Question

Consider the circuit shown in the figure below.



The switch in the circuit is initially closed for a long time, at t = 0 it is opened. Then which of the following statement is true.

(Assume forward drop of a zener diode = 0.7 V).

• A. $\text{It makes a transition from}+5.7V\text{to}-5.7\text{at}t=13\mu \sec$
• B. $\text{It makes a transition from}+5.7Vto-5.7Vatt=14.23\mu sec$
• C. $\text{It makes a transition from}+5.7V\text{at}t=3\mu \sec$
• D. $\text{It makes a transistion from}+5.7V\text{to}-5.7V\text{at}t=1.29\mu \sec .$

Answer 1

The correct option is B $\text{It makes a transition from}+5.7Vto-5.7Vatt=14.23\mu sec$

In the circuit, the capacitor starts charginng from 0 V (as the switch was initially closed) towards the steady state alue of 0 V.
Now, when he switch is flipped open, the capacitor will charge upto 20 V.
$\therefore V_c\left(t\right)=V_c\left(\infty \right)-\left[V_c\right(0)-V_c(\infty \left)\right]e^{-t/RC}$
$RC=1\times {10}^3\times 0.01\times {10}^{-6}=10\mu sec$
$V_c\left(t\right)=20(1-e^{-t/RC})$
Voltage at non-inverting amplifier is obtained as



$V^{+}=V_0\times \frac{100k\Omega }{(10+100)k\Omega }$
$V^{+}=V_0\times \frac{100}{110}=\frac{V_0\times 10}{11}$
$\because$ Initially $V^{-}$ was equal to -10 V, thus $V_0=+5.7V$.
Thus, now capacitor will start charging as soon as the switch is opened.
Thus, $V^{-}=V_C-10V$
or, $V_C=V^{-}+10V$
now, $V^{-}=V^{+}=\frac{5.7V\times 10}{11}[\because$ the op-amp will switch$]$
thus, $V_C=\frac{10\times 5.7}{11}+10$
now, $V_C=20(1-e^{-t/RC})$
$\therefore 20(1-e^{-t/RC})=10+\frac{57}{11}$
$1-e^{-t/RC}=\frac{1}{2}+\frac{57}{220}$
$1-e^{-t/RC}=0.7590$
$1-e^{-t/RC}=0.2409$
$T=14.23\mu \sec$
Hence, the output voltage wave will be,