Question

The circuit shown below works as differential amplifier. Both transistor are of silicon material and have $\beta$ = 100, then the value of $V_0$ if $V_i$ = 5 V will be _____V.
(Take $V_{BE}=0.7Volts$)

• A. 0 V
• B. 12 V
• C. 8 V
• D. 6 V

Answer 1

The correct option is B 12 V

Let us consider $Q_2$ to be off and $Q_1$ is in the active mode, based on the voltage given , then
$I_{E1}=1mA$
$I_{C1}=1mA$ = $\frac{\beta }{1+\beta }{I}_{E1}$
$=\frac{100}{101}\times 1$
= 0.99 mA
and $V_{CBI}=12-5\times 0.99-5$
= 2.05 V
Since $V_{CBI}$ is +ve , $Q_1$ is in active mode.
Assume $V_{BEI}$ = 07 V, $V_E=V_B-V_{BE1}$
= 5 - 0.7 V = 4.3 V
$\because$ $V_E=4.3$ means that
$V_{BE2}=V_{B2}-V_E$
= 2.5 - 4.3 = - 1.8 V
Since, $V_{BE2}$ is negative , our assumptions of $Q_2$ being off is confirmed . As no current flows through $Q_2$
So , $V_0$ = 12 V