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Differential Amplifiers and common mode feedback Differential amplifiers • Cancellation of common mode signals including clock feed-through • Cancellation of even-order harmonics • Increased signal swing Symbol: Two-Stage, Miller, Differential-In, Differential-Out Op Amp peak-to-peak output voltage ≤ 2·OCMR Output common mode range (OCMR) = VDD-VSS - VSDPsat - VDSNsat Two-Stage, Miller, Differential-In, DifferentialOut Op Amp with Push-Pull Output Able to actively source and sink output current Output quiescent current poorly defined Cascode Op Amp with DifferentialOutputs, push-pull output Differential-Output, FoldedCascode OCMR = VDD -VSS - 2VSDP(sat) -2VDSN(sat) Quite limited Two-Stage, Differential Output, Folded-Cascode M11-M13 and M10-M12 provide level shift Common Mode Output Voltage Stabilization Common mode drift at output causes differential signals move into triode region Common Mode feedback • All fully differential amplifier needs CMFB • Common mode output, if uncontrolled, moves to either high or low end, causing triode operation • Ways of common mode stabilization: – external CMFB – internal CMFB Cause of common mode problem Unmatched quiescent currents Vbb I2 Vo1 Vin Vbb=VbbQ+Δ Vin=VinQ Vbb=VbbQ Vo2 I1 Vo1Q Vin=VinQ+ΔVin Vo1 actual Q point M2 is in triode If Vo VOCM I y i.e. Vyy If Vo VOCM I y i.e. Vyy Vxx Ix Vo VOCM Vin Ix(Vo) Iy(Vo) Vyy Iy Vo needneg. feedbackfromVo to V yy Basic concept of CMFB: Vo+ +Vo- Vo+ CM Vo- Dvb 2 measurement Voc CMFB e VoCM + desired common mode voltage Basic concept of CMFB: Vo+ +Vo- Vo+ CM Vo- Dvb 2 measurement Voc CMFB e e VoCM + Find transfer function from e to Voc, ACMF(s) Find transfer function from an error source to Voc Aerr(s) Voc error due to error source: err*Aerr(0)/ACMF(0) example Vb2 CC Vi+ CC Vi- Vo+ VCMFB Vo- Vb1 VCMFB Voc + Vo+ Vo- Example Voc ? ? VoCM Need to make sure to have negative feedback VDD M7A 75/3 150/3 M2A 150/3 M2B BIAS4 300/3 300/3 M13A M13B 1.5pF 1.5pF M7B 75/3 averagerOUT- M3B BIAS3 OUT+ 300/2.25 300/2.25 M6C 20K M3A 20K 300/2.25 300/2.25 75/2.25 INM6AB 75/2.25 IN+ M1A Source M12B M12A 1000/2.25 1000/2.25 follower M1B 200/2.25 BIAS2 M11 150/2.25 M10 CL=4pF 4pF M9A 50/2.25 M9B 50/2.25 M4A 50/2.25 M4B 50/2.25 BIAS1 M8 150/2.25 200/2.25 M5 VSS Folded cascode amplifier Resistive C.M. detectors: Vo+ R1 R2 Vo- Vo Vo , if R1 R2 2 Prob : resistive loading effect. use R1 , R2 very large - difficult to achieve - especially when Vo is at an cascoded node Resistive C.M. detectors: Vo.c. Vo+ Vi+ R1 R1 Vo- Vi- • O.K. if op amp is used in a resistive feedback configuration • & R1 is part of feedback network. • Otherwise, R1 becomes part of g0 & hence reduces AD.C.(v) Buffer Vo+, Vo- before connecting to R1. Voc Vo+ VoR1 R1 Simple implementation: source follower Vo+ Vo.c. * Gate capacitance is your load to Amp. Vo- Why not: Vo+ Vo.c. Vo- * Initial voltage on cap. C1 C2 Vo Vo , if C1 C 2 2 Prob : at high freq. AC diff short Use buffer to isolate Vo node: gate cap is load or resistors Switched cap CMFB Vo+ Vo- Φ1 Φ2 Φ1 VoCM. VoCM. To increase or decrease the C.M. loop gain: e.g. Vo.c. Vo.c.d. Vo.c. Vo.c.d. VC.M.F.B. VC.M.F.B. Another implementation • Use triode transistors to provide isolation & z(s) simultaneously. M1, M2 in deep triode. VGS1, VGS2>>VT Voc Vo+ VoM1 can be a c.s. M2 In that case, circuit above M1, M2 needs to ensure that M1, M2 are in triode. Vo Vo 2 deep triode oper Example: Input state Vo- Vb M1 M2 e.g. Vo+, Vo-≈2V at Q & Vb ≈1V , Then M1&2 will be in deep triode. Vo+ If Vo ,Vo Vo- Vo+ Voc VG1 ,VG 2 Vb1 RM 1 , RM 2 Vb2 VX M1 M2 V X ( I const ) but V X to Vo ,Vo is cascoded C.G. Vo ,Vo Two-Stage, Miller, Differential-In, DifferentialOut Op Amp M10 and M11 are in deep triode Vo++ Vo2 VoCM. VCMFB Vo Vo VCMFB 2 gain can be large Vo+ Vo- Note the difference from the book accommodates much larger VoCM range Small signal analysis of CMFB Example: IB M3 Vo+ M1 M4 IB M2 -Δi Vo- +Δi +Δi M5 +Δi -Δi VCM +Δi VCMFB Δi=0 2Δi -Δi -Δi • Differential Vo: Vo+↓ by ΔVo, Vo-↑ by ΔVo • Common mode Vo: Vo+↑ by ΔVo, Vo-↑ by ΔVo DVo Di g m1 2 1 DVCMFB 2Di g m5 k DVo g m1 g m 2 g m3 g m 4 g m1 k g m5 IB M3 Vo+ M1 VCM M4 IB Vo- M2 +Δi +Δi M5 -Δi -Δi VCMFB Δi=0 2Δi Δi7 M7 + -2Δi M6 - 1 -2Δi gm6 To increase gain : 1 DVG 7 2Di g m6 Di7 DVG 7 g m 7 g m7 2Di g m6 g m7 Di5 2Di1 g m6 g m1 g m 7 1 DVo DVCMFB g m5 g m6 * gain by geometric ratios can be made accurate CMFB loop gain: example Vb2 CC Vi+ CC Vi- Vo+ VCMFB Vo- Vb1 VCMFB Voc + Vo+ Vo- gain from VCMFB Voc Vo Vo 2 -gm5vro2 -gm5vro2gm6 Vo gm5v v Poles: p1 p2 z1 same as before Aocmc g m5 g m 6 g ds 2 g ds6 g ds7 VDD M7A 75/3 150/3 M2A 150/3 M2B BIAS4 300/3 300/3 M13A M13B 1.5pF 1.5pF M7B 75/3 averagerOUT- M3B BIAS3 OUT+ 300/2.25 300/2.25 M6C 20K M3A 20K 300/2.25 300/2.25 75/2.25 INM6AB 75/2.25 IN+ M1A Source M12B M12A 1000/2.25 1000/2.25 follower M1B 200/2.25 BIAS2 M11 150/2.25 M10 CL=4pF 4pF M9A 50/2.25 M9B 50/2.25 M4A 50/2.25 M4B 50/2.25 BIAS1 M8 150/2.25 200/2.25 M5 VSS Folded cascode amplifier Removing the CM measurement Vo+ VoCM VoVCMFB Directly connect Vo+, Vo- to the gates of CMFB diff amp. VDD=+1.65V M11 M12 M3 M4 Vo1 M27 M21 M22 M23 M24 Vo2 M1C M2C IDC=100υA VCM Vi1 M14 M26 M13 M1 M51 M2 M52 -VSS=-1.65V Vi2 M25 CMFB with current feedback M3 M4 Vo+ Vo- M1 IB CM Voc detect M2 VoCM M6 M7 M5 VDD VSS IB desired Q : I 3 I1 ,VoCM 4 2 If Vo Vo by DVo , Voc by DVo Di1 , Di2 by DVo g m 6 (equivalen t to VG1 , VG 2 by DVo )