Question

A transistor is connected in common emitter circuit configuration, the collector supply voltage is 10 V and the voltage drop across a resistor of 1000 $\Omega$ in the collector circuit is 0.6 V. If the current gain factor ($\beta$) is 24, then the base current is _________ $\mu$A. (Round off to the Nearest Integer)

Hint:

In transistor problems, remember the fundamental relationships: $I_E = I_B + I_C$, $\alpha = I_C/I_E$, and $\beta = I_C/I_B$. The problem often gives you enough information to find one current, from which you can find the others using the gain factors.

Answer 1

Correct Answer: 25

First, we find the collector current ($I_C$) using Ohm's law for the collector resistor ($R_C$).
Voltage drop across $R_C$, $V_{R_C} = 0.6$ V.
Resistance of $R_C = 1000 \, \Omega$.
$I_C = \frac{V_{R_C}}{R_C} = \frac{0.6 \text{ V}}{1000 \, \Omega} = 0.6 \times 10^{-3}$ A.
$I_C = 0.6$ mA.
Next, we use the current gain factor ($\beta$) to find the base current ($I_B$). The formula for $\beta$ in a common emitter configuration is:
$\beta = \frac{I_C}{I_B}$
We are given $\beta = 24$.
Rearranging the formula to solve for $I_B$:
$I_B = \frac{I_C}{\beta} = \frac{0.6 \times 10^{-3} \text{ A}}{24}$
$I_B = \frac{6 \times 10^{-4}}{24} = \frac{1}{4} \times 10^{-4} \text{ A} = 0.25 \times 10^{-4}$ A.
The question asks for the answer in microamperes ($\mu$A).
$1 \, \mu\text{A} = 10^{-6}$ A.
$I_B = 0.25 \times 10^{-4} \text{ A} = 25 \times 10^{-6} \text{ A} = 25 \, \mu$A.
The nearest integer is 25.