Basic Opamp Design and Compensation

Hossein Shamsi

1

Two-Stage opamp

2

opamp Gain

3

V3V2V10 AAAA

ds4ds2mV1 rrgA1

Slew Rate

4

C

D5o

CI

maxdtdVSR

1. 100 mV < VDS(SAT) < 200mV2. VDS > 2VDS(SAT)

3. (both Mi and Mj are npmos or pmos)

4.(Transistors are from different types)

Guidelines for Biasing of the Two-Stage opamp

5

j

i

jD

iD

LWLW

II

p

n

p

n

p

n

pD

nD

LWLW

LWLW

II

3

Frequency Response (first-order model)

6

242

2421

11

1

ACrr

ACrr

Cdsds

Cdsdsp

2420 ArrgA dsdsim

1

0

1p

sAsA

It is proven that:C

impta C

gAUGBW 10

Since a first order transfer function is assumed for the Opamp, therefore it does not have stability problem.

Using opamp in closed-loop configurations

7

Hence for the closed-loop amplifier, we have:

tadB

CLA

3

10

Using opamp in closed-loop configurations(Eample#1)

8

tUVstep

After performing an accurate analysis, we have:

step

ta

step

p

o Vss

AVs

AsAA

sV

1

1

111

11

1 0

10

0

0

ta

t

stepo eVA

tv

1,1110

Using opamp in closed-loop configurations(Eample#1)

9

tUVstep

tastept

oo

ta

t

stepo

Vdtdv

dtdv

eVA

tv

0max

0

1,111

times.-settlinglinear-nonandlinearthebothhascircuitTheif

time.-settlinglineartheonlyhascircuitTheif

SRV

SRV

tastep

tastep

Using opamp in closed-loop configurations(Eample#2)

10

tUVstep

Frequency Response (second-order model)

11

Frequency Response (second-order model)

12

Frequency Response (second-order model)

13

C

mz

mp

Cmp C

gCC

gCRRg

7

21

72

2171 ,,1

•As gm7 increases, the separation between the first and second poles increase.•Increasing CC moves the dominant pole to a lower frequency without affecting the second pole.•Hence, the use of a Miller capacitance for compensation is often called “pole-splitting compensation”.•The pole-splitting makes the opamp more stable.

MHz2392πCgω

GHz1.192πCC

gω

kHz1772πωωAω

MHz802πsRadG0.5

2p1m

Cgω

dB53dB

A450RRggA

C

m7z

21

m7p2

p1p10ta

C

m1ta

021m7m10

Example1

14

Consider a two-stage opamp with the following characteristics:

pF2CpF,0.3CpF,0.1Ck15Rk,10RVmA3g,

VmA1g

C21

21

m7m1

a) Calculate A0 , ωp1, ωp2, ωz, and A(s).

96

9

107.5s1

101.1s1

101.5s1

450sA

15

b) Plot the Bode diagram of A(s).c) What is the phase margin if we use the opamp in a closed-loop configuration with feedback coefficient of β=1?

Fّor β=1 PM=66, ωt=525 M Rad/s

16

d) What is the phase margin for β=0.1?

Fّor β=0.1 PM=89, ωt=49.5 M Rad/s

17

Fّor β=0.1 PM=89, ωt=49.5 M Rad/s

Fّor β=1 PM=66, ωt=525 M Rad/s

It is considered that for greater values of β, the stability of the closed-loopsystem decreases.

It is considered that for greater values of β, the bandwidth of the closed-loopsystem increases.

MHz4782πCgω

GHz1.192πCC

gω

kHz3542πωωAω

MHz1602πsRadG1

1p1m

Cgω

dB53dB

A450RRggA

C

m7z

21

m7p2

p1p10ta

C

m1ta

021m7m10

Example2

18

Consider a two-stage opamp with the following characteristics:

pF1CpF,0.3CpF,0.1Ck15Rk,10RVmA3g,

VmA1g

C21

21

m7m1

a) Calculate A0 , ωp1, ωp2, ωz, and A(s).

96

9

107.5s1

102.2s1

103s1

450sA

19

b) What is the phase margin for feedback coefficient of β=1?c) What is the phase margin for feedback coefficient of β=0.1?

Fّor β=1 PM=63, ωt=1.04 G Rad/s

Fّor β=0.1 PM=89, ωt=99 M Rad/s

It is considered that if CC is decreased, the bandwidth of the system is increased while the stability problem is worsened.

20

Fّor β=1 PM=63, ωt=1.04 G Rad/s

Fّor β=0.1 PM=89, ωt=99 M Rad/s

Comparison between the Example1 and Example2

Fّor β=0.1 PM=89, ωt=49.5 M Rad/s

Fّor β=1 PM=66, ωt=525 M Rad/s

Example1: (CC=2 pF)

Example2: (CC=1 pF)

The compensation of the opamp of Example2 is performed better than that of the Example1 because it leads to a higher bandwidth and enough phase margin.

Frequency Response (second-order model)

21

C

mz

mp

Cmp C

gCC

gCRRg

7

21

72

2171 ,,1

•A problem arises due to the right-half-plane zero, ωz . It introduces negative phase shift (phase lag).•The right-half-plane zero makes the stability more difficult.•Fortunately by using the resistor, RC, in series with CC, this problem is mitigated.

Right-Half-Plane Zero Problem

22

Cm

C

zpm

pCm

p

Rg

CCC

gCRRg

7

321

72

2171

11,,,1

There are three approaches to solve the right-half-plane zero problem as follows:

1. zero cancellation2. pole-zero cancellation3. lead compensation

Zero cancellation, Pole-zero cancellation

23

Cm

C

zpm

pCm

p

Rg

CCC

gCRRg

7

321

72

2171

11,,,1

In order to eliminate the right-half0plane zero, we should choose the value of RC as follows:

Pole-zero cancellation is performed if ωp2=ωz. After a few manipulation, the condition for pole-zero cancellation is obtained as follows:

Lead compensation

24

Cm

C

zpm

pCm

p

Rg

Cωω

CCgω

CRRgω

7

321

72

2171

11,,,1

The pole-zero cancellation method is not reliable because C2 is often not known a priori. David Johns recommends the lead compensation method as follows:

C

mtat C

g 1 By using and , the value of RC is given as follows:tz 7.1

171 7.111

7.11

mmmC gggR