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 & Solution

Correct Answer: Option A

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All transistor in the N output current mirror in the figure...

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

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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