The transistor in the common-gate circuit in Figure P4.46 has the same parameters that are given in Problem 4.45. The output resistance \(R_{o}\) is to

be \(500 \Omega\) and the drain-to-source quiescent voltage is to be \(V_{D S Q}=V_{D S}(\) sat \()+0.3 \mathrm{~V}\).

(a) What is the value of \(R_{D}\) ?

(b) What is the quiescent drain current \(I_{D Q}\) ?

(c) Find the input resistance \(R_{i}\).

(d) Determine the small-signal voltage gain \(A_{v}=V_{o} / V_{i}\).

Data From Problem 4.45:-

Figure P4.45 is the ac equivalent circuit of a common-gate amplifier. The transistor parameters are \(V_{T N}=0.4 \mathrm{~V}, k_{n}^{\prime}=100 \mu \mathrm{A} / \mathrm{V}^{2}\), and \(\lambda=0\). The quiescent drain current is \(I_{D Q}=0.25 \mathrm{~mA}\). Determine the transistor \(W / L\) ratio and the value of \(R_{D}\) such that the small-signal voltage gain is \(A_{v}=V_{o} / V_{i}=20\) and the input resistance is \(R_{i}=500 \Omega\).

Transcribed Image Text:

V R 1.2 V = VGSQ Figure P4.46 ww Rp + VDD = 2.2 V