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CMOS Analog Circuits
C fL12: Common Drain Amplifier and Output Stages (23.9.2013)
B Mazhari
G-NumberB. Mazhari, IITK1
B. MazhariDept. of EE, IIT Kanpur
Why do we want one more amplifier configuration?
+3.3V
v0321
vS
R = 1K-3.3V RL 1K
Low output Resistance; Rail-to-Rail voltage swing; Low distortion
High efficiency
G-NumberB. Mazhari, IITK2
How is CS as an output stage?1. We want rail-to-rail swing with minimal distortion
VDD = 3.3V
RD +3 HD
2 (%) 25indsat
vHDV
RD
1k
vO 0+3
-3
225in dsat
HDv V
A R R
VSS = -3 3Vvin
1k o v in m D L inv A v g R R v 2 DSQ
mI
gV V
VSS = -3.3Vin mGSQ T
gV V
211 12 5o DSQ D
HDv I RR R
( ) 0; DDo DSQ
D
VV dc IR
1 12.5QD LR R
211 12 5o DD
D L
HDv VR R
1 12.5D LR R
22
13 3.3 (1 ) 11.361 12 5 D D L
HD H R RR R
G-NumberB. Mazhari, IITK3
1 12.5D LR R
Output will have high distortion
CS as an output stage 2. We want low output Resistance so that outputcharacteristics are insensitive to load.
V = 3 3V
3 3
0.1DR k
RD
VDD = 3.3V
0+3
3.3 33DSQD
I mAR
RD
1k
-3vO
Small output resistancerequires large bias current.
VSS = -3.3Vvin
1k
vOmax1
3.3
V
VDSQvOmax2
~ 0.3 1dsat GSQ T dsatV V V V V V
2( )KP W L Vdsat -3.32( ) ( ) 33
2DSQ GS TNKP W LI V V mA
7333W Is this diagram still valid?
G-NumberB. Mazhari, IITK4
~ 7333WL Large bias current requires very large
Transistor sizes
VDD = 3.3V 1DR k
Suppose we compromise on the output Resistance
RD0
+3
3 3.3 3 3I A
1DR k
1k
-3vO
3.3 3.3DSQD
I mAR
VSS = -3.3Vvin vOmax1
3.3
V
VDSQvOmax2
~ 0.3 1dsat GSQ T dsatV V V V V V Vdsat -3.3dsat GSQ T dsat
W2( ) ( ) 3.3
2DSQ GS TNKP W LI V V mA
~ 733WL
G-NumberB. Mazhari, IITK5
However, the circuit will not give us the required swing 211 12.5o DD
D L
HDv VR R
Output Resistance
VO
RLVS
VO
V
RLVSVS
0V
S
VAV
0
V OS
VAV
S
Av
Avo
0.5 Avo Loading Effect
G-NumberB. Mazhari, IITK6
Ro RL
VinR0
+
RL AVo * Vin RLV0
-
+-
LinV RR
RVAV 00 0L
V VRA A
LRR0 0
0V V
LR R
Low output resistance is desirable and as long as load resistance is muchlarger than output resistance , it is expected that it won’t matter.
G-NumberB. Mazhari, IITK7
However, this is true only from a small-signal point of view
Example
1oR
Si l d i t h l th t t i t d ’t t
G-NumberB. Mazhari, IITK8
Since load resistors are much larger than output resistance, we don’t expectany difference in the two outputs
100mV
G-NumberB. Mazhari, IITK9
The difference between the twocircuits is due to different loadcurrents. In the first case the opamp is
3V
p pnot able to supply 30mA because itsoutput current is limited to 25mA only.In the second case the demand is only15mA. This highlights the importanceof output drive current.
G-NumberB. Mazhari, IITK10
VvOmax1
3.3
VDD = 3.3VV
VDSQvOmax2
31k 0.3mA
Vdsat -3.3
0+3
33
1k-3
3mA
VSS = -3.3Vvin
It is not j st the o tp t resistance b t o tp t c rrent dri e capabilit hich is
G-NumberB. Mazhari, IITK11
It is not just the output resistance but output current drive capability which isalso important. One can have very low output resistance but one may not beable to drive very low output loads because of current limitations.
VDD = 3.3V
3. We want large output current drive capability
DD
≥ 3mA
0+3
33
1k-3
3mA
VSS = -3.3VvinvOmax1
3.3
0.3 3 0 1DmA R k V
VDSQvOmax2
3 0.1DD
mA R kR
33DDVI A
Vdsat -3.3
G-NumberB. Mazhari, IITK12
33DDDSQ
DI mA
R
CS as an output stage 4. We want efficient amplifier where most of thepower drawn from the supply is delivered to the load
3 30.1DR k
VDD = 3.3V
+3 3.3 33DSQD
I mAR
1 T
RD0
-3v
s dd ssP P P 0
1 Tdd DD DDP V I dt
T
P V I
1kvO
( )DD DSQ dI I i Sin t dd DD DSQP V I
SS DSQP V I ( ( ))DD SS DSQP V V I
VSS = -3.3Vvin
(3.3 ( 3.3)) 218S DSQP I mW 2 23 4 5o
LvP mW
ss SS DSQP V I ( ( ))s DD SS DSQP V V I
( ( ))S DSQ 4.52 2L
L LP mW
R R100 2%L
S
PP
G-NumberB. Mazhari, IITK13
Amplifier is highly inefficientS
VDD = 3.3VCurrent source biasing is much better than resistive.
3.3V
+3IO ≥ 3mA v
Vbias2
M2
(W/L)P
0+3
-3
IO3
vO
vS Vbias1
3 3V
M1
(W/L)N
31k 3mA
-3.3V
VSS = -3.3Vvin 3OI mA 3DSQI mA
(3.3 ( 3.3)) 3.3 22SP m mW 2 23 4.5
2 2o
LL L
vP mWR R
2 2L LR R
100 22%LS
PP
1 1
G-NumberB. Mazhari, IITK14
1 1 3.8On p DSQ
R kI
3.3VCS as an output stage
VDD = 3.3V
Vbias2
M2
(W/L)P
RD
DD
0+3
3
vS Vbias1 M1
(W/L)N
vO
RL1k
-3vO
S bias1
-3.3V
1
VSS = -3.3Vvin
33DDDSQ
VI mAR
3DSQI mA
0.1 ; ~ 7333DWR kL
1( ) ~ 1500W
LQDR L
22SP mW218SP mW
~ 22%~ 2%
1 1 3.8On p DSQ
R kI
0.1O DR R k
G-NumberB. Mazhari, IITK15
~ 20vA ~ 17vA
VDD = 3.3V
1kvO
7330/1
0.1k
VSS = -3.3Vvin
1k
3.3V
Vbias2
M2
vO1500/1
750/1
vS Vbias1
-3.3V
M11k
1500/1
G-NumberB. Mazhari, IITK16Positive swing is poor; negative swing is good
3.3V
Vbias2
M2
vO1500/1
750/1
vS Vbias1
-3.3V
M11k
1500/1
3.3V
750/1Vbias2
M2
vO
1k1500/1
vS
Vbias1
-3.3V
M11k
G-NumberB. Mazhari, IITK17Positive swing is good; negative swing is poor
Low output resistanceV 3 3V
V
3.3V
750/1
VDD = 3.3V
Vbias2
M2
v
750/1
vO7330/1
0.1k
vS Vbias1 M1
vO
1k1500/1
v
1k
-3.3VVSS = -3.3Vvin
Resistance looking into drain is large and thus CS amplifier is notreally suitable for obtaining small output resistance
G-NumberB. Mazhari, IITK18
Strategy for obtaining Low output resistance
Resistance looking into source is small !
VDD
vO
RO ~ 1/(gm+gmb)vin RS
Apply input at gate, takeoutput at Source
VSS
G-NumberB. Mazhari, IITK19
p
Small Signal Analysis
vgs
gmvgs
ro
gmbvbs
vi
vO
RS
vin
gm gs
RS
vo
vin RS
m S ov
g R rA
1 1o SR R
g g g g
1 ( )vm mb S o
Ag g R r m mb m mbg g g g
G OG-NumberB. Mazhari, IITK
20
Gain is less than unity ! Output resistance is low !
vO
R v R
iN N m gsi g v
vin RS vin RS
1
RN
1N s
m mbR R
g g
RS
RN
1 ( )m S
vm mb S
g RAg g R
G-NumberB. Mazhari, IITK21
( )m mb Sg g
3 3V
Example
3.3V
Bias GSQ DSQ S SSV V I R V
vO
2/11.85V
( 2 2 )V V V
2( ) [1 ]2N
DSQ GSQ THN n DSQI V V V
vin 330k0 ( 2 2 )THN THN F SBQ FV V V
;V I R V V I R V -3.3V
;SBQ DSQ S DSQ DD DSQ S SSV I R V V I R V
10DSQI A 3.3 1.5 ; 1.85SBQ THN GSQV V V V V V 3.3DSQV V
55.57 / ; 9.67 / ; 6.6m mb og A V g A V r
G-NumberB. Mazhari, IITK22
0.811 ( )
m Sv
m mb S
g RAg g R
1 ~ 14.6o sm mb
R R kg g
3.3V
CD amplifier has good linearity and thus less
vO
2/11.85V
CD amplifier has good linearity and thus lessprone to harmonic distortion
vin 330k
-3.3V
G-NumberB. Mazhari, IITK23
HD2 ~1.6%
Why is distortion so less even though input is so large?
G-NumberB. Mazhari, IITK24
2)5.0( gsmTGSQ
gsmds vgVV
vgi
vO
+ vgs -
2 (%) 25gs
GSQ T
vHD
V V
vin 330kEven though vin is large, distortion is small perhapsbecause vgs is significantly smallerbecause vgs is significantly smaller
G-NumberB. Mazhari, IITK25
Negative feedback helps to reduce distortion
It is interesting that even though vin is pureIt is interesting that even though vin is puresinusoid, vgs is distorted in such a way thattransistor distorts it an opposite manner to giverise to drain current which is distortion free.
vO
+ vgs -
vin 330k
G-NumberB. Mazhari, IITK26
3.3V VDD Vsat1
vi
vO
2/1
330k
1.85VvOmax1
vO 2
VO(dc)vin
-3.3V
330k vOmax2
VSS
G-NumberB. Mazhari, IITK27
Good swing but input sinusoidal amplitude is 4V
Driving low impedance Load
VDD = 3 3VDD 3.3
vO
v R RL = 1K-3V
≥3mAvin RSRL 1K
3mA
≥3mA
VSS =-3.3
G-NumberB. Mazhari, IITK28
3 ( 3.3) 3 100SS
mA RR
3.3 ~ 33
100biasI mA
CD amplifier with current source biasing
VDD = 3.3
v
R 1K
vO-3V
vinRL = 1KIBias
3mA3.3mA
VSS =-3.3
G-NumberB. Mazhari, IITK29
SS
Biasing 3.3DSQI mA
VDD = 3.3
m1V
RL = 1K
vO
m2
m1VBias1
vi
VBias2L
VSS =-3.3
m2vin
(W/L) VGS1(V) VGS2(V)
50 2.6 1.8100 2 3 1 5100 2.3 1.5200 2.06 1.25500 1.85 1.05
G-NumberB. Mazhari, IITK30
700 1.8 11000 1.75 0.94
RVDD = 3.3
1 2
1 1 21 ( )m L o
vm mb L o
g R rA
g g R r
vOm1VBias1
11 11 ( )m L
m mb L
g Rg g R
RL = 1K
O
m2vin
VBias2 1o
m mbR
g g
VSS =-3.3
m2 m mbg g
VDD VSS
vOmax1V (d )
DD Vsat1
If Vo(dc)~0 then it appears that
vOmax2
VO(dc)
V
If Vo(dc) 0 then it appears thatswing within one saturation voltageof supply rails can be achieved.
G-NumberB. Mazhari, IITK31
VSSVsat2
Voltage Swing: Limitations
VDD
vIN m1
1 1O IN GS IN TN satV V V V V V
vin vO1
m2 V (d )+Vm2
VTN1
Vin(dc)+Vinmax1
VSS VDD Vsat
Vsat1
vOmax1
VIN(dc)
sat
vINmax1 vOmax2
VO(dc)
G-NumberB. Mazhari, IITK32VSS
Vsat
vINmax2 VSSVsat2
Biasing 3.3DSQI mA(W/L) VGS1(V) VGS2(V)
VDD = 3.3
1
50 2.6 1.8100 2.3 1.5
RL = 1K
vOm1
VBias2
100 2.3 1.5200 2.06 1.25500 1.85 1.05RL 1K
VSS =-3.3
m2 700 1.8 11000 1.75 0.94
SS
3 33
VTN1=1 53.3 0.3
1.2
VTN1 1.5Vsat1=0.3
vO 1=1 2
0 3
1.84.8
vOmax2
VO(dc)=0vOmax1 1.2
G-NumberB. Mazhari, IITK33
-3.30.3
Vsat2 VSS=-3.3
G-NumberB. Mazhari, IITK34
VTN1=1.5Vsat1=0.3
3
sat1
VO(dc)=-0.9vOmax1=2.1
vOmax2=2.1
VSS=-3.30.3
G-NumberB. Mazhari, IITK35
CD amplifier with P-MOSFET
VDD
V
Source and body are shorted so VT isconstant and smaller
VBias2
vO
(W/L) VSG1(V)
VBias1 RL=1k
O
50 2.52
100 2.04
vinVSS
200 1.69
500 1.39
700 1.31
1000 1.238
1m Lv
g RA One can get gain very close to
G-NumberB. Mazhari, IITK36
11vm L
Ag R
unity in this case
Biasing 3.3DSQI mA (W/L) VSG1(V)
50 2 52VDD
VBias2
50 2.52
100 2.04
200 1 69
R =1kvin
vO
m2200 1.69
500 1.39
700 1 31
VSS
RL=1km1
700 1.31
1000 1.238
3 3VDD=3.3V
V
SS
3.3 0.3
3 65
Vsat2
vOmax1=2.35VO(dc)=0.65
0 3
-0.652.35
3.65
Vsat1=0.45
OvOmax2=2.35
G-NumberB. Mazhari, IITK37
-3.30.3
VTN1=0.86-3
VDD=3.3VVsat2
VDD
V vOmax1=2.35VO(dc)=0.65
vOmax2=2.35
VBias2
vvO
m2
VTN1=0.86
Vsat1=0.45
3
RL=1kvin
m1
CS
-3VSS
G-NumberB. Mazhari, IITK38
Noisem2
iNiN
m1
2 22 2i ii i2 21 21 22 2 2 2
1 1 1 1
f ft tin
m m m m
i ii ieg g g g
Just like a CS stage
G-NumberB. Mazhari, IITK39
Just like a CS stage
Frequency Response
RS ||RLVDD
voCgs
S || L
vO
RS
RRG
gmvgs
Cgd
CdbVBias1
vORG
RL
vinvinVSS
1jjj CR
j
dBf2
13
G-NumberB. Mazhari, IITK40
G gdR C
RCgs
RG
Cgdgmvgs
vovin
CR CdbRS
1db SC Rg R
gsC 1 m Sg R( )
1gs
S Gm S
CR R
g R
1 ( ) gs S dbC R CR C R R
G-NumberB. Mazhari, IITK41
13 ( )
1 1gs S db
dB G gd S Gm S m S
R C R Rg R g R
1( )
szH s K
2( )1
zH s Kgs hs
CgsRG
Cgmvgs
mgs
gzC
vovin
Cgd
CdbRS CdbRS
( )1 1
gssdb gd g s G
m s m S
CRg C C R R Rg R g R
( )1
g sgd db db gs gs gd
R Rh C C C C C C
g R
1 g g g gm sg R
There are in general two poles and a zero all of which will influence 3dBThere are, in general, two poles and a zero, all of which will influence 3dBfrequency if they are close together.
G-NumberB. Mazhari, IITK42
Case-1: Negligible RG
3.3V C R C
330k
v
13
9
2 ( ) 1 1
1 1 10
gs S dbdB G gd S G
m S m S
C R Cf R C R Rg R g R
Hz
VBias1
RGvO
2/1
1.1 10 Hz
There is only one pole in this case since h = 0
vin-3.3V
91 1.1 10p Hz 91.38 10z Hz
G-NumberB. Mazhari, IITK43
3.3V
330k
VRG
330k
vO
2/1
vin
VBias1 2/1
vin-3.3V
RG (MΩ) P1 (GHz) P2 (GHz) Z(GHz)
f3DB(GHz)(est.)
f3DB(GHz)(acc.)( ) ( )
0 1.1 - 1.38 1.1 -0 1 0 89 i0 75 0 89+i0 75 1 38 0 76 1 070.1 0.89-i0.75 0.89+i0.75 1.38 0.76 1.07
1 0.35-i0.1 0.35+i0.1 1.38 0.19 0.249
G-NumberB. Mazhari, IITK44
5 0.047 0.56 1.38 0.043 0.04710 0.023 0.57 1.38 0.022 0.023
CS amplifier
31 1
dBf
2/1
RD = 100K 3.3V
0 14f GH 1 29mgf GH
3 2 ( (1 )) ( )dBS gs gd m D D gd db
fR C C g R R C C
2/1 VO
vin
1.2V100K 3 0.14dBf GHz 1 29
2m
zgd
gf GHzC
1 0 35m gdg Cf GH2 0.35
2m gd
pgs gd gs db gd db
f GHzC C C C C C
3 0.16dBf GHz
G-NumberB. Mazhari, IITK45CD amplifier has a superior frequency response
3.3V
330k
VRG
330k
2/1
vO
v
VBias1 2/1 CL=1pF
vin-3.3V
1 632 ( ) 5.95 10
1 1gs S db
dB G gd S Gm S m S
C R Cf R C R R Hzg R g R
66 10 H 91 38 10 H8
2 3 6 10p Hz 61 6 10p Hz 91.38 10z Hz 2 3.6 10p Hz
With large gate resistance or capacitive load, a dominant pole exists and the
G-NumberB. Mazhari, IITK46
With large gate resistance or capacitive load, a dominant pole exists and thesimple method gives a fairly accurate estimate of 3dB frequency.
Miller’s effect in CD amplifier
VDD
RS
VBi 1
vO
S
RGv
CgsRG
Cgdgmvgs
vi
VBias1 RLvovin
CdbRS vin
VSS
CF
Iif+
-VoVin
+
-
CMi CMo+
-Vin
+
-Vo
Iin Io
1
G-NumberB. Mazhari, IITK47
)11(V
FMo ACC )1( VFMi ACC
CgsRG
gmvgsRG
gmvgsCgsi
RS
vo
RS CgsoS S gso
(1 )gsi gs VC C A 1(1 )gso gsV
C CA
1m s
vg RA
g R
1 m sg R
1gs
gsi
CC
g R
gs
gso
CC
g R
G-NumberB. Mazhari, IITK48
1 m Sg R m Sg R
CgsRG
gmvgsRG
gmvgsCgsi
RS
vo
RS CgsoS S gso
1gs
gsim S
CC
g R
1( )
jzH j K
gsgso
m S
CC
g R
( )1
zH j Kj
p
1( )
(1 ) (1 )H j K
j j
m Sg
1 gs
m
Cz
g 1 2
(1 ) (1 )j jp p
Cmg
1 ( )1
gss G
m S
Cp R R
g R
11 1
gsG
m S
Cp R
g R
1
21
1gs
m S m
Cp
g R g
G-NumberB. Mazhari, IITK49
m Sg
CgsRG
gmvgsRG
gmvgsCgsi
RS
vo
RS CgsoS S gso
gs gs SC C RR ( )
1gs
S GS
CR R
g R
1 1
1( )
Gm S m S m S
gs
Rg R g R g R
CR
1 m Sg R
( )1
gsG
m S m
Rg R g
G-NumberB. Mazhari, IITK50
Summary3.3V VDD
VBi 2CS CDVbias2
M2VBias1
VBias2
R 1k
vO
CS CD
vS Vbias1 M1
vO
1kvin
VBias1
V
RL=1k
-3.3VVSS
1 Low Output resistance requires1. Low Output resistance requireslarge bias current
2 Rail to rail output swing
1. Significantly Lower Outputresistance can be obtained atsame value of bias current2. Rail-to-rail output swing
3. Frequency response suffersf Mill ’ ff d i
2. Swing lower by about a VT drop
from Miller’s effect and isinferior. 3. Good frequency response
G-NumberB. Mazhari, IITK51
Efficiency limited to < 25% for both the stages
Multistage Amplifiers with Negative feedback
Vbias2
3.3V
Vbias2
M2
vO
3 3V
M11kvS
-3.3VRfR1
N i f db k ill h l l h iNegative feedback will help lower the output resistance
G-NumberB. Mazhari, IITK52
Example 3.3V
700/11 99V
M2
vO00/1
700/1IBias=3.3mA
1.99V
-2.31V30 ; 817v oA R
3 3V
M11k
vS
700/1
-3.3V
3.3V
1.99V
M2
700/1
M1k
vO
vS
-100100
M2
700/1 99.94 ; 0.45v oA R
-3.3V
M1S
1k 99k
G-NumberB. Mazhari, IITK53
Ibias = 3.3mA
Ibias = 3.93mAbias
4 5LP mW
G-NumberB. Mazhari, IITK54
sup
4.5 0.1726
Lply
P mWP mW
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