Consider the transconductance feedback amplifier shown in Figure P12.56 with \(R_{D}=1.6 \mathrm{k} \Omega\) and \(R_{L}=248 \Omega\). The transistor parameters are \(V_{T N}=0.5 \mathrm{~V}, V_{T P}=-0.5 \mathrm{~V}, K_{n}=2 \mathrm{~mA} / \mathrm{V}^{2}, K_{p}=10 \mathrm{~mA} / \mathrm{V}^{2}\), and \(\lambda_{n}=\) \(\lambda_{p}=0\). The LED turn-on voltage is \(V_{\gamma}=1.6 \mathrm{~V}\). Assume the LED smallsignal resistance is \(r_{f}=0\). The current source is ideal.

(a) Determine the quiescent values of \(V_{D 1}, I_{D Q 3}\) and \(V_{G 2}\).

(b) Derive the small-signal transconductance function \(A_{g f}=I_{o} / V_{i}\).

(c) Calculate the value of \(A_{g f}=I_{o} / V_{i}\).

Transcribed Image Text:

Vi www V+=3 V RD M1 www Rp M2 M3 D 1Q= 2 mA V = -3 V Figure P12.56 RL