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[SOLVED] Efficiency of active rectifer not true - only 50%?

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minhducpham2009@gmail.com

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I have designed an active bridge rectifier (see attached figure), with 4 Switch
Image 28.png

The circuit rectifies an AC input signal (110VAC) to 200VDC Output

My question is related to the calculation of the power efficiency (n=Pout/Pin) of the rectifier :

The output power is Pout=Idc_out* Vdc_out

The input power is Pin=rms(Iin(t)*Vin(t)).

IN my simulation : I out 2A * 200V(DC) = 400W

and for the input I_in 7.26A( almost sin - because active rectifier) *110V(AC) = 798.6

So the efficiency is appro.. 50% . And I wonder how about 50 % dissipated or I using wrong calculation

Thank you so much
 

minhducpham2009@gmail.com said:
The input power is Pin=rms(Iin(t)*Vin(t)).
Click to expand...

The correct expression for the input power is rather:
Code: 
Pin = (1/T) ∫ ( Iin(t)*Vin(t) )dt
Anyway, the way you're doing seems more like measuring the power rectified, which is under no circumstances related to power lost/transferred power.
 
the efficiency can hardly be worse than that offered by the 4 diodes in the fets, so yes your calculations are wrong, real power on an AC line is not simply Vrms x I rms, unless the load is purely resistive and there is no distortion on the AC sine wave.

Also calculating from the average Vo and the average Io is not quite correct, esp as you have a big cap on the DC bus...
 
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