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  4. A common emitter amplifier is designed with n-p-n transistor (α=0.99). The input impedance is 1 k Ω and load is 10 kΩ. The voltage gain will be

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Q. A common emitter amplifier is designed with $n-p-n$ transistor $(\alpha=0.99)$. The input impedance is $1 \,k \Omega$ and load is $10\, k\Omega$. The voltage gain will be

Semiconductor Electronics: Materials Devices and Simple Circuits

Solution:

Voltage gain = Current gain $\times$ Resistance gain
Current gain $(\beta)=\frac{\alpha}{1-\alpha}=\frac{0.99}{1-0.99}=99$
$\Rightarrow $ Voltage gain $=99 \times \frac{10 \,k \Omega}{1 \,k \Omega}=990$