Consider the circuit shown in Figure 7.25(a). The bias voltages are changed to (V^{+}=3 mathrm{~V}) and (V^{-}=-3

Question:

Consider the circuit shown in Figure 7.25(a). The bias voltages are changed to $V^{+}=3 \mathrm{~V}$ and $V^{-}=-3 \mathrm{~V}$. The load resistor is $R_{L}=20 \mathrm{k} \Omega$. The transistor parameters are $K_{p}=0.1 \mathrm{~mA} / \mathrm{V}^{2}, V_{T P}=-0.6 \mathrm{~V}$, and $\lambda=0$.

(a) Design the circuit such that $I_{D Q}=0.2 \mathrm{~mA}$ and $V_{S D Q}=1.9 \mathrm{~V}$.

(b) Determine the value of $C_{L}$ that produces a corner frequency of $f_{H}=4 \mathrm{MHz}$.

Figure 7.25(a):-