First slide

Transistors

Question

In a n-p-n transistor, 10¹⁰ electrons enter the emitter in 10^-6 s and 4% of the electrons are lost in base. The current transfer ratio will be

Moderate

A

0.98
B

0.97
C

0.96
D

0.94

Solution

Number of electrons reaching the collector,
$n_{C} = \frac{96}{100} \times 10^{10} = 0 .96 \times 10^{10}$
Emitter current, $I_{E} = \frac{n_{E} \times e}{t}$
Collector current, $I_{C} = \frac{n_{C} \times e}{t}$
Thus, current transfer ratio,
$α = \frac{I_{C}}{I_{E}} = \frac{n_{C}}{n_{E}} = \frac{0 .96 \times 10^{10}}{10^{10}} = 0 .96$

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