Question

Two perfectly matched Silicon transistors are connected as shown in the figure. Assuming the $\beta$ of the transistor to be very high and the forward voltage drop in diodes to be 0.7 V, the value of current I is ________mA .

• A. 1.8

Answer 1

The correct option is A 1.8

$\frac{I_{C1}}{I_{C2}}=e^{\left(\frac{V_{BE1}-V_{BE2}}{V_T}\right)}=e^0=1$ $[\because V_{BE1}=V_{BE2}]$

Since $\beta$ for both are same, therefore

$I_{B1}=I_{B2}=I_B$

Applying KVL to loop as shown,

$I=\frac{0-0.7-0.7-(-5)}{2k\Omega }=1.8mA$