In $n-p-n$ transistor circuit, the collector current is 10 mA . If $95 \%$ of the electrons emitted reach…

In $n-p-n$ transistor circuit, the collector current is 10 mA . If $95 \%$ of the electrons emitted reach the collector, then the base current is nearly
  1. 5.3 mA
  2. 53 mA
  3. 35 mA
  4. 0.53 mA

Solution

In n-p-n transistor, $\begin{aligned} & \mathrm{I}_{\mathrm{C}}=10 \mathrm{~mA}, \mathrm{I}_{\mathrm{C}}=95 \% \text { of } \mathrm{I}_{\mathrm{E}}=\frac{95}{100} \mathrm{I}_{\mathrm{E}} \\ & \therefore \mathrm{I}_{\mathrm{E}}=\frac{100 \mathrm{I}_{\mathrm{C}}}{95}=\frac{100 \times 10}{95}=10.53 \mathrm{~mA} \end{aligned}$
Now, $\mathrm{I}_{\mathrm{E}}^{\prime}=\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}$
$\therefore \quad \mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{E}}-\mathrm{I}_{\mathrm{C}}=10.53-10=0.53 \mathrm{~mA}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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