KCET · Physics · Semiconductors
The dc common emitter current gain of a n-p-n transistor is 50, The potential difference applied
across the collector and emitter of a transistor used in \( C E \) configuration is \( V_{C E}=2 V \). If the
collector resistance,
\( R_{C}=4 \mathrm{~K} \Omega \), the base current \( \left(I_{B}\right) \) and the collector current \( \left(I_{C}\right) \) are
- A \( I_{B}=10 \mu \mathrm{A}, I_{C}=0.5 \mathrm{~mA} \)
- B \( I_{B}=0.5 \mu \mathrm{A}, I_{C}=10 \mathrm{~mA} \)
- C \( I_{B}=5 \mu \mathrm{A}, I_{C}=1 \mathrm{~mA} \)
- D \( I_{B}=1 \mu \mathrm{A}, I_{C}=0.5 \mathrm{~mA} \)
Answer & Solution
Correct Answer
(A) \( I_{B}=10 \mu \mathrm{A}, I_{C}=0.5 \mathrm{~mA} \)
Step-by-step Solution
Detailed explanation
Given, current gain,
\[
\begin{array}{l}
\beta=50 ; V_{C E}=2 V ; R_{C}=4 k \Omega \\
\therefore I_{C}=\frac{V_{C E}}{R_{C}}=\frac{2}{4 \times 10^{3}}=0.5 \times 10^{-3} A \\
\Rightarrow I_{C}=0.5 \mathrm{~mA}
\end{array}
\]
\[
\begin{array}{l}
\text { Also, } 85 \beta=\frac{I_{C}}{I_{B}} \\
\Rightarrow I_{B}=\frac{I_{C}}{\beta}=\frac{0.5 \times 10^{-3}}{50}=10^{-5} \mathrm{~A} \\
\Rightarrow I_{B}=10 \mu A
\end{array}
\]
\[
\begin{array}{l}
\beta=50 ; V_{C E}=2 V ; R_{C}=4 k \Omega \\
\therefore I_{C}=\frac{V_{C E}}{R_{C}}=\frac{2}{4 \times 10^{3}}=0.5 \times 10^{-3} A \\
\Rightarrow I_{C}=0.5 \mathrm{~mA}
\end{array}
\]
\[
\begin{array}{l}
\text { Also, } 85 \beta=\frac{I_{C}}{I_{B}} \\
\Rightarrow I_{B}=\frac{I_{C}}{\beta}=\frac{0.5 \times 10^{-3}}{50}=10^{-5} \mathrm{~A} \\
\Rightarrow I_{B}=10 \mu A
\end{array}
\]
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