1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf ·...

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1.0 CMOS OPAMP 1.1. General Opamp Block Diagram A1 -A2 A3 Cc V 1 V 2 V O V in Differential stage gain Inverter stage gain Buffer stage gain Figure 1. Opamp General Block Diagram Note the second stage needs to have a negative gain because of the feedback capacitor. With a positive gain an oscillator or unstable system will be implemented instead. The last stage is a buffer to convert the high output impedance of the inverter stage to a low output impedance. This is needed in developing a good opamp. Ideal opamp requires r in =and r o =0. Figure 2. The Effect of Compensation Capacitance Cc. At the input side, A) 1 ( sC 1 I V ; sC 1 I A) 1 ( V ; sC 1 I AV V ; sC 1 I V V c 1 1 c 1 1 c 1 1 1 c 1 2 1 + = = + = + = This is known as the miller capacitance. The feedback capacitance value appears at the input side of the gain stage with a value magnified by (1+A). At the output side, 1 >> A when ; sC 1 ) A 1 1 ( sC 1 I V ; sC 1 I ) A 1 1 ( V ; sC 1 I A V V ; sC 1 I V V c c 2 2 c 2 2 c 2 2 2 c 2 1 2 = + = = + = + = 1

Transcript of 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf ·...

Page 1: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

1.0 CMOS OPAMP

1.1. General Opamp Block Diagram

A1 -A2 A3

Cc

V1 V2VOVin

Differentialstage gain

Inverterstage gain

Bufferstage gain

Figure 1. Opamp General Block Diagram Note the second stage needs to have a negative gain because of the feedback capacitor. With a positive gain an oscillator or unstable system will be implemented instead. The last stage is a buffer to convert the high output impedance of the inverter stage to a low output impedance. This is needed in developing a good opamp. Ideal opamp requires rin =∞ and ro=0. Figure 2. The Effect of Compensation Capacitance Cc. At the input side,

A)1(sC1

IV ;

sC1

IA)1(V ;

sC1

IAVV ;

sC1

IVV

c1

1

c1

1

c1

11

c1

21

+==

+=

+=

This is known as the miller capacitance. The feedback capacitance value appears at the input side of the gain stage with a value magnified by (1+A). At the output side,

1>>A when ;sC

1

)A11(sC

1IV ;

sC1

I

)A11(V

; sC

1I

AVV

;sC

1I

VV

cc

2

2

c2

2

c2

22

c2

12 =+

==+

=+

=−

1

Page 2: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

Due to Miller effect, the load of the first stage is effectively the compensation capacitance Cc magnified by (1+A2)

A1Vin V1

Cc(1+A2)=Ceq =A2Cc

Figure 3. The Miller Capacitance Loading the First Stage. The gain

o1m2o1m1in

1 ZgZgVV=A1 =−=

where:

Z rds2 / /rds4 / /1

sC1

sC s(1+ A2)C sA2Co1eq eq C C

≅ = ≅

1 1

The voltage gain of the opamp, assuming the output buffer has unity gain (A3=1),

sCcg

sCcgA3

sA2CcgA3A2=A3A2A1

VV

=Av(s) m2m2m2

in

o ==

=

The unity gain bandwidth, wGB is the radian frequency when the gain is 1. That is,

1Ccjw

g)Av(jwAv(s)

GB

m2GB ===

Solving for wGB,

C

m2GB C

gw =

1.2. Slew Rate Determination

2

Page 3: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

VDD

M5

M1 M2

M3 M4

-A2

CcVin- Vin+

V1 V2

Vbias

Differential stage gain

Inverter stage gain

Figure 4. The Differential Gain Stage of the Opamp. Slew rate is the maximum rate at which the output changes when input signals are large. When vin+ >>0 (vin-<<0) is large positive (negative), M1 is on and M2 is off. Hence ISD1=ISD5, since ISD2=0. The current flows through M3 which is then mirrored to M4, therefore ISD1=IDS3=IDS4=ISD5. The current in the compensation capacitor can only flows through M4, since M2 is off and the input impedance of the second stage A2 is ∞. That is ICc=IDS4=ISD5. On the other hand, when vin+<<0 (vin->>0) is large negative (positive), M1 is off and M2 is on. Hence, ISD1=0 which flows through M3 and mirrored to M4, therefore IDS3=IDS4=0. That is, current source ISD5 flows through M2, then to Cc directly, since M4 is off. In both cases, the maximum current that flows through Cc is ISD5. The output voltage of the opamp is approximately equal to v2, since A3≅1.

Cc

2ICcI

tCcq

t)(V

=SR SD1SD5o ==

=d

d

dd

1.3. OpAmp First Order Model

The objective of compensation is to make the opamp to behave as a single pole in the frequency of interest. That is, at least within the unity gain bandwidth frequency, wGB. The desired transfer function is

3

Page 4: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

1

VO

ps1

A=Av(s)

+

where: AVO is the dc gain of the opamp, and p1 is the real axis dominant pole.

DB

w

AVO

wBW=p1 wGB

-20db/dec

Figure 5. Bode Plot of Opamp First Order Model The relation between the unity gain bandwidth, wGB, and the dominant pole p1 can be obtained by the finding the radian frequency when the magnitude of the voltage gain is equal to 1. That is,

1

pw

j

A

pw

j+1

A)Av(jw

1

GB

VO

1

GB

VOGB =≅=

1VOGB pAw ≅ At midband frequency, the gain is approximately given by :

s

ws

pA

ps

A

ps+1

A=Av(s) GB1VO

1

VO

1

VO ==≅

1.4. An Opamp with Feedback

4

Page 5: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

Av(s) VoVin+

B

Figure 6. Ideal Feedback Configuration.

V A V A VO V in V= − Oβ V A AO V V( )1+ = Vinβ

Hence, the closed loop response is given by:

GB

GB

GBCL

ws1

11/sw+1

/swAv(s)+1

Av(s))s(A

ββββ +

===

where: (1/β) is the dc gain of the closed loop system. LG(s)=βAv(s) is the loop gain. The frequency when the closed gain is 3 Db down occurs when the LG becomes unity. That is,

1jww

LG(s)3db-

GB == β

That is, 1 ; www GBGB3db- === ββ 1.5. Settling Time Determination Opamps are compensated to behave as a simple pole system up to a radian frequency of wGB. This is the case whenever the PM of the Opamp is designed for at least 60°. The settling time is the time needed for the output of the opamp to reach a final value to within a predetermined tolerance. This can be determined using the first order model of the opamp given by:

)p

(1

A)(A

1

VOV ss

+=

The unity feedback (β=1) closed loop gain is given by:

5

Page 6: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

1VO

1

VO

VO1VO

VO

1VOVOVO

1

VO

V

VCL

pA1

)A

11(

)A

11(pA1

A11

1

pA)

A11(

1

Ap

1

A)(A1

)(A)(A

ss

sssss

+

+≈

++

+=

++=

++=

+=

The step response is obtained as follows:

)pA

1(

)A

11()U()(A)(V

1VO

1

VOCLO ss

sss+

+==

Taking the inverse Laplace Transform, one gets,

ε−=−=+=∞

−+=

−−

1A

11)A

1(1)(V

u(t))1()A

1(1t)(V

VO

1

VOO

tw1

VOO

GBe

For large AVO,

ε−=∞≈−=

−=−

1)(V)1()t(V

u(t))1(t)(V

Otw

sO

twO

sGB

GB

e

e

Solve for ts in term of ε

)1ln()1ln(pA

1)1ln(w

1)(t1VOGB

s ετ

εεε ===

For a step input the output initially slews, then slowly settles toward its final value, determined by the gain bandwidth product. ε ts (settling time) <1% 4.6τ ≅ 5τ <.5% 5.3τ ≅ 6τ

6

Page 7: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

1.6. Opamp Second Order Model

For compensation of an Opamp, we need at least a second order model

)s/p1)(s/p+(1

A=Av(s)

21

VO

+

DB

w

AVO

p1 wGB

-20db/dec

p2

-40db/dec

Figure 7. Bode Plot of Opamp Second Order Model. For frequency w>>p1, the model reduces to:

)s/p1(s

w)s/p1(s

pA)s/p1)((s/p

AAv(s)

2

GB

2

1VO

21

VO

+=

+=

+≅

The loop gain for this second order model is given by:

)s/p1(s

w=Av(s)=LG(s)

2

GB

β

The radian frequency at 3db down is obtained by setting the magnitude of LG(s) to 1.

7

Page 8: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

1)/pjw1(jw

w=)LG(jw

23db-3db-

GB3db- =

When p2>>w-3db, then

1 ; ww

)p/w(1w

=w 3db-3db-2

23db-3db-

GB ==≅+ βββ

Phase margin is the phase angle between the loop gain phase angle and -180 degree. It is a measure of stability. First let us determine the loop gain phase angle:

)jw/p+(jw)(1

w=LG(jw)

2

GBβ

)w/p(tan-90=LG(jw))( 2

-1−Θ o

The phase margin is computed as follows: )/pw(tan90=(-180)-))LG(jw(=PM 23db-

-13db- −Θ o

PM)tan(90pw 23db- −= o

The closed loop gain is given by:

2

21VOVO

21

CL0CL

s)pp)(A1(

1+sA1

)p/1(1/p+1

AAv(s)+1

Av(s))s(A

βββ

+++

==

where:

1A ;1A+1

AA VO

VO

VOCL0 >>≅=

ββ

Comparing with general second order transfer function:

2

200

VO2

s)(w

1+sQw

1+1

A=s)(H

where: wo is the resonant frequency, and Q is the Q factor. We obtain, 2GB21VO0 pw)pp)(A+(1w ββ ≅=

8

Page 9: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

2

GB

2

1VO

22

1

1VO

21

21VO

21

21VO

21

VO0

pw

ppA

)ppp(

)pA1(

)p(p)pp)(A+(1

)p/1(1/p)pp/()A+(1

Q

)p/1p/1(A1

Qw

βββ

ββ

β

=≅+

+=

+=

+=

++

=

The step response overshoot is a function of Q,

1Q4 2

100=overshoot% −

−π

e That is, when Q=0.5 there is no overshoot (%overshoot=0). In addition, wo = w-3db and PM≈65° PM wp2/wGB Q %overshoot 55° 1.4285 0.925 13.3% 60° 1.733 0.817 8.7% 65° 2.1459 0.717 4.7% 70° 2.748 0.622 1.4% 75° 3.733 0.527 0.008% For no overshoot, select PM=75°. Note that Q is proportional to β. That is the worst case PM occurs when β=1. Thus for opamp compensation where 0<β(s)<=1 is frequency dependent , if the opamp is compensated for β=1, it is guaranteed to be stable for all other β. 1.7. Small Signal Equivalent Circuit

9

Page 10: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

V- V+

Y c (9)(6)

(7)

(1) (2)

(5)

(3) VDD

(4) VSS

M1 M2

M3 M4

M5

M6

M7

Figure 8a. Schematic Diagram of Unbuffered Opamp Circuit.

10

Page 11: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

g ds2

g ds4

g ds6

g ds7

+

-

Vo

+

-

YC

V1a=vid

g ds2

g ds4

g ds6

g ds7

+

-

Vo

+

-

YC

V1a=vid

g m2V

1a

g m6V

1b

V2a=V1b+

-

g m1V

1a

V2a=V1b

+

-

g m6V

1b+

-

V1a=vid

+

-

g m2V

1a

g m6V

1b

V2a=V1b+

-R1 C1 R2 C2V2b=Vo

(a)

(b)

(c)

YC

Vx

Vx

Vx

I3

I2I1

V1 V2

+ +

Y1

Y3

Y2

g m2V

1

VX

g m6V

X

(d)

C2C1

C1

C2

Figure 8. Opamp Small Signal Equivalent Circuit of the First Two Stages. In Figure 8(a) each amplifier stage is replaced by its equivalent circuit. C1 and C2 are the parasitic capacitances of the Opamp at node 6 and 9 respectively. The second stage of an opamp is an inverter stage. The overall gain of the opamp can be made positive, by switching the positive and negative inputs of the differential gain stage. That is the positive input is now the gate of M2. Figure 8(b) shows this by reversing the polarity of the first controlled current source and replacing gm1 by gm2 to indicate M2 gate is the positive input. Figure 8(c) simplify the equivalent circuit by combining the resistances and capacitances.That is,

11

Page 12: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

DS676ds7ds62

SD242ds4ds21

I)(1

gg1R

I)(1

gg1R

λλ

λλ

+=

+=

+=

+=

From Figure 8(c), the port current equations are derived to obtain the Y parameters:

x3m6232x2322xm62

1

)VY-(g)VY(Y)V-(VYVYVgI0I

++=++==

At node xV

231

31

31

m2x

231m2x31

2x31m2x1

x3

x13

2x31m23

VYY

YV

YYg-V

0VY-Vg)VY(Y0)V-(VYVgVY

,Vfor solve and I SubstituteVYI

0)V-(VYVgI

++

+=

=++=++

==++

Substituting to xV 2I

231

3m63231211

31

m2m63

231

31

31

m23m62322

VYY

YgYYYYYYV

YY)gg-(Y

VYY

YV

YYg-)Y-(g)VYY(I

++++

++

=

+

++

++=

The resulting Y-parameter matrix is

323121

31

3m6

31

m2m63

YYYYYY whereYY

YgYY)gg-Y(

00Y

++=∆

++∆

+=

The voltage gain is

3m6

m23m6

L22

21

1

2v Yg

)gY-g(Yy

y-VVA

+∆=

+==

12

Page 13: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

Case1: 0Cor 0,Y c3 ==

21

222

111

YYsCGYsCGY

=∆+=+=

Solving for the voltage gain,

)CsR1()CsR1(GGgg

)sCG()sCG(gg

YYgg

Yg)gY-g(

Yyy

VVA

221121

m6m2

2211

m6m2

21

m6m2

3m6

m23m6

L22

21

1

2v ++

=++

==+∆

=+

−==

)ps-(1)

ps-(1

As)CR+s)(1CR+(1

RRggA

21

V0

2211

21m2m1v ==

where: 21m2m1V0 RRggA =

11

1 CR1p −

=

22

2 CR1p −

=

Case2: C33 sCsCY ==

323221C

2C31232121

3223112211323121

CCCCCC wheres)]sC(CG)C(C[GGG

)sC()sCG()sC()sCG()sCG()sCG(YYYYYY

++=∆∆+++++=

++++++=++=∆

Solving for the voltage gain,

C212

321m6131232

m6

321m6m2

3m62

C31232121

m23m6

3m6

m23m6v

RRs]CRRg)RC(C)RCs[(C1

)gC

s-1(RRgg

Csgs)]sC(CG)C(C[GGG)gsC-g(

Yg)gY-g(

A

∆++++++=

+∆+++++=

+∆=

Compare the denominator with,

21

2

121

2

2121 pp1s

p1s1

pp1s)

p1

p1s(1)

ps-(1)

ps-(1=D(s) +−≅++−=

13

Page 14: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

C21m6C21m61C12C2

1 CRRg1

CRRgR)C(CR)C(C1p −

≅++++

−≅

L21C12L

m6

2

m6

C1C221

Cm62 C C C C C C ;

Cg

Cg

CCCCCCCg-

p ≈>>>>−

=−

=++

C

m61 C

gz =

AV0-20db/dec

-40db/dec

-20db/dec

wp1 p2 zwGB

Figure 9. Bode Plot of Second Order System with RHP zero For good stability a phase margin of at least 60° is desirable This can be achieved as follows:

GB

V01V0GB

w10zA ;pAw

≥∞≈=

Then

14

Page 15: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

m2C

Lm6

GB2

gCC2.2g

orw2.2p

=

=

This is shown below: That is, the transfer function has two poles and one RHP zero. The RHP zero contributes a negative phase angle.

)p

1)(p

1(

)z

1(A)(A

21

V0

V ss

s

s++

−=

The phase angle at the unity gain radian frequency (wGB) is

)p/w(tan)p/w(tan)/w(tan)w(A 2GB1

1GB1

GB1

GBV−−− −−−=∠ z

The phase margin PM is given by:

m2C

Lm6

C

m2

L

m6

GB2

2GB1

2GB1

2GB1

V011-

2GB1

1GB1

GB1

GBV

gCC2.2g

Cg2.2

Cg

2.2wp3.24)p/w(tan

)]p/w(tan907.5[180)]p/w(tan)A(tan)1.0([-tan180 60

)]p/w(tan)p/w(tan)/w(tan[180)w(A180PM

=

=

==

−−−+≈−−+=

=−−−+=∠+=

−−−

−−−

o

ooo

z

To achieve the assumption z=10wGB , the compensation capacitance Cc must be selected properly.

LC

m2C

Lm6

C

m2

L

m6GB2

m2m6C

m2

C

m6GB

0.22CC Combining

gCC2.2g

Cg2.2

Cg

;2.2wp

10gg Cg

10Cg

;10wz

>⇒>>

=⇒==

In addition, LCC

m6

L

m62 CC

Cg

Cgzp <⇒<⇒< . Hence,

LCL CC0.22C <<

Case 3: 33

333

CC33333 sCG

CsGY ;

sC1

G1

sC1

G1

sC1RZ

+=+=+=+=

15

Page 16: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

323121C

323121G

33

3321

2321312C1322313G321

33

3322

33

33112211

323121

CCCCCCGGGGGG

:wheresCG

sCCC)sCCGCCG(G)sCGGCGGC(GGG

sCGCsG

)sC(GsCGCsG

)sC(G)sC)(GsC(G

YYYYYY

++=∆++=∆

++++∆+++∆+

=

+

++

+

++++=

++=∆

{ }

321321

32323131C21

33321m6131232

32m6

3333m6m1

V

321321

32323131C21

321m611223321

32321

3m6

333m63m6m1

3321

2321312C33m61322313G321

m1333m63m6

33

33m6

m133

33m6

3m6

m13m6V

CCCRRRa'CCRRCCRRRRb'

CRCRRg)RC(C)RC[(Cc':where

sa'sb'sc'1

sgC

-CR1RggA

CCCRRRa]CCRRCCRRR[Rb

]CRRgCRCR)CRRR[(c:where

asbscs1GGG

sGg

CG-Cg1Ggg

sCCC)sCCGCCG(G)sCGgCGGCGGC(GGG)s]gCG-C(gG[g

sCGCsG

g

)gsCGCsG

-g(

Yg)gY-g(

A

=++∆=

+++++=

+++

+

=

=++∆=

+++++=

+++

+

=

+++∆++++∆++

=

++∆

+=

+∆=

2.0Opamp Design Specification

16

Page 17: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

M10w=300ul=6.6u

M1w=36.6ul=6.6u

M2w=36.6ul=6.6u

V- V+

M5w=30ul=6.6u

M7w=90ul=6.6u

Cc

M3w=5.4ul-6.6u

M4w=5.4ul-6.6u

M6w=32.4ul=6.6u

M9w=300ul=6.6u

M8w=873.6ul=6.6u

Vo(10)

(9)(6)

(7)

(1) (2)

(8)

(5)

(3) VDD

(4) VSS

IB=100uA

Figure 10 Vdd =+5V, Vss=-5V, Ao>=15,000, GB=2π(5MHz), SR = 10V/uS, Ro ≤ 1.46k -4.5 <=CMR<=3, PM>60° SPICE Parameters: Kn=40uA/V2 , Kp=15uA/V2, λ=0.02 (1) Determine ID5 from the SR specification.

Let Cc=1pF

C

SD5

CI

=SR

uA52uA10

2I

II

333.56)(0.5)-E15(

6)-E10(2VK

I2LW

0.5|-1|3.5--5|V|-V-V|V|-VV10uA12)-6)(1E-(10E)SR)(C(I

SD5SD2SD1

22SD5(SAT)P

SD5

5

TP0biasDDTP0SG5SD5(SAT)

CSD5

====

===

=======

(2) Determine (W/L)1 from the wGB and positive CMR specifications

17

Page 18: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

From wGB specification,

31.4umho5E6)12)(2-E1(wCg

IW/L)(K2C1

Cg=w

GBCm1

SD11PCC

m1GB

===

=

π

57.66)-6)(5E-2(15E

6]-31.4E[IK2)g(

LW 2

SD1P

2m1

1

===

From the positive CMR specification,

G1(max)SD5(SAT)DDSG1

SG1SD5(SAT)DDG1(max)

V-V-VV

V-VVV

=

−=

From the gate bias, Vbias, VSD5 can be determined.

666.26)(0.5)-E15(

6)-E5(2VK

I2LW

0.5|-1|1.5-|V|-VV

5.135.05V-V-VV

22SD1(SAT)p

SD1

1

TP0SG1SD1(SAT)

G1(max)SD5(SAT)DDSG1

===

===

=−−==

To satisfy both specifications select the higher (W/L) ratio. For matching and symmetry, we also choose (W/L)2 = (W/L)1=6.57 (3) Determine (W/L)3=(W/L)4 ratio from negative CMR

422

SSG1(min)N

DS3

3 LW1

(-5))-6)(-4.5-(40E6)-E5(2

)V-V(KI2

LW

====

(4) Determine (W/L)6 from the PM>60 specification. )p/w(tan-z)/w(tan-90PM 2GB

-1GB

-1=

21C221

m6 2

C

m6

C

m2GB CCC since ; pz ;

CCg

p ; Cg

z ; Cgw +><

+===

A pessimistic estimate of PM is obtained by assuming 2pz = . That is,

18

Page 19: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

>

<

−<

−=

<

2PM90tan

gg

2PM90tan

gg

2PM90

ggtan

ggtan290

/Cg/Cg

2tan-90PM

m2m6

m6

m2

m6

m21-

m6

m21-

Cm6

Cm21-

To achieve PM>60,

6-120E6-E122.11760)/2)-tan((90

6E4.31g m6 ≈=−

>

From step 4 VDS6(SAT) =VDS3(SAT) =0.5V

66)(0.5)-E40(

6-E120VKg

LW

DS6(SAT)N

m6

6

===

The current through M6 is given by

uA306)(6)-E40(26)-E120(

W/L)(2Kg

I2

6N

2m6

DS6 ===

For balance condition the current through M6 must be properly ratioed with the current through M3,

30uAuA)5(16I

(W/L)W/L)(

I DS33

6DS6 ===

(5) Determine (W/L)7 from the balance condition and that ISD7=IDS6

16999.15)333.5(uA10uA30W/L)(

II

(W/L) 5SD5

SD77 ≈===

(6) DC gain AVO is computed and compared with the specification:

000,15700,156)-.02)(30E6)(.02-E5)(02.02(.

6)-6)(120E-(31.4E

I)(I)(gg

)gg)(g(ggg

RRggADS676SD242

m6m2

ds6ds5ds4ds2

m6m221m6m2V0

≥=++

=

++=

++==

λλλλ

19

Page 20: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

(7) From the output resistance specification Ro ≤ 1.46k, (W/L)8 can be determined. The output resistance is given by:

SD88PSD88m8

O I(W/L)K21

I21

g1R ===

β

Solving for (W/L)8 , assuming ISD8 = 100uA and Pspice parameter KP=15 uA/V2

1566)-E100(3)+6)(1.46E-E15(2

1IRK2

1(W/L) 2SD8

2OP

8 ===

(8) Determine the (W/L) of the current mirrors. First set up the current source M10 to 100uA using the biasing current source IB. The VGS8 is set up to guarantee it operates at saturation by adding -0.5v beyond its threshold voltage of Vtp=-1v . That is VSG10 = 1.5v. (W/L)10 can now be determined as follows:

33.53(-1))-6)(1.5-(15E

6)-E100(2|)V|V(K

I2W/L)( 22

tpSG10P

SD1010 ==

−=

The rest of current sources are determined by proportionality as follows: ISD9 = 100uA = > (W/L)9 = (W/L)10(ISD9/ISD10) = (53.33)(100uA/100uA)=53.33 This complete the design. The results is summarized in the following table: PAR M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 I(Ua) 5 5 5 5 10 30 30 100 100 100 T P P N N P N P P P P W/L 6.57 6.57 1 1 5.333 6 16 156 53.33 53.33 W(u) 36.792 36.792 5.6 5.6 29.865 33.6 89.6 873.6 298.65 298.65 L(u) 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 Leff(u) 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6

Finding the W, L to the nearest multiple of λ=0.6 length. PAR M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 I(Ua) 5 5 5 5 10 30 30 100 100 100 T P P N N P N P P P P W/L 6.57 6.57 1 1 5.333 6 16 156 53.33 53.33 W(u) 36.6 36.6 5.4 5.4* 30* 32.4* 90* 873.6 300 300 L(u) 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 Leff(u) 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6

20

Page 21: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

* Adjusted to satisfy the balance condition to minimize the input offset voltage, Vos:

6630902

4.54.32

(W/L)W/L)(

2(W/L)W/L)(

5

7

4

6

=

=

=

3.0 Opamp Simulation 3.1. DC Offset Voltage and Differential Gain Determination * Filename="opamp11.cir" * MOS Diff Amp with PMOS Input and Current Mirror Load * Input Signals VID 11 0 DC 0V E- 1 12 11 0 -0.5 E+ 2 12 11 0 0.5 VIC 12 0 DC 0V * Power Supplies VDD 3 0 DC 5VOLT VSS 4 0 DC -5VOLT * Netlist for OpAmp Offset Measurement M1 7 1 5 5 PMOS1 W=36.6U L=6.6U M2 6 2 5 5 PMOS1 W=36.6U L=6.6U M3 7 7 4 4 NMOS1 W=5.4U L=6.6U M4 6 7 4 4 NMOS1 W=5.4U L=6.6U M5 5 8 3 3 PMOS1 W=30U L=6.6U M6 9 6 4 4 NMOS1 W=32.4U L=6.6U M7 9 8 3 3 PMOS1 W=90U L=6.6U M8 4 9 10 10 PMOS1 W=873.6U L=6.6U M9 10 8 3 3 PMOS1 W=300U L=6.6U M10 8 8 3 3 PMOS1 W=300U L=6.6U Cc 9 6 1pF *Bias current IB 8 4 100UA * SPICE Parameters

21

Page 22: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

.MODEL NMOS1 NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL PMOS1 PMOS VTO=-1 KP=15U + GAMMA=0.6 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 *Analysis .OP .DC VID -300uV 300uV .1uV .TF V(10) VID .PROBE .END NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( 1) 0.0000 ( 2) 0.0000 ( 3) 5.0000 ( 4) -5.0000 ( 5) 1.3115 ( 6) -3.4878 ( 7) -3.4878 ( 8) 3.5084 ( 9) -1.1378 ( 10) .1495 ( 11) 0.0000 ( 12) 0.0000

The output voltage Vo=V(10) is 0.145V at quiescent operating point. This non-zero output voltage can be corrected or reduced by applying an input offset voltage, Vos. This offset is determined by finding the value of VID that makes the output

22

Page 23: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

voltage Vo=V(10)=0. From the above graph, this is equal to –8.2484uV. The resulting output voltage from simulation is –2.753E-6V, as shown below: Modify the VID line in the netlist as follow: VID 11 0 DC –8.2484uV NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( 1) 4.124E-06 ( 2)-4.124E-06 ( 3) 5.0000 ( 4) -5.0000 ( 5) 1.3115 ( 6) -3.4864 ( 7) -3.4878 ( 8) 3.5084 ( 9) -1.2881 ( 10)-2.753E-06 ( 11)-8.248E-06 ( 12) 0.0000 The differential gain can be obtained from the DC transfer characteristic by locating two points centered about the bias point of Vid=Vos=-8.2484uV. For accuracy, the two points must be far apart. From the graph:

4E833.16)-(-169.4E-6)-E5.225(

)8356.2(4048.4A vd +=−−

=

This is very closed to the value obtained using TF analysis of 1.810E+4, as shown below: **** SMALL-SIGNAL CHARACTERISTICS V(10)/VID = 1.810E+04 INPUT RESISTANCE AT VID = 1.000E+20 OUTPUT RESISTANCE AT V(10) = 1.342E+03 3.2. Common-mode Voltage Range Determination * Filename="opamp24.cir" * MOS Diff Amp with PMOS Input and Current Mirror Load * Input Signals VID 11 0 DC –8.2484uV E- 1 12 11 0 -0.5 E+ 2 12 11 0 0.5 VIC 12 0 DC 0V * Power Supplies VDD 3 0 DC 5VOLT VSS 4 0 DC -5VOLT * Netlist for Measuring Input CMR M1 7 1 5 5 PMOS1 W=36.6U L=6.6U M2 6 2 5 5 PMOS1 W=36.6U L=6.6U

23

Page 24: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

M3 7 7 4 4 NMOS1 W=5.4U L=6.6U M4 6 7 4 4 NMOS1 W=5.4U L=6.6U M5 5 8 3 3 PMOS1 W=30U L=6.6U M6 9 6 4 4 NMOS1 W=32.4U L=6.6U M7 9 8 3 3 PMOS1 W=90U L=6.6U M8 4 9 10 10 PMOS1 W=873.6U L=6.6U M9 10 8 3 3 PMOS1 W=300U L=6.6U M10 8 8 3 3 PMOS1 W=300U L=6.6U Cc 9 6 1pF * Bias current IB 8 4 100UA * SPICE Parameter .MODEL NMOS1 NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL PMOS1 PMOS VTO=-1 KP=15U + GAMMA=0.6 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 * Analysis .DC VIC -5V 5V .05V .TF V(10) VIC .PROBE .END

24

Page 25: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

The common-mode input range is determined when the output cease to follow the input linearly. At the low end, it determines VG1(min)=-4.4255, and at the high end, it determines VG1(max)=3.1383. These are very closed to the corresponding design specifications of VG1(min=-4.5 and VG1(max)=3. The common-mode gain is obtained by locating two points as far apart in the linear range about the operating point. This is shown in the graph above, and computed as follows:

49373.0)3192.4(1915.3)0677.2(6406.1Acm =

−−−−

=

This is very closed to the value obtained using the TF analysis of 0.4984, as shown below: **** SMALL-SIGNAL CHARACTERISTICS V(10)/VIC = 4.984E-01 INPUT RESISTANCE AT VIC = 1.000E+20 OUTPUT RESISTANCE AT V(10) = 1.342E+03 The common mode reject ration CMRR is calculated as follows:

5.3712549373.0

4E833.1AA

CMRRcm

vd =+

==

3.3. Opamp Frequency Response * Filename="opamp33.cir" * MOS Diff Amp with PMOS Input and Current Mirror Load * Input Signals VID 11 0 DC –8.2484uV AC 1V E- 1 12 11 0 -0.5 E+ 2 12 11 0 0.5 VIC 12 0 DC 0V * Power Supplies VDD 3 0 DC 5VOLT VSS 4 0 DC -5VOLT * Netlist for Frequency Response Measurement M1 7 1 5 5 PMOS1 W=36.6U L=6.6U M2 6 2 5 5 PMOS1 W=36.6U L=6.6U M3 7 7 4 4 NMOS1 W=5.4U L=6.6U M4 6 7 4 4 NMOS1 W=5.4U L=6.6U M5 5 8 3 3 PMOS1 W=30U L=6.6U M6 9 6 4 4 NMOS1 W=32.4U L=6.6U M7 9 8 3 3 PMOS1 W=90U L=6.6U M8 4 9 10 10 PMOS1 W=873.6U L=6.6U M9 10 8 3 3 PMOS1 W=300U L=6.6U M10 8 8 3 3 PMOS1 W=300U L=6.6U

25

Page 26: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

Cc 9 6 1pF CL 10 0 2pF *Bias current IB 8 4 100UA * SPICE Parameters .MODEL NMOS1 NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL PMOS1 PMOS VTO=-1 KP=15U + GAMMA=0.6 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 * Analysis .AC DEC 10 0.1HZ 10000MegHz .TF V(10) VID .PROBE .END

Pspice simulation results:

M16.5f ; 63PM ; 4E810.1A GBVO ==+= o **** SMALL-SIGNAL CHARACTERISTICS V(10)/VID = 1.810E+04 INPUT RESISTANCE AT VID = 1.000E+20

26

Page 27: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

OUTPUT RESISTANCE AT V(10) = 1.342E+03 The theoretical bandwidth and gain bandwidth are calculated as follows:

M3.222

E61402zf

E614012)-E1(

6)-E140(Cg

z

M5)318)(15700(fAf

Hz3182

20002pf

200012)-14E6)(1E6)(5E6)(.7-E140(

1CRRg

1-p

M714.6)-E35)(02.02(.

1I)(

1R

M56)-E5)(02.02(.

1I)(

1R

umho140gumho4.31g

z

C

m6

1V0GB

11

C21m21

6762

2421

m6

m2

===

===

===

===

=−

=≅

=+

=+

=

=+

=+

=

==

ππ

ππ

λλ

λλ

Simulation results are : f1 =258Hz, fGB=5.16M, fz and f2 are difficult to obtain accurately from bode plot. 3.4. Slew Rate Measurement * Filename="opamp44.cir" * MOS Diff Amp with PMOS Input and Current Mirror Load * Input Signal VIN 2 0 PWL(0,-5V 10us,-5V 10.01us,5V 30us, 5V 30.01us,-5V 1s, -5V) *Power Supplies VDD 3 0 DC 5VOLT VSS 4 0 DC -5VOLT * Netlist for Slew Rate Measurement M1 7 1 5 5 PMOS1 W=36.6U L=6.6U M2 6 2 5 5 PMOS1 W=36.6U L=6.6U M3 7 7 4 4 NMOS1 W=5.4U L=6.6U M4 6 7 4 4 NMOS1 W=5.4U L=6.6U M5 5 8 3 3 PMOS1 W=30U L=6.6U M6 9 6 4 4 NMOS1 W=32.4U L=6.6U M7 9 8 3 3 PMOS1 W=90U L=6.6U M8 4 9 10 10 PMOS1 W=873.6U L=6.6U M9 10 8 3 3 PMOS1 W=300U L=6.6U M10 8 8 3 3 PMOS1 W=300U L=6.6U Cc 9 6 1pF * Bias current IB 8 4 100UA

27

Page 28: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

* External Component for Slew Rate Measurement VSH 1 10 DC 0V * SPICE Parameters .MODEL NMOS1 NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL PMOS1 PMOS VTO=-1 KP=15U + GAMMA=0.6 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 *Analysis .TRAN .1ns 40us .PROBE .END

The positive slew rate is the slope of the rising edge of the output. This is computed from the two points on the rising edge. That is,

V/us 96.7085.10931.10

)2985.3(4386.3SR =−−−

=+

Similarly, the new slew rate is computed as follows:

28

Page 29: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

V/us 55.10014.30619.306491.37368.2SR - −=

−−−

=

4.0 Op Amp Design Specification Vdd =+5V, Vss=-5V, Ao>=30,000, GB=2π(5MHz), SR = 10V/uS, Ro ≤ 1.46k -4.5 <=CMR<=3 SPICE Parameters: Kn=40uA/V2 , Kp=15uA/V2, λ=0.02 Increasing Overall DC Gain DC gain AVO is computed and compared with the specification:

000,30700,156)-.02)(30E(.02

6)-(120E6)-E5)(02.02(.

6)-(31.4E

I)(g

I)(g

)gg(g

)g(gg

]R][gRg[AAA

DS676

m6

SD242

m2

ds6ds5

m6

ds4ds2

m22m61m2V2V1V0

≤=

+

+

=

+

+

=

+

+

===

λλλλ

Pspice simulation results:

M16.5f ; 63PM ; 4E810.1A GBVO ==+= o Increasing Overall Gain By Increasing A V1

SD2

2

42

P

SD242

SD22P

SD242

m2V1 I

(W/L))(

2KI)(

I(W/L)2KI)(

gAλλλλλλ +

=+

=+

=

The current I is half the bias current and is independent of ( . That is, the gain can be

adjusted by simply adjusting while remains constant. The gain is proportional to SD2 SD5I

I2W/L)

2W/L)( SD2 V1A

2(W/L) . For example to increase the overall gain by 2, one needs to increase (W/L) by a factor of 4.

21 (W/L) ,

PAR M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 I(Ua) 5 5 5 5 10 30 30 100 100 100 T P P N N P N P P P P

29

Page 30: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

W/L 6.57 6.57 1 1 5.333 6 16 156 53.33 53.33 W(u) 4*36.6 4*36.6 5.4 5.4 30 32.4 90 873.6 300 300 L(u) 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 Leff(u) 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6 Increasing Av1 * Filename="opamp331.cir" * MOS Diff Amp with PMOS Input and Current Mirror Load * Increasing Overall Gain By Increasing W/L of M1 & M2 * Input Signals VID 11 0 DC –8.2484uV AC 1V E- 1 12 11 0 -0.5 E+ 2 12 11 0 0.5 VIC 12 0 DC 0V * Power Supplies VDD 3 0 DC 5VOLT VSS 4 0 DC -5VOLT * Netlist for Frequency Response Measurement M1 7 1 5 5 PMOS1 W={4*36.6U} L=6.6U M2 6 2 5 5 PMOS1 W={4*36.6U} L=6.6U M3 7 7 4 4 NMOS1 W=5.4U L=6.6U M4 6 7 4 4 NMOS1 W=5.4U L=6.6U M5 5 8 3 3 PMOS1 W=30U L=6.6U M6 9 6 4 4 NMOS1 W={32.4U} L=6.6U M7 9 8 3 3 PMOS1 W={90U} L=6.6U M8 4 9 10 10 PMOS1 W=873.6U L=6.6U M9 10 8 3 3 PMOS1 W=300U L=6.6U M10 8 8 3 3 PMOS1 W=300U L=6.6U Cc 9 6 1pF *Bias current IB 8 4 100UA * SPICE Parameters .MODEL NMOS1 NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL PMOS1 PMOS VTO=-1 KP=15U + GAMMA=0.6 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 * Analysis .AC DEC 10 0.1HZ 10000MegHz

30

Page 31: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

.TF V(10) VID

.PROBE

.END Before increasing (W/L) of M1 and M2 **** SMALL-SIGNAL CHARACTERISTICS V(10)/VID = 1.810E+04 INPUT RESISTANCE AT VID = 1.000E+20 OUTPUT RESISTANCE AT V(10) = 1.342E+03 After Increasing (W/L) of M1 and M2 by 4 **** SMALL-SIGNAL CHARACTERISTICS V(10)/VID = 3.590E+04 INPUT RESISTANCE AT VID = 1.000E+20 OUTPUT RESISTANCE AT V(10) = 1.342E+03

Although the DC gain increases the PM decreases.

31

Page 32: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

AV0=3.590E+4; PM=29°; fGB=10.596M The Overall Gain Is Independent of V2A The second stage gain can not be used to increase the overall gain but is used to adjust the phase margin as shown below:

DS6

6

76

N

DS676

DS66N

DS676

m6V2 I

W/L)()(

K2I)(

IW/L)(K2I)(

gA

λλλλλλ +=

+=

+=

Since is determined by current mirror, increasing ( increases I proportionately as shown below:

DS6I 6W/L) DS6

constantI

W/L)(I

(W/L) I

(W/L)W/L)(

IDS4

4

DS6

6DS4

4

6DS6 ==⇒=

That is, increasing or does not increase the overall gain, but g is increased by DS6I 6W/L)( m6

DS66 IW/L)(

DS6I. That is, g is doubled by increasing ( by a factor of 2, and by current mirror

is also increase by a factor of 2. Similarly, ( must be doubled, since I = . The phase

margin, which deteriorates when increasing (W/L) or g , can be compensated by increasing .

m6 6W/L)

m2

7W/L)

2

DS6 DS7I

m6g PAR M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 I(Ua) 5 5 5 5 10 2*30 2*30 100 100 100 T P P N N P N P P P P W/L 6.57 6.57 1 1 5.333 6 16 156 53.33 53.33 W(u) 4*36.6 4*36.6 5.4 5.4 30 2*32.4 2*90 873.6 300 300 L(u) 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 Leff(u) 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6 Pspice simulation results:

M7.9f ; 51.3PM ; 4E590.3A GBVO ==+= o **** SMALL-SIGNAL CHARACTERISTICS V(10)/VID = 3.590E+04 INPUT RESISTANCE AT VID = 1.000E+20 OUTPUT RESISTANCE AT V(10) = 1.342E+03

32

Page 33: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

The phase margin can be increased further by increasing both ( and . The desired phase

margin of 65° is approached when and ( are increased by a factor of 4 6W/L) 7W/L)(

6W/L)( 7W/L) PAR M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 I(Ua) 5 5 5 5 10 4*30 4*30 100 100 100 T P P N N P N P P P P W/L 6.57 6.57 1 1 5.333 6 16 156 53.33 53.33 W(u) 4*36.6 4*36.6 5.4 5.4 30 4*32.4 4*90 873.6 300 300 L(u) 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 Leff(u) 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6 The Overall Gain Is Independent of But Adjust Phase Margin V2A * Filename="opamp331.cir" * MOS Diff Amp with PMOS Input and Current Mirror Load * Increasing Overall Gain By Increasing W/L of M1 & M2 * Input Signals VID 11 0 DC –8.2484uV AC 1V E- 1 12 11 0 -0.5 E+ 2 12 11 0 0.5 VIC 12 0 DC 0V * Power Supplies VDD 3 0 DC 5VOLT VSS 4 0 DC -5VOLT

33

Page 34: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

* Netlist for Frequency Response Measurement M1 7 1 5 5 PMOS1 W={4*36.6U} L=6.6U M2 6 2 5 5 PMOS1 W={4*36.6U} L=6.6U M3 7 7 4 4 NMOS1 W=5.4U L=6.6U M4 6 7 4 4 NMOS1 W=5.4U L=6.6U M5 5 8 3 3 PMOS1 W=30U L=6.6U M6 9 6 4 4 NMOS1 W={4*32.4U} L=6.6U M7 9 8 3 3 PMOS1 W={4*90U} L=6.6U M8 4 9 10 10 PMOS1 W=873.6U L=6.6U M9 10 8 3 3 PMOS1 W=300U L=6.6U M10 8 8 3 3 PMOS1 W=300U L=6.6U Cc 9 6 1pF *Bias current IB 8 4 100UA * SPICE Parameters .MODEL NMOS1 NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL PMOS1 PMOS VTO=-1 KP=15U + GAMMA=0.6 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 * Analysis .AC DEC 10 0.1HZ 10000MegHz .TF V(10) VID .PROBE .END Pspice simulation results:

M17.9f ; 61.5PM ; 4E590.3A GBVO ==+= o **** SMALL-SIGNAL CHARACTERISTICS V(10)/VID = 3.590E+04 INPUT RESISTANCE AT VID = 1.000E+20 OUTPUT RESISTANCE AT V(10) = 1.342E+03

34

Page 35: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

5.0 Phase Margin Correction By Moving the Zero from RHP to LHP The location of zero is given by:

)Rg1(C

1

)Rg1(C

1zC

m6C3

m63 −

=−

=

The zero can be moved from the RHP to the LHP by selecting m6

C g1R >> , with this choice the zero

is approximately given by:

CCCR1-z ≈

A zero in the LHP increase the phase margin rather decrease as in the case of RHP zero. The transfer function of an op amp compensated by resistor R and capacitor connected in series is given by: C CC

35

Page 36: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

=

321

V0

V

ps1

ps1

ps1

zs-1A

s)(A

where:

CC

1C3

21

m62

C21m61

CR1 z

CR1p

CCg

p

CRRg1p

−≈

−≈

+−

−≈

321

21

V0

V p p p since ;

ps1

ps1

zs-1A

s)(A <<<<

+=∠+=

=∠

2

GB1-

1

GB1-GB1-oGBV

o

2

GB1-

1

GB1-GB1-GBV

pw

tanp

wtan

zw

tan180)w(A180PM

pw

tanp

wtan

zw

tan)w(A

With extra phase margin introduced by the LHP zero, the phase margin can be increased without increasing the sizes of M6 and M7. To make sure that the op amp behaves as single order system within the gain bandwidth, . One positions the zero 20% (GBw GB1.2wz = ) over the gain bandwidth. In the

previous example, the non-dominant pole was position at p . Where w is the

original gain bandwidth prior to doubling g which double the gain bandwidth. The new phase margin gain be calculated as follows:

'GBGB w2.2=2 1.1w= '

GB

m2

( ) ( ) ( ) 53.87909.0tanAtan833.tan180

1.1ww

tanp

pAtan

1.2ww

tan180

pw

tanp

wtan

zw

tan180)w(A180PM

1-V0

1-1-o

GB

GB1-

1

1V01-

GB

GB1-o

2

GB1-

1

GB1-GB1-oGBV

o

≈−−+≈

+=

+=∠+=

36

Page 37: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

The value of to place the zero is calculated as follows: CR

k25.136)-x31.4E2(2.1

1g2.11R

Cg2.1w2.1

CR1z

m2C

C

m2GB

CC

===

===

PAR M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 I(Ua) 5 5 5 5 10 35 35 100 100 100 T P P N N P N P P P P W/L 6.57 6.57 1 1 5.333 7 18.665 156 53.33 53.33 W(u) 4*36.6 4*36.6 5.4 5.4 30 37.8 105 873.6 300 300 L(u) 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 Leff(u) 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6 5.6

pF1C k,25.13R CC ==

M10w=300ul=6.6u

M1w=4*36.6ul=6.6u

M2w=4*36.6ul=6.6u

V- V+

M5w=30ul=6.6u

M7w=90ul=6.6u

Cc

M3w=5.4ul-6.6u

M4w=5.4ul-6.6u

M6w=32.4ul=6.6u

M9w=300ul=6.6u

M8w=873.6ul=6.6u

Vo(10)

(9)

(7)

(1) (2)

(8)

(5)

(3) VDD

(4) VSS

Rc(6b)(6a)

IB=100uA

Fig 10a * Filename="opamp33z.cir" * MOS Diff Amp with PMOS Input and Current Mirror Load * Input Signals

37

Page 38: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

VID 11 0 DC -8.2484uV AC 1V E- 1 12 11 0 -0.5 E+ 2 12 11 0 0.5 VIC 12 0 DC 0V * Power Supplies VDD 3 0 DC 5VOLT VSS 4 0 DC -5VOLT * Netlist for Frequency Response Measurement M1 7 1 5 5 PMOS1 W={4*36.6U} L=6.6U M2 6a 2 5 5 PMOS1 W={4*36.6U} L=6.6U M3 7 7 4 4 NMOS1 W=5.4U L=6.6U M4 6a 7 4 4 NMOS1 W=5.4U L=6.6U M5 5 8 3 3 PMOS1 W=30U L=6.6U M6 9 6a 4 4 NMOS1 W={32.4U} L=6.6U M7 9 8 3 3 PMOS1 W={90U} L=6.6U M8 4 9 10 10 PMOS1 W=873.6U L=6.6U M9 10 8 3 3 PMOS1 W=300U L=6.6U M10 8 8 3 3 PMOS1 W=300U L=6.6U Cc 9 6b 1pF Rc 6a 6b {13.25k} *Bias current IB 8 4 100UA * SPICE Parameters .MODEL NMOS1 NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL PMOS1 PMOS VTO=-1 KP=15U + GAMMA=0.6 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 * Analysis .AC DEC 10 0.1HZ 10000MegHz .TF V(10) VID .PROBE .END Pspice simulation results:

M294.10f ; 85.5PM ; 4E590.3A GBVO ==+= o **** SMALL-SIGNAL CHARACTERISTICS V(10)/VID = 3.590E+04

38

Page 39: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

INPUT RESISTANCE AT VID = 1.000E+20 OUTPUT RESISTANCE AT V(10) = 1.342E+03 NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( 1) 4.124E-06 ( 2)-4.124E-06 ( 3) 5.0000 ( 4) -5.0000 ( 5) 1.1562 ( 7) -3.4870 ( 8) 3.5084 ( 9) -1.5153 ( 10) -.2260 ( 11)-8.248E-06 ( 12) 0.0000 ( 6a) -3.4842 ( 6b) -3.4842

6.0 Implementation of Rc with NMOS Transistor Transistor M11 has VDS11=0 since no dc bias current flows through its because of capacitor Cc. Therefore, it is operating at the ohmic or triode region. That is, DSDSTNGSND /2]VV-V-(W/L)[VKI = The resistance value implemented by this transistor is givcen by

39

Page 40: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

[ ]

[ ] 65.1)6.0(2)5(4836.3()6.0(211

2V2VV FSBTOTN

=−−−−++=

−++= φφγ

uu

276276.0

1.65)-(-3.4842)-3)(56(13.25E-E401

)V(VRK1W/L)(

)V(W/L)(VK1

VIrR

TNGSCN

TNGSN

1

DS

DDS10C

==+

=−

=

−=

∂∂

==−

M10

VDD

VSS

M1 M2

M3 M4

M5 M7

M6

M9

M8

CcM11

VDD

(6a) (6b)

(8)

(1) (2)

V- V+

(5)

(7)

(4)

(3)

(9)

(10)

IB

* Filename="opamp33m.cir" * MOS Diff Amp with PMOS Input and Current Mirror Load * Input Signals VID 11 0 DC -8.2484uV AC 1V E- 1 12 11 0 -0.5

40

Page 41: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

E+ 2 12 11 0 0.5 VIC 12 0 DC 0V * Power Supplies VDD 3 0 DC 5VOLT VSS 4 0 DC -5VOLT * Netlist for Frequency Response Measurement M1 7 1 5 5 PMOS1 W={4*36.6U} L=6.6U M2 6a 2 5 5 PMOS1 W={4*36.6U} L=6.6U M3 7 7 4 4 NMOS1 W=5.4U L=6.6U M4 6a 7 4 4 NMOS1 W=5.4U L=6.6U M5 5 8 3 3 PMOS1 W=30U L=6.6U M6 9 6a 4 4 NMOS1 W={32.4U} L=6.6U M7 9 8 3 3 PMOS1 W={90U} L=6.6U M8 4 9 10 10 PMOS1 W=873.6U L=6.6U M9 10 8 3 3 PMOS1 W=300U L=6.6U M10 8 8 3 3 PMOS1 W=300U L=6.6U M11 6b 3 6a 4 NMOS1 W=6U L=27U Cc 9 6b 1pF *Rc 6a 6b {13.25k} *Bias current IB 8 4 100UA * SPICE Parameters .MODEL NMOS1 NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL PMOS1 PMOS VTO=-1 KP=15U + GAMMA=0.6 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 * Analysis .PROBE .AC DEC 10 0.1HZ 10000MegHz .TF V(10) VID .END Pspice simulation results:

M9.11f ; 87.7PM ; 4E590.3A GBVO ==+= o **** SMALL-SIGNAL CHARACTERISTICS V(10)/VID = 3.590E+04 INPUT RESISTANCE AT VID = 1.000E+20 OUTPUT RESISTANCE AT V(10) = 1.342E+03

41

Page 42: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( 1) 4.124E-06 ( 2)-4.124E-06 ( 3) 5.0000 ( 4) -5.0000 ( 5) 1.1562 ( 7) -3.4870 ( 8) 3.5084 ( 9) -1.5137 ( 10) -.2244 ( 11)-8.248E-06 ( 12) 0.0000 ( 6a) -3.4843 ( 6b) -3.4843

42

Page 43: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

7.0 Opamp with ClassAB Output Buffer

M10

VDD

VSS

M1 M2

M3 M4

M5M7

M6

M9

M8CcM11

VDD

(6a) (6b)

(8)

(1) (2)

V- V+

(5)

(7)

(4)

(3)

(9)

(10)

M12

M13

(13)

(14)

k*(W/L)p

(W/L)n

(W/L)n

(W/L)p

k*(W/L)n

I k*I

(W/L)p

IB

Figure 12b k=3 * Filename="op33zab2.cir" * MOS Diff Amp with PMOS Input and Current Mirror Load * Input Signals VID 11 0 DC -8.2484uV AC 1V E- 1 12 11 0 -0.5 E+ 2 12 11 0 0.5 VIC 12 0 DC 0V * Power Supplies VDD 3 0 DC 5VOLT VSS 4 0 DC -5VOLT * Netlist for Frequency Response Measurement M1 7 1 5 5 PMOS1 W={4*36.6U} L=6.6U M2 6a 2 5 5 PMOS1 W={4*36.6U} L=6.6U M3 7 7 4 4 NMOS1 W=5.4U L=6.6U

43

Page 44: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

M4 6a 7 4 4 NMOS1 W=5.4U L=6.6U M5 5 8 3 3 PMOS1 W=30U L=6.6U M6 9 6a 4 4 NMOS1 W={32.4U} L=6.6U M7 14 8 3 3 PMOS1 W={90U} L=6.6U M8 4 9 10 10 PMOS1 W={3*90U} L=6.6U M9 3 14 10 4 NMOS1 W={3*32.4U} L=6.6U M10 8 8 3 3 PMOS1 W=300U L=6.6U M11 6b 3 6a 4 NMOS1 W=6U L=27U M12 9 9 13 13 PMOS1 W=90U L=6.6U M13 14 14 13 4 NMOS1 W=32.4U L=6.6U Cc 9 6b 1pF *Bias current IB 8 4 100UA * SPICE Parameters .MODEL NMOS1 NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL PMOS1 PMOS VTO=-1 KP=15U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 * Analysis .PROBE .DC VID -5V 5V .1V .AC DEC 10 0.1HZ 10000MegHz .TF V(10) VID .END Pspice simulation results:

3E113.1R ; M9.10f ; 86.1PM ; 4E065.3A OGBV0 +==°=+= **** SMALL-SIGNAL CHARACTERISTICS V(10)/VID = 3.065E+04 INPUT RESISTANCE AT VID = 1.000E+20 OUTPUT RESISTANCE AT V(10) = 1.113E+03 NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( 1) 4.124E-06 ( 2)-4.124E-06 ( 3) 5.0000 ( 4) -5.0000 ( 5) 1.1562 ( 7) -3.4870 ( 8) 3.5084 ( 9) -3.5234 ( 10) -2.0125 ( 11)-8.248E-06 ( 12) 0.0000 ( 13) -2.0182

44

Page 45: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

( 14) .6098 ( 6a) -3.4843 ( 6b) -3.4843

45

Page 46: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

8.0 Process and Temperature Independent Opamp Design Using Widlar Current Source

VDD

VSS

(8)

(3)

Rb

MB1

MB2(W/L)p

MB3MB4

(W/L)n4(W/L)n

(W/L)p

(15)

(16)

I I

I

Msnk1(W/L)n

Msrc1(W/L)p

(4)

(5)

+Vsnk

Msnk22(W/L)n

I 2I

Msnk33(W/L)n

3I

+Vsrc

2I 3I

Msrc22(W/L)p

Msrc33(W/L)p

(2)

Figure 12

III D4D3 ==

bGS4GS3 IRVV += Subtracting the threshold voltage (neglecting body effect) from both sides, one obtains: TN0V

b4N3N

bTN0GS3TN0GS3

IRW/L)(K

I2W/L)(KI2

IRV-VV-V

+=

+=

46

Page 47: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

Solve for Rb,

b4

3

3N

R(W/L)(W/L)

1I(W/L)K2

2=

or

34b

m3

b

4

3

m3

4(W/L)(W/L) If ;R1g

R(W/L)(W/L)

12g

==

=

Thus is determined by geometric ratio only. It is independent of power supply voltage, process parameter, and temperature. This is only true to the first order of approximation, since the body effect is neglected. In addition all other transconductance are also stabilized since all transistor currents are derived from the same biasing network. That is,

m3g

r transistochannel-pfor ;gI(W/L)KI(W/L)Kg

r transistochannel-nfor ;gI(W/L)I(W/L)

g

m3D33N

DiiPmi

m3D33

Diimi

=

=

Design a 10uA widlar current source with the minimum transistor sizes for biasing an opamp circuit. Assuming all transistors are of the same length L=6.6u. The minimum width corresponding to (W/L)=1 and 6.0=λ is obtained as follows:

5.4uW

12(0.5)-6.6u

W2LD-LW

LW

eff

===

The minimum W for the widlar current source is 2(5.4u)=10.8u, since the differential amplifier input stage load current is half of the widlar design current value. The minimum pmos transistor (W/L) is calculated to achieve the same transconductance. That is,

30u28.8uu)8.10(15u

u40WKK

W

LW

KK

LW

NP

NP

NeffP

N

Peff

≈===

=

The widlar cu * Filename="op33zab4.cir" * Widlar Source = Supply Independent

47

Page 48: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

* Input Signals * Power Supplies VDD 3 0 DC 5VOLT VSS 4 0 DC -5VOLT VSNK 5 0 DC 0VOLT VSRC 2 0 DC 0VOLT * Netlist for Frequency Response Measurement MB1 15 8 3 3 PMOS1 W=30U L=6.6U MB2 8 8 3 3 PMOS1 W=30U L=6.6U MB3 15 15 4 4 NMOS1 W={10.8U} L=6.6U MB4 8 15 16 4 NMOS1 W={4*10.8U} L=6.6U Msrc1 2 8 3 3 PMOS1 W=30U L=6.6U Msrc2 2 8 3 3 PMOS1 W={2*30U} L=6.6U Msrc3 2 8 3 3 PMOS1 W={3*30U} L=6.6U Msnk1 5 15 4 4 NMOS1 W=10.8U L=6.6U Msnk2 5 15 4 4 NMOS1 W={2*10.8U} L=6.6U Msnk3 5 15 4 4 NMOS1 W={3*10.8U} L=6.6U Rb 16 4 20K * SPICE Parameters .MODEL NMOS1 NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL PMOS1 PMOS VTO=-1 KP=15U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 * Analysis .PROBE .OP .DC VSNK -5V 5V .1V *.DC VSRC -5V 5V .1V .END

48

Page 49: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

49

Page 50: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

NAME MB1 MB2 MB3 MB4 Msrc1 MODEL PMOS1 PMOS1 NMOS1 NMOS1 PMOS1 ID -9.89E-06 -8.70E-06 9.89E-06 8.70E-06 -9.30E-06 VGS -1.46E+00 -1.46E+00 1.50E+00 1.33E+00 -1.46E+00 VDS -8.50E+00 -1.46E+00 1.50E+00 8.37E+00 -5.00E+00 VBS 0.00E+00 0.00E+00 0.00E+00 -1.74E-01 0.00E+00 VTH -1.00E+00 -1.00E+00 1.00E+00 1.11E+00 -1.00E+00 VDSAT -4.59E-01 -4.59E-01 4.99E-01 2.20E-01 -4.59E-01 NAME Msrc2 Msrc3 Msnk1 Msnk2 Msnk3 MODEL PMOS1 PMOS1 NMOS1 NMOS1 NMOS1 ID -1.86E-05 -2.79E-05 1.06E-05 2.11E-05 3.17E-05 VGS -1.46E+00 -1.46E+00 1.50E+00 1.50E+00 1.50E+00 VDS -5.00E+00 -5.00E+00 5.00E+00 5.00E+00 5.00E+00 VBS 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 VTH -1.00E+00 -1.00E+00 1.00E+00 1.00E+00 1.00E+00 VDSAT -4.59E-01 -4.59E-01 4.99E-01 4.99E-01 4.99E-01

9.0 Inferred Biasing

50

Page 51: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

VDD

VSS

(8)

(3)

Rb

MB1

MB2(W/L)p

MB3MB4

(W/L)n4(W/L)n

(W/L)p

(15)

(16)

I I

I

Msnk1(W/L)n

Msrc1(W/L)p

(4)

(5) +Vsnk

Msnk(W/L)n

I I

NOTE: V(5)=V(15)

(6)

Figure 12a * Filename="op33zab5.cir" * Widlar Source = Supply Independent * Input Signals * Power Supplies VDD 3 0 DC 5VOLT VSS 4 0 DC -5VOLT VSNK 6 0 DC 0VOLT * Netlist for Frequency Response Measurement MB1 15 8 3 3 PMOS1 W=30U L=6.6U MB2 8 8 3 3 PMOS1 W=30U L=6.6U

51

Page 52: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

MB3 15 15 4 4 NMOS1 W={10.8U} L=6.6U MB4 8 15 16 4 NMOS1 W={4*10.8U} L=6.6U Msrc1 5 8 3 3 PMOS1 W=30U L=6.6U Msnk1 5 5 4 4 NMOS1 W=10.8U L=6.6U Msnk 6 5 4 4 NMOS1 W=10.8U L=6.6U Rb 16 4 20K * SPICE Parameters .MODEL NMOS1 NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL PMOS1 PMOS VTO=-1 KP=15U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 * Analysis .PROBE .OP .DC VSNK -5V 5V .1V *.DC VSRC -5V 5V .1V .END NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( 3) 5.0000 ( 4) -5.0000 ( 5) -3.5010 ( 6) 0.0000 ( 8) 3.5413 ( 15) -3.5010 ( 16) -4.8260

52

Page 53: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

53

Page 54: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

VDD(3)

Rb

MB3MB4

(W/L)n4(W/L)n(16)

I

MB1

MB2(W/L)p

(W/L)p

I

+

MB3a(W/L)nMB4a

(W/L)n

Msnka(W/L)n

Msnkb(W/L)n

Vsnk

MB1a(W/.L)p

MB2a(W/L)p

(4) VSS

(2)

(15a)

(15b)

(8b)

(15a)

(15b)

8a)

(14)

(13)

(12)

10.0 Cascode Widlar Current Sink

54

Page 55: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

Figure 12b * Filename="op33zab6.cir" * Widlar Source = Supply Independent * Input Signals * Power Supplies VDD 3 0 DC 5VOLT VSS 4 0 DC -5VOLT VSNK 2 0 DC 0VOLT * Netlist for Frequency Response Measurement MB1 13 8b 3 3 PMOS1 W=30U L=6.6U MB1a 15a 8a 13 13 PMOS1 W=30u L=6.6U MB2 8b 8b 3 3 PMOS1 W=30U L=6.6U MB2a 8a 8a 8b 8b PMOS1 W=30U L=6.6U MB3 15b 15b 4 4 NMOS1 W={10.8U} L=6.6U MB3a 15a 15a 15b 4 NMOS1 W=10.8U L=6.6U MB4 14 15b 16 4 NMOS1 W={4*10.8U} L=6.6U MB4a 8a 15a 14 4 NMOS1 W=10.8U L=6.6U Msnka 12 15b 4 4 NMOS1 W=10.8U L=6.6U Msnkb 2 15a 12 4 NMOS1 W=10.8U L=6.6U Rb 16 4 15.5K * SPICE Parameters .MODEL NMOS1 NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL PMOS1 PMOS VTO=-1 KP=15U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 * Analysis .PROBE .OP .DC VSNK -5V 5V .1V .TF V(2) VSNK .END **** SMALL-SIGNAL CHARACTERISTICS V(2)/VSNK = 1.000E+00 INPUT RESISTANCE AT VSNK = 1.489E+09 OUTPUT RESISTANCE AT V(2) = 0.000E+00

55

Page 56: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( 2) 0.0000 ( 3) 5.0000 ( 4) -5.0000 ( 12) -3.4910 ( 13) 3.4907 ( 14) -3.4842 ( 16) -4.8434 ( 8a) 2.0117 ( 8b) 3.5058 ( 15a) -1.3184 ( 15b) -3.4956

56

Page 57: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

11.0 Opamp Design with Process and Temperature Independent Implementation 1

VDD

VSS

M1 M2

M3 M4

M5M7

M6

M9

M8CcM11

VDD

(6a) (6b)

(8)

(1) (2)

V- V+

(5)

(7)

(4)

(3)

(9)

(10)

M12

M13

(13)

(14)

k*(W/L)p

(W/L)n

(W/L)n

(W/L)p

k*(W/L)n

I k*I

(W/L)p

Rb

MB1

MB2(W/L)p

MB3MB4(W/L)n

4(W/L)n

(W/L)p

(15)

(16)

Figure 12c * Filename="op33zab3.cir" * MOS Diff Amp with PMOS Input and Current Mirror Load * Input Signals VID 11 0 DC -8.2484uV AC 1V E- 1 12 11 0 -0.5 E+ 2 12 11 0 0.5 VIC 12 0 DC 0V * Power Supplies VDD 3 0 DC 5VOLT VSS 4 0 DC -5VOLT * Netlist for Frequency Response Measurement M1 7 1 5 5 PMOS1 W={4*36.6U} L=6.6U

57

Page 58: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

M2 6a 2 5 5 PMOS1 W={4*36.6U} L=6.6U M3 7 7 4 4 NMOS1 W=5.4U L=6.6U M4 6a 7 4 4 NMOS1 W=5.4U L=6.6U M5 5 8 3 3 PMOS1 W=30U L=6.6U M6 9 6a 4 4 NMOS1 W={32.4U} L=6.6U M7 14 8 3 3 PMOS1 W={90U} L=6.6U M8 4 9 10 10 PMOS1 W={3*90U} L=6.6U M9 3 14 10 4 NMOS1 W={3*32.4U} L=6.6U M11 6b 3 6a 4 NMOS1 W=6U L=27U M12 9 9 13 13 PMOS1 W=90U L=6.6U M13 14 14 13 4 NMOS1 W=32.4U L=6.6U MB1 15 8 3 3 PMOS1 W=30U L=6.6U MB2 8 8 3 3 PMOS1 W=30U L=6.6U MB3 15 15 4 4 NMOS1 W={10.8U} L=6.6U MB4 8 15 16 4 NMOS1 W={4*10.8U} L=6.6U Rb 16 4 19K Cc 9 6b 1pF * SPICE Parameters .MODEL NMOS1 NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL PMOS1 PMOS VTO=-1 KP=15U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 * Analysis .PROBE .DC VID -5V 5V .1V .AC DEC 10 0.1HZ 10000MegHz .TF V(10) VID .END Pspice simulation results:

3E113.1R ; M43.9f ; 88.5PM ; 4E180.3A OGBV0 +==°=+= **** SMALL-SIGNAL CHARACTERISTICS V(10)/VID = 3.180E+04 INPUT RESISTANCE AT VID = 1.000E+20 OUTPUT RESISTANCE AT V(10) = 1.133E+03 NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( 1) 4.124E-06 ( 2)-4.124E-06 ( 3) 5.0000 ( 4) -5.0000

58

Page 59: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

( 5) 1.1533 ( 7) -3.4964 ( 8) 3.5174 ( 9) -3.5295 ( 10) -2.0277 ( 11)-8.248E-06 ( 12) 0.0000 ( 13) -2.0334 ( 14) .5814 ( 15) -3.4753 ( 16) -4.8170 ( 6a) -3.4936 ( 6b) -3.4936

59

Page 60: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

12.0 Opamp Design with Process and Temperature Independent Implementation 2

VDD

VSS

M1 M2

M3 M4

M5M7

M6

M9

M8

MB1(W/L)p

MB3a(W/L)n

MB3

MB2(W/L)p

MB4a(W/L)n

MB44(W/L)n

CcM11

Rb

(6a) (6b) (9)

(10)

(8)

(1) (2)

V+V-

(7)

(5)

(4)

(3)

(13)

(14)

(15)

(16) (W/L)n

Figure 12a * Filename="opamp33b.cir" * MOS Diff Amp with PMOS Input and Current Mirror Load * Input Signals VID 11 0 DC -8.2484uV AC 1V E- 1 12 11 0 -0.5 E+ 2 12 11 0 0.5 VIC 12 0 DC 0V * Power Supplies VDD 3 0 DC 5VOLT VSS 4 0 DC -5VOLT * Netlist for Frequency Response Measurement M1 7 1 5 5 PMOS1 W={4*36.6U} L=6.6U M2 6a 2 5 5 PMOS1 W={4*36.6U} L=6.6U M3 7 7 4 4 NMOS1 W=5.4U L=6.6U M4 6a 7 4 4 NMOS1 W=5.4U L=6.6U M5 5 8 3 3 PMOS1 W=30U L=6.6U

60

Page 61: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

M6 9 6a 4 4 NMOS1 W={32.4U} L=6.6U M7 9 8 3 3 PMOS1 W={90U} L=6.6U M8 4 9 10 10 PMOS1 W=873.6U L=6.6U M9 10 8 3 3 PMOS1 W=300U L=6.6U M11 6b 13 6a 4 NMOS1 W={20.4U} L=6.6U MB1 13 8 3 3 PMOS1 W=30U L=6.6U MB2 8 8 3 3 PMOS1 W=30U L=6.6U MB3a 13 13 14 4 NMOS1 W=10.8U L=6.6U MB3 14 14 4 4 NMOS1 W=10.8U L=6.6U MB4a 8 13 15 4 NMOS1 W=10.8U L=6.6U MB4 15 14 16 4 NMOS1 W=43.2U L=6.6U Cc 9 6b 1pF *External Bias Resistor Rb 16 4 {17.5K} * SPICE Parameters .MODEL NMOS1 NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL PMOS1 PMOS VTO=-1 KP=15U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6 + TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10 + U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 * Analysis .DC VID -5V 5V .1V .AC DEC 10 0.1HZ 10000MegHz .TF V(10) VID .PROBE .END Pspice simulation results:

3E379.1R ; M65.8f ; 83PM ; 4E790.3A OGBV0 +==°=+= **** SMALL-SIGNAL CHARACTERISTICS V(10)/VID = 3.790E+04 INPUT RESISTANCE AT VID = 1.000E+20 OUTPUT RESISTANCE AT V(10) = 1.379E+03 NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( 1) 4.155E-06 ( 2)-4.155E-06 ( 3) 5.0000 ( 4) -5.0000 ( 5) 1.1520 ( 7) -3.5007 ( 8) 3.5216 ( 9) -1.5420

61

Page 62: 1.0 CMOS OPAMP - Wayne State Universitywebpages.eng.wayne.edu/cadence/ECE7570/doc/OPAMP_ADV.pdf · wo is the resonant frequency, and Q is the Q factor. We obtain, w0 = (1+βAVO) ...

( 10) -.2603 ( 11)-8.310E-06 ( 12) 0.0000 ( 13) -1.3042 ( 14) -3.4895 ( 15) -3.4575 ( 16) -4.8343 ( 6a) -3.4979 ( 6b) -3.4979

[ ] [ ][ ] 64.1)6.0(2)5(4979.3()6.0(211

2VSS-V(6a)2V2V2VV FTOFSBTOTN

=−−−−++=

−++=−++= φφγφφγ

uu

6.64.204.3

1.64)-(-3.4979)-3)(-1.30426(13.25E-E401

)VV(6a))13(V(RK1W/L)(

)V)a6(V)13(W/L)(V(K1

)V(W/L)(VK1

VIrR

TNCN

TNNTNGSN

1

DS

DDS10C

≈=+

=−−

=

−−=

−=

∂∂

==−

62