Question

Consider the circuit shown in the figure below:
Both the transistors are matched with $I_{S1}=I_{S2}=4\times {10}^{-16}A\text{and}{V}_A=\infty .$ If base current for transistor $T_1\text{is}15.625\mu A$ and base current of $T_2$ is very small compared to emitter and collector current, then the values of $\beta$ for the transistors is approximately equal to (Assume $V_T=25.9mV$

• A. 150
• B. 117
• C. 189
• D. 126

Answer 1

The correct option is B 117

Drawing the Thevenin's equivalent circuit of the source, we get

Since the transistors are perfectly matched
$V_{BE1}=V_{BE2}=V_{BE}$
$\therefore 2V_{BE1}=1.6V-5.76\times {10}^3\times 15.625\times {10}^{-6}$
$=1.51V$
Thus, $V_{BE1}=\frac{1.51}{2}=0.755V$
Thus, $I_C=I_s\exp \left(\frac{V_{BE}}{V_T}\right)$
$4\times {10}^{-16}\exp \left(\frac{0.755}{0.0259}\right)$
$=1.828\times {10}^{-3}A$
$\therefore \beta =\frac{I_C}{I_B}=\frac{1.828}{15.625}\times {10}^3=117$