Basic Current Mirrors

电路

• $\mu_n, C_{ox},W$ 一致

• $L$可以不同以提供增益

Current

$$ \begin{aligned} I_{out} &= \frac{1}{2} \mu_n C_{ox} (W/L)1 (V{GS2} - V_{T2})^2 \\ I_{REF} &= \frac{1}{2} \mu_n C_{ox} (W/L)2 (V{GS1} - V_{T2})^2 \end{aligned} $$

$$ \begin{equation} I_{out} = \frac{(W/L)_2}{(W/L)1} \cdot I{REF} \end{equation} $$

Voltage Gain(应用)

• $M_2,M_3$ 构成BCM, $M_1$是CS，$R_L$是$R_{out}$

$$ |A_v| = g_{m1} \frac{(W/L)_3}{(W/L)_2}R_L $$

Cascode Current Mirrors

出发点

• 若考虑CLM

$$ \begin{aligned} I_{out} &= I_{REF} \cdot \frac{(W/L)2}{(W/L)1} \cdot \frac{ 1 + \lambda V{DS1} }{ 1 + \lambda V{DS2} } \end{aligned} $$

• 为了消除CLM的影响，必须使得$V_{DS1} = V_{DS2}$，即$V_X = V_Y$

Cascode Current Mirror

$V_{p, min}$

• 为了让$M_3$工作在Saturation Region

$$ \begin{aligned} V_p &\ge V_N - V_{TH} \\ V_p &\ge V_{GS0} + V_{GS1} - V_{TH} \\ V_p &\ge V_{OV1} + V_{OV0} + V_{TH} \end{aligned} $$

$$ \begin{equation} V_{p,min} = V_{OV1} + V_{OV0} + V_{TH} \end{equation} $$

选择适当的MOSFET

• 使得$\dfrac{(W/L)_3}{(W/L)_0} = \dfrac{(W/L)2}{(W/L)1}$, 这样可以满足$V_X = V_Y$, 从而$I{REF} = I{out}$,

Low Voltage Cascode

电路图

1. (a)中实际上是Cascode Stage Shorted

2. (b)即是Low Voltage Cascode

$V_b$范围

1. 为了让$M_1, M_2$工作在Saturation Region

$$ \begin{aligned} V_X &\ge V_b - V_{TH2} \\ V_{GS1} &\ge V_b - V_{TH2} \\\ \\ V_A &\ge V_{GS1} - V_{TH1} \\ V_{b} - V_{GS1} &\ge V_{GS1} - V_{TH1} \end{aligned} $$

$$ \begin{equation} V_{GS2} + (V_{GS1} - V_{TH1}) \le V_b \le V_{GS1}+ V_{TH2} \end{equation} $$

2. 进一步处理不等式

$$ \begin{equation} V_{GS2} \le V_{TH1} + V_{TH2} \end{equation} $$

改良

• $M_S$ is biased such that $V_{GS s} \approx V_{TH3}$

$$ \begin{aligned} V_{N’} &\approx V_N - V_{TH3} \\ \therefore V_B &\approx V_{N’} - V_{GS3} \\ &= V_{GS0} + V_{GS1} - V_{TH3} - V_{GS3} \\ &= V_{GS1} - V_{TH3} \end{aligned} $$

$$ \begin{equation} V_B \approx V_{GS1} - V_{TH3} \end{equation} $$

题解

• $I_{REF}$工作的最小电压为$0.5V$

$$ \begin{aligned} V_N &= V_{GS0} + V_{GS1} \\ &= \sqrt{\frac{2I_{REF}}{k_{n0}}} - \sqrt{\frac{2I_{REF}}{k_{n1}}} + V_{TH1} + V_{TH0} \\ \because V_{DD} - V_N &= 0.5V \\ \therefore I_{REF} &= \frac{1}{2}\mu_nC_{ox}\frac{(V_{DD} - V_{TH1} - V_{TH2} - 0.5)^2}{(\sqrt{(L/W)_0} + \sqrt{(L/W)_1})^2} \end{aligned} $$

Active Current Mirrors

Voltage Gain

Method I

1. Caculate $G_m$ from (b)

$$ \begin{aligned} |A_v| &= G_m R_{out} \\ G_m &= \frac{I_{out}}{V_{in}} \\ &= \frac{\dfrac{1}{2} g_{m1} V_{in}}{V_{in}} \\ &= \frac{1}{2}g_{m1} \end{aligned} $$

2. Caculate Gain from (c)

$$ \begin{aligned} R_2 &= 2r_{o2} + \frac{1}{g_{m1}} \\ &\approx 2r_{o2} \\ \therefore R_{out} &\approx 2r_{o2} || r_{o4} \\ \therefore |A_v| &\approx \frac{g_{m1}}{2} (2r_{o2} || r_{o4}) \end{aligned} $$

$$ \begin{equation} |A_v| \approx \frac{g_{m1}}{2} (2r_{o2} || r_{o4}) \end{equation} $$

• if $r_{04} \longrightarrow \infin$

$$ \begin{equation} |A_v| = g_{m1} r_{02} \end{equation} $$

Method II

$$ \begin{aligned} A_v &= \frac{V_{out}}{V_{in}} \\ &= \frac{V_{out}}{V_p} \times \frac{V_P}{V_{in}} \\ \frac{V_P}{V_{in}} &= \frac{R_{eq}}{R_{eq} + \dfrac{1}{g_{m1}}} \\ R_{eq} &= \frac{1}{g_{m2}} + \frac{r_{o4}}{g_{m2}r_{o2}} \\ &= \frac{1}{g_{m2}}(1+\frac{r_{o4}}{r_{o2}}) \\ \therefore \frac{V_P}{V_{in}} &= \frac{1 + \dfrac{r_{o4}}{r_{o2}}}{2 + \dfrac{r_{o4}}{r_{o2}}} \\ \frac{V_{out}}{V_p} &= \frac{r_{o2} + \dfrac{1}{g_{m2}}}{R_{eq}} \\ &= \frac{1 + g_{m2}r_{o2}}{1 + \dfrac{r_{o2}}{r_{o4}}} \\ &\approx \frac{g_{m2}r_{o2}}{1 + \dfrac{r_{o2}}{r_{o4}}} \\ \therefore |A_v| &= \frac{g_{m2}}{2} [(2r_{o2}) || r_{o4}] \end{aligned} $$

Large Signal Analysis

实例解析

Small Signal Analysis

电路图

Voltage Gain

1. Method I

• 计算$G_m$

$$ \begin{aligned} I_{D1} &= |I_{D3}| = |I_{D4}| \\ &= \frac{ g_{m 1,2} V_{in}}{ 2 } \\ I_{D2} &= - \frac{ g_{m2} V_{in} }{ 2 } \\ \therefore I_{out} &= g_{m1,2} V_{in} \\ \therefore G_m &= g_{m1,2} \end{aligned} $$

• 计算$R_{out}$

$$ \begin{aligned} I_X &= 2 \frac{ V_X }{ 2r_{o1,2} + \dfrac{ 1 }{ g_{m3} } || r_{o3}} + \frac{ V_X }{ r_{04} } \end{aligned} $$

if $2r_{o1,2} \gg \dfrac{ 1 }{ g_{m3}} || r_{03}$

$$ \begin{aligned} R_{out} &= r_{o1,2} || r_{o4} \end{aligned} $$

• Gain

$$ \begin{equation} |A_v| = g_{m1,2} (r_{o2} || r_{o4}) \end{equation} $$

Common Mode Property

电路图

Gain

• 注意，$r_{o3,4} \gg \dfrac{ 1 }{ g_{m3,4} }$

$$ \begin{equation} \begin{aligned} A_{CM} &\approx \frac{ - \dfrac{ 1 }{ 2g_{m3,4}} || \dfrac{ r_{o3,4} }{ 2 } }{\dfrac{ 1 }{ 2g_{m1,2}} + R_{SS}} \\ &= \frac{ -1 }{ 1 + 2g_{m1,2} R_{SS} } \cdot \frac{ g_{m1,2} }{ g_{m3,4} } \end{aligned} \end{equation} $$

CMMR

$$ \begin{equation} \begin{aligned} CMMR &= | \frac{ A_{DM} }{ A_{CM}} | \\ &= g_{m3,4} (r_{o1} || r_{o2}) ( 1 + g_{m1,2}) \end{aligned} \end{equation} $$

附录

MOSFET参数

• $\mu_n, V_{TH}$ 受温度影响

Headroom Voltage

余量电压是指在电路设计中，为了确保电器或电子设备正常工作，需要提供的超过最低电压要求的额外电压。它表示电路中电源电压与电器或电子设备所需电压之间的差值。