Question

Consider the circuit shown in the figure below:



The value of cut-in voltage $V_{D_1}=V_{D_2}=V_D$ is equal to the forward cut-in voltage of the transistor $V_{BE\left(on\right)}.$ Assume that the value of $\beta$ to be very large, then collector current is equal to

• A. 2.12mA
• B. 1.45mA
• C. 6.55 mA
• D. 3.19mA

Answer 1

The correct option is A 2.12mA



Since, $\beta$ is very large, we can neglect $I_B$
$I_1=I_2$
and $I_C=I_E$
Applying KVL in loop 1, we get,
$0-I_1{R}_1-V_{D1}-V_{D2}-I_2{R}_2+20=0$
$I_1=\frac{-2V_D+20}{R_1+R_2}=\frac{-2V_D+20}{20k\Omega }$
Again applying KVL in loop 2, we get,
$-2V_D-I_2{R}_2+I_C{R}_3+V_{BE}=0$
$\Rightarrow I_C{R}_3=V_D+I_1{R}_2$
$I_C{R}_3=V_D+\frac{-2V_D+20}{20k\Omega }\times 10k\Omega$
$I_C=\frac{10}{4.7}\times {10}^{-3}=2.12mA$