Chapter 14 - Frequency Response

The Frequency response of a circuit is the variation in its behavior with change in signal frequency.

Transfer Function

The transfer function $H(\omega )$$H(\omega)$ of a circuit is the frequency-dependent ratio of a phasor output $Y(\omega )$$Y(\omega)$ to a phasor input $X(\omega )$$X(\omega)$

$H(\omega )=\frac{Y(\omega )}{X(\omega )}$$H(\omega) = \frac{Y(\omega)}{ {X(\omega)}}$

Y and X can either be voltage or current Hence 4 types of transfer function

Voltage gain: $\frac{V_o(\omega )}{V_i(\omega )}$$V_o(\omega) \over V_i(\omega)$

Current gain: $\frac{I_o(\omega )}{I_i(\omega )}$$I_o(\omega) \over I_i(\omega)$

Transfer impedance: $\frac{V_o(\omega )}{I_i(\omega )}$$V_o(\omega) \over I_i(\omega)$

Transfer admittance: $\frac{I_o(\omega )}{V_i(\omega )}$$I_o(\omega) \over V_i(\omega)$

Polynomial expression

$H(\omega )=\frac{N(\omega )}{D(\omega )}$$H(\omega) = \frac{N(\omega)}{ {D(\omega)}}$

N和D分别为由Y和X简化后得到的两个多项式

其中，$N(\omega )=0$$N(\omega) = 0$ 的根被叫做"zeros"($H(\omega )=0$$H(\omega) = 0$)，represented as $j\omega =z_1,z_2,...$$j\omega = z_1,z_2, ...$

其中，$D(\omega )=0$$D(\omega) = 0$ 的根被叫做"poles"($H(\omega )=\mathrm{\infty }$$H(\omega) = \infty$)，represented as $j\omega =p_1,p_2,...$$j\omega = p_1,p_2, ...$

常用s替换jw来避免进行complex algebra，最后再替换回去

Transfer function 实际是由电路的differential equation 进行 Laplace Transform 后得到的式子

Decible Scale

Bel

Used to measure the ratio of two levels of power, called power gain G

$G=\text{Number of bels}=lo{g}_{10}\frac{P_2}{P_1}$$G = \text{Number of bels} = log_{10} \frac{P_2}{P_1}$

Decible - 1/10 of a bel

$G_{dB}=10lo{g}_{10}\frac{P_1}{P_2}$$G_{dB} = 10log_{10} \frac{P_1}{P_2}$

Assume $R_1=R_2$$R_1 = R_2$

$P=V^2/R⟹G_{dB}=10lo{g}_{10}\frac{V_2^2/R_2}{V_1^2/R_1}=20lo{g}_{10}\frac{V_2}{V_1}$$P = V^2 / R \implies G_{dB} = 10log_{10} \frac{V_2^2 / R_2}{V_1^2/R_1} = 20log_{10}\frac{V_2}{V_1}$

$P=I^{2}R⟹G_{dB}=10lo{g}_{10}\frac{I_2^2{R}_2}{I_1^2{R}_1}=20lo{g}_{10}\frac{I_2}{I_1}$$P = I^2R \implies G_{dB} = 10log_{10} \frac{I_2^2 R_2}{I_1^2R_1} = 20log_{10}\frac{I_2}{I_1}$

正是因为平方的关系，Gain的log前面才是20这个比较奇怪的数字

Bode Plots

Bode plots are semilog plots of the magnitude(in decibles) and phase(in degrees) of a transfer function versus frequency.

semilog plot: 一个坐标轴是log scale，另一个是linear

decibles: 指每格增长10倍，所以是semilog plot

bode plots就是两个图，横坐标都是angular frequency in decibles($\omega =2\pi f$$\omega = 2\pi f$)，分别是$H-\omega$$H - \omega$ 和 $\varphi -\omega$$\phi - \omega$ 图像

这两个图分别对应着transfer function $H(\omega )$$H(\omega)$的magnitude 和 phase

使用decible 表示H: $H_{dB}=20lo{g}_{10}H$$H_{dB} = 20log_{10} H$

常见的gain

Magnitude H	$H_{dB}=20lo{g}_{10}H$$H_{dB} = 20log_{10} H$	$P_{out}/P_{in}$$P_{out} / P_{in}$
$1/\sqrt{2}$$1/\sqrt 2$	-3	1/2
1	0	1
$\sqrt{2}$$\sqrt 2$	3	2

Standard Form of Transfer Function

Constant Term - $H(\omega )=K$$H(\omega) = K$

$H=20lo{g}_{10}K,\varphi =0$$H = 20log_{10} K, \phi = 0$ 常量没有虚部，不影响phase，也不随频率变化

Zero/Pole at the origin - ($H(\omega )=(j\omega {)}^N$$H(\omega) = (j\omega)^N$)

For $H(\omega )=j\omega ⟹H=20lo{g}_{10}\sqrt{(j\omega {)}^2}=20lo{g}_{10}\omega ,\varphi =90\mathrm{°}$$H(\omega) = j\omega \implies H = 20log_{10} \sqrt{(j \omega)^2} = 20log_{10}\omega, \phi = 90\degree$

For $H(\omega )j\omega ⟹H=-20lo{g}_{10}\omega ,\varphi =-90\mathrm{°}$$H(\omega) j\omega \implies H = -20log_{10}\omega, \phi = -90\degree$

Generalize 之后，对于$H(\omega )=(j\omega {)}^N⟹H=20NdB/decade$$H(\omega) = (j\omega)^N \implies H = 20N dB/decade$, $\varphi =90N\mathrm{°}$$\phi = 90N\degree$

Simple Zero/Pole - （$H(\omega )=(1+j\omega /z_1{)}^N$$H(\omega) = (1+j\omega/z_1)^N$)

最简单的情况， $H(\omega )=(1+j\omega /z_1)$$H(\omega) = (1+j\omega/z_1)$， 即N= 1时，由复数的性质可得

$H_{dB}=20lo{g}_{10}\sqrt{1^2+(\omega /z_1{)}^2}$$H_{dB} = 20log_{10} \sqrt{1^2 + (\omega/z_1)^2}$

$\varphi =atan(\omega /z_1)$$\phi = atan(\omega/z_1)$

我们用linear approximation来绘制这两个图

当$\omega =z_1$$\omega = z_1$时，$H_{dB}=20lo{g}_{10}\sqrt{2}=3dB$$H_{dB} = 20log_{10} \sqrt 2 = 3dB$, $\varphi =atan(1)=45\mathrm{°}$$\phi = atan(1) = 45\degree$

Generalize之后，对于double zero(N = 2)，或任意N，有

Quadratic Zero/Pole - ($H(\omega )=[1+j2\zeta \omega /{\omega }_k+(j\omega /{\omega }_k{)}^2{]}^N$$H(\omega) = [1+j2\zeta\omega / \omega_k + (j\omega/\omega_k)^2]^N$ )

注意，double pole - $(1+\omega /p_1{)}^2$$(1 + \omega/p_1)^2$ 有别与quadratic pole，重点就在 damping factor - $\zeta$$\zeta$上

?（*具体是什么不知道，长大后在学习）

对于不同的$\zeta$$\zeta$值，绘制出来的图像也不一样，我们使用$\zeta =1$$\zeta = 1$时的图像来进行linear approximation，此时

$H(\omega )=1+j2\omega /{\omega }_k+(j\omega /{\omega }_k{)}^2=(1+j\omega /{\omega }_k{)}^2$$H(\omega) = 1 + j2\omega/\omega_k + (j\omega/\omega_k)^2 = (1+j\omega/\omega_k)^2$

图像与double zero一样，N = 2, H斜率为40dB/decade，$\varphi$$\phi$斜率为90$\mathrm{°}$$\degree$

这个是pole的图像, $H(\omega )=[1+j2\zeta \omega /{\omega }_k+(j\omega /{\omega }_k{)}^2{]}^{-1}$$H(\omega) = [1+j2\zeta\omega / \omega_k + (j\omega/\omega_k)^2]^{-1}$

Zeros cause an increase in slope

Poles cause a decrease

从数学上可以看出来，因为pole在分母上，但为什么会有这种behavior还需要对transfer function有更深入的理解?

Resonance

https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/5bcec4bfba5f2e99754b77509e9e7ab4_resonance_qfactr.pdf

Resonance is a conduction in an RLC circuit in which the capacitive and inductive reactances are equal in magnitude, thereby resulting in a purely resistive impedance (The imaginary part of the transfer function is 0)

Resonance occurs in any system that has a compliex conjugate pair of poles.(?)

The frequency that satifisfies the condition $Im(H({\omega }_0))=0$$Im(H(\omega_0)) = 0$ is called the resonant frequency(${\omega }_0$$\omega_0$)

Series Resonance

任何series circuit都可以变成RLC串联的形式

$Z=H(\omega )=\frac{V_S}{I}=R+j\omega L+\frac{1}{j\omega C}$$Z = H(\omega) = \frac{V_S}{I} = R + j\omega L + \frac{1}{j\omega C}$

$Im(Z)=\omega L-\frac{1}{\omega C}=0⟹{\omega }_{0}L=\frac{1}{{\omega }_{0}C}⟹{\omega }_0=\frac{1}{\sqrt{LC}}rad/s$$Im(Z) = \omega L - \frac{1}{\omega C} = 0 \implies \omega_0 L = {1 \over \omega_0 C} \implies \omega_0 = \frac{1}{\sqrt{LC}} rad/s$

$\omega =2\pi f⟹f_0=\frac{1}{2\pi \sqrt{LC}}Hz$$\omega = 2\pi f \implies f_0 = \frac{1}{2\pi\sqrt{LC}} Hz$

注意频率单位，Hz为频率，弧度为角速度

Half-Power Frequency

The dissipated power is half the maximum value

Average (active) power dissapated by the circuit

active power(P) 指电路中电阻(R)消耗的能量（有功功率）

与之对应的是reactive power(Q) （无功功率）

Apprant Power $S^2=Q^2+P^2$$S^2 = Q^2 + P^2$

在实数域上

$p(t)=V_R(\omega )cos(\omega t)I_R(\omega )cos(\omega t)=V_R{I}_{R}co{s}^2(\omega t)=\frac{1}{2}{V}_R{I}_R$$p(t) = V_R(\omega)cos(\omega t) I_R(\omega)cos(\omega t) = V_RI_Rcos^2(\omega t) = \frac{1}{2} V_RI_R$

$V_R,I_R$$V_R, I_R$ 分别是R的电压和电流函数的amplitude

$P(\omega )=\frac{1}{\pi }{\int }_0^{\pi }co{s}^2(\omega t)dt\ast V_R{I}_R$$P(\omega) = \frac{1}{\pi} \int_0^\pi cos^2(\omega t) dt * V_RI_R$

当resonance时功率最大，此时$I=V_m/R$$I = V_m/R$最大功率为$P_{max}(\omega )=\frac{1}{2}{I}^{2}R=\frac{1}{2}\frac{V_m^2}{R}$$P_{max}(\omega) = \frac{1}{2} I^2R = \frac{1}{2} \frac{V_m^2}{R}$

$P({\omega }_1)=P({\omega }_2)=\frac{V_m^2}{4R}=\frac{(V_m/\sqrt{2}{)}^2}{2R}$$P(\omega_1) = P(\omega_2) = \frac{V_m^2}{4R} = \frac{(V_m/\sqrt 2)^2}{2R}$

因此，当$V=V_m/\sqrt{2}$$V = V_m/\sqrt 2$时得到${\omega }_1,{\omega }_2$$\omega_1,\omega_2$

图中的R代表电阻的R，不是当时电路总阻抗的R值

那么电路总阻抗的R值应该变为$R_{total}=\sqrt{2}R$$R_{total} = \sqrt 2 R$

求解$\sqrt{R^2+(\omega L-\frac{1}{\omega C}{)}^2}=\sqrt{2}R$$\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} = \sqrt 2 R$

${\omega }_1=-\frac{R}{2L}+\sqrt{(\frac{R}{2L}{)}^2+\frac{1}{LC}}$$\omega_1 = -\frac{R}{2L} + \sqrt{(\frac{R}{2L})^2 + \frac{1}{LC}}$

${\omega }_2=\frac{R}{2L}+\sqrt{(\frac{R}{2L}{)}^2+\frac{1}{LC}}$$\omega_2 = \frac{R}{2L} + \sqrt{(\frac{R}{2L})^2 + \frac{1}{LC}}$

${\omega }_0=\sqrt{{\omega }_1{\omega }_2}$$\omega_0 = \sqrt{\omega_1 \omega_2}$

resonant frequency is the geometric mean of the half-power frequencies

Bandwidth B - difference betwee half-power frequencies

$B={\omega }_2-{\omega }_1=\frac{R}{L}$$B = \omega_2 - \omega_1 = \frac{R}{L}$

Quality Factor(Q)

Reactive Power也是Q，但此Q非彼Q

The Quality Factor relates the maximum energy stored to the energy dissipated in the circuit per cycle of oscillation at resonance

$Q=2\pi \frac{\text{Peak energy stored}}{\text{Energy Dissipated}}$$Q = 2\pi \frac{\text{Peak energy stored}}{\text{Energy Dissipated}}$

Peak Energy stored: $E_{max}=\frac{1}{2}{I}_L^{2}L+\frac{1}{2}{V}_C^{2}C$$E_{max} = \frac{1}{2} I_L^2 L + \frac{1}{2} V_C^2 C$

Let $V_C=Asin(\omega t)⟹I_L=C\frac{dv}{dt}=\omega CAcos(\omega t)$$V_C = A sin(\omega t) \implies I_L = C\frac{dv}{dt} = \omega C A cos(\omega t)$

$E_{max}=\frac{1}{2}({\omega }^2{C}^2{A}^{2}co{s}^2(\omega t)+A^{2}si{n}^2(\omega t))$$E_{max} = \frac{1}{2} (\omega^2 C^2 A^2 cos^2(\omega t) + A^2 sin^2(\omega t))$

带入${\omega }_0=\frac{1}{\sqrt{LC}}$$\omega_0 = \frac{1}{\sqrt{LC}}$

$E_{max}=\frac{1}{2}C{A}^2$$E_{max} = \frac{1}{2} CA^2$

$E_{dissipate}={\int }_0^{2\pi /\omega }{I}^{2}Rdt=I_L^{2}R{\int }_0^{2\pi /\omega }co{s}^2(\omega t)dt=I_L^{2}R{\int }_0^{2\pi }\frac{co{s}^2(u)}{{\omega }_0}du=I_{)}^{2}R\pi /{\omega }_0={\omega }_0{C}^2{A}^{2}R\pi$$E_{dissipate} = \int_0^{2\pi/\omega} I^2Rdt = I_L^2R \int_0^{2\pi/\omega} cos^2(\omega t) dt = I_L^2 R \int_0^{2\pi} \frac{cos^2(u)}{\omega_0} du = I_)^2R\pi/\omega_0 = \omega_0C^2A^2R\pi$

$Q=2\pi \ast \frac{\frac{1}{2}C{A}^2}{{\omega }_0{C}^2{A}^{2}R\pi }=\frac{1}{CR{\omega }_0}=\frac{{\omega }_{0}L}{R}$$Q = 2\pi * \frac{\frac{1}{2} CA^2}{\omega_0C^2A^2R\pi} = \frac{1}{CR\omega_0} = \frac{\omega_0 L}{R}$

因此

$B=\frac{R}{L}=\frac{{\omega }_0}{Q}$$B = \frac{R}{L} = \frac{\omega_0}{Q}$

Q与B成反比，因此Q越大，selectivity越小

The selectivity of an RLC circuit is the ability of the circuit to respond to a certain frequency and discriminate against all other frequencies.

$V_C,V_L$$V_C, V_L$实际电压

在谐振电路中，电容和电感上的电压可以远大于输入电压

$|V_L|=|I_L|Z_L=|I|{\omega }_{0}L=\frac{V_m}{R}{\omega }_{0}L=\frac{X_L}{R}=Q{V}_m$$|V_L| = |I_L|Z_L = |I| \omega_0 L = \frac{V_m}{R} \omega_0 L = \frac{X_L}{R}= QV_m$

$|V_C|=|I_C|Z_C=|I|/{\omega }_{0}C=\frac{V_m}{R{\omega }_{0}C}=\frac{X_C}{R}{V}_m=Q{V}_m$$|V_C| = |I_C| Z_C = |I|/\omega_0C = \frac{V_m}{R\omega_0 C} = \frac{X_C}{R} V_m= QV_m$

两个元件上的实际电压都可能大于电源电压

在串联谐振电路中，$X_C=X_L$$X_C = X_L$ （就是之前的${\omega }_{0}L=\frac{1}{{\omega }_{0}C}$$\omega_0 L = {1 \over \omega_0 C}$)

由于电容和电感的特性，电感让电流比电压lead 90，电容让电压比电流lead 90，最后结果就是两个抵消了，电感和电流上电压大小相同，phase相差180度$V_L=-V_C$$V_L = -V_C$

High-Q circuits

When quality factor $Q\ge 10$$Q \geq 10$, we call the circuit High-Q Circuit. For all pratical purposes, the half-power frequencies are symmetrical aronud the resonant frequency, and can be approximated as:

${\omega }_1\approx {\omega }_0-\frac{B}{2}$$\omega_1 \approx \omega_0 - \frac{B}{2}$

${\omega }_2\approx {\omega }_0+\frac{B}{2}$$\omega_2 \approx \omega_0 + \frac{B}{2}$

红色和绿色分别代表$X_C,X_L$$X_C, X_L$

橙色是$H(\omega )$$H(\omega)$ (令R = 1, C = 0.4, L = 1, V_m = 5)，蓝色是$|I|$$|I|$

当$\omega <{\omega }_0$$\omega < \omega_0$时，容性阻抗更大，电路呈现容性

当$\omega >{\omega }_0$$\omega > \omega_0$时，感性阻抗更大，电路呈感性

可以看到这个电流函数其实是不对称的，是由于我们的Q不够大，可以想象当Q足够大时，${\omega }_1,{\omega }_2$$\omega_1, \omega_2$基本对称

Parallel Resonance

一个道理，不过这次改用admittance，之前电压的关系现在变成了电流的关系

$Y=H(\omega )=\frac{I}{V}=\frac{1}{R}+j\omega C+\frac{1}{j\omega L}$$Y = H(\omega) = \frac{I}{V} = \frac{1}{R} + j\omega C + \frac{1}{j\omega L}$

$Im(Y)=\omega C-\frac{1}{\omega L}=0⟹{\omega }_{0}C=\frac{1}{{\omega }_{0}L}⟹{\omega }_0=\frac{1}{\sqrt{LC}}rad/s$$Im(Y) = \omega C - \frac{1}{\omega L} = 0 \implies \omega_0 C = {1 \over \omega_0 L} \implies \omega_0 = \frac{1}{\sqrt{LC}} rad/s$

注意，这里根series不一样，C和L互换了位置，因为Y = 1/Z

电感和电容两个构成回路，对整个电路来说像是断路

$I_L=-I_C$$I_L = -I_C$

$|I_L|=|I_C|=Q{I}_m$$|I_L| = |I_C| = QI_m$

由于$H(\omega )$$H(\omega)$这次与上次不同，half-power frequency, B, Q的计算方式都有所不同

${\omega }_1=-\frac{1}{2RC}+\sqrt{(\frac{1}{2RC}{)}^2+\frac{1}{LC}}$$\omega_1 = -\frac{1}{2RC} + \sqrt{(\frac{1}{2RC})^2 + \frac{1}{LC}}$

${\omega }_2=\frac{1}{2RC}+\sqrt{(\frac{1}{2RC}{)}^2+\frac{1}{LC}}$$\omega_2 = \frac{1}{2RC} + \sqrt{(\frac{1}{2RC})^2 + \frac{1}{LC}}$

$B={\omega }_2-{\omega }_1=\frac{1}{RC}$$B = \omega_2 - \omega_1 = \frac{1}{RC}$

$Q={\omega }_0/B={\omega }_{0}RC=\frac{R}{{\omega }_{0}L}$$Q = \omega_0/B = \omega_0 RC = \frac{R}{\omega_0 L}$

我们可以只记Q的不同，剩下的用Q来表示

$B={\omega }_0/Q$$B = \omega_0/Q$

${\omega }_1,{\omega }_2={\omega }_0\sqrt{1+(\frac{1}{2Q}{)}^2}\pm \frac{{\omega }_0}{2Q}$$\omega_1, \omega_2 = \omega_0 \sqrt{1 + (\frac{1}{2Q})^2} \pm \frac{\omega_0}{2Q}$

Passive Filters

A Filter is a circuit that is designed to pass signals with desired frequencies and reject or attenuate others.

A filter is a passive filter if it consists of only passive elements R, L and C.

A filter is an active filter if it consists of active elements.

Low-pass Filter & High-pass filter

在filter中, cutoff frequency(${\omega }_c$$\omega_c$)对应half-power frequency, also called rolloff frequency

两种filter用的是同一个电路，这个RC电路中，电阻上的电压是高频电压（高频电流流过了capacitor，在电阻产生电压）

电容上是低频电压（低频的信号由于无法跨过电阻在电阻两端产生电压）

Low pass: $H(\omega )=\frac{V_o}{V_i}=\frac{1/j\omega C}{R+1/j\omega C}=\frac{1}{1+j\omega RC}$$H(\omega) = \frac{V_o}{V_i} = \frac{1/j\omega C }{R + 1/j\omega C} = \frac{1}{1+j\omega RC}$

High pass: $H(\omega )=\frac{V_o}{V_i}=\frac{R}{R+1/j\omega C}=\frac{j\omega RC}{1+j\omega RC}$$H(\omega) = \frac{V_o}{V_i} = \frac{R}{R + 1/j\omega C} = \frac{j\omega RC}{1+j\omega RC}$

${\omega }_c=1/RC$$\omega_c = 1/RC$

Band-pass Filter & Band-stop filter

使用resonance circuit，电流的图像刚好符合我们的需求，电阻上的电压是band-pass, 电容和电感上的电压是band-stop

$H(\omega )=\frac{V_o}{V_i}=\frac{R}{R+j\omega L-1/\omega C)}$$H(\omega) = \frac{V_o}{V_i} = \frac{R}{R + j\omega L - 1/\omega C)}$

$H(\omega )=\frac{V_o}{V_i}=\frac{j\omega L-1/\omega C)}{R+j\omega L-1/\omega C)}$$H(\omega) = \frac{V_o}{V_i} = \frac{j\omega L - 1/\omega C)}{R + j\omega L - 1/\omega C)}$

${\omega }_0=1/\sqrt{LC}$$\omega_0 = 1/\sqrt{LC}$

${\omega }_1,{\omega }_2=\pm \frac{R}{2L}+\sqrt{(\frac{R}{2L}{)}^2+\frac{1}{LC}}$$\omega_1, \omega_2 = \pm \frac{R}{2L} + \sqrt{(\frac{R}{2L})^2 + \frac{1}{LC}}$

Band-pass 无法使用一个low-pass串联一个high-pass，因为后面的circuit会load前面的circuit，最终结果会与预期不同。下面可以看到是用op amp可以实现电路隔离。事实上，使用active filter的一大好处就是可以不使用电感（占用PCB空间）

在面对一个未知电路时，如果不知道它是什么类型的filter，可以通过对应上面这个table来决定

Active Filters

Active Filter 特性

1. 可以有大于1的gain，向电路中增加能量

2. 不需要电感

3. passive filter在低频perform poorly

不过Active filters are less reliable and less stable. Pratical limit is about 100kHz, 工作频率要低于次频率。

General first-order active filter (order = number of poles)

First Order Low-Pass Filter

$H(\omega )=\frac{V_o}{V_i}=-\frac{R_f}{R_i}\frac{R_f/j\omega C_f}{R_f+1/j\omega C_f}=-\frac{R_f}{R_i}\frac{1}{1+j\omega C_f{R}_f}$$H(\omega) = \frac{V_o}{V_i} = -\frac{R_f}{R_i} \frac{R_f/j\omega C_f}{R_f + 1/j\omega C_f} = -\frac{R_f}{R_i} \frac{1}{1+j\omega C_f R_f}$

${\omega }_c=\frac{1}{R_f{C}_f}$$\omega_c = \frac{1}{R_fC_f}$

First Order High-Pass Filter

$H(\omega )=-\frac{R_f}{R_i+1/j\omega C_i}=-\frac{j\omega C_i{R}_f}{1+j\omega C_i{R}_i}$$H(\omega) =-\frac{R_f}{R_i + 1/j\omega C_i} = - \frac{j\omega C_iR_f}{1 + j\omega C_iR_i}$

${\omega }_c=\frac{1}{R_i{C}_i}$$\omega_c = \frac{1}{R_iC_i}$

注意，这里不是没有了$-R_f/R_i$$-R_f/R_i$ 的gain，而是无法从分母提出来，但其实当$\omega$$\omega$变大时，$H(\omega )$$H(\omega)$趋向于$-R_f/R_i$$-R_f/R_i$

Band-Pass Filter

$H(\omega )=\frac{V_o}{V_i}=(-\frac{1}{1+j\omega C_{1}R})(-\frac{j\omega C_{2}R}{1+j\omega C_{2}R})(-\frac{R_f}{R_i})=-\frac{R_f}{R_i}\frac{1}{1+j\omega C_{1}R}\frac{j\omega C_{2}R}{1+j\omega C_{2}R}$$H(\omega) = \frac{V_o}{V_i} = (-\frac{1}{1 + j\omega C_1 R})(-\frac{j\omega C_2 R}{1 + j\omega C_2 R}) (-\frac{R_f}{R_i}) = -\frac{R_f}{R_i}\frac{1}{1 + j\omega C_1 R}\frac{j\omega C_2 R}{1 + j\omega C_2 R}$

Low-pass section: ${\omega }_2=\frac{1}{R{C}_1}$$\omega_2 = \frac{1}{RC_1}$

High-pass section: ${\omega }_1=\frac{1}{R{C}_2}$$\omega_1 = \frac{1}{RC_2}$

有了这两个，我们就能算剩下的

${\omega }_0=\sqrt{{\omega }_1{\omega }_2}$$\omega_0 = \sqrt{\omega_1 \omega_2}$, $B={\omega }_2-{\omega }_1$$B = \omega_2 - \omega_1$, $Q={\omega }_0/B$$Q = \omega_0 / B$

使用Standard form化简H得到

$H(\omega )=-\frac{R_f}{R_i}\frac{j\omega /{\omega }_1}{(1+j\omega /{\omega }_1)(1+j\omega /{\omega }_2)}$$H(\omega) = -\frac{R_f}{R_i} \frac{j\omega/\omega_1}{(1+j\omega/\omega_1)(1+j\omega/\omega_2)}$

带入${\omega }_0=\sqrt{{\omega }_1{\omega }_2}$$\omega_0 = \sqrt{\omega_1 \omega_2}$

passband gain : $K=|H({\omega }_0)|=\frac{R_f}{R_i}\frac{{\omega }_2}{{\omega }_1+{\omega }_2}$$K = |H(\omega_0)| = \frac{R_f}{R_i} \frac{\omega_2}{\omega_1 + \omega_2}$

最后的inverter不是为了invert，而是像图中所说的provde gain。事实上，在经过inverter前，信号是跟原来的phase一样的，只是没有$-\frac{R_f}{R_i}$$-\frac{R_f}{R_i}$的gain

Band-Reject Filter (Notch Filter)

$H(\omega )=\frac{R_f}{R_i}(\frac{1}{1+j\omega /{\omega }_2}+\frac{j\omega /{\omega }_1}{1+j\omega /{\omega }_1})=\frac{R_f}{R_i}\frac{(1+j\omega /{\omega }_1{)}^2}{(1+j\omega /{\omega }_2)(1+j\omega /{\omega }_2)}$$H(\omega) = \frac{R_f}{R_i}(\frac{1}{1 + j\omega / \omega_2} + \frac{j\omega/\omega_1}{1 + j\omega/\omega_1}) =\frac{R_f}{R_i}\frac{(1+j\omega/\omega_1)^2}{(1+j\omega/\omega_2)(1 + j\omega / \omega_2)}$

在$\omega$$\omega$趋近于0和无穷大时, $H(\omega )$$H(\omega)$ 趋近于$\frac{R_f}{R_i}$$\frac{R_f}{R_i}$

在${\omega }_0$$\omega_0$时，带入$H(\omega )$$H(\omega)$可得到

$K=|H(\omega )|=\frac{R_f}{R_i}\frac{2{\omega }_1}{{\omega }_1+{\omega }_2}$$K = |H(\omega)| = \frac{R_f}{R_i}\frac{2\omega_1}{\omega_1 + \omega_2}$

Scaling

Magnitude Scaling is the process of increasing all impedances in a network by a factor, the frequency response remaining unchanged

Frequency Scaling is the process of shifting the frequency response of a network up or down the frequency axis while leaving the impedance the same

When scaling together:

${\omega }_0^{\prime }=K_f{\omega }_0$$\omega_0' = K_f \omega_0$

$B={\omega }_2^{\prime }-{\omega }_1^{\prime }=K_f{\omega }_2-K_f{\omega }_1=K_{f}B$$B = \omega_2' - \omega_1' = K_f \omega_2 - K_f \omega_1 = K_f B$

$Q^{\prime }=Q$$Q' = Q$

Diodes

Ideal Diode

名副其实的理想二极管，要么导通(on, turn-on, forward bias)，要么不导通(off, turned-off, reverse bias)

Ideal Model 适合快速分析二极管的通断

Junction Diodes

使用pn结制成的二极管通常会有如下i-v characteristic

Forward-Bias region

$i=I_S(e^{v/V_T}-1)$$i = I_S(e^{v/V_T} - 1)$

$I_S$$I_S$: saturation current/ scale current, strong function of temperature

$V_T$$V_T$: thermal voltage

$I_s$$I_s$ directly proportional to the cross-sectional area of diode

For small signal diodes, $I_s$$I_s$ is in the order of ${10}^{-15}A$$10^{-15} A$. It doubles in value for every $5\mathrm{°}C$$5 \degree C$ rise in temperature

$V_T=\frac{kT}{q}=0.0862T(mV)$$V_T = \frac{kT}{q} = 0.0862T (mV)$

k = Boltzmann's constant = 8.62e-5 eV/K T = absolute temperature q = magnitude of electronic charge = 1.6e-19

Usually use the 20 Celcius, $V_T\approx 25mV$$V_T \approx 25mV$ $V_2-V_1=V_{T}ln\frac{I_2}{I_1}=2.3V_{T}log\frac{I_2}{I_1}$$V_2 - V_1 = V_T ln\frac{I_2}{I_1} = 2.3V_T log\frac{I_2}{I_1}$

Voltage drop decreases about 2mV per 1 Celcius temparature increase. (可以用来做温度计)

Reverse-Bias Region

沿用上面的公式，在v < 0时，可以近似为$i\approx -I_S$$i \approx -I_S$

reverse current主要源于leakage effect，受到pn结横截面积影响，但与$I_S$$I_S$对于温度的相关性不大相同

reverse current doubles for every $10\mathrm{°}C$$10\degree C$ rise in temperature ($I_S$$I_S$ 是$5\mathrm{°}C$$5 \degree C$)

Breakdown Region

二极管的反向击穿，当电流低于Breakdown Voltage $V_{BR}$$V_{BR}$时，电流突然变大

https://www.bilibili.com/video/BV1S34y167Ck/

Avalanche Breakdown 雪崩击穿

电子将共价键中的电子创出来，被创出来的电子又去创其它的，像核裂变一样导致载流子增多

一般用于由较小变化引发较大电流的场景，比如感光器件

Zener Breakdown 齐纳击穿

耗尽层较薄，中间电场较大，电子得到足够能量破坏共价键

一般齐纳击穿所需反向电压较小，$V_{BR}<5V$$V_{BR} < 5V$, 雪崩击穿较大，$V_{BR}>8V$$V_{BR} > 8V$，在中间这里可能会有两种击穿共存

两种击穿都是可逆的，不过雪崩击穿由于电压很大，不控制住电流可能会导致烧掉二极管

稳压二极管 - Zener Diode

通过zener breakdown来钳位电压，适合小功率负载

$V_{ZK}$$V_{ZK}$: The voltage of onset of zener operation, $I_{ZK}$$I_{ZK}$ is the current at that point

在Q点附近（Q点为测试点，测试电压为$V_{ZT}$$V_{ZT}$），i-v几乎呈线性，等效电阻为$r_z$$r_z$， 该直线延长线交x轴于$V_{Z0}$$V_{Z0}$，此时二极管两端电压为$V_Z=V_{Z0}+r_z{I}_Z$$V_Z = V_{Z0} + r_zI_Z$（参考上面等效电路图）

The temperature coefficient of zener voltage $V_Z$$V_Z$ is commonly known as temco, and expressed in $mV/\mathrm{°}C$$mV/\degree C$

Shunt regulators

regulator circuit appears in parallel(shunt) with the load

Diode Models

Ideal model - 上面提到了，一般只用来测通断

Exponential Model

当$V_{DD}>0.5V$$V_{DD} > 0.5V$，此时二极管电流$I>>I_S$$I >> I_S$，因此之前公式可以近似为$I=I_S{e}^{V_D/V_T}$$I = I_Se^{V_D/V_T}$ （后面剪掉的$I_S$$I_S$被忽略了）

由于exponential model不是线性的，所以一般用图像法，或者interative analysis来求得方程组的解

Constant voltage drop model

然而，一般情况下这种分析很耗时，所以观察二极管伏安特性曲线，在$V>0.5V$$V > 0.5V$时开始快速上涨，V 在 0.6-0.8之间时，电流上涨速度很快。因此，我们假定二极管的压降为固定的0.7V

也因为这个性质，可以用二极管来钳位0.7V电压

Small Signal Model

当电路的交流signal很小时，二极管近似工作在线性区域

我们将电压分为直流部分和交流部分，$v_D(t)=V_D+v_d(t)$$v_D(t) = V_D + v_d(t)$

那么可以近似为二极管在Q点的切线上工作

Q点为quiescent point，是只有直流部分时造成的i-v，满足$I_D=I_S{e}^{V_D/V_T}$$I_D = I_Se^{V_D/V_T}$

那么$i_D(t)=I_S{e}^{(V_D+v_D(t))/V_T}=I_D{e}^{v_D(t)/V_T}$$i_D(t) = I_Se^{(V_D + v_D(t))/V_T} = I_D e^{v_D(t) / V_T}$

当$v_D(t)/V_T<<1$$v_D(t)/V_T << 1$时，我们可以对原式进行泰勒展开$i_D(t)\approx I_D(1+\frac{v_D(t)}{V_T})$$i_D(t) \approx I_D(1+\frac{v_D(t)}{V_T})$

这个叫做small-signal approximation，usueally valid when $v_d<5mV$$v_d < 5mV$ ($V_T$$V_T$ is usually 25mV)

此时的等效电阻small-signal resistance $r_d$$r_d$= 1/small-signal conductance = $1/{\left[\frac{\partial i_D}{\partial v_D}\right]}_{i_D=I_D}=V_T/I_D$$1/\left[ \frac{\partial i_D}{\partial v_D} \right]_{i_D = I_D} = V_T/I_D$

Rectifier Circuits

Half-Wave Rectifier

$V_D=0.7V$$V_D = 0.7V$

Peak inverse voltage $PIV=v_s$$PIV = v_s$

一般选元器件时选择50% greater than PIV

Full-Wave Rectifier

需要center-tapped到变压器线圈中间

$V_D=0.7V$$V_D = 0.7V$

$PIV=v_o+v_s=2v_s-v_D$$PIV = v_o + v_s = 2v_s - v_D$，需要承受更大的反向电压

Bridge Rectifier

优势时不用center-tapped到变压器中间，同时相比full-wave只需要一半线圈

缺点是$v_o=v_s-2V_D$$v_o = v_s - 2V_D$，中间有两个diode的压降

假设D1, D2导通，那么$-V_{D4}=v_s-v_{D2},-V_{D3}=v_o+v_{D2}$$-V_{D4} = v_s - v_{D2}, -V_{D3} = v_o + v_{D2}$

$PIV=V_{D4}=V_{D3}=V_s-V_D$$PIV = V_{D4} = V_{D3} = V_s - V_D$

Peak Rectifier

使用一个filter capacitor来将电压维持在最高点

Ripple Voltage

当波形为负时，右半部分构成RC电路，持续放电，因此我们希望$RC>>T$$RC>>T$来保证每个周期之间的压降（ripple voltage$V_r$$V_r$）不会太大，RC电路的特性满足

$v_o=V_p{e}^{-t/RC}$$v_o = V_pe^{-t/RC}$

当R = 0时，电路没有负载，电压会一直稳定在Vp

而在这个rectifier里，t = T

$V_p-V_r=v_o=V_p{e}^{-T/RC}$$V_p - V_r = v_o = V_pe^{-T/RC}$

因为RC >> T，$T/RC<<1$$T/RC << 1$进行泰勒展开近似后得到$V_r=V_p-V_p(1-\frac{T}{RC})=\frac{V_{p}T}{RC}=\frac{V_p}{fRC}=\frac{I_L}{fC}$$V_r = V_p - V_p(1 - \frac{T}{RC}) = \frac{V_pT}{RC} = \frac{V_p}{fRC} = \frac{I_L}{fC}$

Charge

在电容放电时，电流恒定，而放电时间为第一次达到$V_p$$V_p$ 到 第二次碰到正弦波的时间(或者理解为T - 充电时间)

令充电时间 = $\mathrm{\Delta }t$$\Delta t$

$V_{p}cos(\omega \mathrm{\Delta }t)=V_p-V_r$$V_p cos(\omega \Delta t) = V_p - V_r$

因为充电时间相对T来说很短，$\omega \mathrm{\Delta }t=\frac{2\pi \mathrm{\Delta }t}{T}<<1$$\omega \Delta t = \frac{2\pi \Delta t}{T} << 1$，可以使用泰勒近似

$V_p(1-(\omega \mathrm{\Delta }t{)}^2/2)=V_p-V_r⟹\mathrm{\Delta }t=\frac{T}{2\pi }\sqrt{\frac{2V_r}{V_p}}$$V_p(1 - (\omega \Delta t)^2/2) = V_p - V_r \implies \Delta t = \frac{T}{2\pi}\sqrt{\frac{2V_r}{V_p}}$

$Q_{supplied}=i_{C(average)}\mathrm{\Delta }t\approx (i_{D(average)}-I_L)\frac{T}{2\pi }\sqrt{\frac{2V_r}{V_p}}$$Q_{supplied} = i_{C(average)} \Delta t \approx (i_{D(average)} - I_L)\frac{T}{2\pi}\sqrt{\frac{2V_r}{V_p}}$

根据电容特性，并且在充电期间也在持续放电($I_L$$I_L$恒定)

$Q_{lost}=C{V}_r=I_{L}T$$Q_{lost} = CV_r = I_LT$

根据电荷守恒，$Q_{supplied}=Q_{lost}$$Q_{supplied} = Q_{lost}$

$(i_{D(average)}-I_L)\frac{T}{2\pi }\sqrt{\frac{2V_r}{V_p}}=I_{L}T$$(i_{D(average)} - I_L)\frac{T}{2\pi}\sqrt{\frac{2V_r}{V_p}} = I_LT$

$i_{D(average)}=I_L(1+\pi \sqrt{\frac{2V_p}{V_r}})$$i_{D(average)} = I_L(1 + \pi \sqrt{\frac{2V_p}{V_r}})$

根据电路，$i_D=C\frac{d{v}_I}{dt}+i_L$$i_D = C\frac{dv_I}{dt} + i_L$

而充电最快时应为一开始充电时(假定t=0时在电压最高值，$t=-\mathrm{\Delta }t$$t = -\Delta t$)

$v_I=V_{p}cos(\omega t)$$v_I = V_pcos(\omega t)$

$\frac{d{v}_I}{dt}=-V_p\omega sin(\omega t)$$\frac{dv_I}{dt} = -V_p \omega sin(\omega t)$

由上面$Q_{lost}$$Q_{lost}$公式得到$C=\frac{I_{L}T}{V_r}$$C = \frac{I_LT}{V_r}$ 那么 at $t=-\mathrm{\Delta }t$$t = -\Delta t$

$i_{D(max)}=C\frac{d{v}_I}{dt}+i_L=\frac{I_{L}T}{V_r}{V}_p\omega sin(\omega \mathrm{\Delta }t)+I_L$$i_{D(max)} = C\frac{dv_I}{dt} + i_L = \frac{I_LT}{V_r} V_p \omega sin(\omega \Delta t) + I_L$

使用泰勒展开近似

$i_{D(max)}=I_L(1+\frac{V_p}{V_r}\frac{2\pi }{\omega }\omega (\omega \mathrm{\Delta }t))=I_L(1+2\pi \frac{V_p}{V_r}\sqrt{\frac{2V_r}{V_p}})=I_L(1+2\pi \sqrt{\frac{2V_p}{V_r}})$$i_{D(max)} = I_L(1 + \frac{V_p}{V_r} \frac{2\pi}{\omega} \omega (\omega \Delta t)) = I_L(1 + 2\pi \frac{V_p}{V_r}\sqrt{\frac{2V_r}{V_p}}) = I_L(1 + 2\pi \sqrt{\frac{2V_p}{V_r}})$

可以看到当 $V_r<<V_p$$V_r << V_p$时，根号中的数字很大，$2i_{D(average)}\approx i_{D(max)}$$2i_{D(average)} \approx i_{D(max)}$ max current在选择二极管型号时是重要参数

Full-wave peak rectifier

与half-wave推导过程相同，只不过原来的一周期，现在能充电两次，因此现在的周期是T/2

$V_r=\frac{V_p}{2fRC}$$V_r = \frac{V_p}{2fRC}$

$\mathrm{\Delta }t$$\Delta t$相同，但现在$Q_{lost}$$Q_{lost}$为原来的1/2 (每周期冲两次电)

$Q_{lost}^{\prime }=C{V}_r=I_{L}T/2$$Q'_{lost} = CV_r = I_LT/2$

因此两个公式变为

$i_{D(average)}=I_L(1+\pi \sqrt{\frac{V_p}{2V_r}})$$i_{D(average)} = I_L(1 + \pi \sqrt{\frac{V_p}{2V_r}})$

$i_{D(max)}==I_L(1+2\pi \sqrt{\frac{V_p}{2V_r}})$$i_{D(max)} = = I_L(1 + 2\pi \sqrt{\frac{V_p}{2V_r}})$

跟之前的根号项相比为1/2

Peak detector

通过相似的电路，可以得到由正弦波携带的信号

这里面正弦波叫做carrier，可以通过不同的频率来区分不同的信道

Percision Half-Wave Rectifier - Superdiode

使用一个运放来补上二极管的压降，达到理想二极管的伏安曲线

Other Diode Applications

Clamped Capacitor and Bootstrapping (dc restorer)

使用peak rectifier，但取二极管的电压

这样输出电压会是二极管的PIV，而正向导通时会将输出的最低电压拉到0V，因此这个电路能将交流电最低点呀拉高/拉低到0V，使其直流的共模信号为$V_{dc}=V_{pp}/2$$V_{dc} = V_{pp}/2$，因此也叫dc restorer

这个图的$V_I$$V_I$产生的就是方波，不是这个电路转换成的方波

这个电路在PWM Pulsewidth Modulation时会被用来将电压拉到0V以上，然后在后面接上一个RC电路来得到平均电压。

Bootstrapping Circuit

在输出接上一个电源，这样可以把低压拉到任意位置

Voltage Doubler

前半部分是clamped capacitor，后面接上一个peak rectifier，可以使输出的直流电压为$v_o=2v_p$$v_o = 2v_p$

Varactors

在反向区域由于pn结的特性，导致二极管有一定的电容$C_j$$C_j$，而这个容值跟反向电压$V_R$$V_R$相关，因此可以做成voltage-variable capacitors

Photodiodes

在reverse bias状态下工作。当受到光的刺激时，电子获得能量，P区和N区会产生新的电子-空穴对，使得区域少子浓度增加。而少子可以通过耗尽层，导致产生少量的漏电流。

The responsivity R is the reverse current produced per watt of incident light power $i_P=R\cdot P$$i_P = R\cdot P$

在黑暗条件下，会有一个最小漏电流dark current $I_D$$I_D$，同时二极管也有自身的junction capacitance $C_j$$C_j$，与$V_D$$V_D$相关。

不同材质的半导体产生电子-空穴对的能量需求不同，因此会吸收不同波长的光（不同颜色的感光）。通常情况下，direct band gap semiconductors对光更敏感，相比于indirect band gap semiconductors可以用更细的耗尽层来感光（感光时间也更短）。

利用雪崩击穿的photodiodes对光线十分敏感。Biased around their reverse breakdown voltage，因此一个光子足以引发雪崩击穿，产生大量电流。

如果不在reverse bias下工作，那么photodiodes就是太阳能板，受到光照会产生电压

LED

电子从导带落回到价带时会产生能级跃迁，使用不同的本征半导体材料代替硅可以让改变这个能量的大小，也可以让这个过程释放光子(direct band gap semiconductors)而不是热能(indirect band gap semiconductors，电子在落回空穴时会有中间态)

由于材料变化，二极管的压降也随材质改变，光的波长越短，需要能量越多

红色 GaP LED压降1.8V，蓝色 GaIn LED压降2.5-4.0V

归根结底LED是二极管，所以使用时一定要加一个限流电阻防止电流过大（一般commerical LED会有内置电阻）

MOSFETS

结构如图所示

Operation with Zero Gate voltage

正常情况下，Drain 与 Source之间相当于串联了两个pn结二极管，方向相反，因此无法导通(${10}^{12}\mathrm{\Omega }$$10^{12} \ohm$的数量级)

Creating a Channel for current flow

当在G端加上电压后，电子会聚集在氧化物下面，形成一条n沟道。由于n沟道本身导电，此时Drain 与 Source导通

通常使用二极管时会将B和S连接到一起，或者一起接地，因此通常用$V_{GS}$$V_{GS}$ 来替换Gate和Body之间的电压

由于制作芯片时有时无法做到这点，也有专门的公式来用来校准由于B和S电压差造成的误差（Body Effect)

Gate部分构成平行板电容器。刚刚能产生N沟道的电压叫做threshold voltage $V_t$$V_t$，低于这个电压则无法在body中构建n沟道。而Gate上多出来的电压叫做effective voltage / overdrive voltage，满足$V_{GS}-V_t\equiv v_{OV}$$V_{GS} - V_t \equiv v_{OV}$

电容器两端的电荷 $Q=CV=C_{ox}(WL)v_{ov}$$Q = CV = C_{ox} (WL) v_{ov}$

$C_{ox}$$C_{ox}$ - oxide capacitance per unit area $F/m^2$$F/m^2$，$C_{ox}=\frac{{ϵ}_{ox}}{t_{ox}}$$C_{ox} = \frac{\epsilon_{ox}}{t_{ox}}$，t是氧化物的thickness

$Si{O}_2$$SiO_2$ 的permittivity${ϵ}_{ox}=3.9{ϵ}_0=3.9\ast 8.854e-12=3.45e-11F/m$$\epsilon_{ox} = 3.9\epsilon_0 = 3.9*8.854e-12 = 3.45e-11 F/m$

W, L - 沟道的长度和宽度

当$V_{ov}$$V_{ov}$增加时，平行板另一端电荷聚集的更多，一般画图时channel 的宽度表示电子的量，而不是电子实际的位置

Small $v_{DS}$$v_{DS}$

我们想要计算此时从Drain 流向 Source 的电流$i_D$$i_D$

i = electron density on channel surface * drift velocity

electron density: $\frac{|Q|}{L}=C_{ox}W{v}_{ov}$$\frac{|Q|}{L} = C_{ox} W v_{ov}$

drift velocity: ${\mu }_n|E|={\mu }_n\frac{v_{DS}}{L}$$\mu_n |E| = \mu_n \frac{v_{DS}}{L}$

${\mu }_n$$\mu_n$ 为mobility of electrons at the surface of the channel. 根据制作工艺的不同会不一样。

$i_D=[({\mu }_n{C}_{ox})(\frac{W}{L})v_{OV}]v_{DS}$$i_D = [(\mu_nC_{ox})(\frac{W}{L})v_{OV}]v_{DS}$

conductance: $g_{DS}=\frac{v_{DS}}{i_D}=[({\mu }_n{C}_{ox})(\frac{W}{L})v_{OV}]$$g_{DS} = \frac{v_{DS}}{i_D} = [(\mu_nC_{ox})(\frac{W}{L})v_{OV}]$

我们定义process transconductance parameter $k^{\prime }n={\mu }_n{C}_{ox}$$k'n = \mu_n C_{ox}$，因为这两个参数都是process technology 决定的

我们在定义MOSFET transconductance parameter $k_n=k_n^{\prime }(W/L)$$k_n = k'_n(W/L)$ 为process transconductance 乘上 aspect ratio

因此 $g_{DS}=1/r_{DS}=k_n{v}_{OV}$$g_{DS} = 1/r_{DS} = k_nv_{OV}$

以上推导全部仅限于$v_{DS}$$v_{DS}$足够小的情况，形成的n沟道上的电子可以看作均匀排布，下面推导不够小时发生的情况

When $v_{DS}$$v_{DS}$ increase

当$v_{DS}$$v_{DS}$稍微大一些时，Drain一侧由于加大的电压，使Drain与Body之间构成的PN结耗尽层增大，更多自由电子落入空穴中，因此呈现在图像上就是Drain一侧变窄

变窄后channel中的电子数量也相应变少，可以从上图看出，channel的面积代表电子的多少。

我们可以用一个更窄的沟道来平替这个梯形沟道，等价的长方形沟道为$V_{OV}^{\prime }=V_{OV}-\frac{1}{2}{V}_{DS}$$V'_{OV} = V_{OV} - \frac{1}{2}V_{DS}$

我们将新的$V_{OV}^{\prime }$$V'_{OV}$代入原先的公式

$i_D=k_n{v}_{OV}^{\prime }{v}_{DS}=i_D{k}_n(v_{OV}-\frac{1}{2}{v}_{DS})v_{DS}$$i_D = k_n v'_{OV} v_{DS} = i_D k_n (v_{OV}- \frac{1}{2}v_{DS})v_{DS}$

$i_D=k_n[(v_{GS}-V_t)v_{DS}-\frac{1}{2}{v}_{DS}^2]$$i_D = k_n [(v_{GS} - V_t)v_{DS} - \frac{1}{2} v^2_{DS}]$

可以看到伏安特性实际时一个二次函数

Channel Pinch off and Current Saturation - 夹断

还是上面的图，可以看到当$v_{DS}$$v_{DS}$高到一定程度后，channel在Drain端被夹断。Drain所形成的耗尽层导致Drain附近彻底没有电子。然而此时的电压足够大，让PN结工作在反向击穿的状态下。多余的$V_{DS}$$V_{DS}$电压被施加在了PN结两端

我们称此状态为saturation region(saturation mode of operation)，相对应的，未夹断时叫做triode region，从原先晶体管时期得名

The voltage at which saturation occurs $V_{DS(saturation)}=V_{OV}=V_{GS}-V_t$$V_{DS(saturation)} = V_{OV} = V_{GS} - V_t$

diode connected mosfet - 指在电路中直接将Drain和Gate短接的MOSFET总是在Saturation region，因为$V_{DS}=V_{OV}+V_t>V_{OV}$$V_{DS} = V_{OV} + V_t > V_{OV}$

saturation的条件也可以改成$V_{GD}\le V_{OV}$$V_{GD} \leq V_{OV}$

代入之前的公式可以得到此时$i_D=\frac{1}{2}{k}_n{v}_{OV}^2$$i_D = \frac{1}{2} k_n v^2_{OV}$

Channel Length Modulation

通常我们都忽略这个效应，assume$V_{DS}>V_{OV}$$V_{DS} > V_{OV}$时，channel的形状不再改变，因此才有上面的推论。

但实际上，继续加压会导致channel末端远离Drain

多余的电压被$\mathrm{\Delta }L$$\Delta L$部分消耗

之前的saturation时电流公式可以看到，$i_D\propto \frac{1}{L}$$i_D \propto \frac{1}{L}$

因此$v_{DS}$$v_{DS}$的增加会使channel变短，进而增加电流，而$\mathrm{\Delta }L$$\Delta L$与$v_{DS}$$v_{DS}$相关

$\mathrm{\Delta }L$$\Delta L$与$v_{DS}$$v_{DS}$真的是线性相关吗?

我们用$\lambda$$\lambda$作为系数将channel length modulation引入我们的公式

$i_D=\frac{1}{2}{k}_n^{\prime }(\frac{W}{L-\mathrm{\Delta }L})v_{OV}^2=\frac{1}{2}{k}_n^{\prime }(\frac{W}{L})v_{OV}^2(1+\frac{\mathrm{\Delta }L}{L})$$i_D = \frac{1}{2} k'_n(\frac{W}{L-\Delta L}) v^2_{OV} = \frac{1}{2} k'_n(\frac{W}{L}) v^2_{OV}(1 + \frac{\Delta L}{L})$

令$\frac{\mathrm{\Delta }L}{L}=\lambda V_{DS}$$\frac{\Delta L}{L} = \lambda V_{DS}$

$i_D=\frac{1}{2}{k}_n{v}_{OV}^2(1+\lambda v_{DS})$$i_D = \frac{1}{2}k_n v^2_{OV} (1 + \lambda v_{DS})$

因此实际的i-v在saturation region会有一个等效电阻$r_0$$r_0$

$1/r_0=\lambda \frac{1}{2}{k}_n^{\prime }(\frac{W}{L})v_{OV}^2$$1/r_0 = \lambda \frac{1}{2} k'_n(\frac{W}{L}) v^2_{OV}$

$r_0=\frac{1}{\lambda i_D^{\prime }}$$r_0 = \frac{1}{\lambda i'_D}$，这里$i_D^{\prime }$$i'_D$指刚到saturation region的电流

我们一如一个电压$V_A$$V_A$，所有$V_{OV}$$V_{OV}$情况下的saturation region的直线交x轴与点$-V_A=-1/\lambda$$-V_A = -1/\lambda$

因此$V_A=1/\lambda ⟹r_0=\frac{V_A}{i_D^{\prime }}$$V_A = 1/\lambda \implies r_0 = \frac{V_A}{i'_D}$