Question

In case of bipolar transistor $\beta = 45$ . The potential drop across the collector resistance of $1\;{{k\Omega }}$ is $5\;{\text{V}}$ . The base current is approximately
A) $222\;{{\mu A}}$
B) $55\;{{\mu A}}$
C) $111\;{{\mu A}}$
D) $45\;{{\mu A}}$

Answer 1

Hint: From the Ohm’s law the collector current can be calculated from the given resistance and voltage across the collector. And the ratio of the collector current and base current is given. Hence the base current can be easily calculated.

Complete step by step answer:
Given the collector resistance is ${R_C} = 1\;{{k\Omega }}$ and the voltage is $V = 5\;{\text{V}}$.
The two PN junctions of the transistor are the collector junction and the emitter junction. When the emitter junction is forward biased, the collector junction will be reverse biased. In this active region, the base current can control the collector region.
According to Ohm’s law, the voltage drop is proportional to the current. Thus the expression for the resistance is given as,
${I_C} = \dfrac{V}{{{R_C}}}$
Substituting the values in the above expression,
$
 \Rightarrow {I_C} = \dfrac{{5\;V}}{{1 \times {{10}^3}\;\Omega }} \\
 \Rightarrow I_C = 5\;{\text{mA}} \\
$
The current through the collector is $5\;{\text{mA}}$.
We have given the ratio $\beta = 45$.
$\beta $ is the ratio of collector current to the base current. The range of $\beta $ in the NPN transistors has $50 - 200$. The relation between the emitter current and collector current is termed as,
$\alpha = \dfrac{{{I_C}}}{{{I_E}}}$
Where, ${I_E}$ is the emitter current and ${I_C}$ is the collector current.
Thus,
$\beta = \dfrac{{{I_C}}}{{{I_B}}}$
Where ${I_C}$ is the collector current and ${I_B}$ is the base current.
Therefore,
$\Rightarrow 45 = \dfrac{{5\;{\text{mA}}}}{{{I_B}}} $
$\Rightarrow I_B = \dfrac{{5\;{\text{mA}}}}{{45}} $
$\Rightarrow I_B = 0.111\;{\text{mA}} $
$I_B {\text{ = 111}}\,{{\mu A}}$

The base current is ${\text{111}}\,{{\mu A}}$. Hence, the correct answer is option (C).

Note:
The collector current will be greater than the base current. This is because the majority of carriers are flowing through the collector. The base is thinner than the collector.