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

[digitalindustry]

1.  This is not truly a circuit that pulses a voltage (i.e. the drive voltage for the circuit) that goes from zero volts to 12 volts.  What it is is a MOSFET switch that makes a connection and then breaks a connection and goes high-impedance.  There is no "zero volt" signal.  Open-circuit can be way way different from zero volts, especially as you go towards higher frequencies.



MileHigh

this is a relevant point i wanted to talk about but in a different context actually.

@TK but up that garden path there are wonderful things growing ! : D - (no not just psychedelics)

[poynt99]

And yet if we use your calculated average current of 28mA,then 28mA through a 1 ohm resistor would mean that the 1 ohm resistor is dissipating an average of .784mW,and the 94.3 ohm resistor would be dissipating an average 73.93mW,as the average current flowing through them !is! 28mA

So if we use the peak current value through the 94.3 ohm resistor X 20.9%,we get 134mA through 94.3 ohm's = 1.693 watts X 20.9% = 353mW. But if we use the average current through the 94.3 ohm resistor,we get 73.93mW.

If we multiply the peak voltage 12.6 by the average current of 28mA,we get very close to the 351mW you calculated. But now the resistance has to be 450 ohm's across that circuit.



Brad.

Good Grief! 

Based on the above and other posts in this thread, it's astounding that folks still don't understand power calculations and measurements, especially after all that has been explained here over the years, by myself and others.

Number one: The average current through a CVR resistor is only used for calculating the average power of the DC source. The circuit average current can not, and should not be used to calculate the power dissipated in any other circuit component, other than the DC source.

Number two: To calculate the power dissipated by the CVR (or any other series resistor), one must use either the RMS current* (which is vastly different than the average current in a 20% duty-cycle pulsed circuit) or the peak voltage squared over the resistor value times the duty-cycle. [* we can of course also use rms voltage across the resistor]

As in the attached diagram:

Psource = Vsource x Iavg (where Iavg=Vavg/Rcvr)

Pcvr = (Vrms1 x Vrms1) / Rcvr
OR = [(Vp1 x Vp1) / Rcvr] x duty cycle

PRL = (Vrms2 x Vrms2) / RL
OR = [(Vp2 x Vp2) / RL] x duty cycle

[poynt99]

@
Tinman, Gotoluc, Magluvin, Woopy, et al...

I would encourage all of you to print out the above post and the associated diagram and keep in in your lab for reference, preferably somewhere readily in view.

[poynt99]

It should be noted that the following (from the above post, but abbreviated and generalized):

PRx = Vrms² / Rx

Holds for ANY wave form, with any duty cycle.

[EMJunkie]

It should be noted that the following (from the above post, but abbreviated and generalized):

PRx = Vrms² / Rx

Holds for ANY wave form, with any duty cycle.

@Poynt - So you're saying that the example TK gave is wrong? It appears so.

I see you're just trying to help, but the Contempt, any need for it?

For a relatively complicated subject, where many different views are held, some maybe wrong and some may be right, I think treading with light feet, a more gentle way, might be more productive?

Brad does excellent work, always, and does the best he can. So go easy...

   Chris Sykes
       hyiq.org