Question

In a common-base configuration of transistor$\alpha =0.98,I_B=0.02\text{mA},R_L=5\text{k}\Omega$.
Output voltage across load is (approximately)

• A. $6.2\text{V}$
• B. $3.2\text{V}$
• C. $5.2\text{V}$
• D. $4.9\text{V}$

Answer 1

The correct option is D $4.9\text{V}$


Given:
Current gain, $\alpha =0.98$
We know that ,
$\alpha =\frac{I_C}{I_E}$
where ,
$I_C\text{is collector current}$
$I_E\text{is emmiter current}$
$\text{So we get ,}$
$\Rightarrow I_E\left(0.98\right)=I_C$

We know that
$I_E=I_B+I_C$ and
$I_B=0.02\text{mA}$
So, $I_E=1\text{mA}$
Hence, $I_C=0.98\text{mA}$

By ohm's law, voltage is given as:
$V_0=R_L\times I_C$
Since, $R_L=5\text{k}\Omega$
So, $V_0=4.9\text{V}$
Hence, option $\left(B\right)$ is correct.