Common mode feedback for fully differential amplifiers

Differential amplifiers

• Cancellation of common mode signals including clock feed-through

• Cancellation of even-order harmonics

• Double differential signal swing, SNR↑3dB

Symbol:

Two-Stage, Miller, Differential-In, Differential-Out Op Amp

Output common mode range (OCMR) = VDD-VSS - VSDPsat - VDSNsat

peak-to-peak output voltage

≤ 2·OCMR

Common Mode Output Voltage Stabilization

Common mode drift at output

causes differential

signals move into triode

region

Common Mode feedback

• All fully differential amplifier needs CMFB

• Common mode output, if uncontrolled, moves to either high or low end, causing triode operation

• Ways of common mode stabilization:– external CMFB– internal CMFB

Common mode equivalent

Vicm

Vo1cm

VBP

VBN

Vocm

I2

I1

VBN

Vocm

Vo1cm I2

I1

Vo1

Vocm

actual Q point too high

PMOS in triode

Correct Vo1cm

Vo1cm a little low

If Vocm is too high, needs to increase Vo1cm

for single stage, needs positive feedback

Ix

Iy

Vo

Iy(Vo)

Ix(Vo)VO1CM

1 1If i.e. need to

From this single stage's point view, need positive feedbacko o CM x BPV V need I V

But we started by having Vocm too high

need neg. feedback from to o BPV V

Vicm

Vo1cm

VBP

VBN

In both cases, the whole loop has negative loop gain

Vicm

Vo1cm

VBP

VBN

Vo

I2

Vicm

I1

What about single ended?Does it have the same problem?Does it require feedback stabilization?

Vi+

Vo1cm

VBP

VBN

VoI2

Vi- I1

Yes, to all three questions

I3I4

I6

I7

To match I1 and I3, the diode connection provides the single stage positive feedback to automatically generate Vg3.The match between I2 and I4, and I6 and I7 is a two stage problem and requires negative feedback: needs feedback from Vo to Vi-.

Vi+

Vo1cm

VBP

VBN

VoI2

Vi- I1

I3I4

I6

I7

Vi+

Vo1cm

VBP

VBN

VoI2

Vi-

I1

I3I4

I6

I7

All op amps must be used in feedback configuration!

Buffer connection or resistive feedback provides the needed negative feedback

Vo+

Vo-

Fully differential amplifiers are also used in feedback configuration.

Vi+

Vi-

Vp

Vn

, i p p o i n n o

i b i b

V V V V V V V V

R R R R

( ) ( )i i p n p n o o

i b

V V V V V V V V

R R

If amplifier gain is high, is 0,

( ), and ( )

p n

i i o o bo o i i

i b i

V V

V V V V RV V V V

R R R

Hence, differential signal is well defined.

Vo+

Vo-

But when you add the first two equations

Vi+

Vi-

Vp

Vn

, i p p o i n n o

i b i b

V V V V V V V V

R R R R

( ) ( )i i p n p n o o

i b

V V V V V V V V

R R

Solving for ,

( ) ( )

o o

b b io o i i p n

i i

V V

R R RV V V V V V

R R

Since Vp+Vn is undefined, Vo++Vo- is undefined.

You get:

CM

measurement

CMFB

+

-VoCM

Voc

Vo+

Vo-

Vo+ +Vo-

2

desired common mode voltage

Basic concept of CMFB:

evb

CM

measurement

CMFB+

-VoCM

Voc

Vo+

Vo-

Vo+ +Vo-

2

Basic concept of CMFB:

evb e

Find transfer function from e to Voc: ACMF(s)Find transfer function from an error source to Voc: Aerr(s)Voc error due to error source: err*Aerr(0)/ACMF(0)

example

Vb2

Vi+ Vi-

VCMFB Vb1

CC CC

Vo+ Vo-

+-

Vo+

Vo-

VocVCMFB

Voc

VoCM

Example

Need to make sure to have negative feedback

?

?

Source follower

averager

BIAS4

1.5pF 1.5pF

M13A M13B

300/3 300/3

20K 20KOUT+ OUT-

M2A M2B150/3 150/3

BIAS3

300/2.25 300/2.25

IN-

M3A

M3B

IN+

300/2.25 300/2.25

M1A M1B

BIAS2

BIAS1

75/3

M7A

75/3

M7B

M6C 75/2.25

75/2.25

M6AB

M11

150/2.25

M8

150/2.25

200/2.25

200/2.25

M10

M5

CL=4pF 4pFM9A

50/2.25

M4A

50/2.25

M9B

50/2.25

M4B

50/2.25

M12A

1000/2.25

M12B

1000/2.25

VDD

VSS

Folded cascode amplifier

R1 R2Vo-Vo+

node cascodedan at

is when especially-

achieve todifficult -

large very , use

effect. loading resistive :Prob

if,2

21

21

o

oo

V

RR

RRVV

Resistive C.M. detectors:

Resistive C.M. detectors:

Vo.c.

R1 R1Vo-Vo+

Vi-Vi+

Not recommended.The resistive loading kill gain.

• O.K. if op amp is used in a resistive feedback configuration

• & R1 is part of feedback network.

• Otherwise, R1 becomes part of g0 & hence reduces AD.C.(v)

Buffer Vo+, Vo- before connecting to R1.

Vo+ Vo-

Voc

R1 R1

Simple implementation:

source follower

Vo-Vo+Vo.c.

* Gate capacitance is added to your amp load.

Why not:

Vo-Vo+

Vo.c.

* Initial voltage on cap.

C1 C2 short diff C A

freq. high at :Prob

if 21,2

CCVV oo

Use buffer to isolate Vo node:

gate cap is load

or resistors

Switched cap CMFB

Vo+

Vo- VoCM.

VoCM.

Φ2 Φ1Φ1

Vb-replVCMFB

Vo+Vo-

Vb-repl

Small copy of

Small copy of

Small copy of

VCMFB

Vxx

Vyy

In the normal balanced case, when Vo+=Vo- = desired Vocm, Vb-repl is supposed to generate the correct voltage for VCMFB

To increase or decrease the C.M. loop gain:e.g.

Vo.c.d.Vo.c.

VC.M.F.B.

Another implementation• Use triode transistors to provide isolation

& z(s) simultaneously.

Vo-Vo+

Voc

M1 M2

can be a c.s.

M1, M2 in deep triode.

VGS1, VGS2>>VT

In that case, circuit above M1, M2 needs to ensure that M1, M2 are in triode.

2 oo VV

deep triode oper

Example:

M1 M2

Vb

Vo+Vo-

Input stage

e.g. Vo+, Vo-≈2V at Q & Vb ≈1V ,

Then M1&2 will be in deep triode.

Vo+Vo-

Vb1

Vb2

M1 M2

VX

oo

ooX

X

MM

GG

oc

oo

VV

VVV

IV

RR

VV

V

VV

,

,

)(

,

,

,

21

21

C.G. cascoded is

to but

const

If

Two-Stage, Miller, Differential-In, Differential-Out Op Amp

M10 and M11 are in deep triode

Vo++ Vo-

2 VoCM.

VCMFB

Vo+

Vo-

large be can gain

CMFBoo VVV

2

Note the difference from the book

accommodates much larger VoCM range

M1

M3

VCM M4

M2

IB IB

Vo+ Vo-

+Δi

-Δi

-Δi

+Δi+Δi

+Δi -Δi

-ΔiVCMFB

M5

Δi=0 2Δi

Small signal analysis of CMFB

Example:

Differential signal

Common mode signal

• Differential Vo: Vo+↓ by ΔVo, Vo-↑ by ΔVo

• Common mode Vo: Vo+↑ by ΔVo, Vo-↑ by ΔVo

o

m

m

mCMFB

mmmmo

m

Vk

g

gki

gV

ggggV

gi

5

1

5

43211

21

2

M1

M3

VCM M4

M2

IB IB

Vo+ Vo-

-Δi

+Δi+Δi

-ΔiVCMFB

M5

Δi=0 2Δi

M6

M7Δi7

-2Δi+

--2Δi

1

gm6

accurate made be canratiosgeometric by gain*

:gain increase To

om

m

m

mCMFB

m

m

m

mmG

mG

Vg

g

g

gV

g

gii

g

gigVi

giV

6

7

5

1

6

75

6

7777

67

1

12

2

12

CMFB loop gain: example

Vb2

Vi+ Vi-

VCMFB Vb1

CC CC

Vo+ Vo-

+-

Vo+

Vo-

VocVCMFB

Bandwidth of CMFB loop• Ideally, if CM and DM are fully decoupled, CM

only needs to stabilize operating points. CM bandwidth only needs to be wide enough to handle disturbances affecting operating points.

• Practically, there is CMDM conversion. CM loop needs to handle disturbances of bandwidth comparable to DM BW

• But CM loop shares most of CM poles and have additional poles, difficult to achieve similar bandwidth, make CM loop bandwidth a few times low than DM

Here Bandwidth = unity loop gain frequency

Example

Vo1

VBP

VBN

Vo

I2

Vi++Vicm

I1

Vi-+Vicm

VBN VCMFB

CM and DM equivalent circuit Comparison

-½Vid

½Vo1d

VBP

VBN

½Vod

I2

I1

AC GND

Vid

Vo1cm

VBP

VBN

Vocm

I2

I1VCMFB½ M5

ro1=rdsp||rcascode≈rdsp

ro1=rdsp||rdsn≈ ½rdsp

gm = ½gm5gm = gm1

Low frequency pole p1 is about 2X lower in CM; DC gain is change by 2*½gm5/gm1, unity gain frequency gm/CC is changed by ½gm5/gm1, high frequency poles and zeros of DM remain in CM, CM has one additional node at D5

similar or worse PM at unity gain fre

CM DM

To ensure sufficient CMFB loop stability

• CMFB loop gain = CM gain from VCMFB to Voc * gain of CMFB circuit

• To ensure sufficient PM for CMFB loop– Make the DC gain of CMFB circuit to be a few

time less than one– That make the CMFB loop UGF to be a few

times lower than CM gain’s UGF– Make sure the additional pole in the CM gain

and any additional poles from the CMFB circuit to be at higher frequency than DM UGF

CM gain’s additional pole at D5 is given by: -gm1/(Cgs1+½Cdb5)This is close to fT of M1. So at very high fre.

M1f M3f

VCM

M4f M2f

IB IB

Vo+ Vo-

VCMFB

VBP VBP

M5f

If the CMFB circuit below is to be used, then the following needs to be true:

1. IB and M5 sized to give desired VCMFB when Vo+=Vo-=desired2. CMFB circuit DC gain ACMFB=2gm1f/gm5f is small. 3. Pole of CMFB – gm5f/(Cgs5+Cgs5f+Cdb5f+Cdb1,4f) >> ACMFB*UGF of DM

Also, W/L of M1-4f should be small, so that their VEB is large to accommodate Vo+, Vo- swing.

This is consistent with (1.) above

AvDM()

AvCM()

Av() of CMFB circuit

AvCMFBLoop()

80dB

85dB

-15dB

70dB

Example DC gains But does not mean CM Q point can be maintained with -70dB accuracy!

Why?

Because op amp is always used in feedback configuration.DM feedback also kills CM gain, since DM and CM share the same path from vo1 to vo.

Vo1

VBP

VBN

Vo

I2I1

VBN VCMFB

Ri RiRf Rf

Vo1cm

VBP

VBN

Vocm

I2

I1VCMFB½ M5

CM

Vicm

Vicm = -Vocm

gm=gm1/(1+2gm1ro5)-1

1

1 5

5

1 2( ) ( ) ( ) ( )( )

12

m

m oocm vCM CMFB ViCM iCM vCM CMFB vCM ocm

m

g

g rV A V A V A V A V

g

5 5

5 5

( )

( )m o vCM

ocm CMFBm o vCM

g r AV V

g r A

5 5 when large

( ) when small

m ovCMocm

CMFBvCM vCM

g rAV

VA A

AvDM()

AvCM()

Av() of CMFB circuit

AvCMFBLoop()

80dB

85dB

-15dB

70dB

Example DC gains

gm5ro5/ 40dB

New AvCMFBLoop()

New AvCM()

25dB

Is it good enough to stabilize the CM Q points to -25 dB accuracy level?

If not, what can be done?

Increase the effective gm5ro5! That is: use cascode tail current source.

This will improve the CMFB loop gain under DM feedback by about 30 to 35 dB.

Or, increase the gain of CMFB circuit.In doing so, avoid introducing high impedance node, avoid introducing poles near or lower than DM GB.

Source follower

averager

BIAS4

1.5pF 1.5pF

M13A M13B

300/3 300/3

20K 20KOUT+ OUT-

M2A M2B150/3 150/3

BIAS3

300/2.25 300/2.25

IN-

M3A

M3B

IN+

300/2.25 300/2.25

M1A M1B

BIAS2

BIAS1

75/3

M7A

75/3

M7B

M6C 75/2.25

75/2.25

M6AB

M11

150/2.25

M8

150/2.25

200/2.25

200/2.25

M10

M5

CL=4pF 4pFM9A

50/2.25

M4A

50/2.25

M9B

50/2.25

M4B

50/2.25

M12A

1000/2.25

M12B

1000/2.25

VDD

VSS

Folded cascode amplifier

Voc variation range• Voc variation comes from two sources

– Input common mode– Common mode PVT variations

• Vicm induced Voc variation– Find closed-loop Vicm range– Find closed-loop gain from Vicm to Voc– Find contribution to Voc variation

• PVT induced Voc variation– Refer all PVT variations to VBP variation

– Find gain closed-loop from VBP to Voc

– Find contribution to Voc variation

Vo1cm

VBP

VBN

Vocm

I2

I1VCMFB½ M5

CM

V’icm gm=gm1/(1+2gm1ro5)-1

CMFB circuit

Vocm-Des

( ) ( ) ( )ocm vCM CMFB ViCM iCM VBP BPV A V A V A V

( )CMFB VCMFBCircuit ocm des ocmV A V V

' ( )( ' )iCM iCM ocm iCMV V V V

Vicm

Differences between CM and DM loop pole/zero

At input node:

Vi Vicm-Vi Vicm

BIAS4

1.5pF 1.5pF

M13A M13B

300/3 300/3

20K 20KOUT+ OUT-

M2A M2B150/3 150/3

BIAS3

300/2.25 300/2.25

IN-

M3A

M3B

IN+

300/2.25 300/2.25

M1A M1B

BIAS2

BIAS1

75/3 M7A

M6C

M6B

M11

150/2.25

M8

150/2.25

200/2.25

200/2.25

M10

M5

CL=4pF 4pFM9A

50/2.25

M4A

50/2.25

M9B

50/2.25

M4B

50/2.25

M12A

1000/2.25

M12B

1000/2.25

VDD

VSS

M6A

Current feedback eliminate one node in CMFB circuit.

What sizes should M6A-C be in order to maintain the same loop gain?

BIAS5

VDD=+1.65V

-VSS=-1.65V

M11 M12 M3 M4 M26 M27

Vo1 Vo2

M14

IDC=100υAM1C M2C

M13

M1

M2Vi1 Vi2

M51 M52

M21 M22

M25

VCM

M23 M24

VDD

+–

762

65

dsdsds

mmocmc ggg

ggA

CMFB with current feedback

VoCMCM

detect

VocVo+

Vo-

M3 M4

M1 M2

M6 M7

M5

IB

)by , t to(equivalen

by ,

by ,by If

2,

4 :Q desired

21

621

13

oGG

mo

oocooo

SSDDoCM

B

VVV

gVii

VVVVV

VVVI

II