calculating the efficiency of an Inverter

Hi guys,

I have a doubt regarding the efficiency calculations for an inverter (DC-AC). I know that to calculate efficiency i have to divide power-out/power-in and power is i^2*R, V^2/R, etc, but there is something troubling me about the calculations when there is an RMS value. How do I calculate the efficiency of a FullBridge inverter when the input is dc and the output is rms?

Po=Vrms*Irms ? Or is it something like Vpk-pk*Irms?

Let’s say we have an ideal Full Bridge Inverter with an input of 169.7Vdc, with a resistive load, and I want an output of 120Vrms, 1kW. Does that mean that the output current must be 8.3Amps-RMS (8.3*120=1k)? If the inverter is 100% efficient, the input current has to be 1000/169.7=5.9Adc, but 5.9/sqrt(2)=4.17Arms and not 8.3Arms.

Thanks

The DC is already RMS, so at 100% efficiency the DC current should be 5.9 A. All this RMS business is to convert the heating effect of a sinewave to that of DC.
Frank

Thanks for the reply, i get it now.

And, correct me if I'm wrong... The DC input of an inverter will be the peak voltage of the output wave, so if i have 169.7Vdc input, my output will be 120Vrms; and since the power-in must be the same as power-out and 169.7Vdc > 120Vrms, then my output current will be higher than my input current? edit5.9Adc(in) - 8.3Arms(out))?

Yes. Think of it this way :- Imagine a square pulse of amplitude 169.7 V, now draw a half cycle of a sine wave in it, the sinwaves peak just touches the top of the pulse. The start and ending of the two wave forms are the same. If we call this the voltage wave form, then the sinewave bit is clearly smaller the the square bit, so in voltage terms the square pulse has more voltage X time in it. As the conversion is 100% efficient, the the sine wave must have more current X time in it so the watts in both cases are the same.
Frank