GATE EC 2016 Set 1 — Question 49
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Electronic Devices → Carrier Transport → Generation & Recombination
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Question
Consider a silicon sample at , with a uniform donor density , illuminated uniformly such that the optical generation rate is throughout the sample. The incident radiation is turned off at . Assume low-level injection to be valid and ignore surface effects. The carrier lifetimes are and .
The hole concentration at and the hole concentration at , respectively, are
The hole concentration at and the hole concentration at , respectively, are
Correct answer
(A) 1.5 × 10¹³ cm⁻³ and 7.47 × 10¹¹ cm⁻³
Solution
Given:
Donor density (n-type silicon)
Optical generation rate
Hole lifetime
Electron lifetime
Temperature First, find the intrinsic carrier concentration for silicon at . A common value is .
Since low-level injection is assumed, . Therefore, .
.
The total hole concentration at time is .
We need to find :
So, .
Since is negligible compared to , we can approximate:
.Therefore, the hole concentration at is and at is .The correct option is (A).
Donor density (n-type silicon)
Optical generation rate
Hole lifetime
Electron lifetime
Temperature First, find the intrinsic carrier concentration for silicon at . A common value is .
1.Equilibrium hole concentration (): In thermal equilibrium, for an n-type semiconductor:
.2.Excess hole concentration () during steady-state illumination: Just before , the sample is uniformly illuminated, creating excess carriers. The steady-state excess hole concentration is:
3.Hole concentration at (): When the radiation is turned off at , the hole concentration is the sum of equilibrium and excess concentrations:
Since low-level injection is assumed, . Therefore, .
.
4.Hole concentration at (): After the radiation is turned off, the excess carriers decay exponentially with their lifetime. The excess hole concentration at time is:
The total hole concentration at time is .
We need to find :
So, .
Since is negligible compared to , we can approximate:
.Therefore, the hole concentration at is and at is .The correct option is (A).
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