Question

Consider the circuit shown in the figure. Assume base-to-emitter voltage $V_{BE}=0.8$ V and common base current gain $\alpha$ of the transistor is unity

The value of the collector -to-emmitter voltage $V_{CE}$ is

• A. 6 V
• B. 4 V
• C. 3 V
• D. 5 V

Answer 1

The correct option is A 6 V

By taking the Thevenin's equivalent between base and ground nodes, the given circuit can be reduced as follwos

$V_{Th}=\frac{16}{16+44}\times 18V=4.8V,$

$R_{Th}=\left(44k||16k\right)\Omega$ = 11.73 kilo-ohm

Applying KVL in loop 1 we get

$I_E{R}_E=V_{Th}-V_{BE}-I_B{R}_{Th}$

$I_B=0A$

Given, $\alpha =1$

So, $I_E{R}_E=4.8-0.8$

= 4 V

$I_E=\frac{4}{2\times {10}^3}=2mA$

$I_C=I_E=2mA$

$V_{CE}=V_{CC}-I_C{R}_C-I_E{R}_E$

$=18-(2\times 4)-(2\times 2)=6V$