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\title{电子学基础——第九次作业}
\author{LXQ}
\date{2019.12.05}

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\begin{document}

\maketitle

\paragraph{10.51} \label{10.51}
    A student who has a single-ended voltage source constructs the circuit shown in Fig. 10-75, hoping to obtain differential outputs. Assume perfect symmetry but $\lambda = 0$ for simplicity. 

    (b) Viewing $M_1$ as a common-source stage degenerated by the impedance seen at the source of $M_2$, calculate $v_X$ in terms of $v_{in}$.

    (b) Viewing $M_1$ as a source follower and $M_2$ as a common-gate stage, calculate $v_Y$ in terms of $v_{in}$.

    (c) Add the results obtained in (a) and (b) with proper polarities. If the voltage gain is defined as $(v_X - v_Y)/v_{in}$, how does it compare with the gain of differentially driven pairs?

    \begin{figure}[!htb]
        \centering
        \includegraphics[width=0.312\textwidth]{p10-75.png}
        \caption*{Figure 10-75}
    \end{figure}

\paragraph{解}
    (a) $M_2$从源端看入，输入电阻为$\under{g_{m2}}$，而$M_1$为源简并放大器。则
    $$\frac{v_X}{v_{in}} = -\frac{g_{m1}R_D}{1+g_{m1}\cdot \under{g_{m2}}} = -\frac{g_mR_D}{2}$$

    (b) $M_1$为源极跟随器，则$M_1$源端电压即为$v_{s1}=v_{in}$，而$M_2$为共栅放大器，则
    $$\frac{v_Y}{v_{s1}}=g_{m1}R_D$$
    $$\therefore \frac{v_Y}{v_{in}}=g_{m1}R_D$$

    (c) $$\frac{v_X-v_Y}{v_{in}}=-\frac{3}{2}g_mR_D$$
    这个增益是普通差分放大器增益的1.5倍。

\paragraph{10.70} \label{10.70}
    Compute the common-mode rejection ratio of the stages illustrated in Fig. 10-89 and compare the results. For simplicity, neglect channel-length modulation in $M_1$ and $M_2$ but not in other transistors.

    \begin{figure}[!htb]
        \centering
        \includegraphics[width=0.569\textwidth]{p10-89.png}
        \caption*{Figure 10-89}
    \end{figure}

\paragraph{解}
    $$\CMRR = 20\log\left|\frac{A_{vd}}{A_{vc}}\right|$$

    (a) 电路对称，可考虑半边电路。由交流小信号电路中$P$为虚地，则
    $$A_{vd}=-g_{m1}R_D$$
    考虑$A_{vc}$时，可将$M_3$视为$r_{o3}$，进而再半边电路中视为$2r_{o3}$，则$M_1$为源简并放大器：
    $$A_{vc}=\frac{-g_{m1}R_D}{1+2g_{m1}r_{o3}}$$
    则
    $$\CMRR = 20\log(1+2g_{m1}r_{o3})$$

    (b) 同(a)，$P$再交流小信号电路中仍未虚地，则
    $$A_{vd}=-g_{m1}R_d$$
    再考虑$A_{vc}$，将$M_4$视为$r_{o4}$，则$M_3$为源简并放大器，可视为电阻$r=(1+g_{m3}r_{o3})r_{o4}+r_{o3}$
    从而半边电路中可将其视为$2r=2[(1+g_{m3}r_{o3})r_{o4}+r_{o3}]$
    此时$M_1$仍为源简并放大器：
    $$\therefore A_{vc}=\frac{-g_{m1}R_D}{1+2g_{m1}[(1+g_{m3}r_{o3})r_{o4}+r_{o3}]}$$
    $$\therefore \CMRR = 20 \log [1+2g_{m1}((1+g_{m3}r_{o3})r_{o4}+r_{o3})]$$

\end{document}