Download CMFB

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
peak-to-peak
output voltage
≤ 2·OCMR
Output common mode range (OCMR)
= VDD-VSS - VSDPsat - VDSNsat
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
VBP
Vo1cm
Vicm
VBN
I2
Vocm
I1
If Vocm is too high, needs to increase Vo1cm
for single stage, needs positive feedback
Vo1cm
I2
Vo1cm a little low
Correct Vo1cm
Vocm
VBN
I1
Vocm
Vo1
actual Q point
too high
PMOS in triode
If Vo1  Vo1CM  need  I x i.e. need to  VBP
From this single stage's point view, need positive feedback
VBP
Ix
Vo1cm
VO1CM
Vicm
Ix(Vo)
Iy(Vo)
VBN
Iy
Vo
But we started by having Vocm too high
 need neg. feedback from Vo to VBP
In both cases, the whole loop has negative loop gain
VBP
I1
I2
Vo1cm
Vicm
Vo
Vicm
VBN
What about single ended?
Does it have the same problem?
Does it require feedback stabilization?
VBP
I3
Yes, to
all three
questions
I4
Vo1cm
Vi-
I1
VBN
I2
I6
Vo
Vi+
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-.
All op amps must
be used in
feedback
Viconfiguration!
VBP
I3
I4
Vo1cm
I1
I2
I3
Vi-
VBN
Vi+
I4
Vo1cm
I1
Vo
I7
VBN
VBP
I6
I2
I6
Vo
Vi+
I7
Buffer connection
or resistive
feedback provides
the needed
negative feedback
Fully differential amplifiers are also used in
feedback configuration.
Vi   Vp
Ri

Vp  Vo 
Rb
Vi   Vi   (Vp  Vn )
Ri
Vi   Vn Vn  Vo 
,

Ri
Rb

Vi+
Vi-
Vp
Vn
Vp  Vn  (Vo   Vo  )
Rb
If amplifier gain is high, V p  Vn is  0,
Vi   Vi 
(Vo   Vo  )
Rb

, and Vo   Vo    (Vi   Vi  )
Ri
Rb
Ri
Hence, differential signal is well defined.
Vo+
Vo-
But when you add the first
two equations
Vi   Vp
Ri

Vp  Vo 
Rb
Vi   Vn Vn  Vo 
,

Ri
Rb
You get:
Vi   Vi   (Vp  Vn )
Ri
Vi+
Vi-

Vp
Vn
Vo+
Vo-
Vp  Vn  (Vo   Vo  )
Rb
Solving for Vo   Vo  ,
Rb
Rb  Ri
Vo   Vo    (Vi   Vi  ) 
(V p  Vn )
Ri
Ri
Since Vp+Vn is undefined, Vo++Vo- is undefined.
Basic concept of CMFB:
Vo+ +Vo-
Vo+
CM
Vo-
Dvb
2
measurement
Voc
CMFB
e
VoCM
+
desired
common mode
voltage
Basic concept of CMFB:
Vo+ +Vo-
Vo+
CM
Vo-
Dvb
2
measurement
Voc
CMFB
e
e
VoCM
+
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
CC
Vi+
CC
Vi- Vo+
VCMFB
Vo-
Vb1
VCMFB
Voc
+
Vo+
Vo-
Example
Voc
?
?
VoCM
Need to make sure to have negative feedback
VDD
M7A
75/3
150/3 M2A
150/3 M2B
BIAS4
300/3
300/3
M13A
M13B
1.5pF 1.5pF
M7B
75/3
averagerOUT-
M3B
BIAS3
OUT+
300/2.25
300/2.25
M6C
20K
M3A
20K
300/2.25
300/2.25
75/2.25
INM6AB
75/2.25
IN+
M1A
Source
M12B
M12A
1000/2.25
1000/2.25
follower
M1B
200/2.25
BIAS2
M11
150/2.25
M10
CL=4pF
4pF
M9A
50/2.25
M9B
50/2.25
M4A
50/2.25
M4B
50/2.25
BIAS1
M8
150/2.25
200/2.25
M5
VSS
Folded cascode amplifier
Resistive C.M. detectors:
Vo+
R1
R2
Vo-
Vo   Vo 
, if R1  R2
2
Prob : resistive loading effect.
use R1 , R2 very large
- difficult to achieve
- especially when Vo is
at an cascoded node
Resistive C.M. detectors:
Vo.c.
Vo+
Vi+
R1
R1
Vo-
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.
Voc
Vo+
VoR1
R1
Simple implementation:
source follower
Vo+
Vo.c.
Vo-
* Gate capacitance is added to your amp load.
Why not:
Vo+
Vo.c.
Vo-
* Initial voltage on cap.
C1
C2
Vo   Vo 
, if C1  C 2
2
Prob : at high freq.
AC diff short
Use buffer to isolate Vo node:
gate cap is load
or resistors
Switched cap CMFB
Vo+
Φ1
VCMFB
Vo-
Φ2
Φ1
VoCM.
Vb-repl
VoCM.
In the normal balanced case, when Vo+=Vo- = desired Vocm,
Vb-repl is supposed to generate the correct voltage for VCMFB
Vxx
Small copy
of
Vo-
Vo+
Small copy Vb-repl
of
VCMFB
Vyy
Small copy
of
To increase or decrease the C.M. loop gain:
e.g.
Vo.c.
Vo.c.d.
VC.M.F.B.
Another implementation
• Use triode transistors to provide isolation
& z(s) simultaneously.
M1, M2 in deep triode.
VGS1, VGS2>>VT
Voc
Vo+
VoM1
can be a c.s.
M2
In that case, circuit above M1,
M2 needs to ensure that M1, M2
are in triode.
Vo   Vo 
2
deep triode oper
Example:
Input stage
Vo-
Vb
M1
M2
e.g. Vo+, Vo-≈2V at Q & Vb ≈1V ,
Then M1&2 will be in deep triode.
Vo+
If Vo  ,Vo  
Vo-
Vo+
 Voc 
 VG1 ,VG 2 
Vb1
 RM 1 , RM 2 
Vb2
VX
M1 M2
 V X  ( I  const )
but V X to Vo  ,Vo 
is cascoded C.G.
 Vo  ,Vo  
Two-Stage, Miller, Differential-In,
Differential-Out Op Amp
M10 and M11 are in deep triode
Vo++ Vo2
VoCM.
VCMFB
Vo   Vo 
 VCMFB
2
gain can be large
Vo+
Vo-
Note the difference
from the book
accommodates much
larger VoCM range
Small signal analysis of CMFB
Example:
IB
M3
Vo+
M1
M4
IB
M2
-Δi
Vo-
+Δi
+Δi
M5
+Δi
-Δi
VCM
+Δi
VCMFB
Δi=0
2Δi
Differential signal
Common mode signal
-Δi
-Δi
• Differential Vo: Vo+↓ by ΔVo, Vo-↑ by ΔVo
• Common mode Vo: Vo+↑ by ΔVo, Vo-↑ by ΔVo
DVo
Di  g m1
2
1
DVCMFB 
 2Di
g m5
 k  DVo
g m1  g m 2  g m3  g m 4 

g m1 
 k 

g m5 

IB
M3
Vo+
M1
VCM
M4
IB
Vo-
M2
+Δi
+Δi
M5
-Δi
-Δi
VCMFB
Δi=0
2Δi
Δi7 M7
+
-2Δi
M6
-
1
-2Δi
gm6
To increase gain :
1
DVG 7  2Di
g m6
Di7  DVG 7 g m 7
g m7
 2Di
g m6
 g m7 

 Di5  2Di1 
 g m6 
g m1  g m 7 
1 
DVo
DVCMFB 
g m5  g m6 
* gain by geometric ratios  can be made accurate
CMFB loop gain: example
Vb2
CC
Vi+
CC
Vi- Vo+
VCMFB
Vo-
Vb1
VCMFB
Voc
+
Vo+
Vo-
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 CMDM 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
VBP
I1
I2
Vo1
Vo
Vi++Vicm
VBN
Vi-+Vicm
VCMFB
VBN
CM and DM equivalent circuit Comparison
VBP
VBP
CM
Vo1cm
Vocm
VCMFB
ro1=rdsp||rcascode≈rdsp
gm = ½gm5
½Vo1d
I2
Vid
½ M5
DM
VBN
I1
I2
-½Vid
½Vod
AC GND
VBN
I1
ro1=rdsp||rdsn≈ ½rdsp
gm = 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
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.
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-=desired
2. CMFB circuit DC gain ACMFB=2gm1f/gm5f is small.
3. Pole of CMFB – gm5f/(Cgs5+Cgs5f+Cdb5f+Cdb1,4f) >> ACMFB*UGF of DM
VBP
Vo+
M1f
VCM
IB
M3f
VBP
M4f
IB
M2f
Vo-
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
VCMFB
M5f
Example DC gains
85dB
80dB
AvCM(w)
70dB
But does not mean
CM Q point can be
maintained with 70dB accuracy!
AvDM(w)
AvCMFBLoop(w)
w
-15dB
Av(w) of CMFB circuit
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.
VBP
I1
I2
Vo1
Vo
VBN
VCMFB
Ri
Rf
VBN
Rf
Ri
VBP
CM
Vo1cm
I2
Vicm = -bVocm
Vicm
Vocm
-1
gm=gm1/(1+2gm1ro5)
VCMFB
½ M5
Vocm  AvCM (w )VCMFB  AViCM (w )ViCM
Vocm 
g m5 ro 5 AvCM (w )
VCMFB
g m 5 ro 5  b AvCM (w )
VBN
I1
g m1
1  2 g m1ro 5
 AvCM (w )VCMFB 
AvCM (w )( bVocm )
1
g m5
2
Vocm
VCMFB
 g m5 ro5
when AvCM large

b

 A (w ) when A
vCM small
 vCM
Example DC gains
85dB
80dB
AvCM(w)
70dB
New AvCM(w)
AvDM(w)
gm5ro5/b 40dB
25dB
AvCMFBLoop(w)
New AvCMFBLoop(w)
-15dB
Av(w) of CMFB circuit
w
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.
VDD
M7A
75/3
150/3 M2A
150/3 M2B
BIAS4
300/3
300/3
M13A
M13B
1.5pF 1.5pF
M7B
75/3
averagerOUT-
M3B
BIAS3
OUT+
300/2.25
300/2.25
M6C
20K
M3A
20K
300/2.25
300/2.25
75/2.25
INM6AB
75/2.25
IN+
M1A
Source
M12B
M12A
1000/2.25
1000/2.25
follower
M1B
200/2.25
BIAS2
M11
150/2.25
M10
CL=4pF
4pF
M9A
50/2.25
M9B
50/2.25
M4A
50/2.25
M4B
50/2.25
BIAS1
M8
150/2.25
200/2.25
M5
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
VBP
CM
Vo1cm
I2
Vicm
V’icm
Vocm
-1
gm=gm1/(1+2gm1ro5)
VCMFB
½ M5
VBN
I1
CMFB
circuit
Vocm  AvCM (w )VCMFB  AViCM (w )ViCM  AVBP (w)VBP
VCMFB  AVCMFBCircuit (w) Vocmdes  Vocm 
ViCM  V 'iCM  b (w )(Vocm  V 'iCM )
Vocm-Des
Differences between CM and DM loop pole/zero
At input node:
Vi
-Vi
Vicm
Vicm
Current feedback eliminate one node in CMFB circuit.
VDD
75/3
150/3 M2A
M7A
150/3 M2B
BIAS5
BIAS4
300/3
300/3
M13A
M13B
1.5pF 1.5pF
M3B
BIAS3
OUT+
300/2.25
300/2.25
OUT20K
M3A
20K
300/2.25
300/2.25
M6C
INM6A
M6B
IN+
M1A
M1B
M12A
1000/2.25
200/2.25
BIAS2
M11
150/2.25
M10
CL=4pF
4pF
M9A
50/2.25
M9B
50/2.25
M4A
50/2.25
M4B
50/2.25
BIAS1
M8
150/2.25
200/2.25
M5
VSS
What sizes should M6A-C be in order to maintain the same loop gain?
M12B
1000/2.25
VDD=+1.65V
M11
M12
M3
M4
Vo1
M27
M21 M22
M23 M24
Vo2
M1C
M2C
IDC=100υA
VCM
Vi1
M14
M26
M13
M1
M51
M2
M52
-VSS=-1.65V
Vi2
M25
VDD
+
–
Aocmc
g m5 g m6

g ds 2  g ds 6  g ds 7 
CMFB with current feedback
M3
M4
Vo+
Vo-
M1
IB
CM
Voc
detect
M2
VoCM
M6
M7
M5
VDD  VSS
IB
desired Q : I 3   I1 ,VoCM 
4
2
If Vo   Vo   by DVo , Voc  by DVo
Di1 , Di2  by DVo  g m 6
(equivalen t to VG1 , VG 2  by DVo )