Full Wave Rectifier

Thread Starter

ac_dc_1

Joined Jan 27, 2013
74
I am trying to analize this rectifier.I want to determine the expressions of the ouputs and the gain of both Amps and of the montage


VO1= 0V if Vi>0V

For the second OpAmp

Vo2=-R5/R3 Vout 1 - (-R5/R4)Vout for Vin<0 V

and Vo2=-R5/R4*Vin if Vin>0 V


Thanks
 

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Audioguru

Joined Dec 20, 2007
11,248
The circuit works very well when the opamps have a positive and negative power supply.
The first opamp circuit produces only the negative parts of the input. Its level is doubled by the second opamp and added to the input signal then some of the positive cancels the negative and produces full-wave rectification.
 

Thread Starter

ac_dc_1

Joined Jan 27, 2013
74
The circuit works very well when the opamps have a positive and negative power supply.
The first opamp circuit produces only the negative parts of the input. Its level is doubled by the second opamp and added to the input signal then some of the positive cancels the negative and produces full-wave rectification.
So the second opamp has a gain of 2,rigth?
 

Audioguru

Joined Dec 20, 2007
11,248
The second opamp has a gain of 20k/10k (R5/R3=2) for the half-wave rectified signal from the first opamp and it has a gain of 20k/20k (R5/R4=1) for the input signal.

Then when the input signal goes negative one amount the output of the second opamp goes positive two amounts which cancels the negative leaving a signal level of positive one.
 

Thread Starter

ac_dc_1

Joined Jan 27, 2013
74
The second opamp has a gain of 20k/10k (R5/R3=2) for the half-wave rectified signal from the first opamp and it has a gain of 20k/20k (R5/R4=1) for the input signal.

Then when the input signal goes negative one amount the output of the second opamp goes positive two amounts which cancels the negative leaving a signal level of positive one.
So in conclusion Vo=-vi if Vi<0 and Vo=Vi if Vi>0
 

thatoneguy

Joined Feb 19, 2009
6,359
Work out what the voltages would be if the input was +5VDC

Do the same, but with an input of -5VDC

From there, you'll have a good understanding of how the circuit works.
 
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