GATE 2022 EC Q.16: solution and answer

GATE 2022 · ECQ.16 · 1 mark · MCQ

Consider a long rectangular bar of direct bandgap p-type semiconductor. The equilibrium hole density is $1 0^{17} cm^{- 3}$ and the intrinsic carrier concentration is $1 0^{10} cm^{- 3}$ . Electron and hole diffusion lengths are $2 μ m$ and $1 μ m$ , respectively.

The left side of the bar ( $x = 0$ ) is uniformly illuminated with a laser having photon energy greater than the bandgap of the semiconductor. Excess electron-hole pairs are generated ONLY at $x = 0$ because of the laser. The steady state electron density at $x = 0$ is $1 0^{14} cm^{- 3}$ due to laser illumination. Under these conditions and ignoring electric field, the closest approximation (among the given options) of the steady state electron density at $x = 2 μ m$ , is ___________.

(A)
$0.37 \times 1 0^{14} cm^{- 3}$
(B)
$0.63 \times 1 0^{13} cm^{- 3}$
(C)
$3.7 \times 1 0^{14} cm^{- 3}$
(D)
$1 0^{3} cm^{- 3}$

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Answer and solution

ANSWER

The steady-state electron density at x = 2 μm is 0.37 × 10^14 cm^-3, option (A).

Official answer: ASource: Official answer key, GATE 2022 EC (IIT Kharagpur)

In the p-type bar, electrons are minority carriers, so their steady-state spatial profile decays with the electron (minority-carrier) diffusion length $L_{n} = 2 μ m$ — not the hole diffusion length, which is irrelevant to the electron density.
With no electric field and no generation for $x > 0$ , the steady-state minority-carrier diffusion equation reduces to $\frac{d ^{2} Δ n}{d x ^{2}} = \frac{Δ n}{L _{n}^{2}}$ , whose physically bounded solution in a long bar is $Δ n (x) = Δ n (0) e^{- x / L_{n}}$ .
Check the equilibrium (dark) electron density: $n_{0} = \frac{n _{i}^{2}}{N _{A}} = \frac{( 1 0 ^{10} ) ^{2}}{1 0 ^{17}} = 1 0^{3} cm^{- 3}$ , which is negligible next to $1 0^{14} cm^{- 3}$ , so the given $n (0) = 1 0^{14} cm^{- 3}$ is essentially all excess: $n (0) \approx Δ n (0) = 1 0^{14} cm^{- 3}$ .
Evaluate at $x = 2 μ m = L_{n}$ , so the exponent is exactly $- 1$ : $n (2 μ m) = 1 0^{14} \times e^{- 1} = 1 0^{14} \times 0.3679 \approx 0.37 \times 1 0^{14} cm^{- 3}$
This matches option (A).

KEY IDEA

In a long p-type bar, electrons are minority carriers whose steady-state profile away from the injection point decays exponentially over the electron diffusion length $L_{n}$ : $Δ n (x) = Δ n (0) e^{- x / L_{n}}$ .

WATCH OUT FOR

Using the hole diffusion length (1 μm) instead of the electron diffusion length (2 μm) since holes are the majority carrier associated with "p-type," or forgetting to check that the equilibrium electron density $n_{0} = n_{i}^{2} / N_{A}$ is negligible compared to the injected value (option D, $1 0^{3} cm^{- 3}$ , is exactly that distractor).

From the GATE 2022 EC paper. Every question of this paper · Download the full paper · All 9 Carrier transport questions, with notes

GATE 2022 EC Q.16: solution and answer

Answer and solution

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