Question

In a common emitter transistor circuit, the base current is $40\mu A$, then $V_{BE}$ is :-

• A. $2\text{V}$
• B. $0.2\text{V}$
• C. $0.8\text{V}$
• D. Zero

Answer 1

The correct option is B $0.2\text{V}$

Given that $I_B=40\mu A$
Apply KVL
$10V=\left(245\text{k}\Omega \right)\left(40\mu A\right)+V_{BE}$
$V_{BE}=(10-9.8)\text{V}=0.2\text{V}$