Question

Consider the circuit shown in the figure below :



If the value of $\alpha$ = 0.992 and the value of $V_{BE\left(on\right)}=-0.7V,$ then the value of base to collector voltage $V_{BC}$ is equal to

• A. 3.25 V
• B. 4.47 V
• C. -4.47 V
• D. -3.25 V

Answer 1

The correct option is B 4.47 V

The circuit, we obtain the parameter
Emitter voltage, $V_E=V_{EB}=0.7V$
Emitter current, $I_E=\frac{9-0.7}{4\times {10}^3}A=2.075mA$
Collector current, $I_C=\frac{\beta }{1+\beta }{I}_E=\alpha I_E=0.992\times 2.075=2.058mA$



Applying KVL in base-collector loop (loop 1), we get
$0-V_{BC}-2.2\times {10}^3\times I_C+9=0$


$V_{BC}=4.47V$