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

[TinselKoala]

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") ?

The original issue was to account for the power dissipation of the bulb itself, as one of several load elements within a more complex circuit. It was not about the total input power to the circuit. The confusing part was that the bulb in the original circuit, and in my modification, became dimmer when the "average" current through it increased. But as we showed, when the actual power dissipation is considered rather than just the "average" or mean current, the "anomaly" goes away and the bulb's brightness follows the actual average power, as measured and correctly calculated.

[EMJunkie]

For Giggles:

The 1KHz Signal is from a High Side Mosfet Driver. The 1K Resistor is measured to be 1K.

Mean Calculation: Red Channel

V / R = I = 5 / 1000 = 0.005 Amperes

I² * R = P = (0.005² * 1000) * 0.5 (Duty Cycle) = (0.000025 * 1000) * 0.5 (Duty Cycle) = 0.0125 Watts

RMS Calculation: Yellow Channel

V² / R = P = 3.458² / 1000 = 11.957764 / 1000 = 0.011957764 Watts

Now I must have something wrong here? We see: 0.000542236 Watts difference!

   Chris Sykes
       hyiq.org

[EMJunkie]

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.

Yes Sir!!! I could not agree more here!!!

   Chris Sykes
       hyiq.org

[TinselKoala]

For Giggles:

The 1KHz Signal is from a High Side Mosfet Driver. The 1K Resistor is measured to be 1K.

Mean Calculation: Red Channel

I = 5 / 1000 = 0.005 Amperes

P = (0.005² * 1000) * 0.5 (Duty Cycle) = 0.0125 Watts

RMS Calculation: Yellow Channel

P = 3.458² / 1000 = 11.957764 / 1000 = 0.011957764 Watts

Now I must have something wrong here? We see: 0.000542236 Watts difference!

   Chris Sykes
       hyiq.org

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":

Continuous average power ratings are a staple of performance specifications for audio amplifiers and, sometimes, loudspeakers. Continuous average power is derived from the root mean square (RMS) of the AC voltage or current, and is often incorrectly referred to as "RMS power", "RMS watts", or "watts RMS". However, the RMS value of the power waveform is different from the average power value (e.g. 22% higher for a sine wave signal), and doesn't represent anything useful, so these terms should not be used.^{[2][3][4][5][6][7]} The correct term is "continuous average power", which is proportional to the RMS voltage, and specified by the FTC.^[8]

(emphasis mine)

[TinselKoala]

For Giggles:

The 1KHz Signal is from a High Side Mosfet Driver. The 1K Resistor is measured to be 1K.

Mean Calculation: Red Channel

I = 5 / 1000 = 0.005 Amperes

P = (0.005² * 1000) * 0.5 (Duty Cycle) = 0.0125 Watts

RMS Calculation: Yellow Channel

P = 3.458² / 1000 = 11.957764 / 1000 = 0.011957764 Watts

Now I must have something wrong here? We see: 0.000542236 Watts difference!

   Chris Sykes
       hyiq.org

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).