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GATE 2020 EC – Question 43

Electronic Devices · P-N Junction and Semiconductor Devices · 2 marks · Multiple choice

The base of an $npn$ BJT T1 has a linear doping profile $N_B(x)$ as shown below. The base of another $npn$ BJT T2 has a uniform doping $N_B$ of $10^{17}\ \text{cm}^{-3}$. All other parameters are identical for both the devices. Assuming that the hole density profile is the same as that of doping, the common-emitter current gain of T2 is

Diagram for GATE 2020 EC question 43
  1. approximately 2.0 times that of T1.
  2. approximately 0.3 times that of T1.
  3. approximately 2.5 times that of T1.
  4. approximately 0.7 times that of T1.

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Correct answer: (D) approximately 0.7 times that of T1.

Explanation

The current gain of a BJT is inversely proportional to the base Gummel number $\int_0^WN_B(x)\,dx$. For the linear profile falling from $10^{17}$ to $10^{14}\ \text{cm}^{-3}$ this is about $0.5\times10^{17}W$, while the uniform base gives $10^{17}W$. So the gain of T2 is about half that of T1, and of the given options 0.7 is the nearest. The official key awarded this question to all candidates.