Basic Differential Pair
电路
- Bias Current Source $I_{SS}$ : Reducing the Variation of CM level
Qualitative Analysis
$V_{out} - V_{in}$ 图像
$$ \begin{aligned} V_{in1} - V_{in2} &\in (-\infin, + \infin) \end{aligned} $$
共模输入的范围
- 为了使$M_3$工作在饱和区
$$ \begin{aligned} V_P &\ge V_{GS3} - V_{TH3} \\ V_{in,CM} - V_{GS1} &\ge V_{GS3} - V_{TH3} \\ \therefore V_{in,CM} &\ge V_{GS1} + V_{OV3} \end{aligned} $$
- 为了使$M_1,M_2$工作在饱和区
$$ \begin{aligned} V_{D1} &\ge V_{G1} - V_{T1} \\ \therefore V_{in,CM} &\le V_{out1} + V_{TH1} = V_{DD}- \frac{I_{SS}}{2} R_D + V_{TH1} \end{aligned} $$
- $V_{in,CM}$的范围
$$ \begin{align} V_{GS1} + (V_{GS3} - V_{TH3} ) &\le V_{in,CM} \le min \Big[ V_{DD}-\frac{I_{SS}}{2}R_D + V_{TH1, V_{DD}} \Big] \end{align} $$
共模输出的范围
- 最大值
$$ \begin{aligned} V_{out} &\le V_{DD} \end{aligned} $$
- 最小值
- 在Satuartion Region和Triode Region的交界处
$$ \begin{aligned} V_{DS} &= V_{OV} \\ V_D &= V_G - V_{TH} \\ \therefore V_{out} &= V_{in,CM} - V_{TH} \\\ \\ \end{aligned} $$
- $V_{out}$的范围
$$ \begin{equation} V_{in,CM} - V_{TH} < V_{out} < V_{DD} \end{equation} $$
$|A_v| - V_{in,CM}$ 图像
$V_{in,CM}$ | $A_v$ |
---|---|
$(0,V_1)$ | $M_1, M_2$ off |
$(V_1, V_2)$ | All in Satuartion |
$(V_2, \infin )$ | In Triode Region |
Quantitative Analysis
Differential Voltage Gain
- 当$M_1$和$M_2$都工作在Satuartion Region时,有以下分析:
- 推算$\Delta I - \Delta V$
$$ \begin{aligned} V_{in1} - V_{in2} &= V_{G1} - V_{G2} \\ V_{GS} - V_t&= \sqrt{\frac{2I}{k_n}} \\ (V_{in1} - V_{in2})^2 &= \frac{2}{k_n} (I_{SS} - 2\sqrt{I_{D1} I_{D2}}) \\ \frac{1}{2}k_n(V_{in1} - V_{in2})^2 &= I_{SS} - 2\sqrt{I_{D1}I_{D2}} \\ \therefore (I_{D1} - I_{D2})^2 &= I_{SS}^2 - 4I_{D1}I_{D2} \\ &= I_{SS}^2 - (I_{SS} - \frac{1}{2}k_n(V_{in1} - V_{in2})^2)^2 \\ &= I_{SS}k_n(V_{in1} - V_{in2})^2 - \frac{1}{4}k_n^2(V_{in1} -V_{in2})^4 \\ \therefore I_{D1} - I_{D2} &= \frac{1}{2}k_n(V_{in1} - V_{in2})\sqrt{\frac{4I_{SS}}{k_n} - (V_{in1} - V_{in2})^2} \end{aligned} $$
$$ \begin{equation} \Delta I_{D} = \frac{1}{2} k_n\Delta V_{in} \sqrt{\frac{4I_{SS}}{k_n} - \Delta V_{in}^2} \end{equation} $$
- Differential Gain
$$ \begin{equation} G_m = \frac{\partial \Delta I_D}{\partial \Delta V_{in}} = \frac{1}{2} k_n \dfrac{\dfrac{4I_{SS}}{k_n} - 2\Delta V_{in}^2}{\sqrt{\dfrac{4I_{SS}}{k_n} - \Delta V_{in}^2}} \end{equation} $$
- 当$\Delta V_{in} = 0$时
$$ \begin{aligned} G_m &= \sqrt{k_nI_{SS}}\\ |A_v| &= \sqrt{k_nI_{SS}} R_D \\ &= g_mR_D \end{aligned} $$
- 当$\Delta V_{in} = \sqrt{\frac{2I_{SS}}{k_n}}$时
$$ \begin{aligned} G_m &= 0 \\ |A_v| &= 0 \\ V_{GS} &= \sqrt{\frac{I_{SS}}{k_n}} + V_T \\ \end{aligned} $$
- 当$\Delta V_{in} = \sqrt{\frac{4I_{SS}}{k_n}}$时,$M_1, M_2$ NOT in Satuartion Region
Voltage Gain
- 下图中,$M_1$与$M_2$相同,$R_{D1} = R_{D2}$
- 左电路的等效电路 (CS Stage With Source Degeneration)
- 右电路的等效电路 (CG Stage)
$$ \begin{aligned} R_S &= \frac{1}{g_{m2}}, R_T = \frac{1}{g_{m1}} \\ \frac{V_X}{V_{in1}} &= -\frac{R_{D1}}{\dfrac{1}{g_{m1}} + \dfrac{1}{g_{m2}}} \\ \frac{V_Y}{V_{in1}} &= \frac{R_{D2}}{\dfrac{1}{g_{m1}} + \dfrac{1}{g_{m2}}} \\ A_v &=\frac{V_X - V_Y}{V_{in1} - V_{in2}} \\ &= -g_m R_D \end{aligned} $$
对于线性电路,$V_P$不变(虚接地)
- 假设图(a)输入电压以图(b)所示方式变化
$$ \begin{aligned} V_{in1} = V_0 + \Delta V_{in}, V_{in2} &= V_0 - \Delta V_{in} \\ V_1 = V_a + V_1, \Delta V_2 &= V_a - \Delta V_2 \\ \because I_1 + I_2 &= I_T \\ \therefore g_m(\Delta V_1 + \Delta V_2) &= 0 \\ \therefore \Delta V_1 &= - \Delta V_2 \\\ \\ V_{in1} - V_1 &= V_{in2} - V_2 = V_P \\ V_{in1} + \Delta V_{in} - (V_a + \Delta V_1) &= V_{in2} - \Delta V_{in} - (V_a - \Delta V_2) \\ \therefore \Delta V_{in} &= \Delta V_1 \\ \therefore V_P&不变 \end{aligned} $$
- $P$ is a Virtual Ground
Common-Mode Response (Small-Signal Model)
电路图
Common-Mode Gain
$$ \begin{aligned} A_v &= \frac{\dfrac{R_D}{2}}{\dfrac{1}{2g_m} + R_{SS}} \\ G_m &= \frac{g_m}{1 + 2g_m R_{SS}} \end{aligned} $$
Considering the Manufactured Difference
$M_1$ 和 $M_2$ 若不同
- 阻抗为$R_{SS}$
$$ \begin{aligned} V_P &= (g_{m1} + g_{m2})(V_{in,CM} - V_P)R_{ss} \\ \therefore V_P &= \frac{g_{m1} + g_{m2}}{1 + (g_{m1} + g_{m2})R_{SS}} R_{SS} \\ V_X &= g_{m1}R_D(V_{in,CM} - V_P) \\ &= \frac{-g_{m1}}{1 + (g_{m1} + g_{m2})R_{SS}}R_D V_{in,CM} \\ V_Y &= \frac{-g_{m2}}{1 + (g_{m1} + g_{m2})R_{SS}}R_D V_{in,CM} \\
\end{aligned} $$
$$ \begin{equation} A_{CM-DM} = \frac{V_X - V_Y}{V_{in,CM}} = \frac{\Delta g_m R_D}{1 + (g_{m1} +g_{m2}) R_{SS}} \end{equation} $$
- 将$R_{SS}$ 替换为$C_S$
$$ \begin{equation} A_{CM-DM} = \frac{\Delta g_m R_D}{\sqrt{1 + (g_{m1} + g_{m2})^2 \Big|\dfrac{1}{\omega C_S}\Big|^2} } \end{equation} $$
若$R_D$不相同
$R_D$ | $R_D + \Delta R_D$ |
---|
$$ \begin{equation} A_{CM-DM} = \frac{g_m \Delta R_D}{1 + (g_{m1} +g_{m2}) R_{SS}} \end{equation} $$
Differential Gain & CMRR
Differential Gain
- 在电流源被$R_{SS}$替换后(假设$V_{in1} = - V_{in2}$)
$$ \begin{equation} |A_{DM}| = \frac{ R_D }{ 2 } \frac{ g_{m1} + g_{m2} + 4g_{m1} g_{m2} R_{SS} }{ 1 + (g_{m1} + g_{m2} )R_{SS}} \end{equation} $$
CMRR
$$ \begin{equation} \begin{aligned} CMRR &= \frac{A_{DM}}{A_{CM-DM}} \\ &= \frac{ g_{m1} + g_{m2} + 4 g_{m1} g_{m2} R_{SS} }{ 2 \Delta g_m } \\ & \approx \frac{ g_m }{ \Delta g_m } ( 1 + 2g_m R_{SS}) \end{aligned} \end{equation} $$
$g_m$ | $\dfrac{ g_{m1} + g_{m2} }{ 2 }$ |
---|---|
$\approx$ | 假设$g_{m1} \approx g_{m2}$ |
Differential Pair With MOS Loads
Differential Pair With Diode-Connected and Common-Source Loads
Diode-Connected Loads
$$ \begin{equation} \begin{aligned} A_v &= -g_{mN} (\frac{1}{g_{mP}} || r_{oP} || r_{oN}) \\ & \approx -\frac{g_{mN}}{g_{mP}} \\ &= \sqrt{\frac{ k_{nN} }{ k_{nP}}} \end{aligned} \end{equation} $$
- Tradeoff among output voltage swing, voltage gain and CM input range
Current-Source Loads
$$ \begin{equation} A_v = -g_{mN} (r_{oP} || r_{oN}) \end{equation} $$
- Diode-connected Loads需要考虑电阻是因为二极管的非线性特性,而Current Source Loads不需要考虑电阻是因为电流源的线性特性。
Addition of current sources to increase the voltage gain
$$ |A_v| \approx g_{m1} [(g_{m3}r_{o3}r_{o1}) || (g_{m5}r_{o5}r_{o7})] $$
- 电压增益可以近似地表示为晶体管的跨导和输出阻抗的乘积
Gilbert Cell
原理
$$ \begin{aligned} |A_v| &= g_m R_D \\ &= \sqrt{ 2 k_n I_D } R_D \end{aligned} $$
-
$V_{cont} \longrightarrow I_{SS} \longrightarrow I_D \longrightarrow |A_v|$
-
通过电压控制增益
Gilbert Cell
$$ \begin{equation} \begin{aligned} V_{out} &= A_1 V_1 + A_2 V_2 \\ &= R_D(I_{D1} + I_{D4}) - R_D(I_{D2} + I_{D3}) \end{aligned} \end{equation} $$
附录
静态电流损耗
静态电流损耗可能会影响电路的性能,并导致功率损耗和热量产生。为了减少静态电流损耗,可以采取以下措施:
-
选择$\dfrac{1}{g_m}$和$r_o$较小的 MOSFET。
-
将$V_G$保持在最小值,以减少电路总电阻。
-
通过使用负反馈电路来控制 MOSFET 的电流,从而减少静态电流损耗。
Voltage Headroom
电路内部被消耗的,不能转换为Output Voltage Swing的部分