All transistor in the N output current mirror in the figure are matched with a finite gain β and early voltage V_A = ∞. The expression for load current is

A. $$\frac{{{{\text{I}}_{{\text{ref}}}}}}{{\left( {1 + \frac{{\left( {1 + {\text{N}}} \right)}}{{\beta \left( {\beta + 1} \right)}}} \right)}}$$

B. $$\frac{{{{\text{I}}_{{\text{ref}}}}}}{{\left( {1 + \frac{{\text{N}}}{{\beta \left( {\beta + 1} \right)}}} \right)}}$$

C. $$\frac{{\beta {{\text{I}}_{{\text{ref}}}}}}{{\left( {1 + \frac{{\left( {1 + {\text{N}}} \right)}}{{\beta \left( {\beta + 1} \right)}}} \right)}}$$

D. $$\frac{{\beta {{\text{I}}_{{\text{ref}}}}}}{{\left( {1 + \frac{{\text{N}}}{{\beta \left( {\beta + 1} \right)}}} \right)}}$$

Answer: Option A