# 差动放大器作业

\begin{align*} \newcommand{\dif}{\mathop{}\!\mathrm{d}} \newcommand{\belowarrow}[1]{\mathop{#1}\limits_{\uparrow}} \newcommand{\bd}{\boldsymbol} \newcommand{\tx}{\text} \newcommand{\L}{\mathscr{L}} \newcommand{\p}{\partial\,} \end{align*}
1. Describe advantages and drawbacks of differential signals comparing with single-ended signal.
• Higher immunity to environmental noise 更高的环境噪声抑制
• Reject supply noise 更高的电源噪声抑制
• Increase output swing from $V_{DD}-(V_{GS}-V_\tx{TH})$ to $2[V_{DD}-(V_{GS}-V_\tx{TH})]$ 更大的摆幅
• Higher linearity 更高的线性度
• Drawbacks:
• occupy twice area 占两倍面积
2. What is the problem of simple differential circuit? How to solve this problem?
• 问题：如果输入共模电平 Vin,CM 太低,输出会被截断。
• 解决方案：使用差分对电路，并引入电流源$I_{SS}$
• 为什么这样能解决？因为引入 $I_{SS}$ 后，$V_S$ 会随着输入共模电平变化，从而 $V_{GS}$ 始终是一个合适的值。
3. If all symmetry（对称的）, is the small signal gain（小信号增益） of differential pair __ than that of the common-source single stage?
• 要比较这两者，关键是比较 $I_D$
• $I_D$ 相同，则两者相同；$I_{SS}$ 相同，由于差分对的 $I_D=I_{SS}/2$，故比单端的要小。
4. If there is a small mismatch between $M_1$ and $M_2$, how do the parameters of the transistors affect the common mode rejection ratio (CMRR) of a differential pair?
• $\tx{CMRR}\approx\frac{g_m}{\Delta g_m}(1+2g_m R_{SS})$
• 如果 $g_{m1}$ 与 $g_{m2}$ 不匹配，就会导致 CMRR 减小。
5. Write the gain ($A_v$) expression of a differential pair with current-source load (Fig.2). How to increase its gain?
• $A_v = -g_{mN}(r_{ON}\Vert r_{OP})$
• increase $g_{mN}$ (i.e. $\frac{W}{L}$) or $r_{ON}$ or $r_{OP}$ by increasing L1~L4

4.2 Sketch the small-signal differential voltage gain of the circuit shown in Fig. 4.9(a) if $V_{DD}$ varies from 0 to 3V. Assume that $(W/L){1−3}=50/0.5$,$V\tx{in,CM}=1.3\tx{ V}$,and $V_b=1 \tx{ V}$.

• $V_{DD}=0$，$M_1,M_2,M_3$ 处于线性区
• 随着 $V_{DD}$ 增大，$M_3$ 进入饱和，此时 $I_{SS} = \frac{1}{2}\mu_n C_\tx{ox} (\frac{W}{L})(V_{GS3}-V_\tx{TH})^2=0.225 \tx{ mA}$
• 随着 $V_{DD}$ 进一步增大，$M_1,M_2$ 进入饱和区，此时有 $V_\tx{in,CM}=V_{DD}-R_DI_{SS}/2+V_\tx{TH}$，算得 $V_{DD}=0.6+0.1125 R_D \tx{ mV}$

4.13 In Problem 4.11, suppose $W_3=10 \tx{ μm}$, but $W_4=11 \tx{ μm}$. Calculate the CMRR.

A_\tx{DM-DM} = -g_{m} R_D\\ A_\tx{CM-DM} = - \frac{g_m \Delta R_D}{2g_m R_{SS}+1}\\ \therefore \tx{CMRR} = \frac{1+2 g_m R_{SS}}{\Delta R_D / R_D}\\ 其中，\\ g_{m}=\sqrt{2 \mu_n C_\tx{ox} \frac{W}{L} I_D} = 5 \tx{ mS}\\ R_{SS} = \frac{1}{\lambda I_D} = 40 {\rm k\Omega}\\ \begin{aligned} \frac{\Delta R_D}{R_D}&= \frac{1/g_{m3}-1/g_{m4}}{1/g_{m3}}\\ &=1-\frac{g_{m3}}{g_{m4}}\\ &=1-\frac{\sqrt{2\mu_p C_\tx{ox} (W/L)_3 I_D}}{\sqrt{2\mu_p C_\tx{ox} (W/L)_4 I_D}}\\ &=1-\sqrt{\frac{10}{11}}\\ &= 0.0465 \end{aligned}\\ \therefore \tx{CMRR} = 8623

4.18 Assuming that all the transistors in the circuits of Figs. 4.44 and 4.45 are saturated and $λ\neq 0$, calculate the small-signal differential voltage gain of each circuit.

For Fig. 4.44(a), $R_1$ 中间可看作是 0 电位（差分信号的中间），故半边电路为：

\begin{aligned} A_v &= \frac{-\frac{1}{g_{m3}}\Vert r_{O3} \Vert \frac{1}{2} R_1 \Vert r_{O1}}{\frac{1}{g_{m1}}}\\ &= -g_{m1} (\frac{1}{g_{m3}}\Vert r_{O3}\Vert \frac{1}{2} R_1 \Vert r_{O1}) \end{aligned}

For Fig.4.44(e)，半边电路为：

$-\frac{V_X}{r_{O1}}-\frac{g_{m1}V_\tx{in}}{2} = \frac{V_X - \frac{V_\tx{out}}{2}}{R_1} = \frac{V_\tx{out}}{2r_{O3}}+g_{m3} V_X\\ 由后一个等号得：V_X = \frac{(r_{O3}+R_1)V_\tx{out}}{2r_{O3}(1-R_1g_{m3})}\\ 代入原式有：\\ V_\tx{out} (\frac{R_1+r_{O1}}{R_1 r_{O1}}\cdot \frac{r_{O3}+R_1}{r_{O3}(1-R_1g_{m3})}-\frac{1}{R})=-g_{m1} V_\tx{in}\\ 从而：\\ A_v = \frac{V_\tx{out}}{V_\tx{in}} = - \frac{g_{m1}r_{O1}r_{O3}(1-R_1 g_{m3})}{R_1+r_{O1}+r_{O3}+g_{m3}r_{O1}r_{O3}}$

4.19. Consider the circuit shown in Fig. 4.46.

1. Sketch $V_\tx{out}$ as $V_\tx{in1}$ and $V_\tx{in2}$ vary differentially from zero to $V_{DD}$.
2. If $λ=0$, obtain an expression for the voltage gain. What is the voltage gain if $W_{3,4}=0.8W_{5,6}$

(a)

(b) 这题难在 $M_3,M_4$ 组成的交叉结构，我们先计算该结构的输入阻抗：

$I_X = g_{m3} V_2\\ I_X = -g_{m4} V_1\\ \Rightarrow 2I_X = -g_{m3,4}(V_1-V_2)=-g_{m3,4} V_X\\ \Rightarrow \frac{V_X}{I_X} = -\frac{2}{g_{m3,4}}$

\begin{aligned} A_v &= -g_{m1} (\frac{1}{g_{m5}}\Vert \frac{-1}{g_{m3}})\\ &= \frac{g_{m1}}{g_{m5}-g_{m3}} \end{aligned}\\ \tx{要求 } g_{m5}>g_{m3}

$g_m = \mu_p C_\tx{ox} \frac{W}{L} (V_{GS}-V_\tx{TH})\\ \Rightarrow \frac{g_{m3,4}}{g_{m5,6}} = \frac{(W/L)_{3,4}}{(W/L)_{5,6}} = 0.8$

$A_v = -\frac{g_{m1}}{g_{m5}-0.5g_{m5}} = -\frac{5 g_{m1}}{g_{m5}}$