Accurate Measurements on pulsed system's harder than you think.

[verpies]

Yes, yes, but only if they are symmetrical?

In my table, the sawtooth and the pulse train were not symmetrical.

[tinman]

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.

[poynt99]

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.

[EMJunkie]

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 X_L or X_C, which is the Inductive or Capacitive Reactance. If the Inductor is Resonant, then X_L and X_C 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

[digitalindustry]

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?