Question

A common emitter amplifier circuit built using an n-p-n transistor is shown in the figure. Its DC current gain is 250 $R_C=1k\mathrm{\Omega }$ and $V_{cc}=10V$. What is the minimum base current for $V_{CE}$ to reach saturation is:

a. $100\mu A$
b. $7\mu A$
c. $40\mu A$
d. $$10\mu A$$

Answer 1

Hint: The n-p-n transistor is the transistor which is used to when we need to sink the current .Current gain is ratio between collector current and base current so first calculate collector current then use current gain equation to calculate the base current.

Formula used:
The formula for the collector current is given by,
$\Rightarrow i_C={\displaystyle \frac{V_{CC}}{R_C}}$
Where the collector current is $i_C$ the resistance is $R_C$ and the voltage is $V_{CC}$.

Complete step by step answer:
Step1:
At the saturation point the voltage $V_{CE}$ becomes zero.
Given that $V_{CC}=10V$
Therefore current in the collector is given by-
$\Rightarrow i_C={\displaystyle \frac{V_{CC}}{R_C}}$
The value of $V_{CC}=10V$.
$\Rightarrow i_C={\displaystyle \frac{10}{1000}}$
$\Rightarrow i_C=10mA$

Step2:
Now use the current gain equation to calculate the base current.
$\Rightarrow \beta ={\displaystyle \frac{i_C}{i_B}}$
Where $\beta =$ current gain $=250$
Put values in above equation we get,
$\Rightarrow 250={\displaystyle \frac{10}{i_B}}$
$\Rightarrow i_B={\displaystyle \frac{10}{250}}mA$
$\Rightarrow i_B=40\mu A$
The required current in the base is equal to $i_B=40\mu A$.

Hence, the correct answer is option (C).

Additional information:
The configuration in which the emitter is connected between the collector and base is known as a common emitter configuration. The input circuit is connected between emitter and base, and the output circuit is taken from the collector and emitter. Thus, the emitter is common to both the input and the output circuit, and hence the name is the common emitter configuration.

Note: The saturation region allows the transistor to conduct current from the emitter to the collector. With both the base collector junction and the base emitter junction forward-biased, the base current is so strong it exceeds the magnitude at which it can increase the collector current flow. Always keep in mind that at the saturation point $V_{CE}$ becomes zero.