Let us first consider the effects of $R_L$. The peak output current must be $\pm 100 \mu \mathrm{~A}$. In order to meet the $S R$ requirements a current magnitude of $\pm 1 \mathrm{~mA}$ is needed to charge the load capacitance. Since this current is so much larger than the current needed to meet the voltage specification across $R_L$, we can safely assume that all of the current supplied by the inverter is available to charge $C_L$. Using a value of $\pm 1 \mathrm{~mA}, W_1 / L_1$ needs to be approximately $3 \mu \mathrm{~m} / 2 \mu \mathrm{m}$ and $W_2 / L_2, 15 \mu \mathrm{~m} / 2 \mu \mathrm{~m}$.

The small-signal gain of this amplifier is $-8.21 \mathrm{~V} / \mathrm{V}$. This voltage gain is low because of the low output resistance in shunt with $R_L$. The output resistance of the amplifier is $50 \mathrm{k} \Omega$. The zero is located at 1.59 GHz and the pole is located at -11.14 kHz .