Question

Consider the circuit shown in the figure below:

$\text{Both the transistors}{Q}_1\text{and}{Q}_2\text{are matched with}\beta =50\text{. The value of}{V}_{BE}=0.7\text{V, the value of current}{I}_0\text{is equal to \_\_\_\_\_\_mA.}$

• A. 0.241

Answer 1

The correct option is A 0.241
The circuit can be redrawn as,

For current mirror circuit,

$I_{\text{reff}}=\frac{V_{CC}-V_{BE}}{R}=\frac{10-0.7}{37\times {10}^3}=0.251mA$
$\text{now,}{I}_0=\frac{I_{\text{reff}}}{(1+\frac{2}{\beta })}=\frac{I_{\text{reff}}}{(1+\frac{2}{50})}=0.241\text{mA}$