Question

Consider the circuit shown inn the figure below:



Assume the op-amp and the diode $D_1$ and $D_2$ to be ideal, then the transfer charcteristic of the circuit can be given as

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Case -I : When $V_{\text{in}}>-10$ V, then the voltage across diode $D_1$ is positive so diode $D_1$ is in ON state, and therefore the equivalent circuit can be drawn as



$V^{+}=-15\times \frac{10}{15}=-10V$
Due to virtual groud, $V^{+}=V^{-}=-10V$
$\text{and}{V}_0=V^{-}=-10V$
$\therefore V_0=-10V$
Case -II: When $V_{\text{in}}<-10V$
$V_0=+V_{\text{sat}}$
Thus,


$V_0=-\frac{5}{5}\times V_{\text{in}}=-V_{\text{in}}\left(\text{for}{V}_{\text{in}}\right)<-10V$
Alternatively, we can write the equation of the graph by applying KCL at node ${}^{-}$
$\because V^{-}=V^{+}=-10V$
$\frac{-10-V_{\text{in}}}{5k\Omega }+\frac{-10-V_0}{5k\Omega }=0$
$-20-V_{\text{in}}-V_0=0$
$V_0=-V_{\text{in}}-20$