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Author Topic: Accurate Measurements on pulsed system's harder than you think.  (Read 83340 times)

EMJunkie

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #105 on: December 12, 2015, 02:34:36 AM »
;)


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.

tinman

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #106 on: December 12, 2015, 02:49:12 AM »

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.

EMJunkie

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #107 on: December 12, 2015, 03:06:53 AM »
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

EMJunkie

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #108 on: December 12, 2015, 03:48:28 AM »

Re: Heating and Current Shunts, they are relative.

Rated for maximum accuracy at the least Temperature drift.

See Images:

   Chris Sykes
       hyiq.org

tinman

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #109 on: December 12, 2015, 03:52:48 AM »

@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

tinman

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #110 on: December 12, 2015, 04:09:12 AM »
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

EMJunkie

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #111 on: December 12, 2015, 04:20:32 AM »
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

tinman

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #112 on: December 12, 2015, 04:47:09 AM »

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

EMJunkie

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #113 on: December 12, 2015, 04:49:10 AM »
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


EMJunkie

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #114 on: December 12, 2015, 05:02:11 AM »
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

digitalindustry

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #115 on: December 12, 2015, 05:22:11 AM »
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?

TinselKoala

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #116 on: December 12, 2015, 07:28:55 AM »
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.

digitalindustry

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #117 on: December 12, 2015, 07:36:50 AM »
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

TinselKoala

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #118 on: December 12, 2015, 07:42:49 AM »
Sure, ask them. Be sure to let me know when you get an answer.

EMJunkie

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Re: Accurate Measurements on pulsed system's harder than you think.
« Reply #119 on: December 12, 2015, 07:45:12 AM »
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