Power Efficiency calculations for active rectifier

Hi

I have designed an active bridge rectifier (see attached figure), with an active (PMOS) diode in the output controlled from a comparator, to generate a DC voltage. The circuit rectifies an AC input signal (1.5Vp) and generates 1.4V in the output
My question is related to the calculation of the power efficiency (n=Pout/Pin) of the rectifier :
The output power is Pout=Vdc(out)/Rload, which is quite straightforward.
To evaluate the input power i use the formula Pin=rms(Iin(t)*Vin(t)). In this case what should be the Iin(t) and Vin(t): the differential I and V waveforms between the two terminals of the input voltage source, or the I and V waveforms from one terminal of the source (The voltage source is floating, no gnd). In the first case (differential) my efficiency is something like 45%, and i wonder where the rest of the power is dissipated.

I would appreciate any feedback regarding the proper power calculation and also if you have any typical values regarding the power efficiency of the specific circuit.

Thanks

The instantaneous power calculation is OK.
However ,you kindly observe the waveform on a scope .
You will find that there are pinch off and switch ON delays at the zero crossings of the input sine .
The Cgs capacitor (between gate & source) needs to discharge first before the enhanced mode
commences for the Mos Channel to provide a min Rd ON.
This is reason for the powerloss .As such , 27% approx power is not switched at all.
There are 10-12% losses in the MOS P-channel .& 6% in the N channel.
,I will say roughly 60% power should be available to you at the rect. output stages .

Hi,

I simulated similar cicuit with a slight change in the MOS dimensions. But I got around 30% efficiency :-(
I want to implement MOS rectifier for my RFID application.
How to increase the efficiency of the circuit? Please help me out.

I have added my circuit as an image.



Thanks!

moisiad said:

The output power is Pout=Vdc(out)/Rload, which is quite straightforward.

Click to expand...

No. Pout=(Vdc(out))²/Rload

Thanks Erikl for the correction . By mistake i missed the square power.
However the calculation of the input power is still an open issue