Download VDD -VSS

Differential Amplifiers and
common mode feedback
Differential amplifiers
• Cancellation of common mode signals
including clock feed-through
• Cancellation of even-order harmonics
• Increased signal swing
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
Two-Stage, Miller, Differential-In, DifferentialOut Op Amp with Push-Pull Output
Able to actively source and sink output current
Output quiescent current poorly defined
Cascode Op Amp with DifferentialOutputs, push-pull output
Differential-Output, FoldedCascode
OCMR = VDD -VSS - 2VSDP(sat) -2VDSN(sat)
Quite limited
Two-Stage, Differential Output,
Folded-Cascode
M11-M13 and M10-M12 provide level shift
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
Cause of common mode problem
Unmatched quiescent currents
Vbb
I2
Vo1
Vin
Vbb=VbbQ+Δ
Vin=VinQ
Vbb=VbbQ
Vo2
I1
Vo1Q
Vin=VinQ+ΔVin
Vo1
actual Q point
M2 is in triode
If Vo  VOCM  I y i.e.  Vyy
If Vo  VOCM  I y i.e.  Vyy
Vxx
Ix
Vo
VOCM
Vin
Ix(Vo)
Iy(Vo)
Vyy
Iy
Vo
needneg. feedbackfromVo to V yy
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-
• 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.
* Gate capacitance is your load to Amp.
Vo-
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+
Vo-
Φ1
Φ2
Φ1
VoCM.
VoCM.
To increase or decrease the C.M. loop gain:
e.g.
Vo.c.
Vo.c.d.
Vo.c.
Vo.c.d.
VC.M.F.B.
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 state
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, DifferentialOut 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
-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-
gain from VCMFB  Voc 
Vo   Vo 
2
-gm5vro2
-gm5vro2gm6
Vo
gm5v
v
Poles: p1
p2
z1 same as before
Aocmc
g m5 g m 6

g ds 2 g ds6  g ds7 
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
Removing the CM measurement
Vo+
VoCM
VoVCMFB
Directly connect Vo+, Vo- to the gates of
CMFB diff amp.
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
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 )