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CMOS AMPLIFIERS

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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  1. CMOS AMPLIFIERS • Simple Inverting Amplifier • Differential Amplifiers • Cascode Amplifier • Output Amplifiers • Summary

  2. Simple Inverting Amplifiers

  3. Small Signal Characteristics Inverter with diode connection load How do you get better matching?

  4. High gain inverters

  5. 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

  6. Small signal High gain! Especially at low power.

  7. Key to analysis by hand: • Use level 1 or 3 model equations • Use KCL/KVL

  8. Dependence of Gain upon Bias Current

  9. Transfer function of a system input u output y System

  10. 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.

  11. Frequency Response of CMOS Inverters

  12. 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

  13. Total conductance from y to ground: go = gds1 + gds2 KCL at node y: Therefore system pole is:

  14. 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:

  15. Zeros of CMOS Inverters Let vin = u, X=0, VDD = 0, VSS = 0. CGS2, CGD2, CBD1, CBD2, CL are all short gds1, gds2 also short No current in them KCL: Zero is:

  16. Input output transfer function When s=jw0, A(0)  When w∞, A(s)

  17. -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

  18. Unity gain feedback A(s)

  19. 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=UGF,  need high gain bandwidth product

  20. Gain bandwidth product C’L = Ctotal = CGD1+ CGD2+ CBD1+ CBD2+ CL When CL≈ C’L, W↑GB↑, but it saturates, when

  21. Note: If VEB1 and VEB2 are fixed, W1/L1 and W2/L2 must be adjusted proportionally, and they are proportional to DC power.

  22. 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

  23. GB Linear increase region P

  24. For given current or power (current source load) Initially, as W1 increased, GB increases But GB will reach a max, and then drop as W1 increases

  25. NOISE IN MOS INVERTERS

  26. To minimize: • L2 >>L1 • En1 small

  27. For thermal noise

  28. Noise in Push-Pull current source load Inverter

  29. Differential Input, single-ended output single stage Amplifier N-Channel vin- vin+

  30. P-channel

  31. 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

  32. Solving for iD1 and iD2 iD1=iD2=ISS/2 VON1=VON2=(ISS/b)0.5

  33. N-Channel Input Pair Differential Amplifier C.M. Load Simple current reference C.M. Bias

  34. Voltage transfer curve

  35. P-Channel Input Pair Differential Amplifier

  36. Voltage transfer curve

  37. INPUT COMMON MODE RANGE VG1=VG2=ViCM VSDSAT1=VSDSAT2 =VON VD1=VD3= VSS+VT3+VON VG1min=VD1-|VT1| VG1max=VDD- VSD5SAT-|VT1|-VON

  38. Output Range Vomin=Vss+Von4 Vomax=Vicm –|VT2| So what’s the vo range What’s for the N-ch circuit.

  39. SMALL SIGNAL ANALYSIS AV

  40. 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

  41. If not perfectly matched io=aiIC a is a fraction go1≈ gds2 + gds4 iC1 ACM≈ agds5 / 2(gds2 + gds4) CMRR=Av/ACM=2gm1/agds5

  42. Formal detailed analysis

  43. 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

  44. Parasitic Capacitances CT: common mode only CM: mirror cap = Cdg1 + Cdb1 + Cgs3 + Cgs4 + Cdb3 COUT = output cap = Cbd4 + Cbd2 + Cgd2 + CL

  45. 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)

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