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

[verpies]

because a Sinusoidal Wave is calculated: Asine(omega t)

Yes, the sine waveform has many well known metrics

[TinselKoala]

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 V_drop across the 100 ohm (probably 5 or 10 percent tolerance) resistor.
The CH2 measurement is not "Vcc", it is the V_drop 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 V_drop 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 V_drop 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 V_drop across it is accurate. Using those two assumptions we can crosscheck the other results coming from the scope.

Why are you multiplying the V_drop 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.

[verpies]

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.

[verpies]

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") ?

[TinselKoala]

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.