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 very close to unity .

The value of the collector-to-emitter voltage $V_{CE}$ will be

• A. 4 V
• B. 5 V
• C. 6 V
• D. 3 V

Answer 1

The correct option is C 6 V



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

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

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

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 $I_E{R}_E=4.8-0.8=4A$
$I_E=\frac{4}{2\times {10}^3}$ = 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 ) = 6 V