Question 19.29. Determine the voltage gain of each circuit in Fig. 9.59. Assume &mo » 1. Vos VOD MA M5 Voo VOD VOD VIMA VEMA Vino E VEM V63 KM₂ V63 16₂ V63 16 M₂ - Vout M2 V621 Rp V62 M₂ V62 M2 Vinom, Vino 1, Ap Мы Hм, Уыз?НЕ м. Vout O Vout Vout V62 M₂ Vom, Ms (b) (d) (a) Figure 9.59

from the given information,(1) a_{m}=g_{m} and Rout is given by\text { Rout }=\left[\left(R_{p} \| \gamma_{2}\right)+\left(1+g_{m_{2}}\left(R_{p} \| \gamma_{O_{2}}\right) \gamma_{0}\right] \|\left[\gamma_{04}\left(1+g_{m_{3}}^{\gamma_{0}}+\gamma_{3}\right]\right.\right.Hen, as we knowe that, A_{v}=-G m Rout. Then,\text { - } A v=-g_{m_{1}}\left[\left(C R \cdot 11 \gamma_{O_{2}}\right)+\left(1+g_{m}\left(R_{p} \| \gamma_{2}\right)\right) \gamma_{1}\right] \|\left[\gamma_{\alpha_{1}}\left(1+g_{m_{3}}^{25 \gamma_{1}}\right)\right](b) From the given information, G_{m_{1}}=g_{m_{1}} \times\left(\frac{R p l l r_{01}}{\gamma_{1} \| R_{p}+\left(\frac{1}{g_{m_{2}}}\right)}\right) Then, from the given circuit,\begin{array}{l}\left.\left[\gamma_{04}\left(1+g_{m_{3}} \gamma_{3}\right)+\gamma_{3}\right]\right] \\\end{array}C) And, G_{m}=g_{\mathrm{ms}}. from the circuit,\begin{array}{l}\text { Rout }=\left[\left(1+g_{m_{2}} \gamma_{O_{2}}\right)\left(\gamma_{01} 11 r_{0}\right)+\gamma_{0}\right] \|\left[\left(1+g_{m_{3}} \gamma_{0_{3}}\right) \gamma_{0}+\gamma_{0_{3}}\right] \\\text { then } A_{V}=-g_{m s}\left\{\left(1+g_{m_{2}} \gamma_{2}\right)\left(\gamma_{0} \| \gamma_{05}\right)+\gamma_{0_{2}}\right] \|\left[\left(1+g_{m_{3}} \gamma_{0}\right) \gamma_{0}+\gamma_{3}\right]\end{array}(d) a_{m}= arms from the circuit _{\text {(d) }}\begin{array}{l}\text { Rout }=\left[\left(1+g_{m_{2}} \gamma_{2}+\gamma_{0_{2}}\right] \|\left[\left(\gamma_{1} \| r_{o_{4}}\right)\left(1+g_{m_{3}} \gamma_{O_{3}}\right)+\gamma_{O_{3}}\right]\right. \\\text { Then }\end{array}ThenA_{v}=-g_{m s}\left[\left[\left(1+g m_{2} \gamma_{2}\right) \gamma_{0}+\gamma_{2}\right] /\left[\left(r_{0}\left(1 r_{04}\right)\left(1+g_{m_{3}} \gamma_{0_{3}}\right)+\gamma_{0_{3}}\right]\right\}\right. ...