Question

Consider the circuit shown in the figure below.

When the transistor is biased in active region,the value of voltage $|V_{BE}|$ = 0.7 V. Then the value of power dissipated by the transistor is equal to _____ mV.

• A. 5.028

Answer 1

The correct option is A 5.028

Assuming the transistor to be in active region, we get,

$I_C=\alpha I_E=\frac{\beta }{1+\beta }{I}_E=\frac{60}{61}\times 0.61\times {10}^{-3}$

$=0.6mA$

Now, the voltage at $V_C=-10+4.7\times 0.6=-7.18V$

$\text{Value of}{I}_B=\frac{I_C}{\beta }=\frac{0.6}{60}\times {10}^{-3}=1\times {10}^{-5}A=10\mu A$

now, $V_B=I_B{R}_B$

$=10\times {10}^{-6}\times 50\times {10}^5=0.5V$

$\therefore V_E=V_B+V_{EB}=0.5+0.7=1.2V$

$V_{EC}=1.2-(-7.18)=8.38V$

$\therefore \text{Power dissipated}{V}_{EC}\times I_C=8.38\times 0.6\times {10}^{-3}=5.028mW$