DC.
Chris Sykes
hyiq.org
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Yep, I must admit, took me a bit to get it... Little bit of study required.
Chris Sykes
hyiq.org
I really don't see anything out of the norm here now I get it. PW what do you think?
I think accurate Input Measurements could be obtained from "Cap Connected - mean value is wrong" Image, as long as the readings are fixed. Channel 1's Mean Value does not appear to be correct: 3.36 Volts
Channel 2's Mean Value of 144mv does appear to be pretty good.
Chris Sykes
hyiq.org
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Before C2 is connected:
After Q1 switched the Negative side of the Coil to Ground, we will see a Voltage drop across all four components:
1: 1 Ohm Resistor
2: Globe
3: Coil
4: Q1's internal Resistance.
Totaling the Total Battery Voltage. Series Resistance is approximately: R = 126 Ohms. Using Ohms law you can calculate the Power: R x I2 = 1.26 watts approximately.
Introducing C2 (6800uf Capacitor), via SW1, after the Globe (LA1) ensures that the grounding of the above two components:
1: 1 Ohm Resistor
2: Globe
through Components:
3: Coil
4: Q1's internal Resistance.
only occurs after the Capacitor C2 (6800uf Capacitor), charge is below the Voltage Drop of the Coil and Q1, or completely empty of Charge, which will never happen. As a result we see no more Pulsed Waveform, the power through these components is the total power being delivered to C2 (6800uf Capacitor) less the losses of these components.
Series Resistance of Components 3 and 4 is: R = 90 Ohms. The Potential Voltage difference across the 1 Ohm Resistor and the Globe when SW1 is closed is only around 1.844 Volts. Using Ohms law you can calculate the Power: R x I2 = 1.76 watts approximately.
Sorry Brad, but this is the way I see it.
Chris Sykes
hyiq.org
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I really don't see anything out of the norm here now I get it. PW what do you think?
I think accurate Input Measurements could be obtained from "Cap Connected - mean value is wrong" Image, as long as the readings are fixed. Channel 1's Mean Value does not appear to be correct: 3.36 Volts
Channel 2's Mean Value of 144mv does appear to be pretty good.
Chris Sykes
hyiq.org
There are probably spikes that aren't being resolved at that timebase setting. A faster timebase might show these spikes. I think the "mean" calculation in TinMan's shot could be thrown off by these spikes for some reason.
My scope seems to be giving near-correct values for "Average" in spite of the spikes.
My scope does have some baseline offset though; I probably need to run the Self-Cal routine again since the temperature has dropped in here. The readings of the circuit under power should be adjusted for these offsets, probably.
First shot below: baseline offset with no power to my circuit.
Second shot below: traces with power applied, no capacitor.
Third shot below: traces with capacitor connected. You can barely see the spikes at this horizontal timebase.
Fourth shot below: the spikes with cap connected, at a faster timebase.
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There are probably spikes that aren't being resolved at that timebase setting. A faster timebase might show these spikes. I think the "mean" calculation in TinMan's shot could be thrown off by these spikes for some reason.
My scope seems to be giving near-correct values for "Average" in spite of the spikes.
My scope does have some baseline offset though; I probably need to run the Self-Cal routine again since the temperature has dropped in here. The readings of the circuit under power should be adjusted for these offsets, probably.
First shot below: baseline offset with no power to my circuit.
Second shot below: traces with power applied, no capacitor.
Third shot below: traces with capacitor connected. You can barely see the spikes at this horizontal timebase.
Fourth shot below: the spikes with cap connected, at a faster timebase.
Thanks TK,
Can your scope do the math on the spikes? Would be interesting to know what the math is on them.
EDIT: Dur sorry ignore me, was looking for the Math Window...
Chris Sykes
hyiq.org
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Before C2 is connected:
Totaling the Total Battery Voltage. Series Resistance is approximately: R = 126 Ohms. Using Ohms law you can calculate the Power: R x I2 = 1.26 watts approximately.
Introducing C2 (6800uf Capacitor), via SW1, after the Globe (LA1) ensures that the grounding of the above two components:
1: 1 Ohm Resistor
2: Globe
through Components:
3: Coil
4: Q1's internal Resistance.
only occurs after the Capacitor C2 (6800uf Capacitor), charge is below the Voltage Drop of the Coil and Q1, or completely empty of Charge, which will never happen. As a result we see no more Pulsed Waveform, the power through these components is the total power being delivered to C2 (6800uf Capacitor) less the losses of these components.
Series Resistance of Components 3 and 4 is: R = 90 Ohms. The Potential Voltage difference across the 1 Ohm Resistor and the Globe when SW1 is closed is only around 1.844 Volts. R x I2 = 1.76 watts approximately.
Sorry brad, but this is the way I see it.
Chris Sykes
hyiq.org
Then you are seeing things all wrong EMJ.
After Q1 switched the Negative side of the Coil to Ground, we will see a Voltage drop across all four components:
1: 1 Ohm Resistor
2: Globe
3: Coil
4: Q1's internal Resistance.
No you wont,you will see a voltage across all four components when Q1 switches on,as there is no voltage across any when the circuit is open except Q1. As you can see in both mine and TK's scope shot's,there is no voltage across the CVR or bulb when Q1 is open,and the cap disconnected. We are looking at the CVR and the bulb,and that is where the scope probe's are placed.
I ask the same of you that i have asked of others that think there is nothing to it-->post a video without the inductor,and show us how an incandescent bulb can dissipate more power with less total current flowing through it,and less power with more total current flowing through it.
You may also like to look at TK's latest scope shot's,and have a look at the total current flowing through the CVR,and voltage across the bulb with and without the cap connected in the circuit. Are you telling me that the error margin could be this big ?. And dont forget,this is TK carrying out these test,and he knows his way around power measurements.
Brad
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A Circuit to study might be a good start:
Tinman, is this Circuit correct? SW1 being the manual Clip lead connection.
Chris Sykes
hyiq.org
Lets figure out what's going on...
No,that circuit is incorrect.
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Then you are seeing things all wrong EMJ.
No you wont,you will see a voltage across all four components when Q1 switches on,as there is no voltage across any when the circuit is open except Q1. As you can see in both mine and TK's scope shot's,there is no voltage across the CVR or bulb when Q1 is open,and the cap disconnected. We are looking at the CVR and the bulb,and that is where the scope probe's are placed.
I ask the same of you that i have asked of others that think there is nothing to it-->post a video without the inductor,and show us how an incandescent bulb can dissipate more power with less total current flowing through it,and less power with more total current flowing through it.
You may also like to look at TK's latest scope shot's,and have a look at the total current flowing through the CVR,and voltage across the bulb with and without the cap connected in the circuit. Are you telling me that the error margin could be this big ?. And dont forget,this is TK carrying out these test,and he knows his way around power measurements.
Brad
Brad, its your circuit, you know better than me.
Removing the Inductor removes its Impedance thus changing the Circuit.
Is the Circuit correct?
Chris Sykes
hyiq.org
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Something is wrong with the mean calculation!
At 2 volts per division for the Yellow Trace, its not giving you the right figure.
I agree with what Verpies said:
Chris Sykes
hyiq.org
JOKE: Try DC Coupling.... Hahahaha
EMJ
This is where incorrect assumptions start,and i hope you either remove this claim,or correct it.
If you listen to the video,and watch the video carefully,then you will know that i have dropped the channel down 1 division so as to fit the whole wave form in the scopes screen.
So the mean value given is correct,and your red voltage calculation you added to the scope shot is incorrect.
Brad
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No,that circuit is incorrect.
Please elaborate?
Chris Sykes
hyiq.org
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EMJ
This is where incorrect assumptions start,and i hope you either remove this claim,or correct it.
If you listen to the video,and watch the video carefully,then you will know that i have dropped the channel down 1 division so as to fit the whole wave form in the scopes screen.
So the mean value given is correct,and your red voltage calculation you added to the scope shot is incorrect.
Brad
Brad, from Baseline to the indicated Trace Line is only 1.7 volts approx. Have you Changed your baseline value to +2V
Chris Sykes
hyiq.org
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TK
Try an inductor with a core.
I think the results will be better and more evident.
P.S--forgot to ask--no power measurements mentioned on the video?.
Brad
I'm not sure we are actually measuring any true power values yet.
In your original circuit, and my no-Bedini modification, the CH1 voltage drop across the bulb doesn't represent the input voltage to the circuit. The CH2 current is the input current, certainly, and actually the CH1 voltage drop across the bulb must also represent the input current and because of that we could calculate the resistance of the bulb.
That is, if the input current from the one-ohm CVR is, say, 94 mA (because we have 94 mV drop across 1 ohm), and we have 3.4 Vdrop across the bulb, then since both are obviously carrying the same current, we have R = V/I so Rbulb= 3.4V/0.094 A = 36.2 Ohms.
So maybe now we can get the power dissipation in the bulb at that current, since Watts=I2R. So we have for example W = 36.2 x (0.098)2 = 0.347 Watt being dissipated in the bulb. This is with capacitor connected in my circuit.
With cap disconnected and using the "average" values and going through the same calculations, we have average I = 62 mA and average Vdrop = 2.06 V, so the bulb's average resistance is V/I or 2.06/0.062 = 33.2 Ohms. So the average power dissipated in the bulb is 33.2 x (0.062)2 = 0.127 Watt.
What is the conclusion we can draw from this, since the bulb is obviously brighter when the cap is disconnected? Perhaps: Using the "average" values in the way we have been doing is not correct in the case of pulses through a bulb; OR perhaps: Our assumption about the relationship between the bulb's brightness and the "average" power _in the pulses_ is not correct. Or both.
Or maybe my math is just totally screwed.
As far as the inductor goes... it is not necessary to produce the effect. I have eliminated the inductor entirely, along with the long leads I used to connect it, and I've simply placed a 34 Ohm resistor directly on my little breadboard between the bulb and the mosfet Drain, and the effect still happens just as before. There is less spikeyness when the cap is connected but the bulb dims while the "average" current through it goes up, just as before.
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EMJ
This is where incorrect assumptions start,and i hope you either remove this claim,or correct it.
If you listen to the video,and watch the video carefully,then you will know that i have dropped the channel down 1 division so as to fit the whole wave form in the scopes screen.
So the mean value given is correct,and your red voltage calculation you added to the scope shot is incorrect.
Brad
Ah, that's right. EMJ is measuring from the horizontal center graticule line, and your baseline is one division below that, so your mean values are indeed correct, and not thrown off by spikes or whatnot.
In the (nearly) full shot below we can see on the left edge where the channel baselines are positioned. They are both one full division below the centerline. The yellow marker at the screen centerline is the Trigger level marker, isn't it?
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Ah, that's right. EMJ is measuring from the horizontal center graticule line, and your baseline is one division below that, so your mean values are indeed correct, and not thrown off by spikes or whatnot.
In the (nearly) full shot below we can see on the left edge where the channel baselines are positioned. They are both one full division below the centerline. The yellow marker at the screen centerline is the Trigger level marker, isn't it?
I don't believe this is correct, yellow channel baseline is pretty much on the horizontal center graticule line: See Image
Blue Channel Baseline is 1div below the horizontal center graticule line.
Correct me if you have changed the value of this baseline Brad.
Chris Sykes
hyiq.org
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No, both channel baselines are one division below centerline. The Blue channel marker is overlaying the Yellow channel marker so you can't see it, and the yellow marker you do see is the Trigger level indicator, indicating Trigger on CH1 (because it is yellow) and at approx. 2 Volts above the baseline,which puts it just a hair below the screen centerline marker. (See the exact trigger level on the bottom right of the screenshot I've posted.)
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I don't believe this is correct, yellow channel baseline is pretty much on the horizontal center graticule line: See Image
Blue Channel Baseline is 1div below the horizontal center graticule line.
Correct me if you have changed the value of this baseline Brad.
Chris Sykes
hyiq.org
No-the yellow marker for channel 1 is being hidden by the blue marker of channel 2. The yellow marker you see at the base line is the trigger marker,and it is yellow because i am triggering on channel 1. If i was triggering on channel 2,then that marker on the base line would be blue.
Brad
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No, both channel baselines are one division below centerline. The Blue channel marker is overlaying the Yellow channel marker so you can't see it, and the yellow marker you do see is the Trigger level indicator, indicating Trigger on CH1 (because it is yellow) and at approx. 2 Volts above the baseline,which puts it just a hair below the screen centerline marker. (See the exact trigger level on the bottom right of the screenshot I've posted.)
TK, Ok I will take your word for it but certainly looks deceiving. Thinking about this, it is correct and Brads Mean Value is right on Ch1.
Still, this still means we only have a Potential Voltage Difference of 3.374 Volts as opposed to the 8 Volts with the Cap disconnected. Being that Power is V x I when in phase, we can see a clear difference in total power through the Components in question.
8 x 0.096 = 0.768
3.374 x 0.144 = 0.485856
Chris Sykes
hyiq.org
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No-the yellow marker for channel 1 is being hidden by the blue marker of channel 2. The yellow marker you see at the base line is the trigger marker,and it is yellow because i am triggering on channel 1. If i was triggering on channel 2,then that marker on the base line would be blue.
Brad
Yes, certainly not easy to make out from the video. Honest mistake all sorted now.
Chris Sykes
hyiq.org
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I'm not sure we are actually measuring any true power values yet.
In your original circuit, and my no-Bedini modification, the CH1 voltage drop across the bulb doesn't represent the input voltage to the circuit. The CH2 current is the input current, certainly, and actually the CH1 voltage drop across the bulb must also represent the input current and because of that we could calculate the resistance of the bulb.
That is, if the input current from the one-ohm CVR is, say, 94 mA (because we have 94 mV drop across 1 ohm), and we have 3.4 Vdrop across the bulb, then since both are obviously carrying the same current, we have R = V/I so Rbulb= 3.4V/0.094 A = 36.2 Ohms.
So maybe now we can get the power dissipation in the bulb at that current, since Watts=I2R. So we have for example W = 36.2 x (0.098)2 = 0.347 Watt being dissipated in the bulb. This is with capacitor connected in my circuit.
With cap disconnected and using the "average" values and going through the same calculations, we have average I = 62 mA and average Vdrop = 2.06 V, so the bulb's average resistance is V/I or 2.06/0.062 = 33.2 Ohms. So the average power dissipated in the bulb is 33.2 x (0.062)2 = 0.127 Watt.
What is the conclusion we can draw from this, since the bulb is obviously brighter when the cap is disconnected? Perhaps: Using the "average" values in the way we have been doing is not correct in the case of pulses through a bulb; OR perhaps: Our assumption about the relationship between the bulb's brightness and the "average" power _in the pulses_ is not correct. Or both.
Or maybe my math is just totally screwed.
have eliminated the inductor entirely, along with the long leads I used to connect it, and I've simply placed a 34 Ohm resistor directly on my little breadboard between the bulb and the mosfet Drain, and the effect still happens just as before. There is less spikeyness when the cap is connected but the bulb dims while the "average" current through it goes up, just as before.
As far as the inductor goes... it is not necessary to produce the effect.
;) . Indeed TK,the inductor is not needed--just a pulsed system-as the thread title says.
A little info to digest.
To obtain the 22mV across the solar panel resistor(as shown in my last video),using a DC current,i need 186mA @ 5.1v to achieve the 22mV across the solar panel resistor. This is 948mW.
@ TK
Can you place a DMM amp meter in series with the CVR and bulb,and see if that gives you the same value your CVR is giving you in current amount.
Cheers
Brad
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Yes, certainly not easy to make out from the video. Honest mistake all sorted now.
Chris Sykes
hyiq.org
You must have a small or bad monitor EMJ,as it looks clear to me.
Hope this shows how things really are.
Brad
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You must have a small or bad monitor EMJ,as it looks clear to me.
Hope this shows how things really are.
Brad
Hahaha 1080P compared to 240??? Max in the youtube settings is 240P
Maybe you just have better eyes than me? :D
Chris Sykes
hyiq.org
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OK, are we all done chasing the Red Herrings now?
Here's the solution, taken from my present readings of my circuit without inductor. (The calculations take into account the temperature coefficient of resistance since the bulb's actual current and Vdrop are being used to compute its resistance.)
With capacitor connected (bulb dimmer):
The input current from the CH2 CVR is 0.082 A. The Vdrop across the bulb is 2.8 Volts. This is constant wrt time.
So the bulb's resistance is 2.8/0.082 = 34.1 ohms. And the power being dissipated by the bulb is I2R = (0.082)2 x 34.1 = 0.224 Watt.
With capacitor disconnected (bulb brighter):
The input current _during the pulse_ is 0.163 A, and the Vdrop across the bulb _during the pulse_ is 6.32 V. So the resistance of the bulb is 6.32/0.163 = 38.8 ohms.
And the power dissipated _during the pulse_ is therefore I2R = (0.163)2 x 38.8 = 1.03 Watts. BUT the _DUTY CYCLE_ of the pulse is only 31 percent, or 0.31.
Therefore the true _average power_ being dissipated by the bulb is 1.03 W x 0.31 = 0.319 Watt. The thermal lag of the filament smooths out the peak power and produces a nearly constant brightness over the entire time.
Since 0.319 Watt is more than 0.224 Watt, the bulb appears brighter when the capacitor is disconnected.
Please check my math and assumptions, and insert your own values for input current, Vdrop across the bulb, and duty cycle of the pulse train.
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TK, Ok I will take your word for it but certainly looks deceiving. Thinking about this, it is correct and Brads Mean Value is right on Ch1.
Still, this still means we only have a Potential Voltage Difference of 3.374 Volts as opposed to the 8 Volts with the Cap disconnected. Being that Power is V x I when in phase, we can see a clear difference in total power through the Components in question.
8 x 0.096 = 0.768
3.374 x 0.144 = 0.485856
Chris Sykes
hyiq.org
Why do you not see the problem in your own calculations Chris?
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@TinMan:
Putting a DMM as ammeter inline with the CVR input is problematic as the burden of inserting the meter in series can vary with range, meter brand, etc. A better way is to use the DMM as a voltmeter and measure across the CVR just as one does with the scope and convert to current by Ohm's Law. And of course in the case of the pulse train, it will give the same "average" value as the scope indicates, pretty much. But as I've shown, the "average" value of the current and the "average" Vdrop across the bulb are not the correct values to use in the computation of power dissipated by the bulb. They are Red Herrings of the reddest, fishiest kind.
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OK, are we all done chasing the Red Herrings now?
Here's the solution, taken from my present readings of my circuit without inductor. (The calculations take into account the temperature coefficient of resistance since the bulb's actual current and Vdrop are being used to compute its resistance.)
With capacitor connected (bulb dimmer):
The input current from the CH2 CVR is 0.082 A. The Vdrop across the bulb is 2.8 Volts. This is constant wrt time.
So the bulb's resistance is 2.8/0.082 = 34.1 ohms. And the power being dissipated by the bulb is I2R = (0.082)2 x 34.1 = 0.224 Watt.
With capacitor disconnected (bulb brighter):
The input current _during the pulse_ is 0.163 A, and the Vdrop across the bulb _during the pulse_ is 6.32 V. So the resistance of the bulb is 6.32/0.163 = 38.8 ohms.
And the power dissipated _during the pulse_ is therefore I2R = (0.163)2 x 38.8 = 1.03 Watts. BUT the _DUTY CYCLE_ of the pulse is only 31 percent, or 0.31.
Therefore the true _average power_ being dissipated by the bulb is 1.03 W x 0.31 = 0.319 Watt. The thermal lag of the filament smooths out the peak power and produces a nearly constant brightness over the entire time.
Since 0.319 Watt is more than 0.224 Watt, the bulb appears brighter when the capacitor is disconnected.
Please check my math and assumptions, and insert your own values for input current, Vdrop across the bulb, and duty cycle of the pulse train.
TK
Dose your scope have math function?,if so,what value dose that give you as in power value in each scenario ?.
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@TinMan:
Putting a DMM as ammeter inline with the CVR input is problematic as the burden of inserting the meter in series can vary with range, meter brand, etc. A better way is to use the DMM as a voltmeter and measure across the CVR just as one does with the scope and convert to current by Ohm's Law. And of course in the case of the pulse train, it will give the same "average" value as the scope indicates, pretty much. But as I've shown, the "average" value of the current and the "average" Vdrop across the bulb are not the correct values to use in the computation of power dissipated by the bulb. They are Red Herrings of the reddest, fishiest kind.
How dose a scope calculate power using the math trace?.
A x B ?
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Some people always try to make things too complicated. Lol
1. Feed the pulse train into a battery of appropriate size & voltage.
2. Continuously monitor the battery voltage.
3. Keep adding more Loads (maybe light bulbs) to the battery until the battery voltage stays the same.
4. Measure the current to the Loads.
5. Now do a simple calculation of Volts X Amps to arrive at the Wattage that your pulse train is putting out.
6. Really, Really, Really simple folks.
.
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Some people always try to make things too complicated. Lol
1. Feed the pulse train into a battery of appropriate size & voltage.
2. Continuously monitor the battery voltage.
3. Keep adding more Loads (maybe light bulbs) to the battery until the battery voltage stays the same.
4. Measure the current to the Loads.
5. Now do a simple calculation of Volts X Amps to arrive at the Wattage that your pulse train is putting out.
6. Really, Really, Really simple folks.
.
You seem to be missing the entire point of this exercise. Please watch TinMan's first and second videos and my own video on the subject.
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How dose a scope calculate power using the math trace?.
A x B ?
Yes, my scope has Math capability. I think yours does too, doesn't it?
The scope only does what it is told, though. You can multiply two traces, but for the result to be "instantaneous power" the two input traces have to be correct for the task.
OK, by Ohm's Law and going through my calculations above, if you multiply CH1 x CH2 , and CH1 is the voltage drop across the bulb, and CH2 is the current going to the bulb, you do get the Power dissipated in the bulb. This can be done in the Math trace. Then the scope's Measurements function can be used to find the Average value of this Math trace (which accounts for the Duty Cycle.)
So I set up the apparatus and the scope to do this. The results are shown below. You can see that the values are in rough agreement with what I wound up with in my manual calculations above. The differences are due to noise, spikes, offset and rounding errors, and the fact that I probably have a partial cycle on screen at the edge. But we are still very close to what we get with the manual calculation.
I think that what was causing the confusion in the first place was the use of the wrong "average" or "mean" values for the power computation. The average needs to be taken after the multiplication, rather than taking the averages first and then multiplying them.
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OK, are we all done chasing the Red Herrings now?
Here's the solution, taken from my present readings of my circuit without inductor. (The calculations take into account the temperature coefficient of resistance since the bulb's actual current and Vdrop are being used to compute its resistance.)
With capacitor connected (bulb dimmer):
The input current from the CH2 CVR is 0.082 A. The Vdrop across the bulb is 2.8 Volts. This is constant wrt time.
So the bulb's resistance is 2.8/0.082 = 34.1 ohms. And the power being dissipated by the bulb is I2R = (0.082)2 x 34.1 = 0.224 Watt.
With capacitor disconnected (bulb brighter):
The input current _during the pulse_ is 0.163 A, and the Vdrop across the bulb _during the pulse_ is 6.32 V. So the resistance of the bulb is 6.32/0.163 = 38.8 ohms.
And the power dissipated _during the pulse_ is therefore I2R = (0.163)2 x 38.8 = 1.03 Watts. BUT the _DUTY CYCLE_ of the pulse is only 31 percent, or 0.31.
Therefore the true _average power_ being dissipated by the bulb is 1.03 W x 0.31 = 0.319 Watt. The thermal lag of the filament smooths out the peak power and produces a nearly constant brightness over the entire time.
Since 0.319 Watt is more than 0.224 Watt, the bulb appears brighter when the capacitor is disconnected.
Please check my math and assumptions, and insert your own values for input current, Vdrop across the bulb, and duty cycle of the pulse train.
considering you have it set up any chance of a video for the learning?
can i ask how is the duty cycle set in the setup
you are saying the input current is higher when the bulb is brighter you are saying the lower reading was/is due to the averaging measurement TM was taking and the PWM?
edit - ahh i posted this at similar time you posted above so you are saying was due to the average.
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considering you have it set up any chance of a video for the learning?
can i ask how is the duty cycle set in the setup
you are saying the input current is higher when the bulb is brighter you are saying the lower reading was/is due to the averaging measurement TM was taking and the PWM?
edit - ahh i posted this at similar time you posted above so you are saying was due to the average.
I'll probably make a video later this evening. I've been up all night and need to get some sleep now.
The duty cycle (pulse width) and frequency in my No-Bedini-Atoll circuit are set by the Function Generator driving the Gate of the mosfet. In TinMan's original Bedini SSG circuit the frequency and duty cycle are determined by the speed of the rotor and the spacing of the coil-rotor combo and the Base resistance and the characteristics of the switching transistor.
In the first few discussions the "mean" or "average" values of the current were being compared between the Cap and NoCap conditions. (My scope calls it "average" and TinMan's scope calls it "mean" but they both refer to the same thing, and "mean" is actually the more correct term.) Since the Mean current was lower when the bulb was brighter, this seemed strange, but when the Power dissipated in the bulb is actually computed, either by hand using the _correct_ input values (not the averages) or directly by the scope's Math function, the mystery is resolved and it can be seen that the lesser average (or mean) dissipated Power does produce the dimmer bulb.
Now the question becomes how does this happen electrically in the circuit? Why does connecting the capacitor produce less power dissipated by the bulb, and why does this effect depend sensitively on the resistance between the bulb and the switching element?
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Just popping this one in here.
Magnets increasing the efficiency of a pulsed system.
Can you argue with this one?.
https://www.youtube.com/watch?v=tVNABy8fSlI
Brad
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I'll probably make a video later this evening. I've been up all night and need to get some sleep now.
The duty cycle (pulse width) and frequency in my No-Bedini-Atoll circuit are set by the Function Generator driving the Gate of the mosfet. In TinMan's original Bedini SSG circuit the frequency and duty cycle are determined by the speed of the rotor and the spacing of the coil-rotor combo and the Base resistance and the characteristics of the switching transistor.
In the first few discussions the "mean" or "average" values of the current were being compared between the Cap and NoCap conditions. (My scope calls it "average" and TinMan's scope calls it "mean" but they both refer to the same thing, and "mean" is actually the more correct term.) Since the Mean current was lower when the bulb was brighter, this seemed strange, but when the Power dissipated in the bulb is actually computed, either by hand using the _correct_ input values (not the averages) or directly by the scope's Math function, the mystery is resolved and it can be seen that the lesser average (or mean) dissipated Power does produce the dimmer bulb.
Now the question becomes how does this happen electrically in the circuit? Why does connecting the capacitor produce less power dissipated by the bulb, and why does this effect depend sensitively on the resistance between the bulb and the switching element?
ah thank you, yes ive learned here also, as you posted the shots v is optional, but thanks for taking the time originally.
my guess to the last q would be a guess at this stage in saying it is in the nature of the waveform posted when the cap is connected the smoothed pulse and how the filament reacts.
Tm has highlighted a curious effect.
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Just popping this one in here.
Magnets increasing the efficiency of a pulsed system.
Can you argue with this one?.
https://www.youtube.com/watch?v=tVNABy8fSlI
Brad
i agree with this totally, because it is my belief that 'magnets' impart energy.
because 'magnetic' is just 'gravity at frequency' which can then be resolved into fields.
so you can say everything is magnetic or nothing is magnetic and everything is gravity.
the effect is always seen when there is inerta through a field i.e your mag passing the coil.
(or any variant thereof) sys at resonance are no different its just the inertia gets naturally multiplied.
this is why combinations of inductors (electro mag) and perm mags will always work as long as inertia is present (resonate shake, collapse or EMF or all)
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Here is the pulse motor circuit from video 1 and 2,showing scope probe placement as well.
Why for the input power measurement to the entire circuit, the scope probes are not placed like this?:
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Hi TK
Thank's for your input
In your sentences hereunder, can you please elaborate some more why we have to multiply first and average after ? Are the average voltage and current not correct on the scope?
"I think that what was causing the confusion in the first place was the use of the wrong "average" or "mean" values for the power computation. The average needs to be taken after the multiplication, rather than taking the averages first and then multiplying them."
Thank's
Laurent
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In your sentences hereunder, can you please elaborate some more why we have to multiply first and average after ?
Because mathematically: AVERAGE(i1, i2) * AVERAGE(v1, v2) <> AVERAGE(i1*v1, i2*v2).
Take a calculator or a spreadsheet and see for yourself.
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Why do you not see the problem in your own calculations Chris?
@Tinman - My calculations are in agreeance with what you are seeing in the circuit. Are they not?
I think you will find, if you read TK's post, he is saying the same thing (More power more Light, less power less light), his math is likely more precise than mine:
OK, are we all done chasing the Red Herrings now?
Here's the solution, taken from my present readings of my circuit without inductor. (The calculations take into account the temperature coefficient of resistance since the bulb's actual current and Vdrop are being used to compute its resistance.)
With capacitor connected (bulb dimmer):
The input current from the CH2 CVR is 0.082 A. The Vdrop across the bulb is 2.8 Volts. This is constant wrt time.
So the bulb's resistance is 2.8/0.082 = 34.1 ohms. And the power being dissipated by the bulb is I2R = (0.082)2 x 34.1 = 0.224 Watt.
With capacitor disconnected (bulb brighter):
The input current _during the pulse_ is 0.163 A, and the Vdrop across the bulb _during the pulse_ is 6.32 V. So the resistance of the bulb is 6.32/0.163 = 38.8 ohms.
And the power dissipated _during the pulse_ is therefore I2R = (0.163)2 x 38.8 = 1.03 Watts. BUT the _DUTY CYCLE_ of the pulse is only 31 percent, or 0.31.
Therefore the true _average power_ being dissipated by the bulb is 1.03 W x 0.31 = 0.319 Watt. The thermal lag of the filament smooths out the peak power and produces a nearly constant brightness over the entire time.
Since 0.319 Watt is more than 0.224 Watt, the bulb appears brighter when the capacitor is disconnected.
Please check my math and assumptions, and insert your own values for input current, Vdrop across the bulb, and duty cycle of the pulse train.
Less power is applied in total across components 1 and 2 when the Cap is connected compared to when the cap is disconnected. The Current goes up but the voltage goes way down. Less power Less light on the globe.
There is nothing in this circuit that is out of the Norm. Look at the Tesla Switch... What's the Potential Voltage Difference between the Load?
Chris Sykes
hyiq.org
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OK, by Ohm's Law and going through my calculations above, if you multiply CH1 x CH2 , and CH1 is the voltage drop across the bulb, and CH2 is the current going to the bulb, you do get the Power dissipated in the bulb.
...but not the power delivered to the rest of the circuit through that bulb.
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Because mathematically: AVERAGE(i1, i2) * AVERAGE(v1, v2) <> AVERAGE(i1*v1, i2*v2).
Take a calculator or a spreadsheet and see for yourself.
Youp problem here, excuse my ignorance
If on the same pulsed circuit, my scope shows for example : on Chanel 1, an average (mean) voltage of 10 volts and on channel 2, an average (mean) current of 2 A, can i simply assume that the average (mean) power is the multiplication of these 2 average (mean) values, and in this case, the MEAN POWER of the pulse is 10 x 2 = 20 watts ? yes or not ?
Laurent
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Just popping this one in here.
Magnets increasing the efficiency of a pulsed system.
Can you argue with this one?.
https://www.youtube.com/watch?v=tVNABy8fSlI
Brad
@Brad and All
EXCELLENT Demonstration!!!
EVERYONE should watch this and pay very close attention!
Faradays Law of Induction clearly Assists the action in this System!!!
Chris Sykes
hyiq.org
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If on the same pulsed circuit, my scope shows for example : on Chanel 1, an average (mean) voltage of 10 volts and on channel 2, an average (mean) current of 2 A, can i simply assume that the average (mean) power is the multiplication of these 2 average (mean) values, and in this case, the MEAN POWER of the pulse is 10 x 2 = 20 watts ? yes or not ?
Not.
You can assume that only for pure DC.
P.S.
Do you have MS-Excel?
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@Brad and All
EXCELLENT Demonstration!!!
EVERYONE should watch this and pay very close attention!
Faradays Law of Induction clearly Assists the action in this System!!!
Chris Sykes
hyiq.org
Normal Faradays Law of Induction principals apply, for Improvements:
Increasing the Cross Sectional Area of the Coils and Magnets
Increasing the Turns, but not too much...
Increasing the Time Rate of Change (The Frequency)
Back in the day, the Bedini h was pointed out many times... I think many did not see the link?
Any Magnetic Field changing in time is a potential source for Faradays Law of Induction!!!
Chris Sykes
hyiq.org
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Not.
You can assume that only for pure DC.
P.S.
Do you have MS-Excel?
Thank's Verpies
very clear answer
So if i understand you well, i can not use my scope to measure average pulsed power input.
I think the title of this thread is even much harder than Tinman suggested.
1-So why are the scope constructors offers the possibility to get the average voltage of pulsed power if there are not usable ?
2-So have you a good and "simple " way to measure those pulsed average input power with some accuracy ?
finally yes i have MS excel
Laurent
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Because mathematically: AVERAGE(i1, i2) * AVERAGE(v1, v2) <> AVERAGE(i1*v1, i2*v2).
Take a calculator or a spreadsheet and see for yourself.
Thanks Verpies and TK I did not know this! It makes sense! The more averaging over time the more precise the figure would be.
I am happy to put a small program together with all of these little tips and tricks in it if you guys are happy to scrutinise it and test it? This way it can help everyone with their work get an accurate result!
Are you guys keen to participate in this?
Chris Sykes
hyiq.org
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So if i understand you well, i can not use my scope to measure average pulsed power input.
Yes, you can ...by multiplying two traces just like TinselKoala did in these scopeshots (http://overunity.com/16250/accurate-measurements-on-pulsed-systems-harder-than-you-think/msg468209/#msg468209).
1-So why are the scope constructors offers the possibility to get the average voltage of pulsed power if there are not usable ?
Because average voltage still might be useful for other purposes besides power calculation.
2-So have you a good and "simple " way to measure those pulsed average input power with some accuracy ?
If the current is not in phase with voltage then there is not a simple way.
If the current is in phase with voltage then the i&v averages can be multiplied with some scaling.
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Are you guys keen to participate in this?
Will that program compile to machine code like e.g.: C, C++, Pascal, Object Pascal, etc... ?
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Will that program compile to machine code like e.g.: C, C++, Pascal, Object Pascal, etc... ?
Hey Verpies, Yes. I normally use C#, complier is just .net to machine code.
I think if enough work and the right info was gathered and incorporated then this could be very helpful for all out there.
It will, however, not stop the inaccurate input values being supplied.
Chris Sykes
hyiq.org
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Because mathematically: AVERAGE(i1, i2) * AVERAGE(v1, v2) <> AVERAGE(i1*v1, i2*v2).
Take a calculator or a spreadsheet and see for yourself.
For others out there, I have put together a very basic example of what Verpies is explaining.
Verpies, please correct me if I have it wrong!!!
double Answer = Mean(Voltage1, Voltage2) * Mean(Current1, Current2);
double IAnswer = Mean(Voltage1 * Current1, Voltage2 * Current2);
AnswerTextBox.Text = string.Format("{0:0.000} Watts", Answer);
IAnswerTextBox.Text = string.Format("{0:0.000} Watts", IAnswer);
}
private double Mean(double Value1, double Value2)
{
double mean = ((Value1 + Value2) / 2);
return mean;
}
Chris Sykes
hyiq.org
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@Tinman - My calculations are in agreeance with what you are seeing in the circuit. Are they not?
I think you will find, if you read TK's post, he is saying the same thing (More power more Light, less power less light), his math is likely more precise than mine:
Less power is applied in total across components 1 and 2 when the Cap is connected compared to when the cap is disconnected. The Current goes up but the voltage goes way down. Less power Less light on the globe.
There is nothing in this circuit that is out of the Norm. Look at the Tesla Switch... What's the Potential Voltage Difference between the Load?
Chris Sykes
hyiq.org
X x 0.096mA = 0.768mW
0.144mA =Y x 0.485mW
;)
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Just popping this one in here.
Magnets increasing the efficiency of a pulsed system.
Can you argue with this one?.
https://www.youtube.com/watch?v=tVNABy8fSlI
Brad
I am sorry you see what I have done as arguing Brad. I don't. I see it as solving an issue you brought up with sound logic and proof.
TinselKoala verified what I was showing with more evidence.
Facts remain: With C2 disconnected more Power through the Globe (Brighter). With C2 connected less Power through the Globe (Dimmer).
Chris Sykes
hyiq.org
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;)
What is X and what is Y...
This is a bit on the cryptic side...
Chris Sykes
hyiq.org
P.S: Feel free to prove me wrong if you like! Your scope shots tell the story. The Applied Voltage Potential across the components is clearly visible. Also please explain what the differences are in the circuit so we can more precisely see what it is that you said I have wrong.
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What is X and what is Y...
This is a bit on the cryptic side...
Chris Sykes
hyiq.org
P.S: Feel free to prove me wrong if you like! Your scope shots tell the story. The Applied Voltage Potential across the components is clearly visible. Also please explain what the differences are in the circuit so we can more precisely see what it is that you said I have wrong.
X/Y is volt's. It dose not matter what that voltage value is,as we are dealing with an incandescent bulb-a resistor that changes value with an increase in temperature.
Once again,how do you increase the temperature of the bulb?. Like i said,the bulb is doing the opposite to what it should be doing. How do you increase the dissipated power of a resistor ?.
How many times have you seen people using small/low value CVR's to measure high powered system's,and assume that the resistor is ok,as the total average power is below the rated value of that resistor.
Now,put together the outcome and dissipated power of the bulb with my last video(the video showing the experiment with and without the rotor full of magnets.
Brad.
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X/Y is volt's. It dose not matter what that voltage value is,as we are dealing with an incandescent bulb-a resistor that changes value with an increase in temperature.
Once again,how do you increase the temperature of the bulb?. Like i said,the bulb is doing the opposite to what it should be doing. How do you increase the dissipated power of a resistor ?.
How many times have you seen people using small/low value CVR's to measure high powered system's,and assume that the resistor is ok,as the total average power is below the rated value of that resistor.
Now,put together the outcome and dissipated power of the bulb with my last video(the video showing the experiment with and without the rotor full of magnets.
Brad.
@Brad - I completely disagree with this statement!
An Applied Voltage Across a 1 Ohm Resistor of 1 Volt gives 1 Ampere of Current through the Resistor. (1 Watt of Power)
An Applied Voltage Across a 2 Ohm Resistor of 1 Volt gives 0.5 Ampere of Current through the Resistor. (0.5 Watts of Power)
An Applied Voltage Across a 100 Ohm Resistor of 1 Volt gives 0.01 Ampere of Current through the Resistor. (0.01 Watts of Power)
Power = Volts x Amps if Phase Angle = 0. Voltage is a critical part of the Equation!
The Voltage Applied across a Circuit Element is critical and has to be taken into account!!! Look at a Voltage Divider for example! By changing the Circuit Impedance after any component will change the Power that flows through the component. It cant be assumed that it will not, it doesn't work that way!
Introducing the Capacitor did the same thing. Circuit Impedance changed! Voltage across the components in question went down but the Current went up. Total Power through these Components went down! (Thus the Dimmer Bulb) Not Up!
Chris Sykes
hyiq.org
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Re: Heating and Current Shunts, they are relative.
Rated for maximum accuracy at the least Temperature drift.
See Images:
Chris Sykes
hyiq.org
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@Brad - I completely disagree with this statement!
An Applied Voltage Across a 1 Ohm Resistor of 1 Volt gives 1 Ampere of Current through the Resistor. (1 Watt of Power)
An Applied Voltage Across a 2 Ohm Resistor of 1 Volt gives 0.5 Ampere of Current through the Resistor. (0.5 Watts of Power)
An Applied Voltage Across a 100 Ohm Resistor of 1 Volt gives 0.01 Ampere of Current through the Resistor. (0.01 Watts of Power)
The Voltage Applied across a Circuit Element is critical and has to be taken into account!!! Look at a Voltage Divider for example! By changing the Circuit Impedance after any component will change the Power that flows through the component. It cant be assumed that it will not, it doesn't work that way!
Power is Volts x Amps if there is no Phase Angle. Voltage is a critical part of the Equation!
Introducing the Capacitor did the same thing. Circuit Impedance changed! Voltage across the components in question went down but the Current went up. Total Power through these Components went down! (Thus the Dimmer Bulb) Not Up!
Chris Sykes
hyiq.org
Can you raise or lower the voltage across that resistor without raising the current flowing through it?.
Can you increase the temperature of the bulb element without increasing the total current flowing through it? Current is what increases the element temperature,and an increase of temperature means an increase in resistance,but the increase in current must happen before the increase of resistance can exist. We decreased the average current flowing through the bulb,but increased the temperature-thus resulting in an increase of resistance-->but to what extent. Look at the scope shot's again,dose the current trace not tell you that the resistance is rather constant during the on time?-->in fact,is it not opposite to the shape of the trace curve that should exist if the resistance was increasing during that current pulse?. Draw a current trace of a pulse of current that has an increasing resistance during that pulse,and then compare that with the current trace in the scope shot. And remember,we are at low frequencies here.
Brad
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EMJ
Look at the scope shots below.
First is the current trace of a pulsed incandescent bulb
Second is the current trace of the same as above,but with inductor in series--in case anyone thought that the inductor would skew the trace in any way.
Third is a screen shot of the current trace we have in our DUT
See a problem yet?
Brad
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Can you raise or lower the voltage across that resistor without raising the current flowing through it?.
Can you increase the temperature of the bulb element without increasing the total current flowing through it? Current is what increases the element temperature,and an increase of temperature means an increase in resistance,but the increase in current must happen before the increase of resistance can exist. We decreased the average current flowing through the bulb,but increased the temperature-thus resulting in an increase of resistance-->but to what extent. Look at the scope shot's again,dose the current trace not tell you that the resistance is rather constant during the on time?-->in fact,is it not opposite to the shape of the trace curve that should exist if the resistance was increasing during that current pulse?. Draw a current trace of a pulse of current that has an increasing resistance during that pulse,and then compare that with the current trace in the scope shot. And remember,we are at low frequencies here.
Brad
I think we are going around in circles.
Chris Sykes
hyiq.org
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I think we are going around in circles.
Chris Sykes
hyiq.org
I am trying to show you that things are opposite to what they should be.
Look at the scope shots i posted in my last post. The first two show what the current trace should look like across the CVR. But in the scope shot from the DUT,the current wave form is opposite to what it should be. This is showing the bulbs resistance is decreasing during the pulse--not increasing as it should be. As i have been trying to say,the incandescent bulb is showing/doing opposite to what it should be. Measurements being taken from something that dose not act as it should can only lead to measurement error.
Now you have to figure out why the current trace is opposite to what it should be-->then we can start talking about correct measurements.
Brad
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EMJ
Look at the scope shots below.
First is the current trace of a pulsed incandescent bulb
Second is the current trace of the same as above,but with inductor in series--in case anyone thought that the inductor would skew the trace in any way.
Third is a screen shot of the current trace we have in our DUT
See a problem yet?
Brad
@Brad, thanks for your time posting scope shots.
I see, under different Circuit conditions we see different wave forms. But nothing I see is alarming.
What is it that you see that is of importance?
Chris Sykes
hyiq.org
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EMJ
Look at the scope shots below.
First is the current trace of a pulsed incandescent bulb
Second is the current trace of the same as above,but with inductor in series--in case anyone thought that the inductor would skew the trace in any way.
Third is a screen shot of the current trace we have in our DUT
See a problem yet?
Brad
@Brad, if I may suggest, although this is simple, a Circuit when presenting information would be very helpful.
Is this current circuit something like this:
Chris Sykes
hyiq.org
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I am trying to show you that things are opposite to what they should be.
Look at the scope shots i posted in my last post. The first two show what the current trace should look like across the CVR. But in the scope shot from the DUT,the current wave form is opposite to what it should be. This is showing the bulbs resistance is decreasing during the pulse--not increasing as it should be. As i have been trying to say,the incandescent bulb is showing/doing opposite to what it should be. Measurements being taken from something that dose not act as it should can only lead to measurement error.
Now you have to figure out why the current trace is opposite to what it should be-->then we can start talking about correct measurements.
Brad
I entertained this thought in TK measurements.
he stated the resistance of the bulb at the pulse , and calculated that as the over all resistance, however then i thought that might have been accounted for by the duty cycle.
i.e no pulse = no resistance.
EM
you have to admit this is a strange and curious effect, why not lets learn from it if you totally understand it, granted you know more about this than i do, i'll happily tuck away my human pride to learn some more here.
you have to admit when TM posted this you didn't say, 'oh that's simple it's due to the pulse duty cycle average and bulb filament'
if you had said that sentence we could all agree you totally understood it?
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You all are just confusing yourselves with more Red Herrings and misunderstandings. STOP already!
1. By multiplying the values for Current and Voltage at each instant of time, you generate an "instantaneous power curve" that represents the actual power at each instant in your sample set. Phase Angle _does not enter_ into this process! The IP curve is correct and gives the actual power at each instant, no matter the phase angle between V and I.
2. When you tell the scope to compute CH1 x CH2 it is following this process. For each timeslice in its memory it is taking a value from one channel and a value from the other channel, multiplying them, and plotting the resulting value at that timeslice on the Math trace to generate a point on the Instantaneous Power Curve. Then it moves on to the next pair of values for the next timeslice, and repeats until it is done.
3. When the scope computes the "average" or mean of a signal it is doing the following: for each timeslice in the sample set, it is adding the value to the sum of the preceding timeslice values, then is dividing the result by the number of timeslices in the sample set.
So for a constant voltage, it is doing this to find the average voltage over, say, twelve samples:
6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 = 72.0
72.0 / 12 = 6.0
and you get a valid average (mean) value of 6.0 volts across that 12 timeslices.
But for a pulse train with a duty cycle of 1/3 High and 2/3 Low (i.e. 33 percent or 0.33 High), the scope is doing this:
0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 6.0 + 6.0 + 6.0 + 6.0 = 24.0
24 / 12 = 2.0
and 2.0 is the valid average (mean) value for this pulse train of one period in 12 samples.
NOTE that this result is the SAME as what you get by taking the Peak Value (or the "average high" during the pulse ON timeslices) of 6.0 and multiplying it by the Duty Cycle of 0.33.
4. So, let's say you have a set of voltage and current values at a series of timeslices and you want to find the IP curve and the average power. Do you find the average voltage, the average current, then multiply those values to give an average Power? Or do you find the instantaneous VxI at each timeslice to generate an IP value, then average those IP values over the full sample set as in Part 3 above? Let's see what happens if we have a pulse train with ON values of V = 3.0, I = 3.0 and a Duty Cycle of 1/3 ON (33 percent, 0.33 ON) , with a sample size of 12 timeslices.
The first way:
Average voltage is 0+0+0+0+0+0+0+0+3+3+3+3= 12, and 12/12 is 1, so average voltage is 1 volt. This is correct, it is the average voltage.
Average current is 0+0+0+0+0+0+0+0+3+3+3+3= 12, and 12/12 is 1, so average current is 1 amp. This is correct, it is the average current.
Now multiply those averages together and you get 1 Watt "average". This is clearly wrong by inspection, since you can easily see that you have 9 watts during 1/3 of the timeslices and 0 Watts during 2/3 of the timeslices.
The second, correct way:
First we multiply each V and I pair to generate the IP curve values at each timeslice:
0,0,0,0,0,0,0,0,9,9,9,9 and taking the average of those we get
0+0+0+0+0+0+0+0+9+9+9+9= 36, and 36/12 = 3 Watts average.
Also we can see that a constant 9 watts during the ON time x 0.33 duty cycle = 3 Watts average over the full 12-slice (100 percent) period.
5. The error in the first method basically boils down to using the Duty Cycle of 33 percent ON twice in the calculation by computing the separate averages before the multiplication, when it should only be used once, by finding the average after multiplication.
6. The behaviour of a single bulb in a simple circuit can't be directly compared to the behaviour of the bulb in the Bedini SSG circuit because there is a lot of other stuff going on in the Bedini circuit that determines the shape of the pulse displayed on the scope. Those differences are irrelevant when computing the power, though, since the scope's timeslices are very fine indeed and can easily take the pulse shape into account when computing the IP curve values and the subsequent average power.
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EM TM TK and other not to be off topic :
but could we consult with any possible people living on this new planet that was just discovered in our solar system :
https://voat.co/v/Contact/comments/719767
: D
and just think EM was all worried about world war 3 ha ha. (jokes)
we're going to need an accurate pi measurement to get this far out ha ha
: D
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Sure, ask them. Be sure to let me know when you get an answer.
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You all are just confusing yourselves with more Red Herrings and misunderstandings. STOP already!
1. By multiplying the values for Current and Voltage at each instant of time, you generate an "instantaneous power curve" that represents the actual power at each instant in your sample set. Phase Angle _does not enter_ into this process! The IP curve is correct and gives the actual power at each instant, no matter the phase angle between V and I.
2. When you tell the scope to compute CH1 x CH2 it is following this process. For each timeslice in its memory it is taking a value from one channel and a value from the other channel, multiplying them, and plotting the resulting value at that timeslice on the Math trace to generate a point on the Instantaneous Power Curve. Then it moves on to the next pair of values for the next timeslice, and repeats until it is done.
3. When the scope computes the "average" or mean of a signal it is doing the following: for each timeslice in the sample set, it is adding the value to the sum of the preceding timeslice values, then is dividing the result by the number of timeslices in the sample set.
So for a constant voltage, it is doing this to find the average voltage over, say, twelve samples:
6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 = 72.0
72.0 / 12 = 6.0
and you get a valid average (mean) value of 6.0 volts across that 12 timeslices.
But for a pulse train with a duty cycle of 1/3 High and 2/3 Low (i.e. 33 percent or 0.33 High), the scope is doing this:
0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 6.0 + 6.0 + 6.0 + 6.0 = 24.0
24 / 12 = 2.0
and 2.0 is the valid average (mean) value for this pulse train of one period in 12 samples.
NOTE that this result is the SAME as what you get by taking the Peak Value (or the "average high" during the pulse ON timeslices) of 6.0 and multiplying it by the Duty Cycle of 0.33.
4. So, let's say you have a set of voltage and current values at a series of timeslices and you want to find the IP curve and the average power. Do you find the average voltage, the average current, then multiply those values to give an average Power? Or do you find the instantaneous VxI at each timeslice to generate an IP value, then average those IP values over the full sample set as in Part 3 above? Let's see what happens if we have a pulse train with ON values of V = 3.0, I = 3.0 and a Duty Cycle of 1/3 ON (33 percent, 0.33 ON) , with a sample size of 12 timeslices.
The first way:
Average voltage is 0+0+0+0+0+0+0+0+3+3+3+3= 12, and 12/12 is 1, so average voltage is 1 volt. This is correct, it is the average voltage.
Average current is 0+0+0+0+0+0+0+0+3+3+3+3= 12, and 12/12 is 1, so average current is 1 amp. This is correct, it is the average current.
Now multiply those averages together and you get 1 Watt "average". This is clearly wrong by inspection, since you can easily see that you have 9 watts during 1/3 of the timeslices and 0 Watts during 2/3 of the timeslices.
The second, correct way:
First we multiply each V and I pair to generate the IP curve values at each timeslice:
0,0,0,0,0,0,0,0,9,9,9,9 and taking the average of those we get
0+0+0+0+0+0+0+0+9+9+9+9= 36, and 36/12 = 3 Watts average.
Also we can see that a constant 9 watts during the ON time x 0.33 duty cycle = 3 Watts average over the full 12-slice (100 percent) period.
5. The error in the first method basically boils down to using the Duty Cycle of 33 percent ON twice in the calculation by computing the separate averages before the multiplication, when it should only be used once, by finding the average after multiplication.
6. The behaviour of a single bulb in a simple circuit can't be directly compared to the behaviour of the bulb in the Bedini SSG circuit because there is a lot of other stuff going on in the Bedini circuit that determines the shape of the pulse displayed on the scope. Those differences are irrelevant when computing the power, though, since the scope's timeslices are very fine indeed and can easily take the pulse shape into account when computing the IP curve values and the subsequent average power.
Brad, TK and others following - I agree with TK's over all Power Conclusion. We cant be mixing Oranges and Apples if it is just Apples we want to observe.
TK, I think it needs to be said, do you think anything unusual is going on in this circuit?
Chris Sykes
hyiq.org
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Sure, ask them. Be sure to let me know when you get an answer.
will do , they might be interested i have no idea, I'll find out for you.
hopefully you can ask yourself in the near future?
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Brad, TK and others following - I agree with TK's over all Power Conclusion. We cant be mixing Oranges and Apples if it is just Apples we want to observe.
TK, I think it needs to be said, do you think anything unusual is going on in this circuit?
Chris Sykes
hyiq.org
how does one define 'unusual' it's only based on the present set of knowledge any new knowledge is always 'unusual'
TK already said back up there what he thought was 'unusual' re the how around the Cap influence.
TM highlighted an effect where your visual reality is in appearance in opposition to the data.
that was for me at least 'curious'
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how does one define 'unusual' it's only based on the present set of knowledge any new knowledge is always 'unusual'
TK already said back up there what he thought was 'unusual' re the how around the Cap influence.
TM highlighted an effect where your visual reality is in appearance in opposition to the data.
that was for me at least 'curious'
@Digital, I am glad you disagree. Please feel free to prove me wrong.
The Data I presented supports what the circuit is doing. TK's data also supports my Data saying the same thing, Bulb is Dimmer because of Less Power, Bulb is Brighter because of more power!!!
Please feel free to show my work as wrong though and I will happily admit I am wrong.
Chris Sykes
hyiq.org
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how does one define 'unusual' it's only based on the present set of knowledge any new knowledge is always 'unusual'
TK already said back up there what he thought was 'unusual' re the how around the Cap influence.
TM highlighted an effect where your visual reality is in appearance in opposition to the data.
that was for me at least 'curious'
As shown before digitalindustry,the current trace shows the opposite to what we should see.
Both in mine and TK's scope shot's. TK's scope shot shows no curve on his current trace,indicating that the temperature of the globe remains the same during both the on time and off time
My current trace climbs throughout the whole pulse--this is showing a decrease in resistance,and thus a decrease in temperature of the bulb filament<--the opposite to what should be happening.
The last scope shot is a true wave form that should be seen across the CVR when current is passing through an incandescent bulb,but both DUT's show that is not happening-->why?.
Brad
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As shown before digitalindustry,the current trace shows the opposite to what we should see.
Both in mine and TK's scope shot's. TK's scope shot shows no curve on his current trace,indicating that the temperature of the globe remains the same during both the on time and off time
My current trace climbs throughout the whole pulse--this is showing a decrease in resistance,and thus a decrease in temperature of the bulb filament<--the opposite to what should be happening.
The last scope shot is a true wave form that should be seen across the CVR when current is passing through an incandescent bulb,but both DUT's show that is not happening-->why?.
Brad
Maybe there is some inverted/reversed Wave Data because of placement of the Probes and Grounds of the scope. I can see a resemblance: See Image - But I am only guessing with out knowing the placement of your probes in your new scope shots.
Although not as pronounced, this is expected because of the massive circuit impedance change between circuits. I am guessing though and don't know without the correct circuits being presented!
Chris Sykes
hyiq.org
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As shown before digitalindustry,the current trace shows the opposite to what we should see.
Both in mine and TK's scope shot's. TK's scope shot shows no curve on his current trace,indicating that the temperature of the globe remains the same during both the on time and off time
My current trace climbs throughout the whole pulse--this is showing a decrease in resistance,and thus a decrease in temperature of the bulb filament<--the opposite to what should be happening.
The last scope shot is a true wave form that should be seen across the CVR when current is passing through an incandescent bulb,but both DUT's show that is not happening-->why?.
Brad
ah ok, so this tells me we should start to think about what is different between your an TK system based on the current curve.
this is the path to the answer, because now we have two variables.
can you expand on what you think is occurring?
can i ask, the top scope shot (in the post above) this is with the inductor correct? (from your first video)
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@Digital, I am glad you disagree. Please feel free to prove me wrong.
The Data I presented supports what the circuit is doing. TK's data also supports my Data saying the same thing, Bulb is Dimmer because of Less Power, Bulb is Brighter because of more power!!!
Please feel free to show my work as wrong though and I will happily admit I am wrong.
Chris Sykes
hyiq.org
i'm not disagreeing i'm learning, disagreeing is an inefficient but necessary process of learning, but if i disagree i'd base it on something more than the understanding i have now on this, if you read my comment i'm not disagreeing.
but the highlighted current curve is interesting.
if as a result of understanding the difference between TK and TM setup, that brings us to a new understanding you and i both may disagree with your result right?
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ah ok, so this tells me we should start to think about what is different between your an TK system based on the current curve.
this is the path to the answer, because now we have two variables.
can you expand on what you think is occurring?
can i ask, the top scope shot (in the post above) this is with the inductor correct? (from your first video)
Yes,the top shot is with the inductor.
Have you heard of negative,or vacuum energy?.
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Yes,the top shot is with the inductor.
Have you heard of negative,or vacuum energy?.
yes i've heard it called lots of different things, 'cold electricity' etc.
thanks for the info, i will look into this. i have something also to share i will post it here, it may be relevant, i can give a little bit of a 'why' around it but i'm still learning of course.
it's just something to keep in mind because of the relationship that 'energy' systems obviously have to geometry.
some people would/might state there is no relationship however considering your revaluations with Pi the golden ratio and the rotor you are creating this all leads me to believe you do see this relationship to geometry.
after all 'Tau' or t is just the 'turns' in a circle which then represents the degrees of all the different shapes.
as elegantly stated here :
https://www.youtube.com/watch?v=jG7vhMMXagQ
i recommend everyone watch from at least 2m on
so i'll get this shape design that was brought up and i saw/learned about and explain why i think it is interesting. i.e give 'tangible' evidence.
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Hi all
Just a video to expose my problem with those "mean input pulsed power calculation"
https://youtu.be/0QWdtTItfHU
It seems to me that analogic instruments just have the ability to feel the right energy, because they don't calculate, they are feeling the right dissipated power all along the pulse and the mechanical device is able to average that power perfectly, perhaps better than the scope.
Some will say that the inertia of the mechanical device will somehow "brake " the reading, but i think that when the frequency of an event is high enough, the max amplitude should be reached at some point. But this is not the case, and i have tried from 50 Hz up to 100 Hz and the result stays the same.
Voila
Laurent
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Hi Laurent,
Just a quick notice: what if you change channel 1 with channel 2 to check how the shift you presently have on the channel 2 waveform affects the mean value of voltage amplitude? perhaps you would have a more accurate reading for the mean current on channel 1 then?
Can it be a DC shift you have now on channel 2 because of the scope channel 2 input has a DC shift problem?
Gyula
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Hi Laurent,
Just a quick notice: what if you change channel 1 with channel 2 to check how the shift you presently have on the channel 2 waveform affects the mean value of voltage amplitude? perhaps you would have a more accurate reading for the mean current on channel 1 then?
Can it be a DC shift you have now on channel 2 because of the scope channel 2 input has a DC shift problem?
Gyula
Hi Gyula
Thank's for your always very appreciated input
Yes there is a very small difference when i invert probe of channel 1 and probe of channel 2 , which suggest me that i have to invest in brand new probes for my next further experiment.
But this small difference can not explain what i am working on and this 3.85 time LESS energy calculation of the analogic meters in comparison to the "instant power "
showed in the Vcc scope shall be deeply investigated
Laurent
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Hi Laurent,
I am afraid it is not probes problem but input circuit problem of scope channel 2, the problem would stay the same with any new probe I think.
Perhaps in the scope's Service Manual if there is such somewhere on the web, the DC offset adjustment could be figured out whether it has an inner potmeter to readjust the zero line.
EDIT perhaps in the Menu there is a possibility for offset calibration for the input channels, try to go through the Manual.
This scope has a Math function too, so you can multiply channel 1 with channel 2 (of course the result you get on display may also be affected by the small offset problem of channel 2)
Gyula
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Hi all
Just a video to expose my problem with those "mean input pulsed power calculation"
https://youtu.be/0QWdtTItfHU
It seems to me that analogic instruments just have the ability to feel the right energy, because they don't calculate, they are feeling the right dissipated power all along the pulse and the mechanical device is able to average that power perfectly, perhaps better than the scope.
Some will say that the inertia of the mechanical device will somehow "brake " the reading, but i think that when the frequency of an event is high enough, the max amplitude should be reached at some point. But this is not the case, and i have tried from 50 Hz up to 100 Hz and the result stays the same.
Voila
Laurent
Great video Laurent,and shows very clearly what i mean when saying that measuring pulsed systems is not so easy.
There is much more to come yet,and great to see you are seeing such a big difference between your two measurements- in that you now have reason to look a little closer as to why the measurements are 300%+ out from each other.
Brad
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I agree that Laurent made a great clip. The one missing thing is one simple thing: You can check it by working it out on paper, you don't need to rely on any derived meter readings at all, just the basic measurements will suffice. The frequency is low so we can ignore reactive effects.
Let's say the current is when the MOSFET is ON is (12.22 volts/101 ohms) = 121 milliamps.
Let's say the voltage is (12.22 volts x 100/101) = 12.10 volts
The instantaneous power is 12.10 volts x 0.121 amps = 1.46 watts
Let's keep it simple and say the 20% duty cycle is ON for one second and OFF for four seconds. So that 1.46 Joules for every five seconds.
The average power is (1.46 Joules/5 seconds) = 0.292 watts of average power. We calculated this ourselves and know this to be true within the limits of the data we collected.
If we use the scope voltage of 1.26 volts we calculate the average power as 0.315 watts.
If Laurent scoped the voltage output from the battery and took accurate measurements of the resistances of the two resistors and did as accurate a measurement as possible of the duty cycle with his scope we could fine tune our calculation. The voltage output from the battery almost certainly takes a dip when the MOSFET switches on and that is an unknown. We don't know what the voltage drop across the MOSFET is.
The scope method using peak values and factoring in the duty cycle measured 0.337 watts.
The analog meter method measured 0.0875 watts.
The scope averaging method measured 0.0559 watts.
It looks like the scope method using peak values and factoring in the duty cycle is the "correct" value. I put correct in quotations because your calculated values and your measured values have to be much closer together than we see to get satisfaction. Why are you seeing these discrepancies? If you are serious it merits a serious follow-up investigation. How can you be sure if any of your measurements are going to be good if you don't understand exactly what is taking place in this clip?
There are two things to contemplate here that can affect what's going on:
1. This is not truly a circuit that pulses a voltage (i.e. the drive voltage for the circuit) that goes from zero volts to 12 volts. What it is is a MOSFET switch that makes a connection and then breaks a connection and goes high-impedance. There is no "zero volt" signal. Open-circuit can be way way different from zero volts, especially as you go towards higher frequencies.
2. It would be very interesting to see what you would measure with digital multimeters for the voltage and the current measurements. Do not use an in-line ammeter, just use a digital multimeter to measure the average voltage across the one-ohm resistor. Digital multimeters can be deadly accurate at making average measurements because they are integration-based measuring devices.
It's amazing how the simple measurements in this simple circuit do indeed present a challenge to the experimenter. If I was in your shoes, I would investigate this from A to Z and understand exactly what was happening. Do not forget to use your own head, you don't necessarily need to use the full measurement capabilities of your instruments. Just do the basic measurements and then do the calculations on paper. That can be used to double-check what is going on.
MileHigh
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I agree that Laurent made a great clip. The one missing thing is one simple thing: You can check it by working it out on paper, you don't need to rely on any derived meter readings at all, just the basic measurements will suffice. The frequency is low so we can ignore reactive effects.
Let's say the current is when the MOSFET is ON is (12.22 volts/101 ohms) = 121 milliamps.
Let's say the voltage is (12.22 volts x 100/101) = 12.10 volts
The instantaneous power is 12.10 volts x 0.121 amps = 1.46 watts
Let's keep it simple and say the 20% duty cycle is ON for one second and OFF for four seconds. So that 1.46 Joules for every five seconds.
The average power is (1.46 Joules/5 seconds) = 0.292 watts of average power. We calculated this ourselves and know this to be true within the limits of the data we collected.
If we use the scope voltage of 1.26 volts we calculate the average power as 0.315 watts.
If Laurent scoped the voltage output from the battery and took accurate measurements of the resistances of the two resistors and did as accurate a measurement as possible of the duty cycle with his scope we could fine tune our calculation. The voltage output from the battery almost certainly takes a dip when the MOSFET switches on and that is an unknown. We don't know what the voltage drop across the MOSFET is.
The scope method using peak values and factoring in the duty cycle measured 0.337 watts.
The analog meter method measured 0.0875 watts.
The scope averaging method measured 0.0559 watts.
It looks like the scope method using peak values and factoring in the duty cycle is the "correct" value. I put correct in quotations because your calculated values and your measured values have to be much closer together than we see to get satisfaction. Why are you seeing these discrepancies? If you are serious it merits a serious follow-up investigation. How can you be sure if any of your measurements are going to be good if you don't understand exactly what is taking place in this clip?
There are two things to contemplate here that can affect what's going on:
1. This is not truly a circuit that pulses a voltage (i.e. the drive voltage for the circuit) that goes from zero volts to 12 volts. What it is is a MOSFET switch that makes a connection and then breaks a connection and goes high-impedance. There is no "zero volt" signal. Open-circuit can be way way different from zero volts, especially as you go towards higher frequencies.
2. It would be very interesting to see what you would measure with digital multimeters for the voltage and the current measurements. Do not use an in-line ammeter, just use a digital multimeter to measure the average voltage across the one-ohm resistor. Digital multimeters can be deadly accurate at making average measurements because they are integration-based measuring devices.
It's amazing how the simple measurements in this simple circuit do indeed present a challenge to the experimenter. If I was in your shoes, I would investigate this from A to Z and understand exactly what was happening. Do not forget to use your own head, you don't necessarily need to use the full measurement capabilities of your instruments. Just do the basic measurements and then do the calculations on paper. That can be used to double-check what is going on.
MileHigh
Great Video, thanks for sharing Woopy!
I am very impressed MileHigh, very helpful, constructive post! This is the stuff that really does help people advance in their understanding to better see what it is that they are looking at.
Very nice!!!
Chris Sykes
hyiq.org
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A closer look at Woopy's test result's going on the scope shot below.
Calculating by instantaneous power measurements ,and then dividing by 5-as Woopy has a 20% duty cycle.
Version A
Vcc on channel 1 is 12.6v
Vcc on channel 2 is 134mV
CVR is 1 ohm= 134mA
12.6 x 134mA= 1.688 watts / 5 =337.6mW average
Power dissipated by CVR is-->134mV / 5=26.8mV over 1 ohm= .718mW
Version B
Calculating by mean value current.
Vcc on channel 1 is 12.6
V mean channel 2 is 20.4mV over 1 ohm= an average current of 20.4mA
12.6 x 20.4mA = 257.04mW average
Power dissipated by CVR is .416mW
Version C-analog meters.
Amp's = 35mA
Volt's= 2.5 volt's plus the average voltage drop across the CVR of 20mV = 2.52v
2.52v X 35mA = 088mW
Version D
A look at the 100 ohm load resistor.
Scope says Vcc of 12.6v.
12.6v over 100 ohms give us an instantaneous power of 1.587 watts
Divide by 5 for 20% duty cycle,we get an average power dissipation of 317mW
The average current is there for 25.2mA
Version E
Calculating using both mean value's.
2.74v X 24.04mA = 65.87mW.
Version A says our average current should be 26.8mA
Version B (Scope mean value) says average current should be 20.4mA
Analog meter says an average current of 35mA
100 ohm resistor is saying the average current should be 25.2mA
The results show by Woopy's test are as follows
Version A = 337.6mW
Version B = 257.04mW
Version C = 88mW
Version D =317mW <--CVR power dissipation not accounted for here.
Version E =65.87mW
Not one answer the same.
So which one is correct?
The only way to find out is to do a load resistor heat dissipation test with an applied DC current to the load resistor in the DUT,and then compare results against the calculated P/in of that DUT.
Brad
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You are STILL doing it! Leading yourselves "up the creek" by following your wild Red Herrings.
The meters used by Woopy are giving the approximate _average values_ of the voltage and current he is measuring. We know this to be true because 1) Poynt99 has done considerable research and testing to confirm that DMMs are good at this; and 2) The values agree within the error range with what the Scope is calculating. (Don't forget that the Analog ammeter is not even calibrated and cannot be read precisely anyhow.) Multiplying these _average_ values to get some kind of figure and calling that "average Power" in Watts is INCORRECT.
Please review my long, numerical post up above where I take you through the calculations that prove this.
Of course "Accurate Measurements on pulsed system's harder than you think." ... when you use the _incorrect_ procedures for doing the measurement, you use uncalibrated instruments and non-precision sense resistors, and you let yourself chase after Red Herrings. They are actually straightforward, if not exactly "easy", when you use proper instruments, proper procedures and proper sensors, and you have an understanding of the errors and where they come from.
1. The wirewound power resistors Woopy used have a tolerance. They are not guaranteed to be exactly 1.0 ohms or 100.0 ohms. Do they have a "J" printed on them next to the value? This means they could vary by +/- 5 percent from the printed value. Do they have a "K" printed on them? This means +/- 10 percent from the printed value. If you think your resistor is 1.0 ohms but it is actually 0.95 ohms, or even 0.90 ohms, or 1.10 ohms, how does this affect your current _reading_ ?
2. Meters, whether they are analog or digital, have an _insertion loss_ created when they are put into a series circuit as ammeters. In addition, they automatically _average_ the values of pulsed signals, as Poynt99 has shown and as anyone who has experience with moving-coil meters knows. In even more addition, the moving coil ammeter has an unknown calibration and is impossible to read to the precision necessary for this measurement. For this measurement the "average values" from the meters are useless, because as I have demonstrated, using them basically puts the duty cycle into the calculation twice when it should only be used once. (There are other power measurements, like the total input power of the circuit, where meters can be used properly, but that is not the measurement under examination here.)
3. The Duty Cycle of the pulsations is critical when you are trying to do this calculation by hand rather than letting the scope do the instantaneous multiplication and _subsequent_ averaging to get an average Power value. If the Duty Cycle is actually 22 percent instead of 20 percent you are using, what effect does this have on your "average" calculation? Can you see this much difference by eyeballing the trace on the screen?
4. Even the oscilloscope's accuracy can be questioned. Is there a baseline offset that carries through all the measurements, affecting accuracy? Have you adjusted your calculations (and the scope's calculations) to account for any offset your instrument might have? How precise is an 8-bit analog-to-digital converter when measuring voltages to show on the scope's screen?
5. When trying to analyze a circuit you should include everything in your schematic, and you shouldn't leave scope reference leads dangling. In this case Woopy's dangling reference clip is connected, through all the scope's probe cabling shields and chassis, to the same place the other probe's reference clip is connected. And the pulser circuit is also connected to both the mosfet Gate and to the circuit's negative rail (mosfet Source). At very low frequencies these facts may not matter much, but as you increase the operating frequency they will matter more and more, and sloppy technique should not be practiced or it will become habitual.
In short.... averaging before multiplying V and I does NOT yield a valid " average power " result, especially if you are using analog meters. And any result you get from any calculations, done by hand or by machine or by instrument, is subject to the errors in the input values. Know how these errors affect your results! Components have tolerances, meters have offsets, ADCs have limits in precision. Know your instruments, and know how they actually work when calculating things like instantaneous multiplication of traces, signal average values, etc.
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A closer look at Woopy's test result's going on the scope shot below.
Calculating by instantaneous power measurements ,and then dividing by 5-as Woopy has a 20% duty cycle.
Version A
Vcc on channel 1 is 12.6v
Vcc on channel 2 is 134mV
CVR is 1 ohm= 134mA
12.6 x 134mA= 1.688 watts / 5 =337.6mW average
Power dissipated by CVR is-->134mV / 5=26.8mV over 1 ohm= .718mW
Let's use the correct terms and values and see what is actually happening here.
In the first place, the duty cycle is 20.9 percent, or 0.209, not 20 percent.
The CH1 measurement is not "Vcc", it is the Vdrop across the 100 ohm (probably 5 or 10 percent tolerance) resistor.
The CH2 measurement is not "Vcc", it is the Vdrop across the 1 ohm (probably 5 or 10 percent tolerance) resistor.
Let us "assume" that the 1 Ohm resistor is accurate, though.
So we have the voltage drop across the 1 ohm CVR as 134 mV, which translates to 134 mA.
Since this current is flowing through both resistors, we know the current through the 100R resistor is also 134 mA.
But the Vdrop across that resistor is 12.6 volts. What does that tell you about the actual value of that resistor?
R=V/I, so 12.6/0.134 is 94.3 Ohms... just about almost within the 5 percent tolerance and well within the 10 percent tolerance of those _non-precision_ resistors.
But let's just let that slide for the moment.
The power dissipated by the 1 Ohm CVR during the ON time of the pulse, if the readings are accurate, is I2R = (0.134)x(0.134)x1 = 0.0180 Watt. But this is only occurring for 0.209 of the total cycle time, so the _average power_ dissipated by the CVR over the whole 1.0 time is 0.00375 Watt. This is the average power dissipated by the 1 ohm CVR, assuming it is accurately 1 ohm and the scope's voltage value is correct.
The power dissipated by the 94.3 ohm "100 ohm" load resistor, during the ON time of the pulse, is again I2R. We know the current is 0.134 A since we "trust" our one ohm CVR. So the power dissipated by the Load resistor is (0.134)x(0.134)x94.3 = 1.69 Watts. But this is only happening for 20.9 percent of the time (0.209). So the average power dissipated by the Load resistor is 1.69 x 0.209 = 0.353 Watt.
A quick "sanity check" shows that the 1 ohm CVR should be dissipating about 1/100 the power that the 100 ohm Load dissipates, since both are carrying the same current being in series. ("About" because of the tolerance ratings of the resistors. If the 1R is accurate and the scope's voltages are accurate, then the 100R is actually 94.3 Ohms.)
Version B
Calculating by mean value current.
Vcc on channel 1 is 12.6
V mean channel 2 is 20.4mV over 1 ohm= an average current of 20.4mA
12.6 x 20.4mA = 257.04mW average
Power dissipated by CVR is .416mW
The "average current" is 0.134 A x .209 = 0.028 A, not 0.0204 A.
(ETA: Here the scope seems to be making some error. If we believe the "134mA" value and the "20.9" duty cycle value, the 28 mA value is correct, even though the scope is reporting "20.4 mA" as the mean for that channel. More probably, the "1 ohm" resistor is not actually one ohm, so the "134 mA" value is probably incorrect. But my calculations assume that it is exactly one ohm.)
Why are you multiplying the Vdrop across the 100R by the "average current"? What does this value mean? The power dissipated in the 100 R resistor is calculated by I2R, or equivalently V2/R.
This calculation you've made here is totally invalid.
Version C-analog meters.
Amp's = 35mA
Volt's= 2.5 volt's plus the average voltage drop across the CVR of 20mV = 2.52v
2.52v X 35mA = 088mW
Since we "know" that it is not correct to take the average values first and then multiply them, this calculation is invalid.
Version D
A look at the 100 ohm load resistor.
Scope says Vcc of 12.6v.
12.6v over 100 ohms give us an instantaneous power of 1.587 watts
Divide by 5 for 20% duty cycle,we get an average power dissipation of 317mW
The average current is there for 25.2mA
Here you are approximately correct, for the values you are using. The power dissipated in the 100R is
V2/R which indeed results in 1.587 Watts (ignoring the fact that the resistor is actually not 100R.)
The duty cycle is not 20 percent, it is nearly 21 percent, but your 317 mW is correct for the values you are using.
Using the correct values for the resistor (94.3 ohms) and the duty cycle (0.209) we get 1.683 Watts peak
and
(12.6)2/94.3 = 1.683 peak
1.683x0.209 = 0.351 Watts average. Which is in agreement with the values we got by doing Version A _correctly_.
Version E
Calculating using both mean value's.
2.74v X 24.04mA = 65.87mW.
Again, invalid because we do not multiply the means to get an average power. We multiply the actual values to get instantaneous power, then we find the average of _that_ to get average power.
Version A says our average current should be 26.8mA
Version B (Scope mean value) says average current should be 20.4mA
Analog meter says an average current of 35mA
100 ohm resistor is saying the average current should be 25.2mA
The results show by Woopy's test are as follows
Version A = 337.6mW
Version B = 257.04mW
Version C = 88mW
Version D =317mW <--CVR power dissipation not accounted for here.
Version E =65.87mW
Not one answer the same.
So which one is correct?
The only way to find out is to do a load resistor heat dissipation test with an applied DC current to the load resistor in the DUT,and then compare results against the calculated P/in of that DUT.
Brad
You aren't doing it right !!!!!!!!
In the first place, we can completely reject the calculations that start with the average values of voltage and current. It is simply WRONG to try to get an average power value this way!
In the second place, the duty cycle is not 20 percent, it is 20.9 percent according to the scope. This is a significant error you have introduced by using the wrong value for the duty cycle.
In the third place you are going badly wrong in some of your calculations.
The values from correct calculations based on the readings, and assuming that the one-ohm resistor is precisely accurate, are these (rounded to 3 sig digits):
The dutycycle (from the scope reading) is 20.9 percent.
The "100R" resistor is actually 94.3 ohms.
Average Power dissipated in the 1 ohm CVR is 0.00375 Watt.
Average Power dissipated in the 94.3 ohm Load resistor is 0.353 Watt.
Note that the power dissipated in the load resistor is 94 times the power dissipated in the CVR.... as it should be since both are carrying the same current and the Load resistor is 94 times the resistance of the CVR.
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Thank you TK, two very detailed Posts there! This is the stuff that needs to be saved and referenced at times of taking measurements.
You are STILL doing it! Leading yourselves "up the creek" by following your wild Red Herrings.
The meters used by Woopy are giving the approximate _average values_ of the voltage and current he is measuring. We know this to be true because 1) Poynt99 has done considerable research and testing to confirm that DMMs are good at this; and 2) The values agree within the error range with what the Scope is calculating. (Don't forget that the Analog ammeter is not even calibrated and cannot be read precisely anyhow.) Multiplying these _average_ values to get some kind of figure and calling that "average Power" in Watts is INCORRECT.
Please review my long, numerical post up above where I take you through the calculations that prove this.
Of course "Accurate Measurements on pulsed system's harder than you think." ... when you use the _incorrect_ procedures for doing the measurement, you use uncalibrated instruments and non-precision sense resistors, and you let yourself chase after Red Herrings. They are actually straightforward, if not exactly "easy", when you use proper instruments, proper procedures and proper sensors, and you have an understanding of the errors and where they come from.
1. The wirewound power resistors Woopy used have a tolerance. They are not guaranteed to be exactly 1.0 ohms or 100.0 ohms. Do they have a "J" printed on them next to the value? This means they could vary by +/- 5 percent from the printed value. Do they have a "K" printed on them? This means +/- 10 percent from the printed value. If you think your resistor is 1.0 ohms but it is actually 0.95 ohms, or even 0.90 ohms, or 1.10 ohms, how does this affect your current _reading_ ?
2. Meters, whether they are analog or digital, have an _insertion loss_ created when they are put into a series circuit as ammeters. In addition, they automatically _average_ the values of pulsed signals, as Poynt99 has shown and as anyone who has experience with moving-coil meters knows. In even more addition, the moving coil ammeter has an unknown calibration and is impossible to read to the precision necessary for this measurement. For this measurement the "average values" from the meters are useless, because as I have demonstrated, using them basically puts the duty cycle into the calculation twice when it should only be used once. (There are other power measurements, like the total input power of the circuit, where meters can be used properly, but that is not the measurement under examination here.)
3. The Duty Cycle of the pulsations is critical when you are trying to do this calculation by hand rather than letting the scope do the instantaneous multiplication and _subsequent_ averaging to get an average Power value. If the Duty Cycle is actually 22 percent instead of 20 percent you are using, what effect does this have on your "average" calculation? Can you see this much difference by eyeballing the trace on the screen?
4. Even the oscilloscope's accuracy can be questioned. Is there a baseline offset that carries through all the measurements, affecting accuracy? Have you adjusted your calculations (and the scope's calculations) to account for any offset your instrument might have? How precise is an 8-bit analog-to-digital converter when measuring voltages to show on the scope's screen?
5. When trying to analyze a circuit you should include everything in your schematic, and you shouldn't leave scope reference leads dangling. In this case Woopy's dangling reference clip is connected, through all the scope's probe cabling shields and chassis, to the same place the other probe's reference clip is connected. And the pulser circuit is also connected to both the mosfet Gate and to the circuit's negative rail (mosfet Source). At very low frequencies these facts may not matter much, but as you increase the operating frequency they will matter more and more, and sloppy technique should not be practiced or it will become habitual.
In short.... averaging before multiplying V and I does NOT yield a valid " average power " result, especially if you are using analog meters. And any result you get from any calculations, done by hand or by machine or by instrument, is subject to the errors in the input values. Know how these errors affect your results! Components have tolerances, meters have offsets, ADCs have limits in precision. Know your instruments, and know how they actually work when calculating things like instantaneous multiplication of traces, signal average values, etc.
Let's use the correct terms and values and see what is actually happening here.
In the first place, the duty cycle is 20.9 percent, or 0.209, not 20 percent.
The CH1 measurement is not "Vcc", it is the Vdrop across the 100 ohm (probably 5 or 10 percent tolerance) resistor.
The CH2 measurement is not "Vcc", it is the Vdrop across the 1 ohm (probably 5 or 10 percent tolerance) resistor.
Let us "assume" that the 1 Ohm resistor is accurate, though.
So we have the voltage drop across the 1 ohm CVR as 134 mV, which translates to 134 mA.
Since this current is flowing through both resistors, we know the current through the 100R resistor is also 134 mA.
But the Vdrop across that resistor is 12.6 volts. What does that tell you about the actual value of that resistor?
R=V/I, so 12.6/0.134 is 94.3 Ohms... just about almost within the 5 percent tolerance and well within the 10 percent tolerance of those _non-precision_ resistors.
But let's just let that slide for the moment.
The power dissipated by the 1 Ohm CVR during the ON time of the pulse, if the readings are accurate, is I2R = (0.134)x(0.134)x1 = 0.0180 Watt. But this is only occurring for 0.209 of the total cycle time, so the _average power_ dissipated by the CVR over the whole 1.0 time is 0.00375 Watt. This is the average power dissipated by the 1 ohm CVR, assuming it is accurately 1 ohm and the scope's voltage value is correct.
The power dissipated by the 94.3 ohm "100 ohm" load resistor, during the ON time of the pulse, is again I2R. We know the current is 0.134 A since we "trust" our one ohm CVR. So the power dissipated by the Load resistor is (0.134)x(0.134)x94.3 = 1.69 Watts. But this is only happening for 20.9 percent of the time (0.209). So the average power dissipated by the Load resistor is 1.69 x 0.209 = 0.353 Watt.
A quick "sanity check" shows that the 1 ohm CVR should be dissipating about 1/100 the power that the 100 ohm Load dissipates, since both are carrying the same current being in series. ("About" because of the tolerance ratings of the resistors. If the 1R is accurate and the scope's voltages are accurate, then the 100R is actually 94.3 Ohms.)
The "average current" is 0.134 A x .209 = 0.028 A, not 0.0204 A.
(ETA: Here the scope seems to be making some error. If we believe the "134mA" value and the "20.9" duty cycle value, the 28 mA value is correct, even though the scope is reporting "20.4 mA" as the mean for that channel. More probably, the "1 ohm" resistor is not actually one ohm, so the "134 mA" value is probably incorrect. But my calculations assume that it is exactly one ohm.)
Why are you multiplying the Vdrop across the 100R by the "average current"? What does this value mean? The power dissipated in the 100 R resistor is calculated by I2R, or equivalently V2/R.
This calculation you've made here is totally invalid.
Since we "know" that it is not correct to take the average values first and then multiply them, this calculation is invalid.
Here you are approximately correct, for the values you are using. The power dissipated in the 100R is
V2/R which indeed results in 1.587 Watts (ignoring the fact that the resistor is actually not 100R.)
The duty cycle is not 20 percent, it is nearly 21 percent, but your 317 mW is correct for the values you are using.
Using the correct values for the resistor (94.3 ohms) and the duty cycle (0.209) we get 1.683 Watts peak
and
(12.6)2/94.3 = 1.683 peak
1.683x0.209 = 0.351 Watts average. Which is in agreement with the values we got by doing Version A _correctly_.
Again, invalid because we do not multiply the means to get an average power. We multiply the actual values to get instantaneous power, then we find the average of _that_ to get average power.
You aren't doing it right !!!!!!!!
In the first place, we can completely reject the calculations that start with the average values of voltage and current. It is simply WRONG to try to get an average power value this way!
In the second place, the duty cycle is not 20 percent, it is 20.9 percent according to the scope. This is a significant error you have introduced by using the wrong value for the duty cycle.
In the third place you are going badly wrong in some of your calculations.
The values from correct calculations based on the readings, and assuming that the one-ohm resistor is precisely accurate, are these (rounded to 3 sig digits):
The dutycycle (from the scope reading) is 20.9 percent.
The "100R" resistor is actually 94.3 ohms.
Average Power dissipated in the 1 ohm CVR is 0.00375 Watt.
Average Power dissipated in the 94.3 ohm Load resistor is 0.353 Watt.
Note that the power dissipated in the load resistor is 94 times the power dissipated in the CVR.... as it should be since both are carrying the same current and the Load resistor is 94 times the resistance of the CVR.
A 0.1 Ohm Current Sense Resistor would have a much lower Voltage drop Across the resistor and have a much lower over all affect on the circuits, if this helps? Also, why not use the CVR or Current Sense Resistor (CSR) on the Negative Rail? It might be easier to have the Scope Grounds closer to the Negative Terminal of the Power Source. Not saying its wrong, but I always think it is a best practice to minimize any possible Ground Loops.
There is always going to be some Inductance in the Current Sensing Circuit, but trying to minimize it as much as possible is also a best practice: 0.1 ohm, 5 W, ยฑ 1%, Open Element, Through Hole Current Sense Resistor (http://au.element14.com/bourns/pwr4412-2sdr1000f/resistor-metal-strip-0-1-ohm-1/dp/2328369) - Not a bad investment for $3
Again, you can use ohms law to calculate the Ampere Value flowing through the resistor: I = V/R - 0.05 / 0.1 = 0.5 Ampere's as an example.
TK Thanks again for such a detailed post pointing out the traps. Very nice to see such good info here for all!
Chris Sykes
hyiq.org
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So just for grins I went ahead and set up Woopy's circuit and measured it with my scope's Math.
The CVR I used is the same 1 Ohm 1% tolerance Ohmite non-inductive one I used for the previous testing. The 100 Ohm load I used is a 1 Watt metal-film resistor that measured 100.1 ohms on my meter. I used my regulated bench power supply for the power source, and my F43 FG to pulse the mosfet. I set the power supply to give an amplitude of 12.6 V on the Vdrop across the Load resistor, and the FG to give as close to 20 percent ON duty cycle as I could get it. "Ivcr" is the current in the one-ohm CVR, and "LOAD" is the Vdrop across the load resistor. The rest of the readings can be seen on the scopeshot below.
The scope computes the instantaneous power (IP) curve by multiplying the instantaneous values of Icvr and LOAD. It's taking 50 million samples per second, or 500,000 samples per horizontal division at 10ms/div, or 250,000 samples per division per channel, as the "instants" for this operation. _THEN_ the scope is averaging that IP curve, to account for the 20 percent ON duty cycle, and is coming up with the Average Power value displayed in the Measurements section below the traces. Note that the figures are in rough agreement with my calculations using Woopy's numbers done correctly up above. The differences are due to the resistor values, the duty cycle and the mosfet's internal resistance.
@ Woopy, TinMan... don't your scopes have Math capability, so you can do this Ch1 x Ch2 multiplication also?
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Thank you TK, two very detailed Posts there! This is the stuff that needs to be saved and referenced at times of taking measurements.
A 0.1 Ohm Current Sense Resistor would have a much lower Voltage drop Across the resistor and have a much lower over all affect on the circuits, if this helps? Also, why not use the CVR or Current Sense Resistor (CSR) on the Negative Rail? It might be easier to have the Scope Grounds closer to the Negative Terminal of the Power Source. Not saying its wrong, but I always think it is a best practice to minimize any possible Ground Loops.
There is always going to be some Inductance in the Current Sensing Circuit, but trying to minimize it as much as possible is also a best practice: 0.1 ohm, 5 W, ยฑ 1%, Open Element, Through Hole Current Sense Resistor (http://au.element14.com/bourns/pwr4412-2sdr1000f/resistor-metal-strip-0-1-ohm-1/dp/2328369) - Not a bad investment for $3
Again, you can use ohms law to calculate the Ampere Value flowing through the resistor: I = V/R - 0.05 / 0.1 = 0.5 Ampere's as an example.
TK Thanks again for such a detailed post pointing out the traps. Very nice to see such good info here for all!
Chris Sykes
hyiq.org
I have some 0.1 Ohm 1 percent CVRs of the non-inductive type, Ohmite WNBR10FET that I have been using for another project where I can't load the circuit under test very much. They cost a bit less than 3 dollars each (DigiKey lists them at $1.56 each, for example.) These will probably have even less inductance than the straight-wire shunt you've shown, as they are wound by the Aryton-Perry method and come in at under 1 nH at 1MHz. I even have one set up as a "Kelvin" type probe for insertion into a circuit with greatest accuracy. And for measurement of the total power in a circuit like this, one would normally insert it in the negative rail as you said. However that is not how TinMan first used his resistor, nor is Woopy using it that way. For the measurement of an in-circuit load element as we are doing here, there is nothing wrong with the setup we are using.
The 0.1 ohm resistors will of course produce 1/10 the voltage drop for a given current than the 1 ohm units, and for low currents you may be putting yourself in the position of having accurately to measure microvolts of drop. The 0.1 ohm resistor will not disturb the circuit under test as much as the 1.0 ohm resistor will. All of these considerations should be "considered" when choosing a CVR for any particular purpose. It definitely is easiest in most cases to use a simple 1.0 ohm resistor.
You are right to be concerned about groundloops. Most good quality Bench power supplies will not have either of their outputs connected to chassis or Mains ground unless deliberately "strapped" to do so. And many good quality Function Generators, like my F43, can be operated in either mode, fully isolated or with the "Black" BNC shield output conductor connected to chassis (and therefore Mains) ground. Of course I'm using the F43 in isolated mode. So the only other thing to worry about is to insure that both scope Probe references are connected to the same point in the circuit, as mine are.
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Another, all be it very small Loss will be Across the Mosfet itself. If the Mosfet has an RDS(ON) Resistance of 0.550 Ohms (IRF740 for example) the loss in this component is also something to be considered.
Chris Sykes
hyiq.org
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Let's use the correct terms and values and see what is actually happening here.
In the first place, the duty cycle is 20.9 percent, or 0.209, not 20 percent.
The CH1 measurement is not "Vcc", it is the Vdrop across the 100 ohm (probably 5 or 10 percent tolerance) resistor.
The CH2 measurement is not "Vcc", it is the Vdrop across the 1 ohm (probably 5 or 10 percent tolerance) resistor.
Let us "assume" that the 1 Ohm resistor is accurate, though.
So we have the voltage drop across the 1 ohm CVR as 134 mV, which translates to 134 mA.
Since this current is flowing through both resistors, we know the current through the 100R resistor is also 134 mA.
But the Vdrop across that resistor is 12.6 volts. What does that tell you about the actual value of that resistor?
R=V/I, so 12.6/0.134 is 94.3 Ohms... just about almost within the 5 percent tolerance and well within the 10 percent tolerance of those _non-precision_ resistors.
But let's just let that slide for the moment.
The power dissipated by the 1 Ohm CVR during the ON time of the pulse, if the readings are accurate, is I2R = (0.134)x(0.134)x1 = 0.0180 Watt. But this is only occurring for 0.209 of the total cycle time, so the _average power_ dissipated by the CVR over the whole 1.0 time is 0.00375 Watt. This is the average power dissipated by the 1 ohm CVR, assuming it is accurately 1 ohm and the scope's voltage value is correct.
The power dissipated by the 94.3 ohm "100 ohm" load resistor, during the ON time of the pulse, is again I2R. We know the current is 0.134 A since we "trust" our one ohm CVR. So the power dissipated by the Load resistor is (0.134)x(0.134)x94.3 = 1.69 Watts. But this is only happening for 20.9 percent of the time (0.209). So the average power dissipated by the Load resistor is 1.69 x 0.209 = 0.353 Watt.
A quick "sanity check" shows that the 1 ohm CVR should be dissipating about 1/100 the power that the 100 ohm Load dissipates, since both are carrying the same current being in series. ("About" because of the tolerance ratings of the resistors. If the 1R is accurate and the scope's voltages are accurate, then the 100R is actually 94.3 Ohms.)
The "average current" is 0.134 A x .209 = 0.028 A, not 0.0204 A.
as the mean for that channel. More probably, the "1 ohm" resistor is not actually one ohm, so the "134 mA" value is probably incorrect. But my calculations assume that it is exactly one ohm.)
Why are you multiplying the Vdrop across the 100R by the "average current"? What does this value mean? The power dissipated in the 100 R resistor is calculated by I2R, or equivalently V2/R.
This calculation you've made here is totally invalid.
Since we "know" that it is not correct to take the average values first and then multiply them, this calculation is invalid.
Here you are approximately correct, for the values you are using. The power dissipated in the 100R is
V2/R which indeed results in 1.587 Watts (ignoring the fact that the resistor is actually not 100R.)
The duty cycle is not 20 percent, it is nearly 21 percent, but your 317 mW is correct for the values you are using.
Using the correct values for the resistor (94.3 ohms) and the duty cycle (0.209) we get 1.683 Watts peak
and
(12.6)2/94.3 = 1.683 peak
1.683x0.209 = 0.351 Watts average. Which is in agreement with the values we got by doing Version A _correctly_.
Again, invalid because we do not multiply the means to get an average power. We multiply the actual values to get instantaneous power, then we find the average of _that_ to get average power.
You aren't doing it right !!!!!!!!
The values from correct calculations based on the readings, and assuming that the one-ohm resistor is precisely accurate, are these (rounded to 3 sig digits):
The dutycycle (from the scope reading) is 20.9 percent.
The "100R" resistor is actually 94.3 ohms.
Average Power dissipated in the 1 ohm CVR is 0.00375 Watt.
Average Power dissipated in the 94.3 ohm Load resistor is 0.353 Watt.
Note that the power dissipated in the load resistor is 94 times the power dissipated in the CVR.... as it should be since both are carrying the same current and the Load resistor is 94 times the resistance of the CVR.
TK
I was only posting results shown from Woopy's video the way he was showing them.
In the third place you are going badly wrong in some of your calculations.
Im not going wrong in any of !my! calculation's,as these are woopy's calculations-->not mine.
In the first place, we can completely reject the calculations that start with the average values of voltage and current. It is simply WRONG to try to get an average power value this way!
This i know. As i said,these were Woopy's calculations from the video--not mine.
(ETA: Here the scope seems to be making some error. If we believe the "134mA" value and the "20.9" duty cycle value, the 28 mA value is correct, even though the scope is reporting "20.4 mA"
Aint that a hoot.
So much for using a scope to take accurate measurements.
We'll just adjust everything,so as it all adds up.
In the second place, the duty cycle is not 20 percent, it is 20.9 percent according to the scope. This is a significant error you have introduced by using the wrong value for the duty cycle.
Once again,i was going on the duty cycle value Woopy gave us.
Please stop saying that these are my error's-->as they are not.
Brad
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Let's use the correct terms and values and see what is actually happening here.
In the first place, the duty cycle is 20.9 percent, or 0.209, not 20 percent.
The CH1 measurement is not "Vcc", it is the Vdrop across the 100 ohm (probably 5 or 10 percent tolerance) resistor.
The CH2 measurement is not "Vcc", it is the Vdrop across the 1 ohm (probably 5 or 10 percent tolerance) resistor.
Let us "assume" that the 1 Ohm resistor is accurate, though.
So we have the voltage drop across the 1 ohm CVR as 134 mV, which translates to 134 mA.
Since this current is flowing through both resistors, we know the current through the 100R resistor is also 134 mA.
But the Vdrop across that resistor is 12.6 volts. What does that tell you about the actual value of that resistor?
R=V/I, so 12.6/0.134 is 94.3 Ohms... just about almost within the 5 percent tolerance and well within the 10 percent tolerance of those _non-precision_ resistors.
But let's just let that slide for the moment.
The power dissipated by the 1 Ohm CVR during the ON time of the pulse, if the readings are accurate, is I2R = (0.134)x(0.134)x1 = 0.0180 Watt. But this is only occurring for 0.209 of the total cycle time, so the _average power_ dissipated by the CVR over the whole 1.0 time is 0.00375 Watt. This is the average power dissipated by the 1 ohm CVR, assuming it is accurately 1 ohm and the scope's voltage value is correct.
The power dissipated by the 94.3 ohm "100 ohm" load resistor, during the ON time of the pulse, is again I2R. We know the current is 0.134 A since we "trust" our one ohm CVR. So the power dissipated by the Load resistor is (0.134)x(0.134)x94.3 = 1.69 Watts. But this is only happening for 20.9 percent of the time (0.209). So the average power dissipated by the Load resistor is 1.69 x 0.209 = 0.353 Watt.
A quick "sanity check" shows that the 1 ohm CVR should be dissipating about 1/100 the power that the 100 ohm Load dissipates, since both are carrying the same current being in series. ("About" because of the tolerance ratings of the resistors. If the 1R is accurate and the scope's voltages are accurate, then the 100R is actually 94.3 Ohms.)
The "average current" is 0.134 A x .209 = 0.028 A, not 0.0204 A.
(ETA: Here the scope seems to be making some error. If we believe the "134mA" value and the "20.9" duty cycle value, the 28 mA value is correct, even though the scope is reporting "20.4 mA" as the mean for that channel. More probably, the "1 ohm" resistor is not actually one ohm, so the "134 mA" value is probably incorrect. But my calculations assume that it is exactly one ohm.)
Why are you multiplying the Vdrop across the 100R by the "average current"? What does this value mean? The power dissipated in the 100 R resistor is calculated by I2R, or equivalently V2/R.
This calculation you've made here is totally invalid.
Since we "know" that it is not correct to take the average values first and then multiply them, this calculation is invalid.
Here you are approximately correct, for the values you are using. The power dissipated in the 100R is
V2/R which indeed results in 1.587 Watts (ignoring the fact that the resistor is actually not 100R.)
The duty cycle is not 20 percent, it is nearly 21 percent, but your 317 mW is correct for the values you are using.
Using the correct values for the resistor (94.3 ohms) and the duty cycle (0.209) we get 1.683 Watts peak
and
(12.6)2/94.3 = 1.683 peak
1.683x0.209 = 0.351 Watts average. Which is in agreement with the values we got by doing Version A _correctly_.
Again, invalid because we do not multiply the means to get an average power. We multiply the actual values to get instantaneous power, then we find the average of _that_ to get average power.
You aren't doing it right !!!!!!!!
In the first place, we can completely reject the calculations that start with the average values of voltage and current. It is simply WRONG to try to get an average power value this way!
In the second place, the duty cycle is not 20 percent, it is 20.9 percent according to the scope. This is a significant error you have introduced by using the wrong value for the duty cycle.
In the third place you are going badly wrong in some of your calculations.
The values from correct calculations based on the readings, and assuming that the one-ohm resistor is precisely accurate, are these (rounded to 3 sig digits):
The dutycycle (from the scope reading) is 20.9 percent.
The "100R" resistor is actually 94.3 ohms.
Average Power dissipated in the 1 ohm CVR is 0.00375 Watt.
Average Power dissipated in the 94.3 ohm Load resistor is 0.353 Watt.
Note that the power dissipated in the load resistor is 94 times the power dissipated in the CVR.... as it should be since both are carrying the same current and the Load resistor is 94 times the resistance of the CVR.
And yet if we use your calculated average current of 28mA,then 28mA through a 1 ohm resistor would mean that the 1 ohm resistor is dissipating an average of .784mW,and the 94.3 ohm resistor would be dissipating an average 73.93mW,as the average current flowing through them !is! 28mA
So if we use the peak current value through the 94.3 ohm resistor X 20.9%,we get 134mA through 94.3 ohm's = 1.693 watts X 20.9% = 353mW. But if we use the average current through the 94.3 ohm resistor,we get 73.93mW.
If we multiply the peak voltage 12.6 by the average current of 28mA,we get very close to the 351mW you calculated. But now the resistance has to be 450 ohm's across that circuit.
Brad.
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1. This is not truly a circuit that pulses a voltage (i.e. the drive voltage for the circuit) that goes from zero volts to 12 volts. What it is is a MOSFET switch that makes a connection and then breaks a connection and goes high-impedance. There is no "zero volt" signal. Open-circuit can be way way different from zero volts, especially as you go towards higher frequencies.
MileHigh
this is a relevant point i wanted to talk about but in a different context actually.
@TK but up that garden path there are wonderful things growing ! : D - (no not just psychedelics)
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And yet if we use your calculated average current of 28mA,then 28mA through a 1 ohm resistor would mean that the 1 ohm resistor is dissipating an average of .784mW,and the 94.3 ohm resistor would be dissipating an average 73.93mW,as the average current flowing through them !is! 28mA
So if we use the peak current value through the 94.3 ohm resistor X 20.9%,we get 134mA through 94.3 ohm's = 1.693 watts X 20.9% = 353mW. But if we use the average current through the 94.3 ohm resistor,we get 73.93mW.
If we multiply the peak voltage 12.6 by the average current of 28mA,we get very close to the 351mW you calculated. But now the resistance has to be 450 ohm's across that circuit.
Brad.
Good Grief! :o
Based on the above and other posts in this thread, it's astounding that folks still don't understand power calculations and measurements, especially after all that has been explained here over the years, by myself and others.
Number one: The average current through a CVR resistor is only used for calculating the average power of the DC source. The circuit average current can not, and should not be used to calculate the power dissipated in any other circuit component, other than the DC source.
Number two: To calculate the power dissipated by the CVR (or any other series resistor), one must use either the RMS current* (which is vastly different than the average current in a 20% duty-cycle pulsed circuit) or the peak voltage squared over the resistor value times the duty-cycle. [* we can of course also use rms voltage across the resistor]
As in the attached diagram:
Psource = Vsource x Iavg (where Iavg=Vavg/Rcvr)
Pcvr = (Vrms1 x Vrms1) / Rcvr
OR = [(Vp1 x Vp1) / Rcvr] x duty cycle
PRL = (Vrms2 x Vrms2) / RL
OR = [(Vp2 x Vp2) / RL] x duty cycle
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@
Tinman, Gotoluc, Magluvin, Woopy, et al...
I would encourage all of you to print out the above post and the associated diagram and keep in in your lab for reference, preferably somewhere readily in view.
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It should be noted that the following (from the above post, but abbreviated and generalized):
PRx = Vrms2 / Rx
Holds for ANY wave form, with any duty cycle.
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It should be noted that the following (from the above post, but abbreviated and generalized):
PRx = Vrms2 / Rx
Holds for ANY wave form, with any duty cycle.
@Poynt - So you're saying that the example TK gave is wrong? It appears so.
I see you're just trying to help, but the Contempt, any need for it?
For a relatively complicated subject, where many different views are held, some maybe wrong and some may be right, I think treading with light feet, a more gentle way, might be more productive?
Brad does excellent work, always, and does the best he can. So go easy...
Chris Sykes
hyiq.org
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For a NON-Uniform, NON-Symmetrical Wave, taking the Peak Value, taking 0.707 of the Peak Value and calling this the true RMS Value just doesn't seem right to me. What is it I am missing that the scope is doing that I cant see?
I mean, we are measuring the total Area from the Scope Baseline, to Scope Trace, over time.
Chris Sykes
hyiq.org
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It should be noted that the following (from the above post, but abbreviated and generalized):
PRx = Vrms2 / Rx
Holds for ANY wave form, with any duty cycle.
...but only when the load Rx is purely resistive.
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...but only when the load Rx is purely resistive.
Yes, If I am not mistaken, RMS is supposed to be the DC Equivalent Figure given from a Sinusoidal Wave Form! So 0.707 of a Sinusoidal Wave would give the DC Value, of the Sinusoidal Wave Form, so the Area above the Wave, above the 0.707 would fill the missing Trace Area, giving a DC Value?
The wave Brad posted is far from Symmetrical.
See the Image below:
Chris Sykes
hyiq.org
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Yes, If I am not mistaken, RMS is supposed to be the DC Equivalent Figure given from a Sinusoidal Wave Form!
Not only.
The RMS calculation can also be applied to other kinds of waveforms.
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Not only.
The RMS calculation can also be applied to other kinds of waveforms.
Yes, yes, but only if they are symmetrical? I just used the Sinusoidal Wave Form as an example.
Is this right?
Chris Sykes
hyiq.org
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Its an interesting subject!!!
The equation Poynt gave us: PLoad = (Vrms x Vrms) / RLoad - I am not disagreeing with, its just that it wont hold true in all cases. Only because the RMS Equations outside this equation may not be sufficient to give correct values for NON-Symmetrical Waves! As far as I can see.
However, its a good generalized Equation, and holds true for Symmetrical Wave Forms.
As it may be the case that many out there may not understand what is actually being explained here, if I may try to elaborate some, and others please correct me if I am wrong!
The Equation Poynt99 gave us is from the Ohms Law Equation Set: P = V2 / R - See Image below. Ohms Law is your friend!!! Use it!
What Poynt is saying here, is, that the Power P, in Watts, is the product of the Voltage V2, (which is just Voltage x Voltage), taken in the form of RMS (Root Mean Square) value, divided by the Resistance R of the Load.
This equation gives you the Power P in Watts, consumed by the Load Element.
An Example:
Load Resistance R = 10 Ohms
Load Applied Voltage VRMS = 10.5 Volts
Power P = 10.52 / 10 = 110.25 / 10 = 11.025 Watts
Chris Sykes
hyiq.org
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I just got a question from a shy member in a PM:
Q: "So when should we use current RMS values to calculate power and when do we multiply the current by voltage ?"
A: When you have BOTH instantaneous current and instantaneous voltage available, then multiplying their magnitudes and averaging the resulting products is the surest universal method to calculate power.
However, sometimes you don't have BOTH voltage and current available. But in such case, all is not lost yet!
In your question you mention a case when only current measurement is available - "Is it possible to calculate power then?"
The answer is "yes", but with the following provisos:
The current must be in-phase with the voltage, or in other words the current must be proportional to the voltage.
This is tantamount to assuming a purely resistive load because current is proportional to voltage only when flowing through a pure resistor.
In such case you can use the RMS value of the current to calculate power using the formula P=IRMS2R
In another, but similar case, mentioned by Poynt99 - if you only have the voltage measurement which is applied across a pure resistance:
- you can use the RMS value of the voltage to calculate power using the formula P=VRMS2/R
You cannot use the average current (or average voltage) to calculate power because:
Sum of squares <> Square of the sum
or
Integral of squares <> Square of the integral.
An example:
Let's have a square current waveform flowing through a 1ฮฉ resistor: 0A for 1ms and 10A for 1ms.
It is obvious that the average current of such current waveform is 5A because (0A+10A)/2=5A.
What about the power dissipated in the resistor?
It is also obvious that when 0A flows through the resistor then 0W is dissipated in that resistor.
However, when 10A flows through the resistor then 100W is dissipated in that resistor, because P=i2R which calculates to P=10A2*1ฮฉ = 100W.
So we have a 0W pulse for 1ms and 100W pulse for another 1ms., or in other words: 100W half of the time. The average of that is 50W because (0W+100W)/2=50W.
If we used the average current (5A) to calculate power dissipated in that resistor, then we would obtain 25W because P=i2R and that calculates to P=5A2*1ฮฉ = 25W ...which is very wrong.
Bonus Rant:
What about the case when you have BOTH the RMS current and RMS voltage - is it possible to calculate average power from them?
Answer:
Not with an arbitrary load (non-resistive) when the shape of the waveforms and the phase offset between them are not known.
For example, if you know that the shape of the current and voltage is sinusoidal and of the same frequency and the phase shift between them is 30ยบ then you can use the formula P = iRMS * VRMS * cos(30ยบ). ...but this formula will not work with other waveform shapes.
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Bonus Rant:
What about the case when you have BOTH the RMS current and RMS voltage - is it possible to calculate average power from them?
Answer:
Not, with an arbitrary load (non-resistive) when the shape shape of the waveforms is not known and the phase offset between them.
For example, if you know that the shape of the current and voltage is sinusoidal and of the same frequency and the phase shift between them is 30ยบ then you can use the formula P = iRMS * VRMS * cos(30ยบ). ...but this formula will not work for other waveform shapes.
Fantastic Post, thanks for sharing Verpies!
Can you explain why the equation in the above quote will not work across other wave forms?
Thank You!
Chris Sykes
hyiq.org
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Can you explain why the equation in the above quote will not work across other wave forms?
The first thing that comes to mind is the correction term cos(30ยบ) which is related to the sinus shape.
For other shapes, a different correction term would have to be used.
It is important to keep in mind that the method that works for all shapes and I&V phases is the method that multiplies the instantaneous values of current and voltage and then averages the resulting products. You can't go wrong with that.
Addendum:
Err... you can go wrong with that, too ...but by making a conceptual mistake - not a mathematical one.
For example confusing the power dissipated by the light bulb in series with the power supply with the power delivered by the power supply.
It is possible for the bulb to dissipate 10W of power while transferring 1kW of power from the PS to the DUT. How you place your CSR and scope probes determines which power you measure...often inadvertently
In other words, the number of Watts will be correct but it will apply to a different power.
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The first thing that comes to mind is the correction term cos(30ยบ) which is related to the sinus shape.
For other shapes, a different correction term would have to be used.
Ah, of course!
because a Sinusoidal Wave is calculated: Asine(omega t)
The Cosine is the Covariance.
Chris Sykes
hyiq.org
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Yes, yes, but only if they are symmetrical?
In my table, the sawtooth and the pulse train were not symmetrical.
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Quote TK: You aren't doing it right !!!!!!!!
Quote Poynt: PRx = Vrms2 / Rx .Holds for ANY wave form, with any duty cycle.
Quote EMJ: @Poynt - So you're saying that the example TK gave is wrong? It appears so.
Question to Verpies-What about the case when you have BOTH the RMS current and RMS voltage - is it possible to calculate average power from them?
Quote Verpies: Not with an arbitrary load (non-resistive) when the shape of the waveforms and the phase offset between them are not known.
Question from TK--Why are you multiplying the Vdrop across the 100R by the "average current"?
I was not,and do not. Take for example the scope shot below. I use the instantaneous voltage across the resistor !not the supply voltage!,as there is no voltage across the resistor when Q1 is open,and the supply voltage is not always the voltage that is across the resistive load. The voltage to use is the actual voltage across the resistive load,and in the case of the screen shot below,that voltage is 12.2 volt's-not the open supply voltage of 12.4 volt's. If we know exactly what the value of the resistive load is,when can then use ohms law to calculate the dissipated power by that resistive load. As the value of the resistive load is exactly 51.2 ohm's,and the actual voltage is 12.2 volts,then the instantaneous dissipated power is 2.907 watts. As we also know that the exact duty cycle is 30%,then the average power dissipated by the resistive load is 872.1mW.
Then you will notice that (once again) the current from those calculations do not agree with the calculated instantaneous current the scope is showing across the CVR (blue channel).
But it's great to see you guys think it is so easy to measure the power in a pulsed system(see above quotes).
Brad.
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Quote TK: You aren't doing it right !!!!!!!!
Quote Poynt: PRx = Vrms2 / Rx .Holds for ANY wave form, with any duty cycle.
Quote EMJ: @Poynt - So you're saying that the example TK gave is wrong? It appears so.
Question to Verpies-What about the case when you have BOTH the RMS current and RMS voltage - is it possible to calculate average power from them?
Quote Verpies: Not with an arbitrary load (non-resistive) when the shape of the waveforms and the phase offset between them are not known.
Question from TK--Why are you multiplying the Vdrop across the 100R by the "average current"?
If you provide some examples of discrepancies (with the quoted numbers), I may be able to give you an answer.
Then you will notice that (once again) the current from those calculations do not agree with the calculated instantaneous current the scope is showing across the CVR (blue channel).
I don't know to what you are referring. Perhaps some actual numbers would be helpful.
But it's great to see you guys think it is so easy and to measure the power in a pulsed system(see above quotes).
For what you guys are doing here, yes it is easy and straight forward. Follow what is in that post of mine and you will be golden.
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Quote TK: You aren't doing it right !!!!!!!!
Quote Poynt: PRx = Vrms2 / Rx .Holds for ANY wave form, with any duty cycle.
Quote EMJ: @Poynt - So you're saying that the example TK gave is wrong? It appears so.
Question to Verpies-What about the case when you have BOTH the RMS current and RMS voltage - is it possible to calculate average power from them?
Quote Verpies: Not with an arbitrary load (non-resistive) when the shape of the waveforms and the phase offset between them are not known.
Question from TK--Why are you multiplying the Vdrop across the 100R by the "average current"?
I was not,and do not. Take for example the scope shot below. I use the instantaneous voltage across the resistor !not the supply voltage!,as there is no voltage across the resistor when Q1 is open,and the supply voltage is not always the voltage that is across the resistive load. The voltage to use is the actual voltage across the resistive load,and in the case of the screen shot below,that voltage is 12.2 volt's-not the open supply voltage of 12.4 volt's. If we know exactly what the value of the resistive load is,when can then use ohms law to calculate the dissipated power by that resistive load. As the value of the resistive load is exactly 51.2 ohm's,and the actual voltage is 12.2 volts,then the instantaneous dissipated power is 2.907 watts. As we also know that the exact duty cycle is 30%,then the average power dissipated by the resistive load is 872.1mW.
Then you will notice that (once again) the current from those calculations do not agree with the calculated instantaneous current the scope is showing across the CVR (blue channel).
But it's great to see you guys think it is so easy to measure the power in a pulsed system(see above quotes).
Brad.
@Brad - Certainly I do not think that "it is so easy to measure the power in a pulsed system"!!!
I never said that, and to be honest, that's why I am here, to try and learn something!
Lets be honest, we are not even at measuring the Power Input to a Pulsed Circuit!!! We are only measuring the Power through a single Component!
Although there are problems with the solution, I think Poynt has suggested the best solution currently to measure a single Component in a Pulsed DC System. It should be the most accurate over all. TK's method is good, but there is, I believe a Complexity to it that will confuse most.
In saying this, I do like TK's method. He has shown that is can be pretty accurate also.
Resistance in any Resistor that is not purely Resistive is where part of this problem lays! As Verpies has pointed out!!!
Lets take an inductor for example. There are two Resistive components, it is typically written as such: Real Resistance+jImaginary Resistance in Ohms or Real Resistance-jImaginary Resistance in Ohms
+j is inductive ...(current lagging)
-j is capacitive... (current leading)
The Imaginary Resistance is XL or XC, which is the Inductive or Capacitive Reactance. If the Inductor is Resonant, then XL and XC cancel out and the DC Resistance remains. Which is the Real Resistance.
So, the Actual Resistance is again Changing with Time in a Pulsed System. If Resistance R Changes in the Equation: P = V2/R then this presents a massive Accuracy Error!!!
I should point out, this will present itself, typically as a Non-Symmetrical or Non-Uniform Wave Form!!!
Its a pretty complex subject! Not easy!
Chris Sykes
hyiq.org
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If you provide some examples of discrepancies (with the quoted numbers), I may be able to give you an answer.
I don't know to what you are referring. Perhaps some actual numbers would be helpful.
For what you guys are doing here, yes it is easy and straight forward. Follow what is in that post of mine and you will be golden.
no offense but you are kind of coming across like a friend of mine i refer to as 'Homer Simpson'
he doesn't read anything , but then just assumes he is correct, is it because you have the word 'elite' in your user?
it's got a bit of that chess with a pigeon feel.
ha ha
: D
hey i'm sure you have ohms law covered at a higher understanding than i , i don't even know what a volt is measuring, but then the humble diode doesn't adhere to ohms law so as we seem to learn everything shifts as one learns more.
did you watch both the videos, also take into account the 'current' waveform in the original video?
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because a Sinusoidal Wave is calculated: Asine(omega t)
Yes, the sine waveform has many well known metrics
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I have highlighted some items in my analysis just in case some people think that Poynt99 is disagreeing with me, and to address one or two other issues.
Let's use the correct terms and values and see what is actually happening here.
In the first place, the duty cycle is 20.9 percent, or 0.209, not 20 percent.
The CH1 measurement is not "Vcc", it is the Vdrop across the 100 ohm (probably 5 or 10 percent tolerance) resistor.
The CH2 measurement is not "Vcc", it is the Vdrop across the 1 ohm (probably 5 or 10 percent tolerance) resistor.
Let us "assume" that the 1 Ohm resistor is accurate, though.
So we have the voltage drop across the 1 ohm CVR as 134 mV, which translates to 134 mA.
Since this current is flowing through both resistors, we know the current through the 100R resistor is also 134 mA.
But the Vdrop across that resistor is 12.6 volts. What does that tell you about the actual value of that resistor?
R=V/I, so 12.6/0.134 is 94.3 Ohms... just about almost within the 5 percent tolerance and well within the 10 percent tolerance of those _non-precision_ resistors.
But let's just let that slide for the moment.
The power dissipated by the 1 Ohm CVR during the ON time of the pulse, if the readings are accurate, is I2R = (0.134)x(0.134)x1 = 0.0180 Watt.
Note that the Ohm's Law formulations P=V2/R and P=I2R are _equivalent_, they are exactly the same algebraically. With a 1 ohm resistor this is pretty easy to see.
(0.134 V)2/1 ohm = (0.134 A)2x1 ohm
But this is only occurring for 0.209 of the total cycle time, so the _average power_ dissipated by the CVR over the whole 1.0 time is 0.00375 Watt. This is the average power dissipated by the 1 ohm CVR, assuming it is accurately 1 ohm and the scope's voltage value is correct.
The power dissipated by the 94.3 ohm "100 ohm" load resistor, during the ON time of the pulse, is again I2R. We know the current is 0.134 A since we "trust" our one ohm CVR. So the power dissipated by the Load resistor is (0.134)x(0.134)x94.3 = 1.69 Watts. But this is only happening for 20.9 percent of the time (0.209). So the average power dissipated by the Load resistor is 1.69 x 0.209 = 0.353 Watt.
Doing this same calculation using V2/R (the Vdrop across the 94.3 ohm resistor) instead of the current through it:
(12.6)(12.6)/94.3 = 1.683 Watts
multiplying by the duty cycle to 1.683x0.209= 0.3518 Watt
The tiny difference is merely due to the various roundings in the numbers going into the two calculations.
A quick "sanity check" shows that the 1 ohm CVR should be dissipating about 1/100 the power that the 100 ohm Load dissipates, since both are carrying the same current being in series. ("About" because of the tolerance ratings of the resistors. If the 1R is accurate and the scope's voltages are accurate, then the 100R is actually 94.3 Ohms.)
The "average current" is 0.134 A x .209 = 0.028 A, not 0.0204 A.
(ETA: Here the scope seems to be making some error. If we believe the "134mA" value and the "20.9" duty cycle value, the 28 mA value is correct, even though the scope is reporting "20.4 mA" as the mean for that channel. More probably, the "1 ohm" resistor is not actually one ohm, so the "134 mA" value is probably incorrect. But my calculations assume that it is exactly one ohm.)
Here TinMan has emphasized the wrong part of my statement. Don't forget that Woopy used a _non precision_ wirewound power resistor that probably was "J" or even "K" tolerance level and did not report an actual measurement of that resistor. And once again, the "mean" current is simply not useful for our purposes here anyway.
But also, instrument readings should _not_ be taken as automatically correct simply because they are coming from a digital instrument! As I said before, it is important to know your sources of error, to understand how your instruments are calculating their values, and you should always _crosscheck_ the "numbers in boxes" for accuracy and consistency and even sanity. Here we are "assuming" two critical things: that the "1 ohm" CVR is precisely one ohm, and that the scope's measurement of 0.134 Vdrop across it is accurate. Using those two assumptions we can crosscheck the other results coming from the scope.
Why are you multiplying the Vdrop across the 100R by the "average current"? What does this value mean? The power dissipated in the 100 R resistor is calculated by I2R, or equivalently V2/R.
This calculation you've made here is totally invalid.
Since we "know" that it is not correct to take the average values first and then multiply them, this calculation is invalid.
Here you are approximately correct, for the values you are using. The power dissipated in the 100R is
V2/R which indeed results in 1.587 Watts (ignoring the fact that the resistor is actually not 100R.)
The duty cycle is not 20 percent, it is nearly 21 percent, but your 317 mW is correct for the values you are using.
Using the correct values for the resistor (94.3 ohms) and the duty cycle (0.209) we get 1.683 Watts peak
and
(12.6)2/94.3 = 1.683 peak
1.683x0.209 = 0.351 Watts average. Which is in agreement with the values we got by doing Version A _correctly_.
Again, invalid because we do not multiply the means to get an average power. We multiply the actual values to get instantaneous power, then we find the average of _that_ to get average power.
Here I've used the V2/R formulation, again.
You aren't doing it right !!!!!!!!
In the first place, we can completely reject the calculations that start with the average values of voltage and current. It is simply WRONG to try to get an average power value this way!
In the second place, the duty cycle is not 20 percent, it is 20.9 percent according to the scope. This is a significant error you have introduced by using the wrong value for the duty cycle.
In the third place you are going badly wrong in some of your calculations.
Yes, in your Version A calculation you have indeed gone badly wrong. Those are your (TinMan's) calculations based on Woopy's numbers aren't they? The result doesn't even pass the "sanity check" that tells us that the power dissipated in the 1 ohm CVR _must_ be 1/100 the power dissipated in the 100 ohm Load since both are carrying the same current, and power goes _linearly_ as resistance ( remember the two equivalent formulations P=V2/R=I2R). But as we've seen the resistor values are probably not exactly as marked on their labels so this ratio will probably be different in reality, as I've calculated.
For reference here's TinMan's "Version A" again. Note the last line:
Calculating by instantaneous power measurements ,and then dividing by 5-as Woopy has a 20% duty cycle.
Version A
Vcc on channel 1 is 12.6v
Vcc on channel 2 is 134mV
CVR is 1 ohm= 134mA
12.6 x 134mA= 1.688 watts / 5 =337.6mW average
Power dissipated by CVR is-->134mV / 5=26.8mV over 1 ohm= .718mW
Clearly, .718 mW is not anywhere near being 1/100 of 337.6 mW, so sanity check fails.
337.6/.718 = 470.2
The values from correct calculations based on the readings, and assuming that the one-ohm resistor is precisely accurate, are these (rounded to 3 sig digits):
The dutycycle (from the scope reading) is 20.9 percent.
The "100R" resistor is actually 94.3 ohms.
Average Power dissipated in the 1 ohm CVR is 0.00375 Watt. (3.75 mW)
Average Power dissipated in the 94.3 ohm Load resistor is 0.353 Watt. (353 mW)
Note that the power dissipated in the load resistor is 94 times the power dissipated in the CVR.... as it should be since both are carrying the same current and the Load resistor is 94 times the resistance of the CVR.
353/3.75 = 94.13 Sanity check passed.
Ohm's Law is V=IR, and Power = VI.
Doing some algebra,
P=VI
P=(IR)I
P=I2R
and since I=V/R,
P=VI
P=V(V/R)
P=V2/R
Therefore
P = I2R = V2/R
QED.
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Certainly I do not think that "it is so easy to measure the power in a pulsed system"!!!
But it sure helps if the probes are positioned correctly for INPUT power measurement and energy flowing into the DUT ...not for the power dissipation of a light bulb connected in series with the power supply, which can be vastly different.
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From the above it should be obvious that the OUTPUT power can be measured by a brightness of a light bulb's, acting as a load like in the Diag.5 ...but the INPUT power cannot be measured by a brightness of a light bulb connected in series between the PowerSupply and the DUT.
P.S.
Is someone going to ask now, how power measurements can be simplified when an arbitrary DUT is supplied by a constant voltage source (colloquially "by DC") ?
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Quote TK: You aren't doing it right !!!!!!!!
Quote Poynt: PRx = Vrms2 / Rx .Holds for ANY wave form, with any duty cycle.
Quote EMJ: @Poynt - So you're saying that the example TK gave is wrong? It appears so.
Question to Verpies-What about the case when you have BOTH the RMS current and RMS voltage - is it possible to calculate average power from them?
Quote Verpies: Not with an arbitrary load (non-resistive) when the shape of the waveforms and the phase offset between them are not known.
Question from TK--Why are you multiplying the Vdrop across the 100R by the "average current"?
I was not,and do not.
Well, somebody did, in your Version B.
Version B
Calculating by mean value current.
Vcc on channel 1 is 12.6
V mean channel 2 is 20.4mV over 1 ohm= an average current of 20.4mA
12.6 x 20.4mA = 257.04mW average
Power dissipated by CVR is .416mW
Take for example the scope shot below. I use the instantaneous voltage across the resistor !not the supply voltage!,as there is no voltage across the resistor when Q1 is open,and the supply voltage is not always the voltage that is across the resistive load. The voltage to use is the actual voltage across the resistive load,and in the case of the screen shot below,that voltage is 12.2 volt's-not the open supply voltage of 12.4 volt's. If we know exactly what the value of the resistive load is,when can then use ohms law to calculate the dissipated power by that resistive load. As the value of the resistive load is exactly 51.2 ohm's,and the actual voltage is 12.2 volts,then the instantaneous dissipated power is 2.907 watts. As we also know that the exact duty cycle is 30%,then the average power dissipated by the resistive load is 872.1mW.
Then you will notice that (once again) the current from those calculations do not agree with the calculated instantaneous current the scope is showing across the CVR (blue channel).
But it's great to see you guys think it is so easy to measure the power in a pulsed system(see above quotes).
Brad.
Yes, they do agree.
If you have a Vdrop of 12.2 V across a 51.2 ohm resistor, the current is I=V/R= 12.2/51.2 = 0.238 A. If your duty cycle is 30 percent, then the average current is 0.07148 or 71.5 mA. The scope is saying the CH2 mean is 72 mV. Since this is across your 1R CVR, it means 72 mA average. These values agree "perfectly".
Also, I make the "peak" value of the blue trace as slightly over 2.4 vertical divisions, and you are set to 100 mV/div, so the peak instantaneous values are slightly over 240 mA... again in "perfect" agreement with the calculations using your measured V and R for the voltage drop and resistance of your load resistor.
A difference of around one percent is pretty damn "perfect" agreement.
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From the above it should be obvious that the OUTPUT power can be measured by a brightness of a light bulb's, acting as a load like in the Diag.5 ...but the INPUT power cannot be measured by a brightness of a light bulb connected in series between the PowerSupply and the DUT.
P.S.
Is someone going to ask now, how power measurements can be simplified when an arbitrary DUT is supplied by a constant voltage source (colloquially "by DC") ?
The original issue was to account for the power dissipation of the bulb itself, as one of several load elements within a more complex circuit. It was not about the total input power to the circuit. The confusing part was that the bulb in the original circuit, and in my modification, became dimmer when the "average" current through it increased. But as we showed, when the actual power dissipation is considered rather than just the "average" or mean current, the "anomaly" goes away and the bulb's brightness follows the actual average power, as measured and correctly calculated.
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For Giggles:
The 1KHz Signal is from a High Side Mosfet Driver. The 1K Resistor is measured to be 1K.
Mean Calculation: Red Channel
V / R = I = 5 / 1000 = 0.005 Amperes
I2 * R = P = (0.0052 * 1000) * 0.5 (Duty Cycle) = (0.000025 * 1000) * 0.5 (Duty Cycle) = 0.0125 Watts
RMS Calculation: Yellow Channel
V2 / R = P = 3.4582 / 1000 = 11.957764 / 1000 = 0.011957764 Watts
Now I must have something wrong here? We see: 0.000542236 Watts difference!
Chris Sykes
hyiq.org
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But it sure helps if the probes are positioned correctly for INPUT power measurement and energy flowing into the DUT ...not for the power dissipation of a light bulb connected in series with the power supply, which can be vastly different.
Yes Sir!!! I could not agree more here!!!
Chris Sykes
hyiq.org
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For Giggles:
The 1KHz Signal is from a High Side Mosfet Driver. The 1K Resistor is measured to be 1K.
Mean Calculation: Red Channel
I = 5 / 1000 = 0.005 Amperes
P = (0.0052 * 1000) * 0.5 (Duty Cycle) = 0.0125 Watts
RMS Calculation: Yellow Channel
P = 3.4582 / 1000 = 11.957764 / 1000 = 0.011957764 Watts
Now I must have something wrong here? We see: 0.000542236 Watts difference!
Chris Sykes
hyiq.org
The Mean of the (all positive) pulse train is calculated by the value multiplied by the duty cycle. The RMS is calculated by the value multiplied by the square root of the duty cycle. Look back at the table that shows the way that the RMS values are calculated.
Now remember that "RMS Power" is not really a valid quantity, except perhaps in audio loudspeakers.
From the Wiki on "audio power":
Continuous average power ratings are a staple of performance specifications for audio amplifiers and, sometimes, loudspeakers. Continuous average power is derived from the root mean square (https://en.wikipedia.org/wiki/Root_mean_square) (RMS) of the AC voltage or current, and is often incorrectly referred to as "RMS power", "RMS watts", or "watts RMS". However, the RMS value of the power waveform is different from the average power value (e.g. 22% higher for a sine wave signal), and doesn't represent anything useful, so these terms should not be used.[2][3][4][5][6][7] The correct term is "continuous average power", which is proportional to the RMS voltage, and specified by the FTC.[8]
(emphasis mine)
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For Giggles:
The 1KHz Signal is from a High Side Mosfet Driver. The 1K Resistor is measured to be 1K.
Mean Calculation: Red Channel
I = 5 / 1000 = 0.005 Amperes
P = (0.0052 * 1000) * 0.5 (Duty Cycle) = 0.0125 Watts
RMS Calculation: Yellow Channel
P = 3.4582 / 1000 = 11.957764 / 1000 = 0.011957764 Watts
Now I must have something wrong here? We see: 0.000542236 Watts difference!
Chris Sykes
hyiq.org
In addition, do a "sanity check". What is the Mean value of a 5 volt, all positive square wave with a duty cycle of 50 percent? It is 2.5 volts, clearly, by inspection.
Since your scopeshot is showing something different from that as the "mean", you either don't have 5 volts, or you don't have 50 percent duty cycle, or both, or your scope is measuring incorrectly. Garbage in, garbage out.
To me it looks like your voltage isn't quite up to the 5 volt level.
Also, ignoring the false precision in your numbers, we are comparing 0.01196 watts with0.0125 watts. This is a difference of slightly less than 5 percentand is easily accounted for by the inaccuracies in the input values (like the use of 5 volts in the first calculation).
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The Mean of the (all positive) pulse train is calculated by the value multiplied by the duty cycle. The RMS is calculated by the value multiplied by the square root of the duty cycle. Look back at the table that shows the way that the RMS values are calculated.
Now remember that "RMS Power" is not really a valid quantity, except perhaps in audio loudspeakers.
From the Wiki on "audio power":(emphasis mine)
TK - This is Gold!
Thanks for that! So I guess now, we need to get this down in a simple procedure to follow. Step 1, step 2 and so on...
Ohms Law is solid. We need not worry about that. The Math in general is Solid and we need not worry about the Math. Its more so the Wave and the gathering of the Figures.
So,
Scope Mean is best - Always?
Use this Mean value all the time when taking Measurements on Pulsed Waveforms? Even NON-Symmetrical Waveforms?
Chris Sykes
hyiq.org
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Well, somebody did, in your Version B.
Yes, they do agree.
If you have a Vdrop of 12.2 V across a 51.2 ohm resistor, the current is I=V/R= 12.2/51.2 = 0.238 A. If your duty cycle is 30 percent, then the average current is 0.07148 or 71.5 mA. The scope is saying the CH2 mean is 72 mV. Since this is across your 1R CVR, it means 72 mA average. These values agree "perfectly".
Also, I make the "peak" value of the blue trace as slightly over 2.4 vertical divisions, and you are set to 100 mV/div, so the peak instantaneous values are slightly over 240 mA... again in "perfect" agreement with the calculations using your measured V and R for the voltage drop and resistance of your load resistor.
A difference of around one percent is pretty damn "perfect" agreement.
What?-are you saying that all my calculations agree, and are accurate lol.
I now see what your concern with my version B is. This is my typo error, as it should say the load resistor-not the CVR. I really dont know why I typed CVR in stead of the 94.3 ohm load resistor, as I was using the calculated current through the CVR to calculate the power dissipated across the load resistor.
Brad
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TK - This is Gold!
Thanks for that! So I guess now, we need to get this down in a simple procedure to follow. Step 1, step 2 and so on...
Ohms Law is solid. We need not worry about that. The Math in general is Solid and we need not worry about the Math. Its more so the Wave and the gathering of the Figures.
So,
Scope Mean is best - Always?
Use this Mean value all the time when taking Measurements on Pulsed Waveforms?
Chris Sykes
hyiq.org
Now you are oversimplifying. Sometimes the use of RMS values for measurements will be more appropriate, depending on what you are doing.
But as I've said, and as Verpies and Poynt99 have also confirmed, the proper way to do _power_ calculations, if you have the data as you do from a DSO, is to do the instantaneous multiplication of the actual per-sample values of voltage and current, which generates an instantaneous power waveform, then take the _average_ (not RMS) value of that IP waveform. This procedure works for all waveforms, whether complex or not, all load types, and all power factors. All the phase shift and other complications are taken care of by the sample-by-sample calculations performed in the scope, typically at hundreds of thousands of samples per second or more, so the errors caused by this "numerical methods integration" are very small.
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In addition, do a "sanity check". What is the Mean value of a 5 volt, all positive square wave with a duty cycle of 50 percent? It is 2.5 volts, clearly, by inspection.
Since your scopeshot is showing something different from that as the "mean", you either don't have 5 volts, or you don't have 50 percent duty cycle, or both, or your scope is measuring incorrectly. Garbage in, garbage out.
To me it looks like your voltage isn't quite up to the 5 volt level.
Also, ignoring the false precision in your numbers, we are comparing 0.01196 watts with0.0125 watts. This is a difference of slightly less than 5 percentand is easily accounted for by the inaccuracies in the input values (like the use of 5 volts in the first calculation).
I would say, I checked the Voltage with DMM and it is a bit on the low side, Scope is 2.3, DMM shows 2.3.
Chris Sykes
hyiq.org
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I would say, I checked the Voltage with DMM and it is a bit on the low side, Scope is 2.3, DMM shows 2.3.
Chris Sykes
hyiq.org
So turn up your power supply until the scope/DMM reading is 2.5 V and then re-do your power calculations using the scope's new RMS value and let's see how it comes out.
It looks like your Duty Cycle is accurate across the scope screen. I have found that horizontal measurements are usually more accurate than the vertical ones, since there is less noise and you aren't limited to the 8-bit ADC precision levels.
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Now you are oversimplifying. Sometimes the use of RMS values for measurements will be more appropriate, depending on what you are doing.
But as I've said, and as Verpies and Poynt99 have also confirmed, the proper way to do _power_ calculations, if you have the data as you do from a DSO, is to do the instantaneous multiplication of the actual per-sample values of voltage and current, which generates an instantaneous power waveform, then take the _average_ (not RMS) value of that IP waveform. This procedure works for all waveforms, whether complex or not, all load types, and all power factors. All the phase shift and other complications are taken care of by the sample-by-sample calculations performed in the scope, typically at hundreds of thousands of samples per second or more, so the errors caused by this "numerical methods integration" are very small.
Yes, I realise. Possibly a good idea to check with both equations anyway. Its not hard. Just a case of gathering the data which one will be doing anyway.
As one equation partially verifies the other.
Chris Sykes
hyiq.org
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So turn up your power supply until the scope/DMM reading is 2.5 V and then re-do your power calculations using the scope's new RMS value and let's see how it comes out.
It looks like your Duty Cycle is accurate across the scope screen. I have found that horizontal measurements are usually more accurate than the vertical ones, since there is less noise and you aren't limited to the 8-bit ADC precision levels.
Yes, this what I did in the first place, some drift slipped in there.
I should have rechecked it again before posting. Was only five minutes between experiment and taking Pics. Still shows others something else to watch for also.
Chris Sykes
hyiq.org
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For Giggles:
The 1KHz Signal is from a High Side Mosfet Driver. The 1K Resistor is measured to be 1K.
Mean Calculation: Red Channel
V / R = I = 5 / 1000 = 0.005 Amperes
I2 * R = P = (0.0052 * 1000) * 0.5 (Duty Cycle) = (0.000025 * 1000) * 0.5 (Duty Cycle) = 0.0125 Watts
RMS Calculation: Yellow Channel
V2 / R = P = 3.4582 / 1000 = 11.957764 / 1000 = 0.011957764 Watts
Now I must have something wrong here? We see: 0.000542236 Watts difference!
Chris Sykes
hyiq.org
I am still seeing a possible problem if the resistance changes in time, or am I seeing something that is not there?
In all of the Equations we use, Ohm's Law, if R changes during the operation of the circuit, then the Power through the Circuit Element in question will change.
Eliminating the problem, like TK has pointed out, can only be done with a Current Sensing Resistor with absolute minimal Inductance and using this Voltage Drop, Across the Non Inductive Current Sensing Resistor, CSR.
Chris Sykes
hyiq.org
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Actually the Ohm's Law relationships are definitions of each term in terms of the others. So, since the current depends on the resistance (if voltage is constant) then even though the resistance isn't explicitly included in the particular formula you may be using, it is still there hidden in the other variable values. So P=VxI does have the resistance in there, because I = V/R.
Here's where it becomes important to understand how the scope calculates "Average" or mean values. It is adding up values from very short timeslices and then dividing by the number of slices. So even if the resistance (or voltage, or current) may vary during the "on" time of a pulse, the sampling system will catch it, as long as it isn't varying too fast. But at 1 gigasamples per second..... well, you can see that even very small, very fast changes will be caught by the system and will give the true average. This is actually a problem sometimes, since the scope may be including noise or random glitches in the average. Hence, the scopes will have "Bandwidth limiting" that can be selected which will cut down on the presence of this kind of noise in the input samples to the averaging function.
On my Rigol, the 20 MHz bandwidth limiting is indicated by a "B" symbol in the Channel V/Div display at the bottom of the screen.
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Actually the Ohm's Law relationships are definitions of each term in terms of the others. So, since the current depends on the resistance (if voltage is constant) then even though the resistance isn't explicitly included in the particular formula you may be using, it is still there hidden in the other variable values. So P=VxI does have the resistance in there, because I = V/R.
Here's where it becomes important to understand how the scope calculates "Average" or mean values. It is adding up values from very short timeslices and then dividing by the number of slices. So even if the resistance (or voltage, or current) may vary during the "on" time of a pulse, the sampling system will catch it, as long as it isn't varying too fast. But at 1 gigasamples per second..... well, you can see that even very small, very fast changes will be caught by the system and will give the true average. This is actually a problem sometimes, since the scope may be including noise or random glitches in the average. Hence, the scopes will have "Bandwidth limiting" that can be selected which will cut down on the presence of this kind of noise in the input samples to the averaging function.
On my Rigol, the 20 MHz bandwidth limiting is indicated by a "B" symbol in the Channel V/Div display at the bottom of the screen.
Yes, this is the way I see it also, one can be interchanged to work out the other. Each are complimentary.
Ok I see what you mean, the over all Mean Sampling is fast and accurate enough to see any circuit changes most of the time. I must admit, I was not think clearly when posting the last post and really did not need to post it, so I am sorry for the Interlude there.
I think still, the oscilloscope is the most awesome machine we Humans have ever built. We would be lost without them!
Chris Sykes
hyiq.org
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Yes, the scope is awesome. I've often said that the Lathe is the King of Tools, and the Oscilloscope is the King of Test Equipment.
However.... the low-end scopes are not especially good voltmeters. Most of them have 8-bit ADC front-ends, which means you only have 256 voltage levels across the full height of the screen in the best of circumstances and usually even fewer than that. So the precision is not that good. And as far as accuracy is concerned, they are subject to drifts due to ambient temperature, component ageing, and who knows what else.
I've just spent some time checking my scope, comparing manual calculations from the screen data with what the scope comes up with. The three screenshots below show what I've been doing. I used one data capture from the "Woopy" circuit, and used cursors to read off the heights of the CH1, CH2 and Math traces to get values for the manual calculations. And I told the scope to provide Measurements of the Duty Cycle, the CH1 and CH2 RMS, the Math and CH2 Averages.
Here are the results.
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So as you can see the cursors in the CH1 test are showing 12.30 V for the peak values. Multiplying the Peak by the square root of the 19.44 percent Duty Cycle gives
12.30 x sqrt(.1944) = 5.42 Vrms, and the scope is reporting 5.40 Vrms. Not too bad, an error of less than 1 percent.
In the second test with the CH2 readings... the cursors are giving me a value of 128.0 mA for the peak values. Multiplying this by the square root of the Duty Cycle gives 0.1280 x sqrt(.1944) = 0.0564 or 56.4 mA rms, and the scope is reporting 63.1 mA rms. WTF? This is a large error. And for the Mean value, the 0.1280 x .1944 = 24.9 mA, whereas the scope says 30.8 mA. Again, a large error. WTF? Is this due to an inaccurate channel, or bad calibration, bad cursor positioning, or what? Perhaps the channel is just not accurate at very sensitive settings?
In the third test with the Math readings, the cursors give me 1.540 W for the peak value, and factoring in the duty cycle gives 1.540 x 0.1944 = 299.4 mW as the average. And the scope is reporting 305 mW as the average. Again not too bad, about 2 percent error.
These three shots are using the same sample set, I have stopped the scope and am just moving cursors around between the screensaves.
The poor result on CH2 has made me start the Self-Calibrate routine; maybe it will get better, or maybe not.
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Well, I ran the self-cal routine and the CH2 discrepancy did go down, from about 11 percent to around 8 percent. The discrepancies on the other two readings went up slightly to about 2 1/2 percent difference each. I've noticed that the cursor position values seem a little off from what I think they should be, so maybe the error is in the reported cursor positions rather than in the scope's computations themselves.
Yes, I just checked the cursor positions carefully. There is definitely an error in the reported cursor positions.
In the screenshot below, the CH1 marker is exactly on the screen center horizontal marker (offset 0.000). The Cursor B is positioned exactly on this line and reads -100.0 mV. The Cursor A is exactly on the second horizontal line below center and should read -10.00 V but actually reads -10.20. There is a similar error on other channels.
Grrr. I'm reporting this as a bug to Rigol.
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TK - A nice example of how we can see some inaccurate measurements on the Oscilloscope. Thanks for sharing this.
What's the Channel like at higher voltage levels? Does this error margin come down some?
Chris Sykes
hyiq.org
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TK - A nice example of how we can see some inaccurate measurements on the Oscilloscope. Thanks for sharing this.
What's the Channel like at higher voltage levels? Does this error margin come down some?
Chris Sykes
hyiq.org
The error in the cursor position reporting seems to be 1 or 2 ADC counts or maybe vertical pixels at most, regardless of the v/div settings, but depending on where the cursor is positioned vertically on the screen. It's annoying but like many errors in measuring instruments (they all have them) it can be worked around once it is known that it exists and how it behaves.
As far as I can tell the Measurements from the measure menus aren't subject to this kind of error (which is why my computations using the Cursor data didn't agree with the Measurements), but of course they too are limited by the resolution of the 8-bit ADC front end of this and all other low-end DSOs. You could think of this ADC resolution limitation as "rounding error".