Question 10.SE.MCQ.1: For the common-base configuration shown in the following fig......

Wiley Acing the Gate Electronics and Communication Engineering [1079457]

For the common-base configuration shown in the following figure, the value of collector current $(IC)$ is (given that the value of α is 0.95)

(a) 0.9 mA (b) 0.8 mA

(c) 0.947 mA (d) 0.847 mA

1079457-10.1

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The value of load resistance $R_L = 5 kΩ$ The voltage drop across the resistor is 4.5 V.

Therefore, the current flowing through the resistor = $4.5 /\left(5 \times 10^3\right) A =0.9 mA$

The current flowing through the resistor is the collector current.

Therefore, collector current is equal to 0.9 mA.

Ans. (a)

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