In the circuit using Zener diode and ideal opamp with $V_{Z} = 2.5$ V as shown below, an unit step input voltage is applied i.e. $V_{I N} (t) = u (t)$ V. Also it is given the voltage across each of the capacitors is zero then time, t in milliseconds at which the output voltage $V_{o u t}$ crosses - 10 V is ______m sec.

7.5

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Solution

The correct option is A 7.5
For given configuration assuming current I in capacitor branch,

For t > 0, $I = \frac{1 V}{1 k Ω} = 1 m A$

The capacitor charges with constant current I and both $V_{1}$ and $V_{2}$ will increase till $V_{2}$ reaches 2.5 V. At this $V_{2}$ remains at 2.5 V and $V_{1}$ increases with time

$V_{o u t} (t) = - 10 V$

$V_{1} = 7.5 V$

So, $\frac{1}{c} \int_{0}^{t} I d t = \frac{1}{1 \times 10^{- 6}} \int_{0}^{t} 10^{- 3} d t$

$\frac{10^{- 3}}{10^{- 6}} t = 7.5$

$t = 7.5 m s e c$

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