Question

All transistors in the N output mirror shown below are matched with a finite gain $\beta$ and early voltage $V_A=\infty$. The expression for each load current is

• A. $\frac{I_{ref}}{1+\frac{1+N}{\beta (\beta +1)}}$
• B. $\frac{{\beta }_{ref}}{1+\frac{1+N}{\beta +1}}$
• C. $\frac{I_{ref}}{1+\frac{N}{\beta +1}}$
• D. $\frac{\beta I_{ref}}{1+\frac{N}{\beta +1}}$

Answer 1

The correct option is A $\frac{I_{ref}}{1+\frac{1+N}{\beta (\beta +1)}}$



Since transistors $Q_1,Q_2,...Q_N$ are identical, So
$I_{B1}=I_{B2}=...=I_{BN}$
and $I_{CR}=I_{01}=I_{02}=...I_{ON}$
for transistors $Q_R,$
$I_{ref}=I_{CR}+I_{BS}$
$I_{ES}=(\beta +1)I_{BS}$
$\Rightarrow I_{ref}=I_{CR}+\frac{I_{ES}}{\beta +1}$ $..\left(i\right)$
$I_{ES}=I_{BR}+N{I}_B$
$I_{ES}=(N+1)\frac{I_{CR}}{\beta }$ $...\left(ii\right)$
Using (i) and (ii), $I_{ref}=I_{CR}+\frac{(N+1)}{\beta (\beta +1)}{I}_{CR}$
$\Rightarrow I_{CR}=I_{01}=I_{02}=...=\frac{I_{ref}}{1+\frac{(N+1)}{\beta (\beta +1)}}$