模拟集成电路(6)----单级放大器（共源共栅级 Cascode Stage）
大信号分析
对M1
V x ≥ V i n − V T H 1 V x = V B − V G S 2 V B ≥ V i n − V T H 1 + V G S 2 V_{x}\geq V_{in}-V_{TH1}\quad V_{x}=V_{B}-V_{GS2}\\V_{B}\geq V_{in}-V_{TH1}+V_{GS2} Vx≥Vin−VTH1Vx=VB−VGS2VB≥Vin−VTH1+VGS2
对M2
V o u t ≥ V B − V T H 2 V o u t ≥ V i n − V T H 1 + V G S 2 − V T H 2 V o u t ≥ V O D 1 + V O D 2 \begin{aligned}&V_{out}\geq V_{B}-V_{TH2}\\&V_{out}\geq V_{in}-V_{TH1}+V_{GS2}-V_{TH2}\\&V_{out}\geq V_{OD1}+V_{OD2}\end{aligned} Vout≥VB−VTH2Vout≥Vin−VTH1+VGS2−VTH2Vout≥VOD1+VOD2
输入输出电阻
W h e n V i n < V T H 1 , M 1 a n d M 2 a r e o f f , V o u t = V D D , V x = V B − V T H 2 \begin{aligned}&\mathrm{When~}V_{in}<V_{TH1},\mathrm{M}_{1}\mathrm{~and~}\mathrm{M}_{2}\mathrm{~are~off},V_{out}=V_{DD},V_{x}=V_{B}-V_{TH2}\end{aligned} When Vin<VTH1,M1 and M2 are off,Vout=VDD,Vx=VB−VTH2
W h e n V i n > V T H 1 , V o u t d r o p s , − V G S 2 increases, resulting in a drop of V X \begin{aligned}&\mathrm{When~}V_{\mathrm{in}}>V_{\mathrm{TH}1},V_{\mathrm{out}}\mathrm{drops},\\&-V_{GS2}\text{increases, resulting in a drop of }V_{X}\end{aligned} When Vin>VTH1,Voutdrops,−VGS2increases, resulting in a drop of VX
R o u t = [ 1 + ( g m 2 + g m b 2 ) r O 2 ] r O 1 + r O 2 = r O 1 + r O 2 + ( g m 2 + g m b 2 ) r O 2 r O 1 R_{out}=[1+(g_{m2}+g_{mb2})r_{O2}]r_{O1}+r_{O2}\\=r_{O1}+r_{O2}+(g_{m2}+g_{mb2})r_{O2}r_{O1} Rout=[1+(gm2+gmb2)rO2]rO1+rO2=rO1+rO2+(gm2+gmb2)rO2rO1
i f g m r O > > 1 , t h e n R o u t ≈ ( g m 2 + g m b 2 ) r O 2 r O 1 ≈ g m 2 r O 2 r O 1 if \ g_{m}r_{O}>>1,\quad\mathrm{then}\quad R_{out}\approx(g_{m2}+g_{mb2})r_{O2}r_{O1}\approx g_{m2}r_{O2}r_{O1} if gmrO>>1,thenRout≈(gm2+gmb2)rO2rO1≈gm2rO2rO1
R o u t ≈ ( g m 3 + g m b 3 ) r O 3 R S ≈ g m 3 r O 3 R S ‾ R S ≈ g m 2 r O 2 r O 1 → R o u t ≈ g m 3 r O 3 ⋅ g m 2 r O 2 ⋅ r o 1 \begin{aligned}&R_{out}\approx(g_{m3}+g_{mb3})r_{O3}R_{S}\approx\underline{g_{m3}r_{O3}R_{S}}\\&R_{S}\approx g_{m2}r_{O2}r_{O1}\\&\to R_{out}\approx g_{m3}r_{O3}\cdot g_{m2}r_{O2}\cdot r_{o1}\end{aligned} Rout≈(gm3+gmb3)rO3RS≈gm3rO3RSRS≈gm2rO2rO1→Rout≈gm3rO3⋅gm2rO2⋅ro1
小信号分析
增益
A ν = − g m 1 R D A_{\nu}=-g_{m1}R_{D} Aν=−gm1RD
如若是电流源负载
A v = − G m r o u t A_v=-G_mr_{out} Av=−Gmrout
G m ≈ g m 1 G_m{\approx}g_{m1} Gm≈gm1
A ν ≈ − g m 1 ⋅ g m 2 r o 2 r o 1 = − g m 1 r o 1 ⋅ g m 2 r o 2 A_{\nu}\approx-g_{m1}\cdot g_{m2}r_{o2}r_{o1}=-g_{m1}r_{o1}\cdot g_{m2}r_{o2} Aν≈−gm1⋅gm2ro2ro1=−gm1ro1⋅gm2ro2
可以结合课本3.21去理解
g m = 2 I D μ n C o x W L r o = 1 λ I D A i n t = g m r o g_{m}=\sqrt{2I_{\mathrm{D}}\mu_{\mathrm{n}}C_{\mathrm{ox}}\frac{W}{L}}\\r_{\mathrm{o}}=\frac{1}{\lambda I_{\mathrm{D}}}\\A_{\mathrm{int}}=g_{m}r_{\mathrm{o}} gm=2IDμnCoxLWro=λID1Aint=gmro
g m 1 = g m 2 r o l = 4 r o A i n t l = 2 A i n t g_{m1}=\frac{g_{\mathrm{m}}}{2}\\r_{\mathrm{ol}}=4r_{\mathrm{o}}\\A_{\mathrm{intl}}=2A_{\mathrm{int}} gm1=2gmrol=4roAintl=2Aint
g m 2 = g m r o 2 = A i n t r o A int 2 = A int 2 g_{m2}=g_{m}\\r_{\mathrm{o2}}=A_{\mathrm{int}}r_{\mathrm{o}}\\A_{\operatorname{int}2}=A_{\operatorname{int}}^2 gm2=gmro2=AintroAint2=Aint2
用PMOS cascode电流源作负载
r o u t n ≈ g m 2 r o 2 ⋅ r o 1 r o u t p ≈ g m 3 r o 3 ⋅ r o 4 \begin{aligned}r_{outn}&\approx g_{m2}r_{o2}\cdot r_{o1}\\r_{outp}&\approx g_{m3}r_{o3}\cdot r_{o4}\end{aligned} routnroutp≈gm2ro2⋅ro1≈gm3ro3⋅ro4
A ν ≈ − g m 1 ⋅ ( r o u t n ∥ r o u t p ) ≈ − g m 1 ⋅ ( g m 3 r o 3 ⋅ r o 4 ∥ g m 2 r o 2 ⋅ r o 1 ) \begin{aligned}A_{\nu}&\approx-g_{m1}\cdot(r_{outn}\parallel r_{outp})\\&\approx-g_{m1}\cdot(g_{m3}r_{o3}\cdot r_{o4}\parallel g_{m2}r_{o2}\cdot r_{o1})\end{aligned} Aν≈−gm1⋅(routn∥routp)≈−gm1⋅(gm3ro3⋅ro4∥gm2ro2⋅ro1)
V o u t max ≤ V D D − V O D 3 − V O D 4 V_{out\max}\leq V_{DD}-V_{OD3}-V_{OD4} Voutmax≤VDD−VOD3−VOD4
V o u t , m i n ≥ V O D 1 + V O D 2 V_{out,min}\geq V_{OD1}+V_{OD2} Vout,min≥VOD1+VOD2
