Question

The transfer characteristic for precision rectifier circuit shown below is (assume op-amp and diodes are ideal and voltmeter is used to measure output voltage)

• A.
• B.
• C.
• D. None of the above

Answer 1

The correct option is B



$V_{+}=(\frac{5}{6}{V}_i+\frac{20}{6})$

$\Rightarrow V_{+}$ is positive when $V_i\ge -4V_i,V_{+}$ is negative when $V_1<-4V,$
When $V_{+}$ is positive then $D_1$ is off and $D_2$ is on

So,



$V_0=2V_{+}=(\frac{5}{3}{V}_i+\frac{20}{3})$

When $V_{+}$ is negative then $D_1$ is on and $D_2$ is off

So,



Since no current will flow in output branch and oparnp is in negative feedback

So, $V_{+}=V$

and $V_0=V_{+}=\frac{5V_i}{6}+\frac{20}{6}$



So, when $V_{+}$ is positive $V_0=2V_{+}$
$V_{+}$ is negative $V_0=V_{+}$