# Common-Source Stage

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# Common-Source Stage

## Comom-Source Stage with Resistive Load

• $V_\tx{in}=0$, $M_1$ is off, $V_\tx{out}=V_{DD}$
• $V_\tx{in}>V_\tx{TH}$, $M_1$ is in saturation, $V_\tx{out}=V_{DD}-R_D\frac{1}{2} \mu_n C_\tx{ox}\frac{W}{L}(V_\tx{in}-V_\tx{TH})^2$
• $V_\tx{in}>V_\tx{out}+V_\tx{TH}$, $M_1$ is in the triode region, $V_\tx{out}=V_{DD}-R_D \frac{1}{2}\mu_n C_\tx{ox} \frac{W}{L}[2(V_\tx{in}-V_\tx{TH})V_\tx{out}-V_\tx{out}^2]$

We usually ensure that the MOS is in saturation. The small-signal gain is

\begin{aligned} A_v &= \frac{\p V_\tx{out}}{\p V_\tx{in}}\\ &=-R_D \mu_n C_\tx{ox} \frac{W}{L}(V_\tx{in}-V_\tx{TH})\\ &=-g_m R_D \end{aligned}

The small-signal model of Fig 3.4(d) yiels the same result.

And we can rewrite $g_m$ and $R_D$ as

$g_m=\sqrt{2\mu_n C_\tx{ox}\frac{W}{L}I_D}\\ R_D = \frac{V_{RD}}{I_D}$

therefore

\begin{aligned} A_v &= -\sqrt{2\mu_n C_\tx{ox}\frac{W}{L}I_D} \frac{V_{RD}}{I_D}\\ &= -\sqrt{2\mu_n C_\tx{ox}\frac{W}{L}} \frac{V_{RD}}{\sqrt{I_D}} \end{aligned}

How to maxiumize $A_v$?

1. increasing $W/L$
• leads to grater device capacitances
2. increasing $V_{RD}$
• limits the maximum voltage swings
• The maximum output is $V_{DD}$, the mimimum output is $V_\tx{in}-V_\tx{TH}$, ideal $V_{RD}$ should be in the middle of max and min output voltage.
3. decreasing $I_D$
• $R_D$ must increase to abtain the same DC output, leading to a greater time constant at the output node

it seems that 3. is more acceptable.

For large values of $R_D$, the effect of channel-length modulation in $M_1$ becomes significant.

\begin{aligned} A_v &= \frac{\p V_\tx{out}}{\p V_\tx{in}}\\ &= \frac{\p}{\p V_\tx{in}} \left\{ V_{DD}- R_D\frac{1}{2} \mu_n C_\tx{ox}\frac{W}{L}(V_\tx{in}-V_\tx{TH})^2 (1+\lambda V_\tx{out}) \right\}\\ &= - R_D\mu_n C_\tx{ox}\frac{W}{L}(V_\tx{in}-V_\tx{TH}) (1+\lambda V_\tx{out})\\ &\quad -R_D\frac{1}{2} \mu_n C_\tx{ox}\frac{W}{L}(V_\tx{in}-V_\tx{TH})^2 (\lambda \frac{\p V_\tx{out}}{\p V_\tx{in}})\\ &= -R_D g_m - \frac{R_D}{r_O} A_v\\ \therefore A_v &= -g_m \frac{r_O R_D}{r_O+R_D}\\ \frac{1}{r_O}&= \frac{1}{2} \mu_n C_\tx{ox}\frac{W}{L}(V_\tx{in}-V_\tx{TH})^2 \lambda \end{aligned}\\

We can use the small-signal model to get the same result:

$g_m V_1 + V_\tx{out} \cdot (\frac{1}{r_O}+\frac{1}{R_D})=0\\ 代入 V_1 = V_\tx{in} 有\\ g_m V_\tx{in} + V_\tx{out} \cdot (\frac{1}{r_O}+\frac{1}{R_D})=0\\ \Rightarrow A_v = \frac{V_\tx{out}}{V_\tx{in}} = -g_m (r_O \| R_D)$

Calculate the small-signal voltage gain of the circuit. (assuming $M_1$ is biased in saturation)

Current source has an $R_D = \infty$, therefore
$$A_v = -g_m r_O$$
Called the “intrinsic gain 本征增益” of a transistor, this quantity represents the maximum voltage gain that can be achieved using a single device.
The total drain current of $M_1$ is
\begin{aligned} I_{D1}&=\frac{1}{2} \mu_n C_\tx{ox}\frac{W}{L}(V_\tx{in}-V_\tx{TH})^2 (1+\lambda V_\tx{out})\\ &=I_1 \end{aligned}

## CS Stage with Diode-Connected Load

it’s hard to fabricate resistors with the values wanted. Therefore we replace $R_D$ with a diode-connected MOS transistor (Fig.3.10)

With body effect and channel-length modulation (Fig. 3.11), we have

$g_m V_1 + g_{mb} V_{bs} +I_X-\frac{V_X}{r_O}=0\\ 代入 V_1=-V_X, V_{bs}=-V_X有\\ (g_m+g_{mb})V_X+\frac{V_X}{r_O}=I_X\\ \Rightarrow \frac{V_X}{I_X} = \frac{1}{g_m+g_{mb}+1/r_O} \approx \frac{1}{g_m+g_{mb}} \tag{3.21}$

（此处是为了说明 diode-connected MOS 的工作特性就像电阻一样，电阻值是 $1/g_m$、$1/g_{mb}$、$r_O$ 的并联。注意此时 MOS 一定在饱和区）

$$\because V_1=-V_X, V_{bs}=-V_X\\ g_m V_X+g_{mb}V_X=I_X\\ \therefore \frac{V_X}{I_X} = \frac{1}{g_m + g_{mb}} \tag{3.26}$$ 可以看出最后结果和上面忽略了 $r_O$ 的结果是一样的。所以我们认为，只要不考虑 $r_O$，无论 MOS 管怎么连接，源端看进去的电阻都是 $\dfrac{1}{g_m} \Vert \dfrac{1}{g_{mb}}$。（相当于屏蔽了漏端电阻）

$$V_X = I_X Z_L + (I_X - g_m V_X - g_{mb} V_X ) r_O\\ \therefore \frac{V_X}{I_X} = \frac{Z_L+r_O}{1+(g_m+g_{mb})r_O}$$

With negligible channel-length modulation, common-source stage with a diode-connected load has

\begin{aligned} A_v &= -g_{m1} R_D\\ &= - g_{m1}\frac{1}{g_{m2}+g_{mb2}}\\ &= -\frac{g_{m1}}{g_{m2}} \frac{1}{1+\eta} \end{aligned}\\ \tx{where } \eta = g_{mb2}/g_{m2}

Since $g_{m} = \sqrt{\frac{1}{2}\mu_n C_\tx{ox}(W/L)I_D}$,

\begin{aligned} A_v &= - \sqrt{\frac{2\mu_n C_\tx{ox}(W/L)_1I_{D1}}{2\mu_n C_\tx{ox}(W/L)_2I_{D2}}} \frac{1}{1+\eta}\\ &= - \sqrt{\frac{(W/L)_1}{(W/L)_2}} \frac{1}{1+\eta} \end{aligned}

This equation reveals an interesting property: if the variation of $\eta$ with the output voltage is neglected, the gain is independent of the bias currents and voltages (so long as $M_1$ stays in saturation). In otherwords, as the input and output signal levels vary, the gain remains relatively constant, indicating that the input-output characteristic is relatively linear.

（从大信号的角度说明上述内容）$M_2,M_1$ share the same $I_D$, therefore

\begin{aligned} I_{D1}&=I_{D2}\\ \frac{1}{2} \mu_n C_\tx{ox} \left(\frac{W}{L}\right)_1 (V_\tx{in}-V_\tx{TH})^2 &= \frac{1}{2} \mu_n C_\tx{ox} \left(\frac{W}{L}\right)_2 (V_{DD}-V_\tx{out}-V_\tx{TH})^2\\ \sqrt{\left(\frac{W}{L}\right)_1} (V_\tx{in}-V_\tx{TH1}) &=\sqrt{\left(\frac{W}{L}\right)_2} (V_{DD}-V_\tx{out}-V_\tx{TH2}) \end{aligned}

The small-signal gain can also be computed by differentiating both sides with respect to $V_\tx{in}$

$\sqrt{\left(\frac{W}{L}\right)_1} =\sqrt{\left(\frac{W}{L}\right)_2} (-\frac{\p V_\tx{out}}{\p V_\tx{in}}-\frac{\p V_\tx{TH2}}{\p V_\tx{in}})$

where

$\frac{\p V_\tx{TH2}}{\p V_\tx{in}} = \frac{\p V_\tx{TH2}}{\p V_\tx{out}} \frac{\p V_\tx{out}}{V_\tx{in}}$

recall that

$g_{mb}=g_m \left(-\frac{\p V_\tx{TH}}{\p V_{BS}}\right)\\ 代入 V_\tx{TH2}=V_\tx{TH},V_{BS}=-V_\tx{out}\\ \Rightarrow \frac{\p V_\tx{TH2}}{\p V_\tx{out}} = \frac{g_{mb}}{g_m}=\eta$

therefore

$\frac{\p V_\tx{out}}{\p V_\tx{in}}=-\sqrt{\frac{(W/L)_1}{(W/L)_2}} \frac{1}{1+\eta}$

• $V_\tx{in}<V_\tx{TH1}$，这样的话，$I_{D1}$ 为 0，$V_\tx{out}$ 应该为 $V_{DD}$，但实际中会有寄生电容，会减缓 $V_\tx{out}$ 趋向 $V_{DD}$，所以我们一般认为 $V_\tx{out}=V_{DD}-V_\tx{TH2}$ （此时 $M_2$ 有点像三极管，$V_{DD}$ 经过 gs 压降后到 $V_\tx{out}$）
• $V_\tx{in}>V_\tx{TH1}$，就是上面讨论的情况
• $V_\tx{in}-V_\tx{TH1}>V_\tx{out}$，此时 $M_1$ 进入可变电阻区，输入输出非线性

\begin{aligned} I_{D1}&=\vert I_{D2} \vert\\ \mu_n \left(\frac{W}{L}\right)_1(V_{GS1}-V_{TH1})^2 &=\mu_p \left(\frac{W}{L}\right)_2(V_{GS2}-V_{TH2})^2\\ \mu_n \left(\frac{W}{L}\right)_1(V_\tx{in1}-V_{TH1})^2 &=\mu_p \left(\frac{W}{L}\right)_2(V_{DD}-V_\tx{out}-V_{TH2})^2\\ \end{aligned}

$A_v = -\sqrt{ \frac{\mu_n(W/L)_1}{\mu_p (W/L)_2}} \frac{1}{1+\eta}\\ = \frac{\vert V_{GS2}-V_\tx{TH2} \vert}{V_{GS1}-V_\tx{TH1}} \frac{1}{1+\eta}$

$A_v = -g_{m1}(r_{O1} \Vert \frac{1}{g_{m2}}\Vert r_{O2})$

$$\vert A_v \vert = \frac{g_{m1}}{g_{m2}}$$

$$\vert A_v \vert = \sqrt{ \frac{\mu_n(W/L)_1}{\mu_p (W/L)_2}}\\ = \frac{\vert V_{GS2}-V_\tx{TH2} \vert}{V_{GS1}-V_\tx{TH1}}$$

$$\vert A_v \vert = \frac{\mu_n (W/L)_1(V_{GS1}-V_\tx{TH1})}{\mu_p (W/L)_2\vert V_{GS2}-V_\tx{TH2} \vert}$$

## CS Stage with Current-Source Load

According the relationship $A_v = -g_m R_D$, we need to increase the load impedance of the CS stage. However, increasing the load resistance translates to a large dc drop across the load, thereby limiting theoutput voltage swing.

A solution to that is to replace the load with a current source, which we have discussed before.

The gain is given by is

$A_v = -g_{m1} (r_{O1}\|r_{O2})$

（补充：为什么 M2 的小信号模型是$r_{O2}$ 而不是像 diode-connected 那样？因为 $r_O$ 的定义就是 $\dfrac{\p V_{DS}}{\p I_D}$，并且 $V_{GS2}=V_b-V_{DD}$、$V_{SB2}=V_{DD}$ 是恒定的，可看作 AC ground，所以不存在 $g_m,g_{mb}$）

$r_O=\frac{1}{\lambda I_D}$

$g_{m1}r_{O1}=\sqrt{2 \left(\dfrac{W}{L}\right)_1 \mu_n C_\tx{ox} I_D} \dfrac{1}{\lambda I_D}\\ 且 \lambda \propto \frac{1}{L_1}$

$g_m \propto \sqrt{L_1}$，所以 $L_1 \uparrow \Rightarrow A_v \uparrow$

$$V_{GS}-V_\tx{TH} \propto 1/\sqrt{(W/L)}$$

$$V_{DS}>V_{GS}-V_\tx{TH}$$ 从而 $L_2$ 增大会导致（相同电流下） $V_{DS}$ 增大。

## CS Stage with Active Load

current-source load 中，PMOS serves as a constant current source，如果我们把 PMOS 也当作 amplifier 用，就叫做 Active Load，这种接法也叫 complementary CS stage

$-(g_{m1}+g_{m2})V_\tx{in} \cdot (r_{O1}\Vert r_{O2}) = V_\tx{out}\\ \Rightarrow A_v = -(g_{m1}+g_{m2})(r_{O1}\Vert r_{O2})$

## CS Stage with Triode Load

A MOS device operating in the deep triode region behaves as a resistor andcan therefore serve as the load in a CS stage. The resistor value can be calculated as

$R_\tx{on2} = \frac{1}{\mu_p C_\tx{ox}(W/L)_2 (V_{DD}-V_b-\vert V_\tx{THP} \vert )}$
• 缺点：$R_\tx{on2}$ 会随工艺和温度变化
• 优点：消耗的电压裕度小于 Diode-connect（$V_\tx{out,max}=V_{DD}$）

## CS Stage with Source Degeneration

$I_D$ 与 overdrive voltage $(V_{DS}-V_\tx{TH})$ 之间是非线性关系，这就导致输出是非线性的。之前我们说过可以用 diode-connected load 来修正非线性效应，这里我们介绍另一种方法：增加 $R_S$

\begin{aligned} A_v &= \frac{\p V_\tx{out}}{\p V_\tx{in}}\\ &=\frac{\p (V_{DD}-I_DR_D)}{\p V_\tx{in}}\\ &=-\frac{\p I_D}{\p V_\tx{in}}R_D\\ &=-G_m R_D \end{aligned}

\begin{aligned} G_m &= \frac{\p I_D}{\p V_\tx{in}}\\ &= \frac{\p I_D}{\p V_{GS}} \frac{\p V_{GS}}{\p V_\tx{in}}\\ \frac{\p V_{GS}}{\p V_\tx{in}}&=\frac{\p }{\p V_\tx{in}}(V_\tx{in}-I_DR_S)\\ &=1-R_S \frac{\p I_D}{\p V_\tx{in}}\\ &=1-R_S G_m\\ \frac{\p I_D}{\p V_{GS}}&=g_m \end{aligned}

$G_m = (1-R_SG_m)g_m\\ \Rightarrow G_m = \frac{g_m}{1+g_mR_S}$

\begin{aligned} A_v &= -G_m R_D\\ &=\frac{-g_mR_D}{1+g_mR_S}\\ &=\frac{-R_D}{1/g_m+R_S} \end{aligned}

$G_m$ or $A_v$ is degenerated by a factor of $1+g_mR_S$. 我们可以这样记忆 $A_v$，分子 $R_D$ 是从 $V_{DD}$ 往下看的电阻，而分母 $1/g_m+R_S$ 则是从 GND 往上看的电阻。

With body effect and channel-length modulation considered, we have (Fig. 3.24)

$V_X = (\frac{V_X}{R_S}-g_mV_1-g_{mb}V_{bs}) r_O\\ \tx{where } V_1=V_\tx{in}-V_X\\ V_{bs}=V_X\\ V_X=I_\tx{out}R_S$

\begin{aligned} I_\tx{out} &= g_m V_1 - g_{mb} V_X -\frac{I_\tx{out}R_S}{r_O}\\ &= g_m(V_\tx{in}-I_\tx{out}R_S)+g_{mb}(-I_\tx{out}R_S)-\frac{I_\tx{out}R_S}{r_O} \end{aligned}\\ \Rightarrow I_\tx{out} = \frac{g_mV_\tx{in}}{1+g_m R_S+g_{mb}R_S+\frac{R_S}{r_O}}\\ \begin{aligned} G_m&=\frac{I_\tx{out}}{V_\tx{in}}\\ &=\frac{g_m r_O}{R_S+r_O+(g_m+g_{mb})R_S r_O}\\ &=\frac{g_m/R_S}{1/r_O+1/R_S+g_m+g_{mb}} \end{aligned}

Typically，$(g_m+g_{mb})r_O\gg 1$，所以我们会忽略分母中的 $R_S$ 项，从而

$G_m \approx \frac{g_m}{1+(g_m+g_{mb})R_S}$

$G_m$ or $A_v$ is degenerated by a factor of $1+(g_m+g_{mb})R_S$（之前是 $1+g_m R_S$）

• $R_S=0$（饱和区）
• $I_D=\frac{1}{2}\mu_n C_\tx{ox} (W/L)(V_\tx{in}-V_\tx{TH})^2$
• $g_m=\mu_n C_\tx{ox}(W/L)(V_\tx{in}-V_\tx{TH})$
• $R_S\neq0$（饱和区）
• $I_D=\frac{1}{2}\mu_n C_\tx{ox} (W/L)(V_\tx{in}-I_D\cdot R_S-V_\tx{TH})^2$
• 😀“易”算出：
• $I_D=\dfrac{1}{R_S} \left( V-\dfrac{1 \pm \sqrt{1- 4kR_SV}}{2kR_S}\right)$，其中，$k=\frac{1}{2}\mu_n C_\tx{ox}(W/L)$，$V=V_\tx{in}-V_\tx{TH}$
• $G_m=\dfrac{1}{1/g_m+(1+g_{mb}/g_m)R_S}$，$V_\tx{in}\uparrow \Rightarrow g_m \uparrow \Rightarrow G_m=1/R_S$

Fig. 3.25(a) 对应的是 Resistive load，Fig. 3.25(b) 对应的是 degenerated laod. 可以看出，在 $V_\tx{in}$ 较大时，degenerated load 的 $G_m$ 趋向于一个恒定值 $1/R_S$ 说明输入输出近似于线性的。

$V_X = (I_X-g_mV_1-g_{mb}V_{bs})r_O+I_XR_S\\ \tx{where } V_1=V_{bs}=-I_XR_S$

\begin{aligned} R_\tx{out} &= \frac{V_X}{I_X}\\ &= [1+(g_m+g_{mb})R_S]r_O+R_S\\ &= [1+(g_m+g_{mb})r_O]R_S+r_O\\ &= R_S+r_O+(g_m+g_{mb})r_OR_S \end{aligned}

$\Delta I = \frac{\Delta V_{R_S}}{R_S}= \Delta V \frac{\frac{1}{g_m+g_{mb}}\Vert R_S}{\frac{1}{g_m+g_{mb}}\Vert R_S+r_O} \Big/R_S\\ =\Delta V \frac{1}{[1+(g_m+g_{mb})]R_Sr_O+R_S}$ $R_\tx{out}=\frac{\Delta V}{\Delta I}=\frac{1}{[1+(g_m+g_{mb})R_S]r_O+R_S}$

\begin{aligned} I_{r_O} &= -\frac{V_\tx{out}}{R_D}-(g_m V_1 + g_{mb}V_{bs})\\ &= - \frac{V_\tx{out}}{R_D}-\left[g_m\left(V_\tx{in}+V_\tx{out}\frac{R_S}{R_D}\right)+g_{mb}V_\tx{out}\frac{R_S}{R_D}\right] \end{aligned}

\begin{aligned} V_\tx{out}&=I_{r_O} r_O-\frac{V_\tx{out}}{R_D}R_S\\ &=\left\{ - \frac{V_\tx{out}}{R_D}-\left[g_m\left(V_\tx{in}+V_\tx{out}\frac{R_S}{R_D}\right)+g_{mb}V_\tx{out}\frac{R_S}{R_D}\right] \right\} r_O-\frac{V_\tx{out}}{R_D}R_S\\ &=-r_O\frac{V_\tx{out}}{R_D}-g_m r_O V_\tx{in} +(g_m+g_{mb})r_OV_\tx{out}\frac{R_S}{R_D}-V_\tx{out}\frac{R_S}{R_D} \end{aligned}

$A_v = \frac{V_\tx{out}}{V_\tx{in}}=\frac{-g_mr_OR_D}{R_D+R_S+r_O+(g_m+g_{mb})R_Sr_O}$

\begin{aligned} A_v &= - \frac{g_mr_O}{R_S+r_O+(g_m+g_{mb})R_Sr_O}\cdot \frac{1}{\frac{1}{R_D}+\frac{1}{R_S+r_O+(g_m+g_{mb})R_Sr_O}}\\ &=- \frac{g_mr_O}{r_O+[1+(g_m+g_{mb})r_O]R_S}\cdot \frac{1}{\frac{1}{R_D}+\frac{1}{r_O+[1+(g_m+g_{mb})r_O]R_S}}\\ &=-G_m(R_\tx{out}\Vert R_D) \end{aligned}

# Summary

$A_v$, $\eta=\lambda=0$ $-g_m R_D$ $- g_{m1}\dfrac{1}{g_{m2}+g_{mb2}}$

$-g_{m1} (r_{O1}|r_{O2})$ 不存在 $R_D=R_\tx{on2}$ $-\dfrac{1}{1/g_m+R_S}R_D$
$A_v$, $\eta,\lambda\neq 0$ $-g_m (r_O \Vert R_D)$

• $g_m = \mu_n C_\tx{ox} \frac{W}{L}(V_\tx{in}-V_\tx{TH})$
• $r_O=\dfrac{1}{\lambda I_D}$
• $g_{mb}=g_m \dfrac{\gamma}{2\sqrt{2\Phi_F+V_{SB}}}$

$A_v = \frac{-i_d \cdot \frac{1}{g_{m2}}}{i\cdot \frac{1}{g_{m1}}}=\frac{\frac{1}{g_{m2}}}{\frac{1}{g_{m1}}}=-\frac{g_{m1}}{g_{m2}}$

$A_v = \frac{R_D}{1/g_m+R_S}$

$A_v = \frac{R_D \Vert \{ r_O+[1+(g_m+g_{mb})r_O]R_S\}}{R_S+\frac{R_D+r_O}{1+(g_m+g_{mb})r_O}}$