CBSE 2026 · Region 2 · Set 1 · Q29 · 4 marks

A p-type or n-type semiconductor can be converted into a p-n junction by doping it with suitable impurity. The motion of majority charge carriers causes diffusion current across the junction while the barrier electric field causes motion of minority carriers for drift current. In case of unbiased diode, the diffusion and drift currents are equal. This equilibrium is disturbed by the biasing batteries. Diodes, therefore, allow currents in one direction. This property of diode is used in making rectifiers.

(i)

Silicon is doped with which of the following to obtain p-type semiconductor?

(A)

Phosphorus

(B)

Arsenic

(C)

Boron

(D)

Antimony

(ii)

A semiconductor has an electron concentration of $\displaystyle 5 \times 10^{22} \mathrm{~m}^{-3}$. The concentration of holes is (given $\displaystyle \mathrm{n}_{\mathrm{i}}=1.5 \times 10^{16} \mathrm{~m}^{-3}$ )

(A)

$\displaystyle 5 \times 10^{22} \mathrm{~m}^{-3}$

(B)

$\displaystyle 1.5 \times 10^{6} \mathrm{~m}^{-3}$

(C)

$\displaystyle 9 \times 10^{8} \mathrm{~m}^{-3}$

(D)

$\displaystyle 4.5 \times 10^{9} \mathrm{~m}^{-3}$

(iii)

During forward biasing of a p-n junction diode, the

(A)

current is mainly due to drifting of majority carriers.

(B)

current is mainly due to drifting of minority carriers.

(C)

diffusion and drift currents are equal.

(D)

current is of the order of $\displaystyle 1$ A.

(iv)OR(iv)

The threshold voltage for silicon diode is about

(A)

0.$\displaystyle 2$ V

(B)

0.$\displaystyle 5$ V

(C)

0.$\displaystyle 7$ V

(D)

1.$\displaystyle 5$ V

CBSE Class 12 Physics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.