MOS基本电路的特性

这几天突然开始需要用到模电知识，却发现我已经不能闭眼写出这些基本电路的性质了，于是决定复习一下。感觉《模拟电路原理》开卷是件错误的事情，最该背的全都没背（）反而是微光这种应该开卷。 ## Mos器件模型  长沟道MosFET的本征电容模型如上。其中 $$ g_m = \mu C_{\mathrm{ox}}\frac{W}{L}V_{DS} $$ $$ r_o = \frac{1}{\mu C_{\mathrm{ox}}\frac{W}{L}(V_{GS}-V_t-V_{DS})} $$ ## CS组态电路计算  CS最简单了，直接闭眼写出 $$ A_v = g_m(r_o\parallel R_L) $$ 显然$R_{out}=r_o$，且低频下$R_{in}\approx \infty$. ## CG组态电路计算  结合 $v_{gs}=-v_i$ 暴力列出节点电压方程 $$ \begin{bmatrix} \frac{1}{R_S}+s(C_{gs}+C_{sb})+\frac{1}{r_o} & -\frac{1}{r_o}\\ -\frac{1}{r_o} & \frac{1}{R_L}+s(C_{gd}+C_{db}) +\frac{1}{r_o} \end{bmatrix}\begin{bmatrix} v_i\\ v_o \end{bmatrix}=\begin{bmatrix} i_i-(g_{mb}+g_m)v_i\\ (g_{mb}+g_m)v_i \end{bmatrix} $$ 令$C_{gs}+C_{sb}=C_S$, $C_{gd}+C_{db}=C_L$,$g_m+g_{mb}=g_m'$ 则 $$ \begin{bmatrix} \frac{1}{R_S}+sC_{S}+g_m'+\frac{1}{r_o} & -\frac{1}{r_o}\\ -\frac{1}{r_o}-g_m' & \frac{1}{R_L}+sC_{L}+\frac{1}{r_o} \end{bmatrix}\begin{bmatrix} v_i\\ v_o \end{bmatrix}=\begin{bmatrix} i_i\\ 0 \end{bmatrix} $$ 利用**克莱默法则**解出 $$ \begin{aligned} v_o&=\frac{i_i\cdot(g_m'+\frac{1}{r_o})}{(\frac{1}{R_S}+sC_{S}+g_m'+\frac{1}{r_o})\cdot(\frac{1}{R_L}+sC_{L}+\frac{1}{r_o})-(-\frac{1}{r_o})\cdot(-\frac{1}{r_o}-g_m')}\\ &\underbrace{=}_{r_o\to\infty\,,C_L\to 0}\frac{i_i\cdot(g_m')}{(\frac{1}{R_S}+sC_{S}+g_m')\cdot(\frac{1}{R_L})}\\ i_o&=\frac{v_o}{R_L}=\frac{i_i\cdot(g_m')}{(\frac{1}{R_S}+sC_{S}+g_m')}\\ \frac{i_o}{i_i}&=\frac{g'_mR_S}{1+g'mR_S+sC_SR_S}\\ &=\frac{g'_mR_S}{1+g'mR_S}\cdot\frac{1}{1+s\frac{R_SC_S}{1+g'_mR_S}}\underbrace{\approx}_{g'_mR_S\gg 1}\boxed{\frac{1}{1+s\frac{C_S}{g'_m}}} \end{aligned} $$ **输入阻抗**同理： $$ \begin{aligned} R_{in}&=\left.\frac{v_i}{i_i}\right|_{R_S\to\infty}=\frac{\frac{1}{R_L}+sC_{L}}{(sC_{S}+g_m'+\frac{1}{r_o})\cdot(\frac{1}{R_L}+sC_{L}+\frac{1}{r_o})-(-\frac{1}{r_o})\cdot(-\frac{1}{r_o}-g_m')}\\ &=\frac{1}{(sC_{S}+g_m'+\frac{1}{r_o})-(-\frac{1}{r_o})\cdot\frac{(-\frac{1}{r_o}-g_m')}{\frac{1}{R_L}+sC_{L}}}\\ &\underbrace{=}_{r_o\gg\frac{1}{g'_m}\,,C_L\to 0}\frac{1}{(sC_{S}+g_m')-\frac{1}{r_o}\cdot\frac{g_m'}{\frac{1}{R_L}}}\\ &=\frac{R_L+r_o}{g'_mr_o}\cdot\frac{1}{sC_S\frac{R_L+r_o}{g'_mr_o}+\frac{R_L+r_o}{g'_mr_o}\cdot (g_m'-\frac{g'_mR_L}{r_o})}\\ &=\frac{R_L+r_o}{g'_mr_o}\cdot\frac{1}{sC_S\frac{R_L+r_o}{g'_mr_o}+\frac{R_L+r_o}{g'_mr_o}\cdot g_m'(\frac{r_o-R_L}{r_o})}\\ &\underbrace{=}_{r_o^2\gg R_L^2}\boxed{\frac{R_L+r_o}{g'_mr_o}\cdot\frac{1}{sC_S\frac{R_L+r_o}{g'_mr_o}+1}} \end{aligned} $$ **低频**下又近似为 $$ \boxed{R_{in}\approx \frac{R_L+r_o}{g'_mr_o}\approx\frac{1}{g'_m}} $$ **输出阻抗**类似，只要利用 $$ R_{out}\approx\left.R_L\right|_{v_o=\frac{1}{2} v_{o,R_L=\infty}} $$ 代入计算 $$ \frac{i_i\cdot(g_m'+\frac{1}{r_o})}{(\frac{1}{R_S}+sC_{S}+g_m'+\frac{1}{r_o})\cdot(\frac{1}{R_{out}}+sC_{L}+\frac{1}{r_o})-(-\frac{1}{r_o})\cdot(-\frac{1}{r_o}-g_m')}=\frac{1}{2}\frac{i_i\cdot(g_m'+\frac{1}{r_o})}{(\frac{1}{R_S}+sC_{S}+g_m'+\frac{1}{r_o})\cdot(sC_{L}+\frac{1}{r_o})-(-\frac{1}{r_o})\cdot(-\frac{1}{r_o}-g_m')} $$ 比较分母系数得到 $$ (\frac{1}{R_S}+sC_{S}+g_m'+\frac{1}{r_o})\cdot(sC_{L}+\frac{1}{r_o})-(-\frac{1}{r_o})\cdot(-\frac{1}{r_o}-g_m')=(\frac{1}{R_S}+sC_{S}+g_m'+\frac{1}{r_o})\cdot(1/R_{out}) $$ $$ (\frac{1}{R_S}+sC_{S}+g_m'+\frac{1}{r_o})\cdot(\frac{1}{r_o})-(\frac{1}{r_o})\cdot(\frac{1}{r_o}+g_m')=(\frac{1}{R_S}+sC_{S}+g_m'+\frac{1}{r_o})\cdot(\frac{1}{R_{out}}) $$ $$ \begin{aligned} (\frac{1}{r_o})-(\frac{1}{r_o})\cdot\frac{\frac{1}{r_o}+g_m'}{\frac{1}{R_S}+sC_{S}+g_m'+\frac{1}{r_o}}&=\frac{1}{R_{out}}\\ \frac{1}{r_o}\cdot\frac{\frac{1}{R_S}+sC_S}{\frac{1}{R_S}+sC_{S}+g_m'+\frac{1}{r_o}}&=\frac{1}{R_{out}}\\ \underbrace{\Rightarrow}_{DC;s\to0}\frac{1}{r_o+R_s+g_m'r_oR_S}&=\frac{1}{R_{out}} \end{aligned} $$ 一般来说近似认为 $$ \boxed{R_{out}=(1+g_m'r_o)R_S} $$ ## CD组态电路计算  老规矩 $$ \begin{bmatrix} s(C_{gs}+C_{gd}+C_{gb}) & -sC_{gs}\\ -sC_{gs}& s(C_{gs}+C_L+C_{sb})+\frac{1}{r_o}+\frac{1}{R_L} \end{bmatrix} \begin{bmatrix} v_i\\v_o \end{bmatrix}= \begin{bmatrix} i_i\\ g_m(v_i-v_o)-g_{mb}v_o \end{bmatrix} $$ $$ \begin{bmatrix} s(C_{gs}+C_{gd}+C_{gb}) & -sC_{gs}\\ -sC_{gs}-g_m& s(C_{gs}+C_L+C_{sb})+\frac{1}{r_o}+\frac{1}{R_L}+g'_m \end{bmatrix} \begin{bmatrix} v_i\\v_o \end{bmatrix}= \begin{bmatrix} i_i\\ 0\\ \end{bmatrix} $$ 再次利用**克莱默法则**求解得到 $$ \begin{bmatrix} v_i\\ v_o \end{bmatrix}=\frac{\begin{bmatrix} s(C_{gs}+C_L+C_{sb})+\frac{1}{r_o}+\frac{1}{R_L}+g'_m & sC_{gs}\\ sC_{gs}+g_m&s(C_{gs}+C_{gd}+C_{gb}) \end{bmatrix}\begin{bmatrix}i_i\\0\end{bmatrix}}{s(C_{gs}+C_{gd}+C_{gb})\cdot(s(C_{gs}+C_L+C_{sb})+\frac{1}{r_o}+\frac{1}{R_L}+g'_m)-(-sC_{gs})(-sC_{gs}-g_m)} $$ 令$R_L' = r_o\parallel R_L\parallel \frac{1}{g_{mb}}$ , $C_L'=C_L+C_{sb}$,则 $$ \begin{bmatrix} v_i\\ v_o \end{bmatrix}=\frac{\begin{bmatrix} s(C_{gs}+C_L')+1/R_L'+g_m & sC_{gs}\\ sC_{gs}+g_m&s(C_{gs}+C_{gd}+C_{gb}) \end{bmatrix}\begin{bmatrix}i_i\\0\end{bmatrix}}{s(C_{gs}+C_{gd}+C_{gb})\cdot(s(C_{gs}+C_L')+1/R_L'+g_m)-(-sC_{gs})(-sC_{gs}-g_m)} $$ $$ v_i = \frac{ (s(C_{gs}+C_L')+1/R_L'+g_m) i_i}{s(C_{gs}+C_{gd}+C_{gb})\cdot(s(C_{gs}+C_L')+1/R_L'+g_m)-(-sC_{gs})(-sC_{gs}-g_m)} $$ $$ v_o = \frac{ (sC_{gs}+g_m) i_i}{s(C_{gs}+C_{gd}+C_{gb})\cdot(s(C_{gs}+C_L')+1/R_L'+g_m)-(-sC_{gs})(-sC_{gs}-g_m)} $$ $$ A_v(s)=\frac{v_o}{v_i}=\frac{sC_{gs}+g_m}{s(C_{gs}+C_L')+1/R_L'+g_m}=\frac{g_m}{g_m+\frac{1}{R_L'}}\cdot \frac{1\textcolor{red}{+s\frac{C_{gs}}{g_m}}}{1+s\frac{C_{gs}+C_{L}'}{g_m+\frac{1}{R_L'}}} $$ 可见这里有一个**RHP零点**。通过改变零点和极点的相对位置，可以改变其频率响应特性。 对于输入阻抗，只要利用上述结果 $$ \begin{aligned} Z_{in}&=\left.\frac{v_i}{i_i}\right|_{R_S=0}\\ &=\frac{ (s(C_{gs}+C_L')+1/R_L'+g_m) }{s(C_{gs}+C_{gd}+C_{gb})\cdot(s(C_{gs}+C_L')+1/R_L'+g_m)-(-sC_{gs})(-sC_{gs}-g_m)}\\ &=\frac{1}{ s(C_{gs}+C_{gd}+C_{gb})-sC_{gs}\textcolor{red}{A_v(s)} }\\ &\underbrace{=}_{A_v(s)\approx 1}\frac{1}{s(C_{gd}+C_{gb})} \end{aligned} $$ 该输入阻抗十分大。对于**没有源阻抗时**的输出阻抗，利用 $$ Z_{out}=\left.Z_L\right|_{v_o=\frac{1}{2}v_{o, open}\approx\frac{1}{2}v_{i}} $$ $$ \begin{aligned} \frac{1}{2}&\frac{ (sC_{gs}+g_m) i_i}{s(C_{gs}+C_{gd}+C_{gb})\cdot(s(C_{gs}+C_{sb})+g_m+1/r_o+g_{mb})-(-sC_{gs})(-sC_{gs}-g_m)}\\ &=\frac{ (sC_{gs}+g_m) i_i}{s(C_{gs}+C_{gd}+C_{gb})\cdot(s(C_{gs}+C_{sb}+C_L)+1/r_o+1/R_{out}+g_{mb}+g_m)-(-sC_{gs})(-sC_{gs}-g_m)} \end{aligned} $$ 比较分母得到 $$ \begin{aligned} &s(C_{gs}+C_{gd}+C_{gb})\cdot(sC_{out}+1/R_{out})\\ &=s(C_{gs}+C_{gd}+C_{gb})\cdot(s(C_{gs}+C_{sb})+g_m+1/r_o+g_{mb})\\ &-(-sC_{gs})(-sC_{gs}-g_m) \end{aligned} $$ $$ \frac{1}{Z_{out}}=s(C_{gs}+C_{sb})+g_m+1/r_o+g_{mb}-\frac{sC_{gs}(sC_{gs}+g_m)}{s(C_{gs}+C_{gd}+C_{gb})} $$ 忽略$r_o$影响， $$ \boxed{Z_{out}=s(C_{gs}+C_{sb})\parallel\frac{1}{g_m+g_{mb}}} $$  对于有$R_S$的情况，先戴维南一下  **依然节点电压**： $$ \begin{bmatrix} s(C_{gs}+C_{gd}+C_{gb})+\frac{1}{R_S} & -sC_{gs}\\ -sC_{gs}-g_m& s(C_{gs}+C_L+C_{sb})+\frac{1}{r_o}+\frac{1}{R_L}+g'_m \end{bmatrix} \begin{bmatrix} v_1\\v_o \end{bmatrix}= \begin{bmatrix} v_i/R_S\\ 0\\ \end{bmatrix} $$ 继续利用 $$ Z_{out}=\left.Z_L\right|_{v_o=\frac{1}{2}v_{o, open}\approx\frac{1}{2}v_{i}} $$ 注意$v_o$分子上只有原始矩阵的12元素$-sC_{gs}$和$R_L$无关,因此只需要对比行列式即可: $$ \frac{1}{2}\cdot\frac{1}{\Delta_{open}}=\frac{1}{\Delta}\Rightarrow \Delta = 2\Delta_{open} $$ $$ \begin{aligned} (s(C_{gs}+C_{gd}+C_{gb})+\frac{1}{R_S})&\cdot(s(C_{gs}+C_{out}+C_{sb})+\frac{1}{r_o}+\frac{1}{R_{out}}+g'_m)\\ -( -sC_{gs})\cdot(-sC_{gs}-g_m)&=2\left((s(C_{gs}+C_{gd}+C_{gb})+\frac{1}{R_S})\right.\\ \cdot(s(C_{gs}+C_{sb})&\left.+\frac{1}{r_o}+g'_m)-(-sC_{gs})\cdot(-sC_{gs}-g_m)\right) \end{aligned} $$ $$ (s(C_{gs}+C_{gd}+C_{gb})+\frac{1}{R_S})(sC_{out}+\frac{1}{R_{out}})=(s(C_{gs}+C_{gd}+C_{gb})+\frac{1}{R_S})\cdot(s(C_{gs}+C_{sb})+\frac{1}{r_o}+g'_m)-(-sC_{gs})\cdot(-sC_{gs}-g_m) $$ $$ Y_{out}=sC_{out}+\frac{1}{R_{out}}=s(C_{gs}+C_{sb})+\frac{1}{r_o}+g'_m-\frac{sC_{gs}(sC_{gs}+g_m)}{s(C_{gs}+C_{gd}+C_{gb})+\frac{1}{R_S}} $$ $$ \begin{aligned} Y_{out}&=s(C_{gs}+C_{sb})+\frac{1}{r_o}+g_{mb}+g_m-\frac{sC_{gs}(sC_{gs}+g_m)}{s(C_{gs}+C_{gd}+C_{gb})+\frac{1}{R_S}}\\ &\underbrace{=}_{C_S:=C_{gd}+C_{gb}} s(C_{gs}+C_{sb})+\frac{1}{r_o}+g_{mb}+g_m-\frac{sC_{gs}(sC_{gs}+g_m)}{s(C_{gs}+C_{S})+\frac{1}{R_S}}\\ &=s(C_{gs}+C_{sb})+\frac{1}{r_o}+g_{mb}+g_m\left(1-\frac{sC_{gs}(s\frac{C_{gs}}{g_m}+1)}{s(C_{gs}+C_{S})+\frac{1}{R_S}}\right)\\ &=s(C_{gs}+C_{sb})+\frac{1}{r_o}+g_{mb}+g_m\left(1-\frac{sC_{gs}R_S(s\frac{C_{gs}}{g_m}+1)}{s(C_{gs}+C_{S})R_S+1}\right)\\ &=s(C_{gs}+C_{sb})+\frac{1}{r_o}+g_{mb}+g_m\left(\frac{s(C_{gs}+C_{S})R_S+1-sC_{gs}R_S(s\frac{C_{gs}}{g_m}+1)}{s(C_{gs}+C_{S})R_S+1}\right)\\ &=s(C_{gs}+C_{sb})+\frac{1}{r_o}+g_{mb}+g_m\left(\frac{sC_{S}R_S+1-sC_{gs}R_S(s\frac{C_{gs}}{g_m})}{s(C_{gs}+C_{S})R_S+1}\right)\\ &=sC_{sb}+\frac{1}{r_o}+g_{mb}+g_m\left(\frac{sC_{S}R_S+1-sC_{gs}R_S(s\frac{C_{gs}}{g_m})}{s(C_{gs}+C_{S})R_S+1}+s\frac{C_{gs}}{g_m}\right)\\ &=sC_{sb}+\frac{1}{r_o}+g_{mb}+g_m\left(\frac{sC_{S}R_S+1-\sout{sC_{gs}R_S(s\frac{C_{gs}}{g_m})}+{s}\frac{C_{gs}}{g_m}+\sout{s\frac{C_{gs}}{g_m}R_SsC_{gs}}+s\frac{C_{gs}}{g_m}R_SsC_{S}}{s(C_{gs}+C_{S})R_S+1}\right)\\ &=sC_{sb}+\frac{1}{r_o}+g_{mb}+g_m\left(\frac{sC_{S}R_S+1+{s}\frac{C_{gs}}{g_m}+s^2\frac{C_{gs}}{g_m}R_SC_{S}}{s(C_{gs}+C_{S})R_S+1}\right)\\ &=sC_{sb}+\frac{1}{r_o}+g_{mb}+g_m\left(\frac{(1+sC_SR_S)(1+sC_{gs}/g_m)}{s(C_{gs}+C_{S})R_S+1}\right)\\ \end{aligned} $$ 因此显然有 $$ Z_{out}=\frac{1}{Y_{out}}=\boxed{\frac{1}{sC_{sb}+\frac{1}{r_o}+g_{mb}+g_m\left(\frac{(1+sC_SR_S)(1+sC_{gs}/g_m)}{s(C_{gs}+C_{S})R_S+1}\right)}} $$