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CMOS AMPLIFIERS. Simple Inverting Amplifier Differential Amplifiers Cascode Amplifier Output Amplifiers Summary. Simple Inverting Amplifiers. Small Signal Characteristics. Inverter with diode connection load. How do you get better matching?. High gain inverters.

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## CMOS AMPLIFIERS

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**CMOS AMPLIFIERS**• Simple Inverting Amplifier • Differential Amplifiers • Cascode Amplifier • Output Amplifiers • Summary**Small Signal Characteristics**Inverter with diode connection load How do you get better matching?**Current source load or push-pull**• Refer to book for large signal analysis • Must match quiescent currents in PMOS and NMOS transistors • Wider output swing, especially push-pull • Much high gain (at DC), but much lower -3dB frequency (vs diode load) • About the same GB • Very power dependent**Small signal**High gain!**Key to analysis by hand:**• Use level 1 or 3 model equations • Use KCL/KVL**Transfer function of a system**input u output y System**When u(s) = 0, y(s) satisfies:**These dynamics are the characteristic dynamics of the system. The roots of the coefficient polynomial are the poles of the system. When y(s) = 0, u(s) satisfies: These dynamics are the zero dynamics of the system. The roots of the coefficient polynomial are the zeros of the system.**Poles of CMOS Inverters**Let vin = 0, x = 0, VDD = 0, VSS = 0. CGS1, CGS2, CBS1, CBS2 are all short y CGD1, CGD2, CBD1, CBD2, CL in parallel C’L = Ctotal = CGD1+ CGD2+ CBD1+ CBD2+ CL**Total conductance from y to ground:**go = gds1 + gds2 KCL at node y: Therefore system pole is:**Zeros of CMOS Inverters**Let vin = x = u, VDD = 0, VSS = 0. CGD1, CGD2, are in parallel, CBD1, CBD2, CL are all short gds1, gds2 also short No current in them KCL: Zero is:**Input output transfer function**When s=jw0, A(0) When w∞, A(s)**-3dB frequency of closed loop**=b*GB A0=gm/go Acl=1/b 0 dB |p1|= g0/CL’ Unity gain frequency =|A0p1| =GB =gm/CL’ |z1| =gm/Cgd =GB*CL’/Cgd**Unity gain feedback**A(s)**If a step input is given, the output response is**In the time domain: Final settling determined by A0 need high gain Settling speed determined by A0p1=GB, need high gain bandwidth product**Gain bandwidth product**C’L = Ctotal = CGD1+ CGD2+ CBD1+ CBD2+ CL When CL≈ C’L, W↑GB↑, but it saturates, when**Note:**If VEB1 and VEB2 are fixed, W1/L1 and W2/L2 must be adjusted proportionally, and they are proportional to DC power.**Therefore:**P is proportional to W1, W2 CL constant, but C(W1,W2) proportional to W1, W2 When C(W1, W2) << CL, GB proportional to P When C(W1,W2)CL or >CL, GB saturates**GB**Linear increase region P**Differential Input, single-ended output single stage**Amplifier N-Channel vin- vin+**Large Signal Eq. in a N-channel Differential pair**=0.5b1(VGS1-VT)2 =(2ID1/b1)0.5 iD1=0, when iD2=ISS and VGS2=VT+(2ISS/b)0.5**Solving for iD1 and iD2**iD1=iD2=ISS/2 VON1=VON2=(ISS/b)0.5**N-Channel Input Pair Differential Amplifier**C.M. Load Simple current reference C.M. Bias**INPUT COMMON MODE RANGE**VG1=VG2=ViCM VSDSAT1=VSDSAT2 =VON VD1=VD3= VSS+VT3+VON VG1min=VD1-|VT1| VG1max=VDD- VSD5SAT-|VT1|-VON**Output Range**Vomin=Vss+Von4 Vomax=Vicm –|VT2| So what’s the vo range What’s for the N-ch circuit.**Common Mode Equivalent Circuit, with perfect match**iC1=VIC/(1/gm1 +2rds5) ro1≈1/gm3 ACM≈ 1/ 2rds5gm3 iC1 CMRR=Av/ACM=2gm1gm3/(gds4+gds2)/gds5**If not perfectly matched**io=aiIC a is a fraction go1≈ gds2 + gds4 iC1 ACM≈ agds5 / 2(gds2 + gds4) CMRR=Av/ACM=2gm1/agds5**SLEW RATE: the limit of the rate of change of the output**voltage C’Ldvo/dt=i4-i2 Max |CLdvo/dt|=ISS ISS ISS Slew Rate = ISS/C’L 0 ISS Output swing: Vosw GB frequency: fGB vo(t)=Voswsin(2pfGBt) Max dvo/dt =Vosw2pfGB To avoid slewing: ISS > C’L Vosw2pfGB**Parasitic Capacitances**CT: common mode only CM: mirror cap = Cdg1 + Cdb1 + Cgs3 + Cgs4 + Cdb3 COUT = output cap = Cbd4 + Cbd2 + Cgd2 + CL**Impedances**• rout = rsd2 || rds4 = 1 / (gds2 + gds4) • rM = 1/gm3 || rds3 || rds1 ≈ 1/ gm3 • Hence the output node is the high impedance node • When vi=0, slowest discharging node is output node with dominant pole p1 = -1/(C’outrout), where C’out = Cout+ Cgd4 • Approximate transfer function AV(s) = AV/(s/p1─1)**When vG1=vG2=0(AC)**KCL at D1: KCL at D2:**Gain bandwidth product**• Gain AV(0) = gm1 / (gds2 + gds4) • Bandwidth ≈ |p1| ≈ (gds2 + gds4) / C’out • GBW ≈ gm1 / C’out • gm1 = {2*ID1mCoxW1/L1}½ • increase gm1 increase GBW • increase W1 increase GBW • But C’out has Cdb2 and Cgd2 W1 • Once Cdb2 and Cgd2 become comparable to CL, increasing W1 reduces GBW

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