In a \(p\) - \(n-p\) transistor working as a common base amplifier, when the current gain is 0.96 and…
In a \(p\) - \(n-p\) transistor working as a common base amplifier, when the current gain is 0.96 and emitter current is \(7.2 \mathrm{~mA}\), the base current is
\(0.4 \mathrm{~mA}\)
\(0.2 \mathrm{~mA}\)
\(0.29 \mathrm{~mA}\)
\(0.35 \mathrm{~mA}\)
Solution
In common base amplifier for \(p-n-p\) transistor, current gain, \(\alpha=0.96\)
Emitter current, \(I_E=7.2 \mathrm{~mA}\)
We know that,
\(\begin{aligned}
& \alpha & =\frac{I_C}{I_E} \Rightarrow 0.96=\frac{I_C}{7.2} \\
\Rightarrow \quad & I_C & =0.96 \times 7.2=6.912 \mathrm{~mA}
\end{aligned}\)
\(\therefore\) Base current,
\(\begin{aligned}
I_B & =I_E-I_C=7.2-6.912 \\
& =0.288 \mathrm{~mA} \simeq 0.29 \mathrm{~mA}
\end{aligned}\)