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Author Topic: Accurate Measurements on pulsed system's harder than you think.  (Read 53211 times)

Offline TinselKoala

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #75 on: December 11, 2015, 12:35:28 PM »
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.

Free Energy | searching for free energy and discussing free energy


Offline tinman

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #76 on: December 11, 2015, 12:41:46 PM »
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?

Offline TinselKoala

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #77 on: December 11, 2015, 12:47:55 PM »
@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.

Free Energy | searching for free energy and discussing free energy

Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #77 on: December 11, 2015, 12:47:55 PM »
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Offline tinman

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #78 on: December 11, 2015, 12:53:05 PM »
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 ?.

Offline tinman

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #79 on: December 11, 2015, 01:00:16 PM »
@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 ?

Free Energy | searching for free energy and discussing free energy

Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #79 on: December 11, 2015, 01:00:16 PM »
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Offline FatBird

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #80 on: December 11, 2015, 04:17:47 PM »
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.


                                                                                                                     .

Offline TinselKoala

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #81 on: December 11, 2015, 04:21:37 PM »
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.


Free Energy | searching for free energy and discussing free energy

Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #81 on: December 11, 2015, 04:21:37 PM »
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Offline TinselKoala

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #82 on: December 11, 2015, 04:42:25 PM »
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.

Offline digitalindustry

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #83 on: December 11, 2015, 04:44:53 PM »
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.

Free Energy | searching for free energy and discussing free energy

Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #83 on: December 11, 2015, 04:44:53 PM »
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Offline TinselKoala

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #84 on: December 11, 2015, 05:26:56 PM »
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?
 

Offline tinman

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #85 on: December 11, 2015, 05:31:32 PM »
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

Free Energy | searching for free energy and discussing free energy

Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #85 on: December 11, 2015, 05:31:32 PM »
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Offline digitalindustry

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #86 on: December 11, 2015, 06:12:31 PM »
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.

Offline digitalindustry

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #87 on: December 11, 2015, 06:21:32 PM »
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)

Offline verpies

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #88 on: December 11, 2015, 07:11:41 PM »
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?:

Offline woopy

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #89 on: December 11, 2015, 07:21:25 PM »
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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