GATE 2022 EC Q.56: solution and answer

GATE 2022 · ECQ.56 · 2 marks · NAT

A $p$ -type semiconductor with zero electric field is under illumination (low level injection) in steady state condition. Excess minority carrier density is zero at $x = \pm 2 l_{n}$ , where $l_{n} = 1 0^{- 4}$ cm is the diffusion length of electrons. Assume electronic charge, $q = - 1.6 \times 1 0^{- 19}$ C. The profiles of photo-generation rate of carriers and the recombination rate of excess minority carriers ( $R$ ) are shown. Under these conditions, the magnitude of the current density due to the photo-generated electrons at $x = + 2 l_{n}$ is ________ mA/cm $^{2}$ (rounded off to two decimal places).

GATE 2022 EC question 56: A bar labelled 'Semiconductor' spans x = -2l_n to +2l_n. — Figure from the GATE 2022 EC paper. It is part of the question.

Answer and solution

ANSWER

The current density due to photo-generated electrons at $x = + 2 l_{n}$ has magnitude $q (1 0^{20}) l_{n} e^{- 1} \approx 0.59 mA/cm^{2}$ .

Official answer: 0.57 to 0.61Source: Official answer key, GATE 2022 EC (IIT Kharagpur)

Read the two profiles from the figure. The generation rate is $G (x) = 1 0^{20} cm^{- 3} s^{- 1}$ for $∣ x ∣ < l_{n}$ and $0$ for $l_{n} < ∣ x ∣ < 2 l_{n}$ . The recombination rate is $R (x) = 1 0^{20} e^{- ∣ x ∣/ l_{n}} cm^{- 3} s^{- 1}$ for $∣ x ∣ < l_{n}$ and $0$ for $l_{n} < ∣ x ∣ < 2 l_{n}$ . It drops to zero as a step at $x = \pm l_{n}$ and does not decay all the way to $\pm 2 l_{n}$ .
Write the steady-state continuity equation for electrons, with zero field and $q = 1.6 \times 1 0^{- 19} C$ as a magnitude: $0 = \frac{1}{q} \frac{d J _{n}}{d x} + G (x) - R (x) \Rightarrow \frac{d J _{n}}{d x} = q [R (x) - G (x)]$
Use symmetry. $G$ and $R$ are even in $x$ , and the boundary conditions $n^{'} (\pm 2 l_{n}) = 0$ are symmetric too, so the excess electron profile $n^{'} (x)$ is even. The diffusion current $J_{n} = q D_{n} \frac{d n ^{'}}{d x}$ is then odd in $x$ , which gives $J_{n} (0) = 0$ .
Integrate from $0$ to $2 l_{n}$ : $J_{n} (2 l_{n}) = J_{n} (0) + q \int_{0}^{2 l_{n}} [R (x) - G (x)] d x = q \int_{0}^{2 l_{n}} [R (x) - G (x)] d x$
Total recombination per unit area on the right half. $R$ is nonzero only on $0 < x < l_{n}$ : $\int_{0}^{2 l_{n}} R d x = \int_{0}^{l_{n}} 1 0^{20} e^{- x / l_{n}} d x = 1 0^{20} l_{n} (1 - e^{- 1})$
Total generation per unit area on the right half: $\int_{0}^{2 l_{n}} G d x = \int_{0}^{l_{n}} 1 0^{20} d x = 1 0^{20} l_{n}$
Take the difference: $\int_{0}^{2 l_{n}} (R - G) d x = 1 0^{20} l_{n} [(1 - e^{- 1}) - 1] = - 1 0^{20} l_{n} e^{- 1}$ Physically, the electrons generated in $0 < x < 2 l_{n}$ but not recombined there, $1 0^{20} l_{n} e^{- 1}$ per unit area per second, must leave through $x = 2 l_{n}$ . Between $l_{n}$ and $2 l_{n}$ we have $G = R = 0$ , so the current is constant in that region.
So $∣ J_{n} (2 l_{n}) ∣ = q \times 1 0^{20} \times l_{n} \times e^{- 1}$ , with $q = 1.6 \times 1 0^{- 19} C$ and $l_{n} = 1 0^{- 4} cm$ : $∣ J_{n} ∣ = (1.6 \times 1 0^{- 19}) (1 0^{20}) (1 0^{- 4}) (0.3679) A/cm^{2}$
Compute: $(1.6 \times 1 0^{- 19}) (1 0^{20}) = 16$ , and $16 \times 1 0^{- 4} = 1.6 \times 1 0^{- 3}$ . Then $1.6 \times 1 0^{- 3} \times 0.3679 = 5.886 \times 1 0^{- 4} A/cm^{2}$ .
Convert: $5.886 \times 1 0^{- 4} A/cm^{2} = 0.5886 mA/cm^{2} \approx 0.59 mA/cm^{2}$ .

KEY IDEA

In steady state with no field, the continuity equation says $d J_{n} / d x = q (R - G)$ . Because the profiles are symmetric, $J_{n} (0) = 0$ . So the current leaving through $x = 2 l_{n}$ is $q$ times the carriers generated but not recombined between $0$ and $2 l_{n}$ , per unit area per second. You get it by integrating $G - R$ ; you never need to solve the diffusion equation for $n^{'} (x)$ .

WATCH OUT FOR

The usual error is assuming $R = 1 0^{20} e^{- ∣ x ∣/ l_{n}}$ continues to $x = \pm 2 l_{n}$ . The figure shows $R$ stepping down to zero at $x = \pm l_{n}$ . Integrating $R$ out to $2 l_{n}$ gives $e^{- 2}$ in place of $e^{- 1}$ and the wrong answer $0.22 mA/cm^{2}$ . A second trap is using $q = - 1.6 \times 1 0^{- 19}$ C with its sign and reporting a negative value: the question asks for the magnitude.

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.56: solution and answer

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