In a PNP transistor working as a common-base amplifier, current gain is 0.96 and emitter current is 7.2 mA. The base current is (a) 0.4 mA (b) 0.2 mA (c) 0.29 mA (d) 0.35 mA
Added by Ashu G.
Step 1
96, Ie = 7.2 mA Using the formula, we get: 0.96 = Ic / 7.2 Ic = 0.96 * 7.2 Ic = 6.91 mA Show more…
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In a $P N P$ transistor working as a common-base amplifier, current gain is $0.96$ and emitter current is $7.2 \mathrm{~mA}$. The base current is (A) $0.4 \mathrm{~mA}$ (B) $0.2 \mathrm{~mA}$ (C) $0.29 \mathrm{~mA}$ (D) $0.35 \mathrm{~mA}$
The current gain of a transistor in a common emitter configuration is $40 .$ If the emitter current is $8.2 \mathrm{~mA}$, then base current is (a) $0.02 \mathrm{~mA}$ (b) $0.2 \mathrm{~mA}$ (c) $2.0 \mathrm{~mA}$ (d) $0.4 \mathrm{~mA}$
Electronic Devices
Round 1
A transistor operating in a common base configuration has forward current gain factor, $\alpha=0.99 .$ If the emitter current changes by $1 \mathrm{~mA}$, then the changes in the base current will be : (a) $100 \mathrm{~mA}$ (b) $0.01 \mathrm{~mA}$ (c) $0.99 \mathrm{~mA}$ (d) $99 \mathrm{~mA}$
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