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First of all look at my circuit.

Full Wave Precision Rectifier

If we assume a sinusoidal signal as input Vin then the signals V01 and V02 should be like the signals shown in the picture. From the two signal how can we get the V0 like that?

Another question is that why we take the value of resistances (marked with red pen in the picture) R/2 and R?

Actually I can't understand the full operation of the circuit. Can anyone help me?

Raihan Khalil
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  • First of all, [learn how to draw a good schematic](http://electronics.stackexchange.com/a/28255/11683). At a minimum, each component should have a unique reference designator, so that we can talk about it without having to use an elaborate description of where it is. – Dave Tweed May 14 '15 at 11:31
  • Can you please help me with any link or something?? – Raihan Khalil May 14 '15 at 12:19

1 Answers1

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The first part of the circuit, whose output is \$v_{o1}\$, is a halfwave precision rectifier.

For the positive half of the input, the diode D2 will be forward biased and D1 will be reverse biased. So it acts as an inverting amplifier with unity gain. So \$v_{o1}\$ will be inverted version of input for positive half cycle. ie, \$v_{01} = -v_{in}\$ for positive half of input.

For the negative half, diode D1 will be forward biased and D2 will be reverse biased and hence \$v_{o1}\$ will be zero for negative half cycle. ie, \$v_{01} = 0\$ for negative half of input

The second part of the circuit is a summing amplifier whose output is given by: $$v_o = -\left(\frac{R}{R/2}v_{o1} + \frac{R}{R}v_{o2}\right) = -(2v_{o1}+v_{o2})$$

where \$v_{o2}\$ is the input.

For positive half of input, \$v_{o1}=-v_{in}\$ and \$v_{o2}=v_{in}\$. Hence \$v_o = v_{in}\$
For negative half of input, \$v_{o1}=0\$ and \$v_{o2}=v_{in}\$. Hence \$v_o = -v_{in}\$

And this is the input output relation for a full-wave rectifier. The resistor values R and R/2 are chosen deliberately to obtain this input-output relationship.

nidhin
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