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1. |
All transistor in the N output current mirror in figure given below are matched with a finite gain β and early voltage VA = ∞. The expression for load current is |
A. | \(\frac{{{I_{ref}}}}{{\left( {1 + \frac{{\left( {1 + N} \right)}}{{\beta \left( {\beta + 1} \right)}}} \right)}}\) |
B. | \(\frac{{{I_{ref}}}}{{\left( {1 + \frac{N}{{\beta \left( {\beta + 1} \right)}}} \right)}}\) |
C. | \(\frac{{\beta {I_{ref}}}}{{\left( {1 + \frac{{\left( {1 + N} \right)}}{{\beta \left( {\beta + 1} \right)}}} \right)}}\) |
D. | \(\frac{{\beta {I_{ref}}}}{{\left( {1 + \frac{N}{{\beta \left( {\beta + 1} \right)}}} \right)}}\) |
Answer» B. \(\frac{{{I_{ref}}}}{{\left( {1 + \frac{N}{{\beta \left( {\beta + 1} \right)}}} \right)}}\) | |