Question

In the circuit shown in figure, the base current is, $I_B=10\mu \text{A}$ and the collector current is, $I_C=5.2\text{mA}$. The transistor is in which mode?

• A. Active
• B. Saturation
• C. Cut-off
• D. Inverted

Answer 1

The correct option is B Saturation


Applying Kirchhoff's law along $ABMNP$, we get

$V_{BE}+I_B{R}_1=V_C$

$\Rightarrow V_{BE}=V_C-I_B{R}_1$

$=5.5-10\times {10}^{-6}\times 500\times {10}^3=0.5\text{V}.....\left(1\right)$

Similarly, applying Kirchhoff's law along $AEBCNP$,

$V_{CE}+I_C{R}_2=V_C$

$\Rightarrow V_{CE}=V_C-I_C{R}_2$

$=5.5-5.2\times {10}^{-3}\times {10}^3=0.3\text{V}...\left(2\right)$

From $\left(1\right)$, $V_{BE}=V_B-V_E=0.5\text{V}$

From $\left(2\right)$, $V_{CE}=V_C-V_E=0.3\text{V}$

$\therefore V_{BC}=V_B-V_C=0.5-0.3=0.2\text{V}$

This is a $\text{n-p-n}$ transistor and $V_{BE},V_{BC}$ both are positive i.e $V_B>V_E$ and $V_B>V_C$ or emitter base and collector-base both are forward biased.

Hence, the transistor is in saturation mode.

Thus, option $\left(B\right)$ is right.