Analog Integrated Circuits (Jieh Tsorng Wu)

Analog Integrated Circuits(Jieh Tsorng Wu)

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Unformatted text preview: d1 V id IM d2 V V i2 M1 M2 I I • RD1 = RD2 ≡ RD . IM • Neglect ro. • RSS → ∞. SS SS VSS Vi d ≡ Vi 1 − Vi 2 Id d ≡ Id 1 − Id 2 Vod ≡ Vo1 − Vo2 = −Id d RD Assume M1 and M2 are in the saturation region, Id 1 1 = k (Vgs1 − Vt )2 2 Differential Gain Stages Id 2 1 = k (Vgs2 − Vt )2 2 7-5 W k = µCox L Analog ICs; Jieh-Tsorng Wu Source-Coupled Pair Large-Signal Behavior Summing currents at the common source node, we have Id 1 + Id 2 = ISS ⇒ Id 1 ISS Id d + = 2 2 Id 2 ISS Id d − = 2 2 The gate voltages can be written as Vgs1 = Vt + 2Id 1 Vgs2 = Vt + K 2Id 2 K The differential input voltage is Vi d = Vgs1 − Vgs2 = 2Id 1 k − 2Id 2 k = 2 k Id 1 − Id 2 Squaring Vi 2 d Differential Gain Stages 2 2 I + I − 2 Id 1Id 2 = I− = k d1 d2 k SS 7-6 2 2 ISS − Id d Analog ICs; Jieh-Tsorng Wu Source-Coupled Pair Large-Signal Behavior Rearrange, then we have Id d k = Vi d 2 4ISS k − Vi 2 d and Id 1 ISS Id d + = 2 2 Id 2 ISS Id d − = 2 2 Define VIM as the differential input voltage at which one of the MOST is turned off, i.e., ISS k = VIM 2 4ISS 2 − VIM k Differential Gain Stages ⇒ VIM = 2ISS = k 7-7 2 (Vov 1 )|Vi d =0 = 2 (Vov 2 )|Vi d =0 Analog ICs; Jieh-Tsorng Wu Small-Signal Analysis of Differential Amplifiers v o1 v o2 v i1 v o1 v o2 v i2 v id 2 v id 2 v ic The differential and common-mode signals are defined as vi d ≡ vi 1 − vi 2 vi 1 + vi 2 vi c ≡ 2 1 1 vi 1 = vi c + vi d vi 2 = vi c − vi d 2 2 vo1 A11 A12 vi 1 = vo2 A21 A22 vi 2 Differential Gain Stages vod ≡ vo1 − vo2 voc vo1 + vo2 ≡ 2 1 1 vo1 = voc + vod vo2 = voc − vod 2 2 vod Ad m Acd m vi d = voc Ad cm Acm vi c 7-8 Analog ICs; Jieh-Tsorng Wu Small-Signal Analysis of Differential Amplifiers The voltage gain are defined as A11 vo1 = vi 1 vi 2 =0 A12 vo1 = vi 2 vi 1 =0 A21 vo2 = vi 1 vi 2 =0 A22 vo2 = vi 2 vi 1 =0 The differential and common-mode gains are Differential-Mode Gain = Ad m Common-Mode Gain = Acm vod = vi d voc = vi c A11 − A12 − A21 + A22 = 2 vi c =0 A11 + A12 + A21 + A22 = 2 vi d =0 Differential-Mode-to-Common-Mode Gain = Ad cm voc = vi d A11 − A12 + A21 − A22 = 4 vi c =0 Common-Mode-to-Differential-Mode Gain = Acd m vod = vi c = A11 + A12 − A21 − A22 Differential Gain Stages 7-9 vi d =0 Analog ICs; Jieh-Tsorng Wu Small-Signal Analysis of Differential Amplifiers • Usually want to sense vi d while rejecting vi c, thus want Ad m Acm, Acd m, Ad cm . • The common-mode-rejection ratio is defined as CMRR ≡ Ad m Acm • In a perfectly balanced circuit, Acmd = Ad cm = 0. However, in practice, these transfer functions are not zero because of component imbalances. • The ratio Ad m/Acd m is important because it indicates the extent to which a commonmode input corrupts the differential output. Differential Gain Stages 7-10 Analog ICs; Jieh-Tsorng Wu Emitter-Coupled Pair Differential-Mode Half Circuit VCC v + od 2 RC VCC RC Q1 RC v − od 2 Q2 RS RS vi d 2 − iE E vod 2 Q1 E + + RE E vi d 2 RS + vi d 2 VEE ve = 0 Differential Gain Stages ⇒ Ad m vod vod /2 rπ (ro RC ) = = = −gm vi d rπ + RS v i d /2 7-11 Analog ICs; Jieh-Tsorng Wu Emitter-Coupled Pair Common-Mode Half Circuit VCC VCC RC RC RC voc voc Q1 voc ix Q2 Q1 RS vi c 2RE E VEE ix = 0 2RE E RS vi c RS vi c 2RE E VEE ⇒ Acm gmRC RC voc = =− ≈− 2g m R E E vi c 2RE E 1+ αo Differential Gain Stages 7-12 Analog ICs; Jieh-Tsorng Wu Emitter-Coupled Pair Input Resistances Rid 2 Rid Rid 2 vi1 vi1 Ric vi1 vi1 Ric 2 Ric Assume RS = 0 and rb = 0. When vi c = 0, ib1 = −ib2 ≡ ibd , Differential-Mode Input Resistance = Ri d ≡ vi d ibd vi c =0 = 2rπ When vi d = 0, ib1 = ib2 ≡ ibc, vi c Common-Mode Input Resistance = Ri c ≡ ibc In general ib1 = + Differential Gain Stages vi d Ri d + vi c vi d =0 ib2 = − Ri c 7-13 = rπ + 2RE E (βo + 1) ≈ 2βoRE E vi d Ri d + vi c Ri c Analog ICs; Jieh-Tsorng Wu Emitter-Coupled Pair Frequency Response |Ad m| Differential-Mode VCC RC Common-Mode VCC |Acm | RC vod 2 voc Q1 zE CMRR = Q1 RS vi c |Ad m | |Acm | RS vi d 2 ω p1 2RE E CE 2 ω Differential Gain Stages 7-14 Analog ICs; Jieh-Tsorng Wu Emitter-Coupled Pair Frequency Response • Using the Miller approximation, the differential Response can be written as vod Ad m(0) ≈ Ad m(s) = vi d 1 − s/p1 Ad m(0) = −gmRC rπ RS + rb + rπ 1 p1 = − Ct [(RS + rb) rπ ] Ct = Cπ + Cµ(1 + gmRC ) • Because RE E is usually large, the common-mode response is typically dominated by the time constant at the tail node of the pair. RC voc Acm(s) = ≈− ≈ Acm(0) 1 − s/zE vi c ZE (s) Acm(0) = − Differential Gain Stages RC 2RE E 7-15 ZE (s) = zE = − 1 1 2RE E +s CE 2 2RE E = 1 + sCE RE E 1 RE E CE Analog ICs; Jieh-Tsorng Wu Emitter-Coupled Pair Input Offset Voltage and Current VCC VCC R C1 R C2 RC Vo Q1 V RC Vo OS Q2 Q1 Q2 Vi Vi I I EE VEE OS 2 I EE VEE Circuit with No Mismatches • VOS and IOS is equal to the value of VID = VI 1 − VI 2 and IBD = IB1 − IB2 that must be applied to the input to drive VOD = 0. Differential Gain Stages 7-16 Analog ICs; Jieh-Tsorng Wu Emitter-Coupled Pair Input Offset Voltage For BJTs in the forward-active region, IC = IS eVBE /UT VBE = UT ln 2 IC IS = A IS qni Dn G (VCB ) The output condition is VOD = −(IC1RC1 − IC2 RC2) = 0 ⇒ IC1 IC2 = RC2 RC1...
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This note was uploaded on 03/26/2013 for the course EE 260 taught by Professor Choma during the Winter '09 term at USC.

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