Question

The figure below shows a circuit and its input voltage $V_{i}$ as function of time $t$.

Assuming the diodes to be ideal, which of the following graphs depicts the output voltage $V_{0}$ as function of time $t$ ?

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

As input voltage $V_{i}$ falls from $0$ to $-4 V$, diode $D_{1}$ is in reverse bias and it does not conducts.
As input voltage $V_{i}$ falls from $0$ to - $3 V$, diode $D_{2}$ is in reverse bias and it does not conducts. So, potential drop $V_{0}=V_{i}$
As input voltage drops below $-3 V$, diode $D_{2}$ is in forward bias and it conducts and bypass the excess current.
So, $V_{0}=-3 V$ when $-4 \leq V_{i} \leq-3 V$.
As $V_{i}$ increases from $0$ to $1 V$, diode $D_{1}$ remains in reverse bias and $D_{2}$ is also in reverse bias. So, $V_{0}=V_{i}$ as $0 \leq V_{i} \leq 1$
When $V_{i}$ increases from $0$ to $4 V$, diode $D_{1}$ conducts while diode $D_{2}$ is in reverse bias.
So, $V_{0}=1 V$ as $V_{i}>1 V$