Power Efficiency calculations for active rectifier

[moisiad]

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

 

[vimalkhanna]

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 .

 

[shaikss]

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!

 

[erikl]

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

No. Pout=(Vdc(out))²/Rload

 

[moisiad]

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