Consider the transistor circuit shown in the figure below :

The transistor is a silicon with $V_{B E} = 0.7 V$ when transistor conducts and $V_{C E (s a t)} = 0.2 V$ If the transistor enters into saturation region. The value of $R_{B}$ was suppose to be $300 k Ω$ but due to the fault in rheostate, the actual value of the resistor was 50% less than the value which was assumed then the value of current $I_{C}$ is equal to _______mA.
4.9

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Solution

The correct option is A 4.9
Now, $R_{B} = R_{B} - 50 % R_{B} = 300 - 0.5 \times 300 = 150 k Ω$
$∴ I_{B} = \frac{10 - 0.7}{150} \times 10^{- 3} = \frac{9.3}{150} m A = 62 μ A$
$∴ I_{C} = β I_{B} = 6.2 m A$
But for $I_{C} = 6.2 m A$
$V_{C E} = V_{C C} - I_{C} R_{C} = - 2.4 V$
Thus, the transistor is in saturation region
$∴ I_{C} = \frac{10 - 0.2}{2} \times 10^{- 3} = 4.9 m A$

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