ISRO EC 2023 Q.43: solution and answer

ISRO 2023 · ECQ.43 · 1 mark · MCQ

Operating Frequency limitation for a Bipolar semi-conductor device is due to

(A)
Junction Parasitic
(B)
Electron Mobility
(C)
both (a) and (b)
(D)
None of above

See the question in the paper

Answer and solution

ANSWER

The operating frequency of a bipolar device is limited by both junction parasitic capacitances and electron mobility, which is option (c).

Official answer: CSource: ISRO's revised and final answer key of 16 February 2024

The upper frequency limit of a BJT is set by two things: how fast carriers cross the device, and how fast its internal capacitances can be charged and discharged.
Junction parasitics: the base-emitter capacitance $C_{π}$ (diffusion plus depletion) and the collector-base capacitance $C_{μ}$ form low-pass networks with the source and base resistances. Because of them the current gain falls at high frequency. So option (a) is a genuine limit.
Carrier mobility: for a uniform base of width $W_{B}$ , the base transit time is $τ_{b} = \frac{W _{B}^{2}}{2 D _{n}}$ , and the Einstein relation gives $D_{n} = μ_{n} V_{T}$ . A lower electron mobility gives a longer transit time and so a lower cut-off $f_{T} = \frac{1}{2 π ( τ _{b} + τ _{e} + \dots )}$ . So option (b) is also a genuine limit.
Both effects act together to fix the upper frequency of operation, so (a) alone and (b) alone are each incomplete, and (d) is wrong.
The correct choice is (c), both (a) and (b).

KEY IDEA

A BJT's high-frequency limit has two parts: the RC delay of the junction capacitances $C_{π}$ and $C_{μ}$ , and the carrier transit time across the base, which depends on mobility through $D_{n} = μ_{n} V_{T}$ .

WATCH OUT FOR

Picking only one cause. Many students name only the junction capacitances, or only the transit time, when both contribute to the cut-off frequency.

From the ISRO EC 2023 paper. Every question of this paper · The full paper (PDF, ISRO) · All 1 ISRO BJT question, with notes