Question: 42: The collector current in a common base amplifier using \mathrm{n-p-n} transistor is 24 mA . If 80 \% of the electrons released by the emitter is accepted by the collector, then the base current is numerically:
(1) 6 mA and leaving the base
(2) 3 mA and leaving the base
(3) 6 mA and entering the base
(4) 3 mA and entering the base
Answer: Option (3)
Explanation:
In a common base amplifier, the current gain \alpha is defined as the ratio of collector current to emitter current.
\alpha=\frac{I_C}{I_E}It is given that 80\% of the electrons released by the emitter are accepted by the collector.
\alpha=0.8The collector current is given as:
I_C=24\ \mathrm{mA}Using the relation for \alpha, the emitter current is:
I_E=\frac{I_C}{\alpha}=\frac{24}{0.8}=30\ \mathrm{mA}The base current is the difference between emitter current and collector current.
I_B=I_E-I_C I_B=30-24=6\ \mathrm{mA}In an \mathrm{n-p-n} transistor, the base current enters the base region.
Hence, the base current is 6\ \mathrm{mA} and it enters the base.
Therefore, the correct option is (3).