stability of operational amplifiersextras.springer.com/2006/978-0-387-25746-4/chapter_05.pdf ·...

Willy Sansen 10-05 051

Stability ofOperational amplifiers

Willy Sansen

KULeuven, ESAT-MICASLeuven, Belgium

[email protected]


Table of contents

• Use of operational amplifiers

• Stability of 2-stage opamp

• Pole splitting

• Compensation of positive zero



Operational amplifiers do operations

+- vOUT

RF

R1

R2

R3

v1

v2

v3

RF

vOUT- = +R1

v1

R2

v2

R3

v3+

Requires High gainHigh speedLow noiseLow power

Opamp specs : Voltage gain is largeDifferential input voltage ≈ 0Input current = 0Bandwidth is highGainbandwidth GBW is very, very high


Single-ended or fully differential ?


Voltage input or current input ?

Voltage inputCurrent output

Current input Current output


Classification

Operational amplifier

Opamp OTA OCA CM ampOperational

Transconduct.amplifier

OperationalCurrent amplifier

CurrentMode

amplifier

+-

+-

+-

+-

Av =vOUTvIN

Ag =iOUTvIN

Ai =iOUTiIN

Ar =vOUTiIN

Av = = Ag RL = Ai = ArRLRS

1RS

GBW


Feedback configurations

+- vOUT

R2

R1vIN

Av = -R1

R2

+-

vOUT

R2R1

vIN+-

vOUT

vIN

Av = 1 +R1

R2

RIN = R1 RIN = ∞ RIN = ∞

Av = 1


Integrator

|Av|

f

f

-20 dB/dec

φ (Av)

-90o

0o

90o

+- vOUT

RvIN

C

-90o

Av = fp = 1

2π RC

fp

fj

1


Low-pass filter

|Av|

fp f

f

-20 dB/dec

φ (Av)

-90o

0o

90o

+- vOUT

R2

R1vIN

C

R1

R2

-90o

Av = fp = 1

2π R2C

fp

f

Av0

(1 + j )

Av0Av0 = -


High-pass filter

+- vOUT

R2

R1vIN

L

R1

R2 Av = Av0

fp

f(1 + j )

Av0 = -

|Av|

fp f

f

20 dB/dec

φ (Av)

-90o

0o

90o

-90o

fp = R2

2π L

fp

fj

Av0


High-pass filter

+-

vOUT

R2R1

vIN

Av0 = 1 +R1

R2

C

|Av|

φ (Av) fz f

f

-90o

0o

20 dB/dec

90o

Av = AV0 (1 + j ) fz = 1

2π RCfz

f R = R1//R2 = R1+R2

R1R2

Av0


Low-pass filter with finite attenuation

+-

vOUT

R2R1

vIN

Av0 = 1 +R1

R2

|Av|

φ (Av) fz f

f

-90o

0o

-20 dB/dec

90o

fz = 1

2π RCR = R1+R2

Av0C

Av = Av0

fz

fj

fz

f(1 + j )


Exchange of gain and bandwidth

-180o

-90o

0o

Ao

1

|A|

f1 f

Loop gain (1+T)Ao open loop gain

Ac closed loop gain

GBW Ao f1 =

Ac

f1c Ac f1c =

Ac

f1c

Ac f1c =

GBW

45o

45o

fφA


Open- and closed-loop gain

G

H

Σ

if the loop gain GH = T >> 1

vε = vIN - H vOUT

1 + GH≈ 1

H

vIN vOUTvε

vOUT = G vε

vOUT

vIN= G

P. Gray, P.Hurst, S.Lewis, R. Meyer: Design of analog integrated circuits,4th ed., Wiley 2001

Ac =


What makes an opamp an opamp ?

Operational amplifier :Single-pole amplifierHigh impedance = high gainExchange Gain-BandwidthStable for all gain values

Wideband amplifier :Multiple-pole amplifierLow impedances at nodesWide BandwidthStable for one gain only

+vout

vin

CL

vin

vout


Single-pole system

-180o

-90o

0o

Ao

Ac=1

|A|

f1f

Ao open loop gain

Closed loop gain Ac = 1

φA

f

-20 dB/dec

open loop closed loop

GBW

PM phase marginPM

loop gain


Two-pole system

-180o

-90o

0o

Ao

Ac=1

|A|

f1 f2 fφA

Ao open loop gain

Closed loop gain Ac = 1-20 dB/dec

-40 dB/dec

open closed loop

vIN+-

vOUTloop gain

PM phase marginPM

GBW


Higher loop gain gives less PM

-180o

-90o

0o

Ao

Ac

1

|A|

f1 f2 f

loop gain Ao open loop gain

Ac closed loop gain

PM phase marginPM

φA open



-180o

-90o

0o

Ao

Ac

1

|A|

f1 f2 f


Ac closed loop gain

PM phase marginPM

φA open



-180o

-90o

0o

Ao

Ac=1

|A|

f1 f2 f


Ac closed loop gain

PM phase marginPM

Worst case for Ac = 1

φA open


Increase PM by increasing f2 : low f2

-180o

-90o

0o

Ao

Ac=1

|A|

f1 f2 f

PM ≈ 0o

φA


fopen


Increase PM by increasing f2

-180o

-90o

0o

Ao

Ac=1

|A|

f1 f2 f

PM ≈ 45o

φA


fopen


Set PM by setting f2 ≈ 3 GBW

-180o

-90o

0o

Ao

Ac=1

|A|

f1 f2 f

PM ≈ 70o

φA


f

GBW f2 ≈ 3 GBW

open


Calculate PM for f2 ≈ 3 GBW

Open loop gain A =

Closed loop gain Ac = ≈

Ao

(1 + j )(1 + j )ff1

ff2

A1+A

1

1 + j + j2 f2GBW f2

fGBW

1

1 + j 2ζ + j2 f2

fr2

ffr

≈

ζ is the damping (=1/2Q)fr is the resonant frequency

vIN+-

vOUT

Ac = 1

AH = 1


Relation PM, damping and f2/GBW

fr = GBW f2

f2GBW

PM (o) = 90o - arctan = arctanf2

GBWf2GBW

f2GBW

ζ =

0.5 27 0.35 3.6 2.3

12PM (o)

1 45 0.5 1.25 1.31.5 56 0.61 0.28 0.732 63 0.71 0 0.373 72 0.87 0 0.04

Pf (dB) Pt (dB)


Amplitude response vs frequency

ζ = Q = 0.7

Pf = 1

2 ζ 1 - ζ2

Pf


Amplitude response vs time

0.1

1

2

0.7

0.4

ζ = Q = 0.7VIN VOUT

Pt = - π ζ

1 - ζ21 + e

Pt


Table of contents



• Pole splitting




Generic 2-stage opamp

+- vOUT

RL

vIN1

vIN2

Cc

+- 1gm1 gm2

CL

Av = gm11

jω Cc

|Av| = 1 fnd =gm2

2π CLGBW = gm1

2π Cc

f1

GBW


Generic 2-stage opamp

+- vOUT

RL

vIN1

vIN2

Cc

+-

Cn1

1gm1 gm2

CL

Av = gm11

jω Cc

|Av| = 1 fnd =gm2

2π CLGBW = gm1

2π Cc

1

1 +Cc

Cn1

f1

GBW


Elementary design of 2-stage opamp

GBW = gm12π Cc

fnd = 3 GBW =gm2

2π CL

1

1 +Cc

Cn1

≈ 0.3

GBW = 100 MHz for CL = 2 pF

Solution: choose Cc = 1 pF

gm2gm1

≈ 4Cc

CL

Larger current in 2nd stage !


Table of contents



• Pole splitting




Generic 2-stage opamp : Miller OTA

+- vOUT

RL

vIN1

vIN2

Cc

+-

Cn1

1gm1 gm2

CL

gm1(vIN2-vIN1)

-

+

-

+1 vOUTCc

Rn1

vn1

Cn1 gm2vn1RL CL

Av0 = - Av1Av2

Av1 = gm1Rn1

Av2 = - gm2RL


Generic two-stage opamp

+- vOUT

RL

vIN1

vIN2

Cc

+-

Cn1

1gm1 gm2

CL

Av0 = - Av1Av2

Av1 = gm1Rn1

Av2 = gm2RL

Av = Av0

1 - sCcgm2

1 + (Rn1Cn1+Av2Rn1Cc+RLCL)s + Rn1RLCCs2

Rn1

CC = Cn1Cc + Cn1CL + CcCL


Approximate poles and zeros

A = A0 1 - cs

1 + a s + b s2

Zero s = 1c

Pole s1 = - 1a

s2 = - if s2 >> s1ab


Miller OTA : pole splitting with Cc

1k 1M Hz

1k 1M

1pF

0.1pF

Cc

|Av|10fF

1

10

100

1000

f

f

fd

fnd

fz

Av0

BW

GBW

Pole splitting

0.1

fd = 2π Av2Rn1Cc

1

fz = 2π Cc

gm2

Hz

Pole splitting for high Cc :

is a positive zero !

1pF

10fF


Effect of positive zero

Av = Av0

-180o

-90o

0o

f1 f2 f

180o

φAf

1 + j f / f2

1 + j f / f1Av = Av0

1 - j f / f21 + j f / f1

-180o

-90o

0o

f1 f2 fφAf

|Av ||Av |

For phase, a positive

zerois like a negative

pole !!!

Negative zero Positive zero


Miller OTA : pole splitting with gm2

1k Hz

1k 1M

10µS

gm2

|Av|0.1µS

1

10

100

1000

f

f

fd fnd fz

BW

Pole splitting

0.1

fd = 2π Av2Rn1Cc

1

fz = 2π Cc

gm2

Hz

Pole splitting for high gm2 :

is a positive zero !0.1µS

250 µS10 µS

1µS 1 µS

0.1 µS

250 µS10 µS

1µS

1M

GBW


Pole splitting by ...

gm2gm1

≈ 4Cc

CL

or gm2 Cc ≈ 4 gm1 CL

both gm2 Cc


Table of contents



• Pole splitting




Positive zero because feedforward

+-

vOUT

RL

vIN1

vIN2

Cc

+-

Cn1

gm1 gm2

CL

vOUT

RL

vIN1

vIN2

Cc

+-

Cn1

gm1

CL

Miller effectIs feedback

Feedforward

Cut !


Cut feedforward through Cc - 1

+- vOUT

RL

Cc

+-

Cn1

gm1 gm2

CL

1 1vOUTCc

Voltagebuffer

Sourcefollower

Ref. Tsividis, JSSC Dec.76, 748-753



+- vOUT

RL

Cc

+-

Cn1

gm1 gm2

CL

1

vOUTCc1

Currentbuffer

CascodeRef. Ahuja, JSSC Dec 83, 629-633

vOUTCc1

Currentbuffer

Cascode


Compensation with cascodes

IB

+1

vin M1

CL+2

+3vout

CcIB

+1

vin M1

+2

+3vout

Cc

CL



+- vOUT

RL

Cc

+-

Cn1

gm1 gm2

CL

11 vOUT

CcRc

vOUTCc

Rc1

fz = 2π Cc (1/gm2 - Rc)

1Rc = 1/gm2 No zeroRc > 1/gm2 Negative zero

3

Ref. Senderovics, JSSC Dec 78, 760-766


Negative zero compensation

fz = -2π Cc Rc

1

Rc >> 1/gm2

fz = 3 GBW Rc = 3 gm1

1

Final choice :

< Rc <1

gm2

13gm1


Exercise of 2-stage opamp

GBW = 50 MHz for CL = 2 pFFind IDS1; IDS2 ; Cc and Rc !

Choose Cc = 1 pF > gm1 = 2π CcGBW = 315 µSIDS1 = 31.5 µA & 1/gm1 ≈ 3.2 kΩ

fnd = 150 MHz > gm2 = 2π CL4GBW = 8gm1 = 2520 µSIDS2 = 252 µA & 1/gm2 ≈ 400 Ω

400 Ω < Rc < 1 kΩ : Rc = 1/√2.5 ≈ 400√2.5 ≈ 640 Ω ± 60%


Table of contents



• Pole splitting




1-stage CMOS OTA

vOUT

CLM1

gm1

vIN

GBW =gm1

2π CL


2-stage Miller CMOS OTA

vOUT

CLM2

CC

M1

gm1 gm2

vIN

GBW =gm1

2π CC

fnd1 =gm2

2π CL

fnd1 = 3 GBW


3-stage Nested Miller CMOS OTA

vOUT

CLM2 M3

CD

CC

M1

gm1 gm2 gm3

vIN

GBW =gm1

2π CC

fnd1 =gm2

2π CD

fnd2 =gm3

2π CL

fnd1 = 3 GBW fnd2 = 5 GBW


Nested Miller with differential pair

Huijsing, JSSC Dec.85, pp.1144-1150

CDCC

gm1

gm2 gm3


Relation between the fnd’s

10987654322

3

4

5

6

7

8

9

10

PM60

PM65

PM70

PM with two fnd's

PM = 90o

- arctan( )

- arctan( )GBWfnd2

GBWfnd1

GBWfnd1

GBWfnd2


Relation fnd’s and power

0

1

2

3

4

5

6

7

8

9

10

0 1 2 3 4 5 6 7 8 9 10

GBWfnd1

GBWfnd2

PM = 60o

Totalcurrent2I1+2I2+I3


Elementary design of 3-stage opamp

GBW = gm12π CC

fnd1 = 3 GBW =gm2

2π CD

gm3gm1

≈ 5CC

CL

Even larger current in output stage !

fnd2 = 5 GBW =gm3

2π CL

gm2gm1

≈ 3

Choose CD ≈ CC !


Exercise of 3-stage opamp

GBW = 50 MHz for CL = 2 pFFind IDS1; IDS2 ; IDS3 ; CC and CD !

Choose CC = CD = 1 pF > gm1 = 2π CCGBW = 315 µSIDS1 = 31 µA

fnd1 = 150 MHz > gm2 = 2π CD3GBW = 3gm1 = 945 µSIDS2 = 95 µA

fnd2 = 250 MHz > gm3 = 2π CL5GBW = 10gm1 = 3150 µSIDS3 = 315 µA


Comparison 1, 2 & 3 stage designs

GBW = 50 MHz for CL = 2 pF

Three stages : Choose CC = CD = 1 pFIDS1 = 31 µA IDS2 = 95 µA IDS3 = 315 µA

ITOT = 2IDS1 + 2IDS2 + IDS3 = 567 µA

Two stages : Choose CC = 1 pFIDS1 = 31 µA IDS2 = 252 µA ITOT = 2IDS1 + IDS2 = 314 µA

Single stage : IDS1 = 31 µA ITOT = 2IDS1 = 62 µA


Table of contents



• Pole splitting



stability of operational amplifiersextras.springer.com/2006/978-0-387-25746-4/chapter_05.pdf ·...

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