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GATE 2016 EC – Question 49

Electronic Devices · Carrier Transport · 2 marks · Multiple choice

Consider a silicon sample at T = 300 K, with a uniform donor density $N_d = 5 \times 10^{16}$ cm$^{-3}$, illuminated uniformly such that the optical generation rate is $G_{opt} = 1.5 \times 10^{20}$ cm$^{-3}$s$^{-1}$ throughout the sample. The incident radiation is turned off at $t = 0$. Assume low-level injection to be valid and ignore surface effects. The carrier lifetimes are $\tau_{p0} = 0.1$ μs and $\tau_{n0} = 0.5$ μs.

The hole concentration at $t = 0$ and the hole concentration at $t = 0.3$ μs, respectively, are

  1. $1.5 \times 10^{13}$ cm$^{-3}$ and $7.47 \times 10^{11}$ cm$^{-3}$
  2. $1.5 \times 10^{13}$ cm$^{-3}$ and $8.23 \times 10^{11}$ cm$^{-3}$
  3. $7.5 \times 10^{13}$ cm$^{-3}$ and $3.73 \times 10^{11}$ cm$^{-3}$
  4. $7.5 \times 10^{13}$ cm$^{-3}$ and $4.12 \times 10^{11}$ cm$^{-3}$

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Correct answer: (A) $1.5 \times 10^{13}$ cm$^{-3}$ and $7.47 \times 10^{11}$ cm$^{-3}$

Explanation

In steady state the holes, which are the minority carriers in this $n$-type sample, are generated at the rate $G_{opt}$ and recombine with lifetime $\tau_{p0}$, so $p(0) = G_{opt}\tau_{p0} = 1.5 \times 10^{20} \times 0.1 \times 10^{-6} = 1.5 \times 10^{13}$ cm$^{-3}$. After the light is turned off, the excess holes decay as $e^{-t/\tau_{p0}}$, so at $t = 0.3$ μs the concentration is $1.5 \times 10^{13} \times e^{-3} = 7.47 \times 10^{11}$ cm$^{-3}$.