# 噪声

\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*}

# 噪声的统计学性质

## 平均功率

$\tx{average power} P_\tx{av} = \lim_{T\rightarrow \infty} \frac{1}{T} \int_{-T/2}^{+T/2} \frac{v^2(t)}{R_L} \dif t\\$

（To be more rigorous, $v^2(t)$ should be replaced by $v(t)\cdot v^*(t)$）

$P_\tx{av} = \lim_{T\rightarrow \infty} \frac{1}{T} \int_{-T/2}^{+T/2} v^2(t) \dif t$

## Noise Spectrum

Power spectral density (PSD) $S_x(f)$ shows how much power the signal carries at each frequency. $S_x(f)$ is defined as the average power carried by $x(t)$ in a one-hertz bandwidth around $f$，根据这个定义，$S_x(f)$ 的单位为 ${\rm V^2/Hz}$

$S_Y(f)=S_x(f)\vert H(f) \vert^2$

$P_\tx{av} = \int_{-\infty}^{+\infty} S_x(f) \dif f$

## Amplitude Distribution

Distribution of the amplitude, also called the “probability density function” (PDF), is defined as

$p_x(x)\dif x = \tx{probability of } x<X<x+\dif x$

where $X$ is the measured value of $x(t)$ at some point in time.

## Correlated and Uncorrelated Sources

\begin{aligned} P_\tx{av} &= \lim_{T\rightarrow \infty} \frac{1}{T} \int_{-T/2}^{+T/2} [x_1(t)+x_2(t)]^2 \dif t\\ &=\lim_{T\rightarrow \infty} \frac{1}{T} \int_{-T/2}^{+T/2} x_1^2(t) \dif t + \lim_{T\rightarrow \infty} \frac{1}{T} \int_{-T/2}^{+T/2} x_2^2(t) \dif t\\ &\quad + \lim_{T\rightarrow \infty} \frac{1}{T} \int_{-T/2}^{+T/2} 2x_1(t)x_2(t) \dif t\\ &= P_\tx{av1}+P_\tx{av2}+\lim_{T\rightarrow \infty} \frac{1}{T} \int_{-T/2}^{+T/2} 2x_1(t)x_2(t) \dif t \end{aligned}

## Signal-to-Noise Ratio

$\tx{SNR} = \frac{P_\tx{sig}}{P_\tx{noise}}\\ 或 =10\log (P_\tx{sig}/P_\tx{noise})$

## Noise Analysis Procedure

1. Identify the sources of noise (e.g., resistors and transistors) and write down the spectrum of each
2. Find the transfer function from each noise source to the output (as if the source were a deterministicsignal)
3. Utilize the theorem $S_Y(f)=S_x(f)\vert H(f)\vert^2$ to calculate the output noise spectrum contributed by each noise source. (The input signal is set to zero.)
4. Add all of the output spectra, paying attention to correlated and uncorrelated sources.

# Types of Noise

• electronic noise
• thermal noise
• flicker noise (1/f noise)
• environmantal noise

## Thermal Noise

Resistor Thermal Noise

• 产生原因：The random motion of electrons in a conductor
• one-sided spectral density：$S_v(f)=4kTR, f\geq 0$
• $k=1.38\times10^{-23}$ J/K is the Boltmann constant
• 模型：
• 无噪声的电阻与一个 $\overline{V_n^2}=4kTR$ 的电压源串联
• 无噪声的电阻与一个 $\overline{I_n^2}=4kT/R$ 的电流源并联

$$\frac{V_\tx{out}}{V_R}(s) = \frac{1}{RCS+1}$$

\begin{aligned} S_\tx{out}(f) &= S_v(f)\left| \frac{V_\tx{out}}{V_R}(j\omega)\right|^2\\ &=4kTR\frac{1}{4\pi^2 R^2 C^2 f^2 + 1} \end{aligned}

\begin{aligned} P_{n,\tx{out}} &= \int_0^{\infty} \frac{4kTR}{4\pi^2 R^2 C^2 f^2 + 1} \dif f\\ &=\frac{2kT}{\pi C} \int_0^\infty \frac{1}{x^2 + 1} \dif x\\ &=\frac{2kT}{\pi C} \tan^{-1} x\big\vert_0^\infty\\ &=\frac{kT}{C} \end{aligned} 这说明输出的噪声功率与电阻无关。因为当电阻增大时，虽然噪声也会增大，但传输函数减小（即带宽减小），最终噪声功率并没有变。

MOSFETs

• 产生原因：the noise generated in the channel
• 模型：
• 在 SD 两边并联上电流源 $\overline{I_n^2} = 4kT\gamma g_m$
• 在 $r_O$ 上串联上电压源 $\overline{V_n^2}=(4kT\gamma g_m)r_O^2$
• 注：
• 式中的 $\gamma$ 不是体效应的 $\gamma$。对于长沟道；$\gamma=2/3$，对于短沟道，这个值会变大。并且还与源漏电压有一定关系。一般我们认为 $\gamma\approx 1$
• $r_O$ 是等效电阻，不产生噪声

\begin{aligned} \overline{I_n^2} &= \lim_{n\rightarrow \infty}\sum_{i=1}^n \dfrac{g_m^2 i^2}{n^2}4kT\dfrac{R_G}{n}\\ &= g_m^2 \cdot 4kTR_G \lim_{n\rightarrow\infty} \sum_{i=1}^n \frac{i^2}{n^3}\\ &= g_m^2 \cdot 4kTR_G \cdot \lim_{n \rightarrow \infty} \frac{n(n+1)(2n+1)}{6n^3}\\ &= g_m^2 \cdot 4kTR_G \cdot \frac{1}{3} \end{aligned}

## Filcker Noise

• 产生原因：沟道与 ${\rm SiO_2}$ 边界表面的悬挂键断裂时，导致漏极电流有噪声
• 模型：$\overline{V_n^2}=\dfrac{K}{C_\tx{ox}WL}\cdot \dfrac{1}{f}$
• $K$ 取决于工艺，数量级在 $10^{-25} {\rm V^2F}$
• $C_\tx{ox}$ 也取决于工艺

$4kT\gamma g_m = \frac{K}{C_\tx{ox}WL}\cdot\frac{1}{f}\cdot g_m^2\\ \Rightarrow f_C=\frac{K}{\gamma C_\tx{ox} WL} g_m \frac{1}{4kT}$

$f_C$ 称为 corner frequency. 虽然这个值也与 $g_m$ 有关，但我们可以近似认为它只与 $L$ 有关，nanomoter transistors 中大概是 10~50 MHz。当高于该频率时，可以忽略 fliker noise.

# 噪声在电路中的表示

MOS 管包括热噪声与 1/f 噪声，电阻有热噪声
$$\overline{I_{n,\tx{th}}^2}=4kT\gamma g_m\\ \overline{I_{n,1/f}^2}=Kg_m^2/(C_\tx{ox}WLf)\\ \overline{I_{n,R_D}^2}=4kT/R_D$$

\begin{aligned} \overline{V_{n,\tx{out}}^2}&=(\overline{I_{n,\tx{th}}^2}+\overline{I_{n,1/f}^2}+\overline{I_{n,R_D}^2})R_D^2\\ &= \left( 4kT\gamma g_m + Kg_m^2/(C_\tx{ox}WLf) + \frac{4kT}{R_D}\right) R_D^2 \end{aligned}

\begin{aligned} \overline{V_{n,\tx{in}}^2}&=\frac{\overline{V_{n,\tx{out}}^2}}{A_v^2}\\ &=4kT \left( \frac{\gamma}{g_m}+\frac{1}{g_m^2 R_D} \right)+\frac{K}{C_\tx{ox}WL} \frac{1}{f} \end{aligned}

• 求输入噪声电压功率（Fig. 7.34a）：设上一级阻抗为 $0$，求出输出电压功率，然后再求输入电压功率
• 求输入噪声电流功率（Fig. 7.34b）：设上一级阻抗为 $\infty$，求出输出电压功率，然后再求输入电流功率

$V_{n,X} = \frac{Z_\tx{in}}{Z_\tx{in}+Z_S}V_{n,\tx{in}}+\frac{Z_\tx{in}Z_S}{Z_\tx{in}+Z_S}I_{n,\tx{in}}$

\begin{aligned} V_{n,X} &= \frac{Z_\tx{in}}{Z_\tx{in}+Z_S} \frac{V_{n,\tx{out}}}{A_v} + \frac{Z_\tx{in}Z_S}{Z_\tx{in}+Z_S} \frac{V_{n,\tx{out}}}{A_v Z_\tx{in}}\\ &= \frac{V_{n,\tx{out}}}{A_v}\\ \Rightarrow &V_{n, \tx{out}} = A_v V_{n,X} = V_{n,\tx{out}} \end{aligned}

$\tx{if }\vert Z_S \vert^2 \ll \frac{\overline{V_{n,\tx{in}}^2}}{\overline{I_{n,\tx{in}}^2}}\\ \tx{then the input-referred noise current can be neglected}$

# 单极放大器的噪声

$\overline{V_{n,\tx{out}}} = -\frac{\overline{V_{n,\tx{out}}}}{Z_L}Z_S+\left( -\frac{\overline{V_{n,\tx{out}}}}{Z_L}-\frac{\overline{V_{n,\tx{out}}}}{Z_L} Z_S(g_m+g_{mb})- \overline{I_{n}}\right)\cdot r_O\\ \Rightarrow \overline{V_{n,\tx{out}}} = \frac{-r_OZ_L\overline{I_n}}{Z_L+Z_S+r_O+(g_m+g_{mb})Z_Sr_O}$

$\overline{V_{n,\tx{out}}}=\overline{V_{n}} A_v\\ A_v = \frac{-g_mr_OZ_L}{Z_L+Z_S+r_O+(g_m+g_{mb})Z_Sr_O}\\ \Rightarrow \overline{V_{n,\tx{out}}} = \frac{-r_OZ_L\overline{I_n}}{Z_L+Z_S+r_O+(g_m+g_{mb})Z_Sr_O}$

## CS Stage

$\overline{V_{n,\tx{in}}^2} = 4kT \left( \frac{\gamma}{g_m}+\frac{1}{g_m^2 R_D} \right)+\frac{K}{C_\tx{ox}WL} \frac{1}{f}$

$\begin{cases} \overline{I_{n,\tx{th}}^2}=4kT\gamma g_m\\ \overline{I_{n,1/f}^2}=Kg_m^2/(C_\tx{ox}WLf)\\ \overline{I_{n,R_D}^2}=4kT/R_D \end{cases}\\ \Downarrow \tx{divided by } g_m^2\\ \begin{cases} \overline{V_\tx{Thev1}^2}=4kT\gamma/g_m\\ \overline{V_\tx{Thev2}^2}=K/(C_\tx{ox}WLf)\\ \overline{V_\tx{Thev3}^2}=4kT/(g_m^2 R_D) \end{cases}$ $\overline{V_{n,\tx{in}}^2} = \overline{V_\tx{Thev1}^2}+\overline{V_\tx{Thev2}^2}+\overline{V_\tx{Thev3}^2}$

Calculate the input-referred thermal noise voltage of the amplifier shown in Fig. 7.42(a)

the thermal noise of $M_1$ and $M_2$ are uncorrelated, therefore we write
\overline{V_{n,\tx{out}}^2} = 4kT(\gamma g_{m1}+\gamma g_{m2})(r_{O1}\Vert r_{O2})^2\\ \begin{aligned} \overline{V_{n,\tx{in}}^2} &= \frac{\overline{V_{n,\tx{out}}^2}}{g_{m1}^2(r_{O1}\Vert r_{O2})^2}\\ &= 4kT \gamma \left( \frac{1}{g_{m1}}+\frac{g_{m2}}{g_{m1}^2} \right) \end{aligned}
The equation above reveals the dependence of $\overline{V_{n,\tx{in}}^2}$ upon $g_{m1}$ and $g_{m2}$, confirming that $g_{m2}$ must be minimized because $M_2$ serves as a current source rather than a transconductor.

• 增大 $g_{m1}$，即
• 增大 $I_D$——trade-off: greater power dissipation
• 增大 $W$——trade-off: larger input and output capacitance
• 增大 $R_D$——trade-off: limiting the voltage headroom and lowering the speed
• 增大 $WL$，同时保持 $W/L$ 不变——trade-off: the device capacitances increase

## CG Stage

$\left( 4kT\gamma g_m + \dfrac{4KT}{R_D} \right) R_D^2 = \overline{V_{n,\tx{in}}^2} (g_m+g_{mb})^2 R_D^2\\ \Rightarrow \overline{V_{n,\tx{in}}^2} = \dfrac{4kT(\gamma g_m+1/R_D)}{(g_m+g_{mb})^2}$

$\overline{I_{n,\tx{in}}^2} = \frac{4kT}{R_D}=\overline{I_{n,R_D}^2}$

With the input shorted to ground, we have

$\overline{V_{n1,\tx{out}}^2} = \frac{1}{C_\tx{ox} f} \left[ \frac{g_{m1}^2K_N}{(WL)_1}+\frac{g_{m3}^2K_P}{(WL)_3} \right] (r_{O1}\Vert r_{O3})^2\\ A_v = (g_{m1}+g_{mb1})(r_{O1}\Vert r_{O3})\\ \Rightarrow \overline{V_{n,\tx{in}}^2} = \frac{1}{C_\tx{ox} f} \left[ \frac{g_{m1}^2K_N}{(WL)_1}+\frac{g_{m3}^2K_P}{(WL)_3} \right] \frac{1}{(g_{m1}+g_{mb1})^2}$

With the input open, we have

$\overline{V_{n2,\tx{out}}^2} = \frac{1}{C_\tx{ox} f} \left[ \frac{g_{m2}^2K_N}{(WL)_2}+\frac{g_{m3}^2K_P}{(WL)_3} \right] R_\tx{out}^2\\ \overline{I_{n,\tx{in}}^2} = \frac{1}{C_\tx{ox} f} \left[ \frac{g_{m2}^2K_N}{(WL)_2}+\frac{g_{m3}^2K_P}{(WL)_3} \right]$

## Source Followers

$\overline{V_{n,\tx{out}}^2} = \overline{I_{n2}^2} \left( \frac{1}{g_{m1}}\Vert \frac{1}{g_{mb1}} \Vert r_{O1} \Vert r_{O2} \right)^2\\ A_v = \frac{g_{m1}}{g_{m1}+g_{mb1} + 1/r_{O1} + 1/r_{O2}}\\ \Rightarrow \overline{V_{n2,\tx{in}}^2}= \frac{\overline{V_{n,\tx{out}}^2}}{A_v^2}=4kT\gamma \frac{g_{m2}}{g_{m1}^2}$

\begin{aligned} \overline{V_{n,\tx{in}}^2} &= \overline{V_{n1,\tx{in}^2}}+\overline{V_{n2,\tx{in}^2}}\\ &=4kT\gamma \left( \frac{1}{g_{m1}}+\frac{g_{m2}}{g_{m1}^2} \right) \end{aligned}

$\overline{V_{n,\tx{out}}^2}=\overline{V_{n,1}^2}(g_{m1}R_\tx{out})^2+\overline{V_{n,2}^2}(g_{m2}R_\tx{out})^2$

$\overline{V_{n,\tx{in}}^2} = \overline{V_{n,1}^2}+\overline{V_{n,2}^2} \frac{g_{m2}^2}{g_{m1}^2}$

## Cascode Stage

\begin{aligned} \overline{V_{n,\tx{in}}^2} &= \frac{\overline{I_{n,\tx{out}}^2}}{g_{m1}^2}\\ &= 4kT \left( \frac{\gamma}{g_{m1}}+\frac{1}{g_{m1}^2R_D} \right) \end{aligned}

# 差分放大电路的噪声

We can model the overall noise as depicted in Fig. 7.55(b). For low-frequency operation, $\overline{I_{n,\tx{in}}^2}$ is negligible.

To calculate the thermal component of $\overline{V_{n,\tx{in}}^2}$, we first obtain the total output noise with the inputs shorted together (Fig. 7.56a). Since the noise sources in the circuit are uncorrelated, we simply derive the effect of each source individually (Fig. 7.56b). the contribution of In1 is obtained by first reducing the circuit to that in (Fig. 7.56c). With neglecting channel-length modulation, we can prove that half of $I_{n1}$ flows through $R_{D1}$ and the other half through $M_2$ and $R_{D2}$ (这个我是真不知道怎么证明😥，目前唯一的解释就是 $M_1$ 的源端看进去的电阻也等于 $1/(g_{m2}+g_{mb2})$，这样一半电流往下，另一半则流过 $M_1$).

Thus, the differential output noise due to M1 is equal to

$V_{n,\tx{out}}\Big\vert_{M1}=\frac{I_{n1}}{2} R_{D1}+\frac{I_{n1}}{2} R_{D2}$

$\overline{V_{n,\tx{out}}^2}\Big\vert_{M1}=\overline{I_{n1}^2}R_D^2$

$\overline{V_{n,\tx{out}}^2}\Big\vert_{M2}=\overline{I_{n2}^2}R_D^2$

\begin{aligned} \overline{V_{n,\tx{out}}^2} &= \left( \overline{I_{n1}^2}+\overline{I_{n2}^2} \right)R_D^2 + 2(4kTR_D)\\ &=8kT(\gamma g_m R_D^2+R_D) \end{aligned}

$\overline{V_{n,\tx{in}}^2}=8kT \left( \frac{\gamma}{g_m}+\frac{1}{g_m^2 R_D} \right)$

$\overline{V_{n,\tx{in}}^2}=8kT \left( \frac{\gamma}{g_m}+\frac{1}{g_m^2 R_D} \right)+\frac{2K}{C_\tx{ox}WL}\frac{1}{f}$

\begin{aligned} \Delta I_{D1}-\Delta I_{D2} &= g_m \Delta V_\tx{in}\\ &= \sqrt{2\mu_n C_\tx{ox} \frac{W}{L}\left( \frac{I_{SS}+I_n}{2} \right)} \Delta V_\tx{in}\\ &\approx \sqrt{2\mu_n C_\tx{ox} \frac{W}{L} \frac{I_{SS}}{2}} \Delta V_\tx{in} \left( 1+\frac{I_n}{2 I_{SS}} \right) \Delta V_\tx{in}\\ &=g_{m0} \left( 1+\frac{I_n}{2 I_{SS}} \right) \Delta V_\tx{in} \end{aligned}

• 对于 $M_3$，它的噪声电流主要流经自己（即 $1/g_{m3}$），然后产生噪声电压 $4kT\gamma/g_{m3}$，再通过 $M_4$ 放大，得到输出噪声电流 $4kT \gamma g_{m4}^2 / g_{m3}$
• $M_4$ 的噪声电流直接流到输出端，等于 $4kT \gamma g_{m4}$

$\overline{V_{n,\tx{in}}^2} = 8kT\gamma \left(\frac{1}{g_{m1,2}}+\frac{g_{m3,4}}{g_{m1,2}^2}\right)$

# Noise Bandwith

$V_0^2 \cdot B_n = \int_0^\infty \overline{V_{n,\tx{out}}^2} \dif f$