Topic: Analog Electronics
All transistor in the N output current mirror in figure given below are matched with a finite gain $\beta$ and early voltage $V_A=\infty$. The expression for load current is
- A. $\frac{I_{ref}}{(1+\frac{(1+N)}{\beta(\beta+1)})}$
- B. $\frac{I_{ref}}{(1+\frac{N}{\beta(\beta+1)})}$
- C. $\frac{\beta I_{ref}}{(1+\frac{(1+N)}{\beta(\beta+1)})}$
- D. $\frac{\beta I_{ref}}{(1+\frac{N}{\beta(\beta+1)})}$
Answer: A. $\frac{I_{ref}}{(1+\frac{(1+N)}{\beta(\beta+1)})}$
Explanation: $I_{ref}=I_C+\frac{(N+1)I_B}{(\beta+1)}$.
$I_{ref}=I_B(\beta+\frac{(N+1)}{\beta+1})$.
$I_B=\frac{I_{ref}}{(\beta+\frac{(N+1)}{\beta+1})}$.
$I_1=\beta I_B=\frac{\beta I_{ref}}{(\beta+\frac{(N+1)}{\beta+1})}=\frac{I_{ref}}{(1+\frac{(N+1)}{\beta(\beta+1)})$.