Modified Precision Full Wave Rectifier

By modifying the Precision Full Wave Rectifier circuit, we get some gain for PFWR. The following circuit diagram shows the modified Precision Full Wave Rectifier.

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In positive half cycle of applied ac input signal, output of op-ampA1 is negative. so diode D2 is reversed biased. Thus op-amp A1 works as an inverting amplifier with gain=(-R/R1 )
Therefore voltage at point ‘A’ (output of op-amp A1) is given as
VA=(-R/R1 ) Vin
The non-inverting terminal of op-amp A2 is also virtually grounded, so op-amp A2 also works as an inverting amplifier. Therefore the output voltage Vo is given as
∴Vo=-R/R [-R/R1 Vin ]
∴Vo=(R/R1 ) Vin
In negative half cycle of applied ac input signal, output of op-amp A1 is positive .so diode D2 is forward biased and diode D1 is reverse biased. Thus op-amp A1 works as an inverting amplifier.
Let V is the output of op-amp A1.
The current flowing through feedback path of op-amp A2& A1 is given as
(Vo-V)/R=(V-0)/2R
∴Vo=V/2+V
∴Vo=V[1/2+1]
∴V= 2/3 Vo
Applying KCL at node present at inverting terminal of op-amp A1, we get
(Vin-0)/R1 +(V-0)/R+(V-0)/2R=0
Vin/R1 +V/R+V/2R=0
Vin/R1 +V[1/R+1/2R]=0
Substituting the equation of ‘V’ in above equation as follows
Vin/R1 +2/3 Vo [(2R+R)/(2R)R]=0
Vin/R1 +2/3 Vo [3R/(2R)R]=0
Vin/R1 +2/3 Vo [3/((2R) )]=0
Vin/R1 +Vo/R=0
∴Vo=-(R/R1 ) Vin
But in negative half cycle input magnitude is negative therefore we get,
∴Vo=-(R/R1 )(-Vin )
∴Vo=(R/R1 ) Vin
Thus in both the half cycles of applied ac input signal, output remains same with gain (R/R1 )

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