duty cycle watts

If you have a load that is "on" for 20 uS, off for 3.5 mS, which is 2.3 volts when on, zero volts off, 10mA on and zero volts off, and the current and voltage are in phase, how many watts is the load?
Thanks again, Rich

 

I guess 2.3 is a dc voltage, am i correct?

what is your load?

 

Calculate for DC conditions and divide by the duty cycle.

 

So I don't need to use RMS?

 

Only if the applied voltage is AC. It's somewhat unusual to specify a duty cycle for AC.

 

It's a square wave, so: [(amplitude)^2]^1/2 = amplitude. That is, the RMS of a square wave is the amplitude. John

 

The power, which is Joules/sec when on is 2.3*10mA = 23mW.
So the energy when on is 23mW*20us = 460nJ
The energy when off is 0nJ.

The average power for the load is the total Joules divided by the total time.
460nJ / 10020us = 45.9e-6W or about 46uW.

 

How do I arrive at "10020us"?

 

I thought the RMS voltage of a rectangular wave was:
SQRT(duty cycle) * Peak voltage
As in:
SQRT (.005714) * 2.3 = .0756 * 2.3 = .17388 volts
and the RMS current of the .57% duty cycle was:
SQRT (.005714) * .01 = .000756
so the power would be .000756*.17388 = .0001314 watts or 131uW

 

Where did your formula come from, particularly the 0.005714 component?

I refer you to Horowitz and Hill, page 18, Figure 1.23. Clearly duty cycle enters into the calculation of the power, but the mean amplitude of a square wave pulse is its amplitude and the square root of the amplitude squared is also just the amplitude. John

 

Sorry, I screwed that up.
On power is 2.3V*10mA = 23mW. Off power is 0mW.
For one cycle, the total energy consumed is 23mW*20us = 460nJ
This is averaged over one cycle, which is 20us + 3.5ms = 3520us.
So average power is 460nJ/3520us = 0.13mW

Sorry about that.

 

You can use I * E * DutyCycle.

Yes, duty cycle is part of the solution.

20 uS on and 3.5 mS off is 20 uS/3.52 mS or 0.005681818

2.3 V * 10 mA is 0.023W

W * Duty Cycle is 130.681 uW