GATE 2023 EC Q.56: solution and answer

GATE 2023 · ECQ.56 · 2 marks · NAT

In an extrinsic semiconductor, the hole concentration is given to be $1.5 n_{i}$ where $n_{i}$ is the intrinsic carrier concentration of $1 \times 1 0^{10} cm^{- 3}$ . The ratio of electron to hole mobility for equal hole and electron drift current is given as ________ (rounded off to two decimal places).

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Answer and solution

ANSWER

The ratio of electron to hole mobility is $μ_{n} / μ_{p} = 2.25$ .

Official answer: 2.20 to 2.30Source: Official answer key, GATE 2023 EC (IIT Kanpur)

Given: hole concentration $p = 1.5 n_{i}$ , with $n_{i} = 1 \times 1 0^{10} cm^{- 3}$ .
Mass-action law (valid for any non-degenerate extrinsic semiconductor in equilibrium): $n p = n_{i}^{2} \Rightarrow n = \frac{n _{i}^{2}}{p} = \frac{n _{i}^{2}}{1.5 n _{i}} = \frac{n _{i}}{1.5}$ .
Drift current densities under the same applied field $E$ : $J_{n} = q n μ_{n} E$ and $J_{p} = q p μ_{p} E$ .
Equal drift currents: $J_{n} = J_{p} \Rightarrow n μ_{n} = p μ_{p} \Rightarrow \frac{μ _{n}}{μ _{p}} = \frac{p}{n}$ .
Substitute: $\frac{p}{n} = \frac{1.5 n _{i}}{n _{i} /1.5} = 1.5 \times 1.5 = 2.25$ .
Hence $\frac{μ _{n}}{μ _{p}} = 2.25$ .

KEY IDEA

In an extrinsic semiconductor, electron and hole concentrations obey the mass-action law $n p = n_{i}^{2}$ , so knowing one concentration (here $p = 1.5 n_{i}$ ) fixes the other. Equal drift current densities for electrons and holes (same applied field $E$ ) then give a direct relation between $n$ , $p$ and the mobility ratio.

WATCH OUT FOR

Inverting the ratio — computing $μ_{p} / μ_{n}$ instead of $μ_{n} / μ_{p}$ — or forgetting to apply the mass-action law and instead assuming $n = p$ , which would (incorrectly) give a ratio of 1.

From the GATE 2023 EC paper. Every question of this paper · Download the full paper · All 9 Carrier transport questions, with notes

GATE 2023 EC Q.56: solution and answer

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