JEE Main 2026 — Semiconductors Question with Solution

If $A=0$ and $B=0$: Both diodes are forward biased. The potential at $C$ is pulled down to approximately $0V$ (logic 0).
If $A=0$ and $B=1$: Diode $D_1$ is forward biased, pulling the potential at $C$ down to $0V$ (logic 0). Diode $D_2$ is reverse biased.
If $A=1$ and $B=0$: Diode $D_2$ is forward biased, pulling the potential at $C$ down to $0V$ (logic 0). Diode $D_1$ is reverse biased.
If $A=1$ and $B=1$: Both diodes are reverse biased. No current flows through the resistor $R$, so the potential at $C$ remains at $V_{dc}=5V$ (logic 1).
The truth table is:

A	B	C
0	0	0
0	1	0
1	0	0
1	1	1

This truth table corresponds to the AND gate logic operation $C=A\cdot B$.
Therefore, the circuit functions as an AND gate.