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

[EMJunkie]

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

[tinman]

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

[TinselKoala]

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.

[EMJunkie]

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

[TinselKoala]

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.