# Cascode Stage

\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*}

# Cascode Stage

• 对 $M_1$ 来说，负载是从 $M_2$ 源端看进去的电阻，故 $A_{v1}=-\dfrac{g_{m1}}{g_{m2}}$
• 对 $M_2$ 来说，$A_{v2}=g_{m1}R_D$

1. 两管工作在饱和区：
• $V_\tx{in}>V_\tx{TH1}$ 且 $V_\tx{in}-V_\tx{TH1} \leq V_X$
• $V_b-V_X>V_\tx{TH2}$ 且 $V_b-V_\tx{TH2} \leq V_\tx{out}$
2. 电压间满足一定关系：
• $V_X=V_b-V_{GS2}$
• $V_\tx{in}=V_{GS1}$

\begin{aligned} V_\tx{out} &\geq V_\tx{in} - V_\tx{TH1}+V_{GS2}-V_\tx{TH2}\\ &=(V_{GS1}-V_\tx{TH1})+(V_{GS2}-V_\tx{TH2}) \end{aligned}

$I_{D2} = g_{m1}V_\tx{in} \frac{g_{m2}+g_{mb2}}{\frac{1}{r_{O1}}+(g_{m2}+g_{mb2})}$

$(g_{m2}+g_{mb2})$ 为从 $M_2$ 源端看进去的电导。从而增益为：

$A_v = g_{m1} \frac{g_{m2}+g_{mb2}}{\frac{1}{r_{O1}}+(g_{m2}+g_{mb2})} R_D$

\begin{aligned} R_\tx{out} &= [1+(g_{m2}+g_{mb2})r_{O2}]r_{O1}+r_{O2}\\ &\approx (g_{m2}+g_{mb2})r_{O2}r_{O1} \end{aligned}

$V_X$ is higher than $V_Y$ by $\Delta V$. Calculate the resulting difference between $I_{D1}$ and $I_{D2}$ if $λ≠0$.

(a) We have
\begin{aligned} I_{D1}-I_{D2} &= \frac{1}{2} \mu_n C_\tx{ox} \frac{W}{L} (V_b-V_\tx{TH})^2 (\lambda V_{DS1}-\lambda V_{DS2})\\ &=\frac{1}{2} \mu_n C_\tx{ox} \frac{W}{L} (V_b - V_\tx{TH})^2 (\lambda \Delta V) \end{aligned}
(b) Cascode 起到分压的作用
\begin{aligned} \Delta V_{PQ} &= \Delta V \frac{r_{O1}}{[1+(g_{m3}+g_{mb3})r_{O3}]r_{O1}+r_{O3}}\\ &\approx \frac{\Delta V}{(g_{m3}+g_{mb3})r_{O3}} \end{aligned}
Thus,
$$I_{D1}-I_{D2}=\frac{1}{2} \mu_n C_\tx{ox} \frac{W}{L} (V_b - V_\tx{TH})^2 \frac{\lambda \Delta V}{(g_{m3}+g_{mb3})r_{O3}}$$
In other words, cascoding reduces the mismatch between $I_{D1}$ and $I_{D2}$ by a factor of $(g_{m3} + g_{mb3})r_{O3}$.（个人觉得这就是简单的加大电阻来减小电流变化🧐）

The shielding property of cascodes diminishes if the cascode device enters the triode region. because
$$I_{D2}=\frac{1}{2}\mu_n C_\tx{ox} \left(\frac{W}{L}\right)_2\left[ 2(V_{b2}-V_P-V_\tx{TH2})(V_X-V_P)-(V_X-V_P)^2 \right]$$
As $V_X$ decreases, $V_P$ also drops, so that $I_{D2}$ remains constant.

\begin{aligned} V_D&=V_S+V_{DS}\\ &=V_S+r_Og_mV_S\\ &=(1+r_Og_m)V_S \end{aligned}

# Folded Cascode

$I_1$ 用于提供直流偏置，显然 $\vert I_{D1}\vert+I_{D2}=I_1$，如果 $I_{D1}$ 变小，就会使得 $I_{D2}$ 变大，从而改变 $V_\tx{out}$

• $V_\tx{in}>V_{DD}-\vert V_\tx{TH1} \vert$，$M_1$ is off and $M_2$ carries all of $I_1$. yielding $V_\tx{out}=V_{DD}-I_1R_D$
• $V_\tx{in}<V_{DD}-\vert V_\tx{TH1} \vert$, $M_1$ turns on in saturation, giving
$I_{D2}=I_1-\frac{1}{2}\mu_p C_\tx{ox} \left( \frac{W}{L} \right)_1 (V_{DD}-V_\tx{in}-\vert V_\tx{TH1} \vert)^2$
• $I_{D2}$ decrease further as $V_\tx{in}$ drops. If $I_{D2}=0$, $V_\tx{out}=V_{DD}$. The $V_\tx{in}$ at this point is denoted as $V_\tx{in1}$
$\frac{1}{2}\mu_p C_\tx{ox} \left( \frac{W}{L} \right)_1 (V_{DD}-V_\tx{in1}-\vert V_\tx{TH1} \vert)^2=I_1$

Thus,

$V_\tx{in1} = V_{DD}-\sqrt{\frac{2I_1}{\mu_p C_\tx{ox}(W/L)1}} - \vert V_\tx{TH1} \vert$