Two identical enhancement NMOS transistors M_1 and M_2 form a large-signal current mirror

Two identical enhancement NMOS transistors \(M_1\) and \(M_2\) form a large-signal current mirror. Both have \(V_{tn}=0.70\,\mathrm{V}\) and \(k_n=0.50\,\mathrm{mA}/\mathrm{V}^2\). Sources and bodies of both devices are grounded. Gate current is zero. \(M_1\) is diode-connected: its gate is shorted to its drain, and an ideal current source \(I_{\mathrm{REF}}=80.0\,\mu\mathrm{A}\) enters that drain-gate node. The gate of \(M_2\) is connected to the gate of \(M_1\). The drain of \(M_2\) is held at a voltage \(v_{D2}\) by an ideal source. Drain currents \(i_{D1}\) and \(i_{D2}\) enter the respective drains. In saturation with channel-length modulation, \[ i_D=\frac12 k_n(v_{GS}-V_{tn})^2(1+\lambda v_{DS}). \] When \(\lambda=0\) the factor \((1+\lambda v_{DS})\) is omitted. Saturation of either device still requires \(v_{DS}\ge v_{GS}-V_{tn}\) for that device. (a) Take \(\lambda=0\). Compute \(v_{GS}\) of the pair. Determine the smallest \(v_{D2}\) for which \(M_2\) is in saturation, and the corresponding \(i_{D2}\) for every \(v_{D2}\) in saturation. (b) Take \(\lambda=0.030\,\mathrm{V}^{-1}\). Write the equation that determines \(v_{GS}\) from \(M_1\), solve it, and then compute \(i_{D2}\) at \(v_{D2}=1.00\,\mathrm{V}\) and at \(v_{D2}=4.00\,\mathrm{V}\). For each of those two drain voltages, check whether \(M_2\) is in saturation. (c) With the \(\lambda>0\) model, form the ratio \(i_{D2}/I_{\mathrm{REF}}\) at \(v_{D2}=4.00\,\mathrm{V}\). Explain, using only the large-signal formulae, why the ratio is not identically one.

The diode-connected reference sets a single \(v_{GS}\) shared by both gates; \(M_2\)’s current then follows from that \(v_{GS}\) and from \(v_{D2}\).