模拟集成电路(3)----单级放大器（共源极）
放大是模拟电路的基本功能
- 大多数自然模拟信号太小而无法处理
- 需要足够的信噪比
理想的放大器
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线性：无限的幅度和频率范围
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输入阻抗无限大
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输出阻抗无限小
共源放大器
共源放大器就是将源极接AC ground。
一般我们对三点进行分析：
- 直流摆幅有多大（饱和区）
- 小信号的增益
- 输入输出的阻抗
电阻负载
大信号分析
- V i n < V T H , c u t o f f V_{in}<V_{TH},\mathrm{~cut~off} Vin<VTH, cut off
V o u t = V D D V_{out}=V_{DD} Vout=VDD
- V i n − V T H ≤ V o u t , saturation V_{in}-V_{TH}\leq V_{out,}\text{saturation} Vin−VTH≤Vout,saturation(一般只考虑饱和区)
V o u t = V D D − I d ⋅ R D = V D D − μ n C o x 2 W L ( V i n − V T H ) 2 ⋅ R D \begin{aligned}&V_{out}=V_{DD}-I_{d}\cdot R_{D}\\&=V_{DD}-\frac{\mu_{n}C_{ox}}{2}\frac{W}{L}(V_{in}-V_{TH})^{2}\cdot R_{D}\end{aligned} Vout=VDD−Id⋅RD=VDD−2μnCoxLW(Vin−VTH)2⋅RD
- V i n − V T H > V o u t , triode V_{in}-V_{TH}>V_{out,}\text{triode} Vin−VTH>Vout,triode
小信号增益
大信号的斜率就是小信号的增益
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V_{out}=V_{DD}-\frac{\mu_{n}C_{ox}}{2}\frac{W}{L}\left(V_{in}-V_{TH}\right)^{2}\cdot R_{D}
Vout=VDD−2μnCoxLW(Vin−VTH)2⋅RD
A v = ∂ V o u t ∂ V i n = − μ n C o x W L ( V i n − V T H ) ⋅ R D A_{v}=\frac{\partial V_{out}}{\partial V_{in}}=-\boxed{\mu_{n}C_{ox}\frac{W}{L}(V_{in}-V_{TH})}\cdot R_{D} Av=∂Vin∂Vout=−μnCoxLW(Vin−VTH)⋅RD
框出的部分即为跨导
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A_v=\frac{\partial V_{out}}{\partial V_{in}}=-\boxed{g_m}\cdot R_D
Av=∂Vin∂Vout=−gm⋅RD
发现，不同的
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Vin的值会影响增益的值，
- 小信号等效电路
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\begin{cases} \begin{aligned}&\nu_{out}=-i_{d}R_{D}\\&i_{d}=g_{m}\nu_{in}\end{aligned} \end{cases}
{νout=−idRDid=gmνin
A ν = ν o u t ν i n = − g m R D A_{\nu}=\frac{\nu_{out}}{\nu_{in}}=-g_{m}R_{D} Aν=νinνout=−gmRD
- 考虑沟道长度调制效应
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\begin{cases} \begin{aligned}&\nu_{out}=-i_{d}(R_{D}\parallel r_{o})\\&i_{d}=g_{m}\nu_{in}\end{aligned} \end{cases}
{νout=−id(RD∥ro)id=gmνin
根据常识有
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R_D \ll r_o
RD≪ro
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\begin{aligned}A_{\nu}&=-g_{m}\cdot(R_{D}\parallel r_{o})\\&\approx-g_{m}\cdot R_{D}\end{aligned}
Aν=−gm⋅(RD∥ro)≈−gm⋅RD
输入输出阻抗
r i n = ν i n i i n = ∞ r_{in}=\frac{\nu_{in}}{i_{in}}=\infty rin=iinνin=∞
r o u t = r o ∥ R D ≈ R D r_{out}=r_{o}\parallel R_{D}\approx R_{D} rout=ro∥RD≈RD
V D S = V i n 1 − V T H V_{\mathrm{DS}}=V_{\mathrm{in}1}-V_{\mathrm{TH}} VDS=Vin1−VTH
V i n 1 − V T H = V D D − μ n C o x 2 W L ( V i n 1 − V T H ) 2 ⋅ R D V_{\mathrm{in1}}-V_{\mathrm{TH}}=V_{\mathrm{DD}}-\frac{\mu_{\mathrm{n}}C_{\mathrm{ox}}}{2}\frac{W}{L}(V_{\mathrm{in1}}-V_{\mathrm{TH}})^{2}\cdot R_{\mathrm{D}} Vin1−VTH=VDD−2μnCoxLW(Vin1−VTH)2⋅RD
可以得到 V i n 1 V_{in1} Vin1是 R D R_D RD的一个函数。
R D R_D RD越大会导致 V i n 1 V_{in1} Vin1越小
二极管接法负载
在M1和M2的电流是一样的，于是我们可以列出如下等式：
1 2 μ n C o x ( W L ) 1 ( V i n − V T H 1 ) 2 = 1 2 μ n C o x ( W L ) 2 ( V D D − V o u t − V T H 2 ) 2 \begin{aligned}&\frac{1}{2}\mu_{n}C_{ox}\left(\frac{W}{L}\right)_{1}\left(V_{in}-V_{TH1}\right)^{2}\\&=\frac{1}{2}\mu_{n}C_{ox}\left(\frac{W}{L}\right)_{2}\left(V_{DD}-V_{out}-V_{TH2}\right)^{2}\end{aligned} 21μnCox(LW)1(Vin−VTH1)2=21μnCox(LW)2(VDD−Vout−VTH2)2
( W L ) 1 ( V i n − V T H 1 ) = ( W L ) 2 ( V D D − V o u t − V T H 2 ) \sqrt{\left(\frac{W}{L}\right)_{1}}(V_{in}-V_{TH1})=\sqrt{\left(\frac{W}{L}\right)_{2}}(V_{DD}-V_{out}-V_{TH2}) (LW)1(Vin−VTH1)=(LW)2(VDD−Vout−VTH2)
可得 V i n V_{in} Vin和 V o u t V_{out} Vout几乎是一个线性关系，如果两个晶体管的 V T H V_{TH} VTH不变，那么可以认作是线性关系。
由于有电容的存在，所以 V o u t V_{out} Vout并不是直接变大。
大信号分析
小信号增益
( W L ) 1 ( V i n − V T H 1 ) = ( W L ) 2 ( V D D − V o u t − V T H 2 ) \sqrt{\left(\frac{W}{L}\right)_{1}}(V_{in}-V_{TH1})=\sqrt{\left(\frac{W}{L}\right)_{2}}(V_{DD}-V_{out}-V_{TH2}) (LW)1(Vin−VTH1)=(LW)2(VDD−Vout−VTH2)
( W L ) 1 = ( W L ) 2 ( − ∂ V o u t ∂ V i n − ∂ V T H 2 ∂ V i n ) \sqrt{\left(\frac{W}{L}\right)_1}=\sqrt{\left(\frac{W}{L}\right)_2}(-\frac{\partial V_{out}}{\partial V_{in}}-\boxed{ \frac{\partial V_{TH2}}{\partial V_{in}}}) (LW)1=(LW)2(−∂Vin∂Vout−∂Vin∂VTH2)
框住的为
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M2的体效应
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\frac{\partial V_{TH2}}{\partial V_{in}}=\frac{\partial V_{TH2}}{\partial V_{out}}\cdot\frac{\partial V_{out}}{\partial V_{in}}=\eta\cdot\frac{\partial V_{out}}{\partial V_{in}}
∂Vin∂VTH2=∂Vout∂VTH2⋅∂Vin∂Vout=η⋅∂Vin∂Vout
得到增益：
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A_{\nu}=\frac{\partial V_{out}}{\partial V_{in}}=-\sqrt{\frac{\left(W/L\right)_{1}}{\left(W/L\right)_{2}}}\cdot\frac{1}{1+\eta}
Aν=∂Vin∂Vout=−(W/L)2(W/L)1⋅1+η1
小信号模型
小信号模型增益
i x = v x / r o + g m v 1 v 1 = v x } → r e q = v x i x = r o ∥ 1 g m ≈ 1 g m \begin{aligned}&i_{x}=v_{x}/r_{o}+g_{m}v_{1}\\&v_{1}=v_{x}\end{aligned}\biggr\}\to r_{eq}=\frac{v_{x}}{i_{x}}=r_{o}\parallel\frac{1}{g_{m}}\approx\frac{1}{g_{m}} ix=vx/ro+gmv1v1=vx}→req=ixvx=ro∥gm1≈gm1
- 考虑体效应
i x = v x r o + ( g m 2 + g m b 2 ) v x i_x=\frac{v_x}{r_o}+(g_{m2}+g_{mb2})v_x ix=rovx+(gm2+gmb2)vx
r e q = v x i x = r o ∥ 1 g m 2 + g m b 2 ≈ 1 g m 2 + g m b 2 = 1 ( 1 + η ) g m 2 r_{eq}=\frac{v_{x}}{i_{x}}=r_{o}\parallel\frac{1}{g_{m2}+g_{mb2}}\approx\frac{1}{g_{m2}+g_{mb2}}=\frac{1}{(1+\eta)g_{m2}} req=ixvx=ro∥gm2+gmb21≈gm2+gmb21=(1+η)gm21
- 用小信号的方法计算增益
A v = − g m 1 ⋅ ( r e q ∥ r o 1 ) ≈ − g m 1 ⋅ r e q A_v=-g_{m1}\cdot(r_{eq}\parallel r_{o1})\approx-g_{m1}\cdot r_{eq} Av=−gm1⋅(req∥ro1)≈−gm1⋅req
A v = − g m 1 g m 2 ⋅ 1 1 + η A_{v}=-\frac{g_{m1}}{g_{m2}}\cdot\frac{1}{1+\eta} Av=−gm2gm1⋅1+η1
对于PMOS
A v = − g m 1 g m 2 A v = − μ n ( W / L ) 1 μ p ( W / L ) 2 A_{v}=-\frac{g_{m1}}{g_{m2}}\\A_{v}=-\sqrt{\frac{\mu_{n}(W/L)_{1}}{\mu_{p}(W/L)_{2}}} Av=−gm2gm1Av=−μp(W/L)2μn(W/L)1
输入输出电阻
r i n = ∞ r_{in}=\infty rin=∞
r o u t = r o 1 ∥ r o 2 ∥ 1 g m 2 ( 1 + η ) ≈ 1 g m 2 ( 1 + η ) \begin{aligned}r_{out}&=r_{o1}\parallel r_{o2}\parallel\frac{1}{g_{m2}(1+\eta)}\\&\approx\frac{1}{g_{m2}(1+\eta)}\end{aligned} rout=ro1∥ro2∥gm2(1+η)1≈gm2(1+η)1
r i n = ∞ r_{in}=\infty rin=∞
r o u t = r o 1 ∥ r o 2 ∥ 1 g m 2 ≈ 1 g m 2 \begin{aligned}r_{out}&=r_{o1}\parallel r_{o2}\parallel\frac{1}{g_{m2}}\\&\approx\frac{1}{g_{m2}}\end{aligned} rout=ro1∥ro2∥gm21≈gm21
电流源负载
一般我们的电流源会用mos管实现，例如pmos
如下是pmos作电流源负载：
M1小信号模型如下：
所以总的小信号模型就是在
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A_{v}=-g_{m}\cdot(r_{o1}\parallel r_{o2})\\r_{in}=\infty\quad r_{out}=r_{o1}\parallel r_{o2}
Av=−gm⋅(ro1∥ro2)rin=∞rout=ro1∥ro2
电流源负载和电阻负载进行对比：
所以电流源负载可实现小电流实现大增益。
通用的CS分析方法
v i n → i d = g m v i n v → i → i d → v o u t = − i d r o u t i → v ν o u t v_{in}\xrightarrow{i_{d}=g_{m}v_{in}}_{v\to i}\to i_{d}\xrightarrow{v_{out}=-i_{d}r_{out}}_{i\to v}\nu_{out} vinid=gmvinv→i→idvout=−idrouti→vνout
v o u t = − i d r o u t = − g m v i n r o u t A v = v o u t / v i n = − g m r o u t v_{out}=-i_{d}r_{out}=-g_{m}v_{in}r_{out}\\A_{v}=v_{out}/v_{in}=-g_{m}r_{out} vout=−idrout=−gmvinroutAv=vout/vin=−gmrout
r o u t = r O ∥ r L o a d r_{out}=r_{O}\parallel r_{Load} rout=rO∥rLoad
有源负载的共源极
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\frac{v_{\mathrm{out}}}{v_{\mathrm{in}}}=-(g_{\mathrm{ml}}+g_{\mathrm{m2}})(r_{\mathrm{ol}}\parallel r_{\mathrm{o2}})
vinvout=−(gml+gm2)(rol∥ro2)
带源极负反馈的共源级
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Assuming \lambda=\gamma=0
Assumingλ=γ=0
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\begin{aligned} I_{d}& =\frac{1}{2}\mu_{n}C_{ox}\frac{W}{L}\big(V_{gs}-V_{TH}\big)^{2} \\ &=\frac12\mu_nC_{ox}\frac WL(V_{in}-R_SI_d-V_{TH})^2 \end{aligned}
Id=21μnCoxLW(Vgs−VTH)2=21μnCoxLW(Vin−RSId−VTH)2
等效跨导如下：
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G_m=\frac{\partial I_d}{\partial V_{in}}
Gm=∂Vin∂Id
G m = ∂ I d ∂ V i n = μ n C o x W L ( V i n − R S I d − V T H ) ( 1 − R S G m ) G_{m}=\frac{\partial I_{d}}{\partial V_{in}}=\boxed{\mu_{n}C_{ox}\frac{W}{L}(V_{in}-R_{S}I_{d}-V_{TH})}(1-R_{S}G_{m}) Gm=∂Vin∂Id=μnCoxLW(Vin−RSId−VTH)(1−RSGm)
框住的部分是
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G_{m}=g_{m}(1-R_{S}G_{m})\longrightarrow \quad G_{m}=\frac{g_{m}}{1+g_{m}R_{S}}
Gm=gm(1−RSGm)⟶Gm=1+gmRSgm
A ν = − G m R D = − g m R D 1 + g m R S A_{\nu}=-G_{m}R_{D}=-\frac{g_{m}R_{D}}{1+g_{m}R_{S}} Aν=−GmRD=−1+gmRSgmRD
I f R s is large enough → G m ≈ 1 / R s , A v = R D / R s \mathrm{If~}R_s\text{ is large enough }\to G_m{\approx}1/R_s,A_v{=}R_D/R_s If Rs is large enough →Gm≈1/Rs,Av=RD/Rs
小信号分析
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v_{1}=v_{in}-v_{x}\quad\nu_{x}=-v_{bs}=R_{S}i_{out}
v1=vin−vxνx=−vbs=RSiout
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