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 $-4V$, diode $D_1$ is in reverse bias and it does not conducts.
As input voltage $V_i$ falls from $0$ to - $3V$, diode $D_2$ is in reverse bias and it does not conducts. So, potential drop $V_0=V_i$
As input voltage drops below $-3V$, diode $D_2$ is in forward bias and it conducts and bypass the excess current.
So, $V_0=-3V$ when $-4\le V_i\le -3V$.
As $V_i$ increases from $0$ to $1V$, diode $D_1$ remains in reverse bias and $D_2$ is also in reverse bias. So, $V_0=V_i$ as $0\le V_i\le 1$
When $V_i$ increases from $0$ to $4V$, diode $D_1$ conducts while diode $D_2$ is in reverse bias.
So, $V_0=1V$ as $V_i>1V$