Common mode feedback for fully differential amplifiers
description
Transcript of Common mode feedback for fully differential amplifiers
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