Question

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

• A. $2V$
• B. $0.2V$
• C. $0.8V$
• D. zero

Answer 1

Answer: (B)

$0.2V$

Given, $I_B=40\mu A=40\times {10}^{-6}A$

In the given circuit, the emitter part is grounded, i.e., at zero potential. Therefore, we can use Kirchhoff’s Voltage law-

${\text{V}}_{\text{CC}}-\left(245k\Omega \right).I_B-{\text{V}}_{\text{BE}}=0$

${\text{V}}_{\text{BE}}=10-\left(245\times {10}^3\times 40\times {10}^{-6}\right)$

${\text{V}}_{\text{BE}}=\left(10-9.8\right)\text{Volt}=0.2\text{Volt}$