In a transistor, when the emitter current changes by 9.85 $\mathrm{mA}$, the collector current changes to $9…
In a transistor, when the emitter current changes by 9.85 $\mathrm{mA}$, the collector current changes to $9.5 \mathrm{~mA}$. Then the base current is
- $0.05 \mathrm{~mA}$
- $0.85 \mathrm{~mA}$
- $0.8 \mathrm{~mA}$
- $0.35 \mathrm{~mA}$
Solution
$\begin{aligned}
& \mathrm{I}_E=9.85 \mathrm{~mA} \\
& \mathrm{I}_{\mathrm{C}}=9.5 \mathrm{~mA} \\
& \mathrm{I}_E=\mathrm{I}_B+\mathrm{I}_{\mathrm{C}}
\end{aligned}$
Base current, $I_B=I_E-I_C$
$=9.85-9.5=0.35 \mathrm{~mA}$
Asked in: AP EAMCET 2023 (15 May Shift 2)
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