Determine the voltage gain of each circuit in Fig. 9.59. Assume

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Vot \left.=-g_{2} V_{2}, h_{D}, 1, \ldots 1\right)(1(0,+1) 1 \cdotsfromern (2) \operatorname{lgg} s_{2}=-\frac{g m_{1}}{g m_{c}} \lg _{2} s_{2}Now romen"(2), \quad v_{n}=\operatorname{lgg}_{2}+\frac{g m_{2}}{g m_{2}} v_{g s_{2}}=\left(\left(1+\frac{g m_{2}}{g m_{2}}\right) v_{g}\right..n (1,1+1+1)^{2}+\left(1+g m_{2} / g m_{2}\right)so V_{\text {out }},-g_{n_{2}}\left(\frac{\operatorname{vin}}{1+g m_{2} / g m_{2}}\right), R_{D}A_{v}=\frac{v_{o u t}}{v_{i n}}=-\left(\frac{m_{1} \times g m_{2}}{g_{m_{2}}+g_{n}}\right) R_{D}(d)\begin{array}{l}\operatorname{ly}_{1} \sim d_{2} v_{y_{1}}\left\{k_{0}\right. \\\& g_{2} g_{s_{2}}=-g m_{2} \theta_{g_{2}} \\\text { h. } \\\text { vgig } \\\mathrm{gm}_{2} \mathrm{gr}_{2} \\c_{2} \square \\\text { of } \theta_{g_{2}}=-\frac{g_{m_{1}}}{g_{m_{2}}} v_{g_{2}} \\\Rightarrow v_{m_{n}}=v v_{g_{1}}-\frac{g m_{1}}{g_{m_{2}}} v_{y_{1}} \\\Rightarrow v_{n}=\operatorname{gg}_{2}\left(L-\frac{g m_{1}}{g m_{2}}\right) \\\end{array}Sine v_{\text {out }}=-g_{m_{1}} v_{g_{2}} R_{D}\begin{array}{l}\left.=-g m_{2} h_{D} \times \frac{V m_{1}}{\left(1-g m_{1} / g m_{2}\right.}\right)=-\left(\frac{g m_{2} g m_{2}}{g m_{2}-g m_{2}}\right) R_{D} v n \\A v=-\left(\frac{g m_{2} g m_{2}}{g m_{2}-g m_{2}}\right) R_{D}\end{array}Thesfor, v_{i n}=-v_{g_{1}}+\left(\frac{g_{m_{1}}}{g_{m_{2}}}\right) v_{g_{1}}=v_{g_{1}}\left(\frac{g_{m_{1}}}{g_{m_{2}}}-1\right)\Rightarrow \lg _{2}=\mathrm{km} /\left(g_{\frac{m_{2}}{}}^{g_{m_{2}}}\right)Now KChat node Vout,\because V_{\text {out }}=\mathrm{gg}_{3}<\mathrm{gm}_{3} \mathrm{Vg}_{3}+g_{m_{2}} \mathrm{gg}_{2}=0\begin{array}{l}\Rightarrow v_{\text {out }} g_{m_{3}}+g m_{1} \times \frac{v_{n_{n}}}{\left(\frac{g_{m_{1}}-g_{m_{2}}}{g_{m_{2}}}\right)}=0 \\\Rightarrow v_{\text {aut }} g_{m_{3}}=\frac{-g m_{1} g m_{2}}{\left(g_{m_{2}}-g m_{2}\right)} v_{\text {in }} \\\Rightarrow A V=\frac{V_{\text {but }}}{V n_{n}}=\frac{-g m_{1} g_{m_{2}}}{g_{m_{3}}\left(g_{m_{2}}-g_{m_{2}}\right)} \\\end{array} ...