efficiency of rectifier.....weird result

Thread Starter

youdontknowme

Joined May 30, 2017
15
Hi,

For my wireless power transfer, the picture shows the secondary side (Receiving).
I'm trying to measure the efficiency of full bridge rectifier by using "eff = (v2*i2) / (v1*i1)" and i'm getting very wrong result.
Voltage is measured with oscilloscope and current with multimeter.

v2 = 17.982v (dc coupled) & i2 = 0.22a (dc) hence p2 = 3.956w

v1 = 32.5v (ac coupled) & i1 = 0.0615a (ac) hence p1 = 1.998w
and eff value would be 1.9799

what am i doing wrong ??

R1 is the resistor load, 83ohm.
L1 is the secondary coil.

for diodes, i'm using 1N5711 schottky diode.
frequency is at 53khz.


thanks





Capture.JPG
 
Last edited:

bertus

Joined Apr 5, 2008
22,278
Hello,

The capacitor C1 will act as a wattless voltage dropper.
The value will influence the output voltage.

Bertus
 

BobTPH

Joined Jun 5, 2013
9,003
This is how over unity devices created: misunderstanding how AC power is calculated. You cannot get average power by multiplying average voltage by average current. You have to integrate the product over the entire period. The phase of the two makes the difference.

Bob
 

DickCappels

Joined Aug 21, 2008
10,187
Do you suppose the fact that a 1N5711 is only rated at 15 ma continuous current (or average current in your case) has anything to do with the problem?

1583498954690.png
Consider the 1N5819, it was designed for the kind of circuit you have.
 
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