Consider the current gain amplifier shown in Figure P12.46. The transistor parameters are \(K_{n}=K_{p}=10 \mathrm{~mA} / \mathrm{V}^{2}, V_{T N}=1 \mathrm{~V}, V_{T P}=-1 \mathrm{~V}\), and \(\lambda_{n}=\lambda_{p}=0\). The LED turn-on voltage is \(V_{\gamma}=1.6 \mathrm{~V}\). Assume the LED small-signal resistance is \(r_{f}=0\).

(a) Determine the quiescent currents \(I_{D Q 1}\) and \(I_{D Q 2}\).

(b) Show that the small-signal current gain is given by

\[A_{i}=\frac{I_{o}}{I_{i}}=\frac{-g_{m 2} R_{D 1}}{1+\frac{1}{g_{m 1}\left(R_{F}+R_{D 2}\right)}+\frac{g_{m 2} R_{D 1} R_{D 2}}{R_{F}+R_{D 2}}}\]

(c) Calculate the value of \(A_{i}\).

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V+= 10 V M VG = 7.6 V RDI = 52592 ww M2 Figure P12.46 LED ww Rp=500 2 RD2= 250