Stability of Operational amplifiers

Transcription

1 Stability o Operational ampliiers Willy Sansen KULeuven, ESAT-MICAS Leuven, Belgium willy.sansen@esat.kuleuven.be Willy Sansen

2 Table o contents Use o operational ampliiers Stability o 2-stage opamp Pole splitting Compensation o positive zero Stability o 3-stage opamp Willy Sansen

3 Operational ampliiers do operations v v 2 v 3 R R 2 R 3 R F - v OUT v OUT R F v - = R v 2 R2 v 3 R3 Requires High gain High speed Low noise Low power Opamp specs : Voltage gain is large Dierential input voltage 0 Input current = 0 Bandwidth is high Gainbandwidth GBW is very, very high Willy Sansen

4 Single-ended or ully dierential? Willy Sansen

5 Voltage input or current input? Voltage input Current output Current input Current output Willy Sansen

6 Classiication Opamp OTA OCA CM amp Operational ampliier Operational Transconduct. ampliier Operational Current ampliier Current Mode ampliier A v = v OUT v IN A g = i OUT v IN A i = i OUT i IN A r = v OUT i IN A v = = A g R L R = A L i = A R r S GBW R S Willy Sansen

7 Feedback conigurations v IN R - R 2 v OUT R v IN - R 2 v OUT v IN - v OUT R 2 A v = - R R 2 A v = R A v = R IN = R R IN = R IN = Willy Sansen

8 Integrator C A v -20 db/dec v IN R - v OUT φ (A v ) 90 o 0 o -90 o A v = p = j p 2π RC Willy Sansen

9 Low-pass ilter C R 2 A v A v0-20 db/dec v IN R - v OUT φ (A v ) 90 o 0 o p -90 o A v0 = - R 2 R A v0 A v = p = ( j ) p 2π R 2 C Willy Sansen

10 High-pass ilter L R 2 A v A v0 20 db/dec v IN R - v OUT φ (A v ) 90 o 0 o p A v0 = - R 2 R A v = A v0 j p ( j ) p -90 o p = R 2 2π L Willy Sansen

11 High-pass ilter v IN - v OUT A v 20 db/dec C R R 2 A v0 φ (A v ) 90 o 0 o z R 2-90 o A v0 = R A v = A V0 ( j ) z z = 2π RC R R 2 R= R //R 2 = R R 2 Willy Sansen

12 Low-pass ilter with inite attenuation v IN - v OUT A v -20 db/dec R 2 R C A v0 φ (A v ) 90 o 0 o z R 2 A v0 = R A v = A v0 ( j ) j z z -90 o z = 2π RC R= R R 2 Willy Sansen

13 Exchange o gain and bandwidth A A o A c Loop gain (T), A()=A o /(j/ ) Ao open loop gain A c closed loop gain A c φ A 0 o -90 o -80 o GBW A o = c A c c = c 45 o 45 o A c c = GBW Willy Sansen

14 Open- and closed-loop gain v IN Σ v ε G v OUT H v ε =v IN -Hv OUT v OUT = G v ε A c = v OUT v IN = G GH H i the loop gain GH = T >> P. Gray, P.Hurst, S.Lewis, R. Meyer: Design o analog integrated circuits, 4th ed., Wiley 200 Willy Sansen

15 What makes an opamp an opamp? v out C L v out v in v in Operational ampliier : Single-pole ampliier High impedance = high gain Exchange Gain-Bandwidth Stable or all gain values Wideband ampliier : Multiple-pole ampliier Low impedances at nodes Wide Bandwidth Stable or one gain only Willy Sansen

16 Single-pole system A A o -20 db/dec A o open loop gain Closed loop gain A c = loop gain A c = φ A 0 o open loop GBW closed loop phase shit -90 o -80 o PM PM phase margin Willy Sansen

17 Two-pole system A A o -20 db/dec A o open loop gain Closed loop gain A c = A c = φ A 0 o -90 o -80 o loop gain GBW -40 db/dec 2 open closed loop PM v IN - v OUT PM phase margin Willy Sansen

18 Higher loop gain gives less PM A A o A c loop gain A o open loop gain A c closed loop gain φ A 0 o open 2-90 o -80 o PM PM phase margin Willy Sansen

19 Higher loop gain gives less PM A A o A c loop gain A o open loop gain A c closed loop gain φ A 0 o open 2-90 o -80 o PM PM phase margin Willy Sansen

20 Higher loop gain gives less PM A A o loop gain A o open loop gain A c closed loop gain A c φ A 0 o open 2-90 o -80 o PM PM phase margin Willy Sansen

21 Higher loop gain gives less PM A A o loop gain A o open loop gain A c closed loop gain A c = φ A 0 o open 2 Worst case or A c = -90 o -80 o PM PM phase margin Willy Sansen

22 Increase PM by increasing 2 : low 2 A A o Closed loop gain A c = A c = φ A 0 o open 2-90 o -80 o PM 0 o Willy Sansen

23 Increase PM by increasing 2 A A o Closed loop gain A c = A c = φ A 0 o open 2-90 o -80 o PM 45 o Willy Sansen

24 Set PM by setting 2 3 GBW A A o Closed loop gain A c = A c = φ A 0 o open GBW GBW -90 o -80 o PM 70 o Willy Sansen

25 Calculate PM or 2 3 GBW Open loop gain A = A o ( j )( j ) 2 v IN - A c = A v OUT H = A Closed loop gain A c = A j j 2 2 GBW GBW 2 ζ is the damping (=/2Q) r is the resonant requency j 2ζ j 2 2 r r 2 Willy Sansen

26 Relation PM, damping and 2 /GBW r = GBW 2 PM ( o ) = 90 o GBW - arctan = arctan 2 2 GBW 2 GBW PM ( o ) ζ = 2 2 GBW P (db) P t (db) Willy Sansen

27 Amplitude response vs requency P ζ = Q = 0.7 P = 2 ζ - ζ 2 Willy Sansen

28 Amplitude response vs time P t V IN 0.7 V OUT ζ = Q = P t = e - π ζ - ζ 2 Willy Sansen

29 Table o contents Use o operational ampliiers Stability o 2-stage opamp Pole splitting Compensation o positive zero Stability o 3-stage opamp Willy Sansen

30 Generic 2-stage opamp v IN v IN2 - g m C c - g v OUT m2 A v = g m jω C c GBW R L C L g m A v = GBW = nd = g m2 2π C c 2π C L Willy Sansen

31 Generic 2-stage opamp v IN v IN2 - g m C n C c - g v OUT m2 R L C L A v = g m jω C c GBW g m A v = GBW = nd = g m2 2π C c 2π C L Cc C n Willy Sansen

32 Elementary design o 2-stage opamp GBW = g m 2π C c nd = 3 GBW = g m2 2π C L C n Cc g m2 C L { 0.3 g m 4 Cc Larger current in 2nd stage! GBW = 00 MHz or C L = 2 pf Solution: choose C c = pf Willy Sansen

33 Table o contents Use o operational ampliiers Stability o 2-stage opamp Pole splitting Compensation o positive zero Stability o 3-stage opamp Willy Sansen

34 Generic 2-stage opamp : Miller OTA C c v IN v IN2 - g m C n g m (v IN2 -v IN ) - g v OUT m2 R L C L C c v OUT A v0 = - A v A v2 A v = g m R n R n v n C n - g m2 v n R L C L - A v2 = - g m2 R L Willy Sansen

35 Generic two-stage opamp v IN v IN2 - g m R n C n C c - g v OUT m2 R L C L A v0 = - A v A v2 A v = g m R n A v2 = g m2 R L C - c s g m2 A v = A v0 (R n C n R C A n c v2 R n C c R L C L R C )s R L c n R L CCs 2 CC = C n C c C n C L C c C L Willy Sansen

36 Approximate poles and zeros A = A 0 - cs a s b s 2 Zero s = c Pole s = - a s 2 = - i s 2 >> s a b Willy Sansen

37 Miller OTA : pole splitting with C c C c pf 0.pF 0F A v 000 A v0 BW d z Pole splitting nd k M Hz 0F Pole splitting or high C c : d = 2π Av2 R n C c g m pf k GBW M Hz z = 2π Cc is a positive zero! Willy Sansen

38 Eect o positive zero Negative zero Positive zero j / 2 - j / 2 A v = A v0 A v = A v0 j / j / A v A v For phase, a positive zero is like a negative pole!!! φ A 0 o 2 φ A 0 o 2-90 o -80 o -90 o -80 o 80 o Willy Sansen

39 Miller OTA : pole splitting with g m2 g m2 0µS µs 0.µS A v µs 0 µs µs 0. µs 250 µs 0 µs µs 0.µS Pole splitting d nd z BW k M Hz GBW Pole splitting or high g m2 : d = 2π Av2 R n C c g m2 z = 2π Cc is a positive zero! 0. k M Hz Willy Sansen

40 Pole splitting by... g m2 C L g m 4 Cc or g m2 C c 4 g m C L both g m2 C c Willy Sansen

41 Table o contents Use o operational ampliiers Stability o 2-stage opamp Pole splitting Compensation o positive zero Stability o 3-stage opamp Willy Sansen

42 Positive zero because eedorward v IN v IN2 - g m C n C c - g m2 R L v OUT C L Miller eect Is eedback v IN v IN2 - g m C n C c R L v OUT C L Feedorward Cut! Willy Sansen

43 Cut eedorward through C c - - g m C n C c - g v OUT m2 R L C L C c v OUT Voltage buer Source ollower Re. Tsividis, JSSC Dec.76, Willy Sansen

44 Cut eedorward through C c -2 - g m C n C c - g v OUT m2 R L C L C c v OUT Current buer Cascode C c v OUT Re. Ahuja, JSSC Dec 83, Current buer Cascode Willy Sansen

45 Compensation with cascodes 3 C c I B v out 3 I B v out 2 C L 2 C L v in M v in M C c Willy Sansen

46 Cut eedorward through C c -3 - g m C n C c - g v OUT m2 R L 3 C L R c C c v OUT R c C c v OUT z = 2π Cc (/g m2 -R c ) R c = /g m2 R c > /g m2 No zero Negative zero Re. Senderovics, JSSC Dec 78, Willy Sansen

47 Negative zero compensation R c >> /g m2 z = 3 GBW Final choice : z = - 2π Cc R c R c = 3gm < R c < g m2 3g m Willy Sansen

48 Exercise o 2-stage opamp GBW = 50 MHz or C L = 2 pf Find I DS ; I DS2 ; C c and R c! Choose C c = pf > g m = 2π C c GBW = 35 µs I DS = 3.5 µa & /g m 3.2 kω nd = 50 MHz > g m2 = 2π C L 4GBW = 8g m = 2520 µs I DS2 = 252 µa & /g m2 400 Ω 400 Ω < R c < kω :R c Ω ± 60% Willy Sansen

49 Table o contents Use o operational ampliiers Stability o 2-stage opamp Pole splitting Compensation o positive zero Stability o 3-stage opamp Willy Sansen

50 -stage CMOS OTA GBW = g m 2π C L v OUT v IN M C L g m Willy Sansen

51 2-stage Miller CMOS OTA GBW = g m 2π C C C C v OUT nd = g m2 2π C L v IN M M2 C L g m g m2 nd = 3 GBW Willy Sansen

52 3-stage Nested Miller CMOS OTA C C GBW = g m 2π C C C D v OUT nd = g m2 2π C D v IN M M2 g m g m2 g m3 M3 C L nd2 = g m3 2π C L nd = 3 GBW nd2 = 5 GBW Willy Sansen

53 Nested Miller with dierential pair C C C D g m g m2 g m3 Huijsing, JSSC Dec.85, pp Willy Sansen

54 Relation between the nd s nd GBW 0 9 PM with two nd's PM60 PM65 PM70 PM = 90 o GBW - arctan( ) nd 5 4 GBW - arctan( ) nd nd2 GBW Willy Sansen

55 Relation nd s and power nd GBW Total current 2I 2I 2 I 3 PM = 60 o nd2 GBW Willy Sansen

56 Elementary design o 3-stage opamp GBW = g m 2π C C nd = 3 GBW = g m2 2π C D Choose C D C C! nd2 = 5 GBW = g m3 2π C L g m2 g m 3 g m3 C L g m 5 CC Even larger current in output stage! Willy Sansen

57 Exercise o 3-stage opamp GBW = 50 MHz or C L = 2 pf Find I DS ; I DS2 ; I DS3 ; C C and C D! Choose C C = C D = pf > g m = 2π C C GBW = 35 µs I DS = 3 µa nd = 50 MHz > g m2 = 2π C D 3GBW = 3g m = 945 µs I DS2 = 95 µa nd2 = 250 MHz > g m3 = 2π C L 5GBW = 0g m = 350 µs I DS3 = 35 µa Willy Sansen

58 Comparison, 2 & 3 stage designs GBW = 50 MHz or C L = 2 pf Single stage : I DS = 3 µa I TOT = 2I DS = 62 µa Two stages : Choose C C = pf I DS = 3 µa I DS2 = 252 µa I TOT = 2I DS I DS2 = 34 µa Three stages : Choose C C = C D = pf I DS = 3 µa I DS2 = 95 µa I DS3 = 35 µa I TOT = 2I DS 2I DS2 I DS3 = 567 µa Willy Sansen

59 Table o contents Use o operational ampliiers Stability o 2-stage opamp Pole splitting Compensation o positive zero Stability o 3-stage opamp Willy Sansen