In a semiconductor p-n diode, the doping concentrations on p-side and n-side are \(10^{15}\) atoms/\(\mathrm{cm}^{3}\) and \(10^{18}\) atoms \(/\mathrm{cm}^{3}\), respectively. Which one of the following statement is true?

(B) The depletion region width is more on p-side compared to that in n-side

(A) Widths of depletion region on either side of the interface are equal

(C) The depletion region width is more on n-side compared to that in p-side

(D) No depletion region forms because of unequal doping concentrations on p and n-sides

Semiconductors — JEE Main · 2026 · 05 Apr 2026 · Shift 2 Question with Solution

In a semiconductor p-n diode, the doping concentrations on p-side and n-side are

1 0^{15}

atoms/

cm^{3}

and

1 0^{18}

atoms

/ cm^{3}

, respectively. Which one of the following statement is true?

(A)Widths of depletion region on either side of the interface are equal
(B)The depletion region width is more on p-side compared to that in n-side✓ Correct
(C)The depletion region width is more on n-side compared to that in p-side
(D)No depletion region forms because of unequal doping concentrations on p and n-sides

Answer: (B)

Solution

The depletion region holds equal and opposite charge on its two sides, so

N_{p} x_{p} = N_{n} x_{n}

, where

x_{p}

and

x_{n}

are the depletion widths and

N_{p}

N_{n}

the doping concentrations. With

N_{p} = 1 0^{15} c m^{- 3}

and

N_{n} = 1 0^{18} c m^{- 3}

\frac{x _{p}}{x _{n}} = \frac{N _{n}}{N _{p}} = 1 0^{3}

. The lightly doped p-side therefore extends a thousand times farther than the n-side. The depletion region width is greater on the p-side than on the n-side.

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Semiconductors

Solution

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