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5/21/2018 Chapter 18
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Willy Sansen 10-05 181
Distortionin elementary transistor circuits
Willy Sansen
KULeuven, ESAT-MICAS
Leuven, Belgium
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Why distortion ?
mc
f
mc
f
CM3
Non-linearity :
distortion
Ch1 Ch2 Ch1 Ch2
Mixing up channels !!!
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Table of contents
Definitions : HD, IM, intercept point, ..
Distortion in a MOST Single-ended amplifier
Differential amplifier
Distortion in a bipolar transistor
Reduction of distortion by feedback
Distortion in an opamp
Other cases of distortion and guide lines
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Linear distortion
vIN
t
vOUT
t
vOUT
vIN
f
High-pass filter
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Linear distortion
vIN
t
vOUT
t
vOUT
vIN
f
Low-pass filter
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Non-linear distortion
vIN
vOUT
t
t
vIN
vOUT
QRounded or compressed
Quiescent point
Expanded
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Soft and hard non-linearity
vIN
vOUT
t
tvIN
vOUT
Q
Soft non-lin.
vIN
vOUT
t
t
vIN
vOUT
Q
Hard non-lin.
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Non-linearity : by power series
vIN
vOUT
t
tvIN
vOUT
Q
Soft non-lin.
vOUT = a0 +
a1vIN + a2vIN2 + a3vIN
3 + ...
vIN(t) vOUT(t)
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How to find a0, a1, a2, a3, ...
y = a0 + a1u + a2u2 + a3u
3 +
a0 = y a1 =dy
du u = 0
a3 =
d3y
du3 u = 06
1
u = 0
a2 =
d2y
du2 u = 02
1
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Definition of harmonic distortion HD
y = a0 + a1u + a2u2 + a3u
3 +
With u = U cos t cos2 x = 1/2 ( 1 + cos 2x)cos3 x = 1/4 ( 3 cos x + cos 3x)
y = a0 + a1u + a2u2
+ a3u3
+ = a0 +
(a1+ a3U2) U cos t + U2 cos 2t + U3 cos 3t
HD2 = U
4
a22
a34
3
2
1
4
1a2a1
a3a1
HD3 = U2
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Amplitude HD versus input signal
U
HD10 %
1 %
0.1 %
- 20 dB
- 40 dB
- 60 dB
0.01 0.1 1
HD2
HD3
slope 1
slope 2
Low-distortion
region :slopes 1 and 2
Noise
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HD of a resistor
HD
(dB)
Output Voltage (Vptp)
slope 1
slope 1
slope 2
slope 2
THD = HD22 + HD3
2 +
HD2
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Output spectrum
Freq. (Hz)
Output
amplitude
(dB)HD2 HD3
2f 3ff
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Definition of intermodulation distortion IM
y = a0 + a1u + a2u2 + a3u
3 +
with u = U (cos 1t + cos 2t )
y = a0 +
IM2 = 2 HD2 = U
IM2 at 1 2
IM3 at 21 2 and 1 22
a2a1
a3a14
3IM3 = 3 HD3 = U
2
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IM components
f2f1-f2 f1 f2 2f2-f19 10 11 12
2f1 f1+f2 2f220 21 22
3f1 2f1+f2 2f2+f1 3f230 31 32 33 MHz
0 f2-f10 1
IM3IM
3IM
3IM
3
IM2IM2
HD2 HD2 HD3 HD3
F F
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IM components
f2f1-f2 f1 f2 2f2-f19 10 11 12
2f1 f1+f2 2f220 21 22
3f1 2f1+f2 2f2+f1 3f230 31 32 33 MHz
0 f2-f10 1
IM3IM
3IM
3IM
3
IM2IM2 HD2
HD2 HD3 HD3
a1U
2 x or 6 dB
3 x or 9.5 dB
a3U33
4 a3U31
4
a2U2a2U
2a2U
21
2
IM3 3 x larger !
Next to Fs !
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Output spectrum of amplifier for IM
Output
amplitude
(dB)
Freq. (MHz)
IM3IM3
Ref.: J.Silva-Martinez, Kluwer 1993
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Amplitude IM3 versus input signal
Vin (dB)Noise
Vout (dB)
a1Vin
4
3a3Vin
3
VNout
IM3
IMFDR3
DRN
IP3
1 dB -1 dB
Compression
point
IP3 is the IM3Intercept point
SFDR3
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Relation IP3 and IM3
Vin
Vout
a1Vin
4
3a3Vin
3
VNout
IM3
IMFDR3
DRN
IP3
1 dB IP3 is Vin
where IM3 = F
IP3 =a1
a33
4
= VinIM
3
1
= VindB - IM3dB2
1
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Vin
Vout
a1Vin
4
3
a3Vin3
VNout
IM3
IMFDR3
DRN
IP3
1 dB
Relation IMFDR3 and IP3
IMFDR3 = max DR
@ VNout = 43 a3Vin3
IMFDR3 =
a1
a33
4 1
VNin2
3
=IP3
VNin( )
2/3
= (IP3dB-VNindB)3
2
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Vin
Vout
a1Vin
4
3
a3Vin3
VNout
IM3
IMFDR3
DRN
IP3
1 dB
= IP3dB - 9.6 dB
-1 dB is x 0.891
Vin1dBc = 0.109a1a33
4
Vin1dBc = 0.109 IP3
The IP3 and -1 dB compression point
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Relationship exercise
Vin
Vout
a1Vin
4
3a3Vin
3
VNout
IM3
IMFDR3
DRN
IP3
1 dBa1 = 20 a3 = 0.4
Vin = 0.45 VRMS or 6 dBm
IP3 = 6 + 25 = 31 dBm
IM3 = 0.3 % or -50 dB
VNin = 30 VRMS (-78 dBm)
IMFDR3 = 119 = 73 dB32
Vin1dBc = 21 dBm
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Definition of crossmodulation distortion CM
y = a0 + a1u + a2u2 + a3u
3 +
with u = U cos 1t + U (1 + mc cos ct ) cos 2t
CM3 = mc U2 = mc IM34
3
mc
f
mc
f
CM3
a3a1
1 2 1 2
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Table of contents
Definitions : HD, IM, intercept point, ..
Distortion in a MOST
Single-ended amplifier
Differential amplifier
Distortion in a bipolar transistor
Reduction of distortion by feedback
Distortion in an opamp
Other cases of distortion and guide lines
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iDS = K (vGS - VT)2
W
L
IDS + ids = K (VGS + vgs - VT)2
IDS is the DC component
iDS is the DC + ac component
ids is the ac component
Ids is the amplitude of the ac component
Distortion in a single-MOST amplifier
K = K
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DC and ac components
vIN
iDS
t
tvIN
Q
IDS : DC component
iDS : DC + ac component
ids : ac componentIds : amplitude of the ac
component
Ids
IDSiDS
ids
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IDS = K (VGS - VT)2
W
L
IDS + ids = K (VGS + vgs - VT)2
ids = K (VGS + vgs - VT)2 - K (VGS - VT)2
ids = 2K (VGS - VT) vgs + K vgs2
Distortion in a single-MOST amplifier
K = K
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ids = 2K (VGS - VT) vgs + K vgs2
or ids = g1vgs + g2vgs2 + g3vgs3 +
g1 = 2K (VGS - VT)
g2 = K
g3 = 0
IM2 = Vgs =g2g1
& IM3 = 0Vgs
2(VGS-VT)
Coefficients a1, a2, a3 by comparison
WL
K = K
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Normalized current swing
ids = 2K (VGS - VT) vgs + K vgs2 iDS = K (vGS - VT)
2
or y = a1u + a2u2 + a3u
3 + ..
y = =
ids
IDS
y = = u + u2
Ids
IDS
2 vgs
VGS - VT+
1
4
2 vgs
VGS - VT( )2
1
4
Vgs
(VGS - VT)/2U =
y is the relative current swing !
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Numerical example
The peak value of Vgs
is Vgsp
= 100 mV
(then VgsRMS = 100 /2 = 71 mVRMS)
if VGS-VT = 0.5 V then Vgsp/[ 2(VGS-VT] = 0.1
gives IM2 = 10 % (HD2 = 5 %) & IM3 = 0
The relative current swing U = 0.1/0.25 = 0.4 !
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In general
ids = gmvgs + K2gmvgs2
+ K3gmvgs3
+govds + K2govds
2 + K3govds3 +
gmbvbs + K2gmbvbs2 + K3gmbvbs
3 +
K2gm&gmbvgsvbs + K3,2gm&gmbvgs2vbs
+ K3,gm&2gmbvgsvbs2 +
+
K3gm&gmb&govgsvdsvbs
More coefficients a1, a2, a3 ...
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Distortion of a MOST diode
iDS = K (vDS - VT)2
y = =ids
IDS
y = = u + u2Ids
IDS
2 vds
VDS - VT
+1
4
2 vds
VDS - VT
( )2
1
4
Vds
(VDS - VT)/2U =
Same as for a MOST transistor amplifier !
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The zero HD3 point for smaller L
gm
VT
wi
vs
VGSgm
gm
IDS = gm
gmsat = WCoxvsat
HD3 = 0 at VGS = VT ?
VGS
si
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Derivatives of gm
Ref. Fager JSSC Jan. 2004, 24-33
W = 60 m
L = 0.6
m
VDS = 2 V
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A differential pair is symmetrical
vOd
vIN
t
t
0
RLIB
rounded:
compressed
vId
symmetrical:
no 2nd order
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y = =iOd
IB
vId
VGS-VT
1
4
vId
VGS-VT
( )2
IB
VGS - VT
1 -
vId is the differential input voltage
iOd is the differential output current (gmvId) or
twice the circular current gmvId/2
IB
is the total DC current in the pair
Note that gm = = K (VGS - VT)
Distortion in MOST differential pair
W
L
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y = = U 1 - U2IOd
IB
1
4 VId
VGS - VT
U =IM3 = U
23
32
U - U31
8
1 - x 1 -x
2
Distortion in MOST differential amplifier
y = =iOd
IB
vId
VGS
-VT
1
4
vId
VGS-VT
( )21 -
IP3 = 4 (VGS - VT) 3.3 (VGS - VT)2
3
IM2 = 0
U is the relative current swing
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Distortion in linear region
VDS1 = RDID0.2 V
IDS1 = 1VDS1(VGS1-VT)
gm1 = 1VDS1 is constant
Ref. Alini,JSSC, Dec.92, pp.1905-1915
Low distortion !
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Table of contents
Definitions : HD, IM, intercept point, ..
Distortion in a MOST
Single-ended amplifier
Differential amplifier
Distortion in a bipolar transistor
Reduction of distortion by feedback
Distortion in an opamp
Other cases of distortion and guide lines
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ICE = IS exp( )
ICE + ice = IS exp ( )
VBE
kTe/qVBE + vbe
kTe
/q
1 + y = exp( )vbe
kTe/q
exp (u) = 1 + u + + + if u
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is the non-linear equation
y u + + + ...u2
2
u3
6U =
Vbe
kTe/q
y is the relative current swing !
a1 = 1
a2 = 1/2
a3 = 1/6
IM2 = U =
IM3 = U2 = ( ) 2
a2a1
Vbe
kTe
/q
1
2
3
4
a3a1
Vbe
kTe/q
1
8
Distortion in a bipolar transistor amplifier
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Numerical example
1. Relative current swing is 10 %
yp = 0.1 gives IM2 = 5 % (HD2 = 2.5%)IM3 = 0.125 % (HD3 = 0.04 %)
As a result Vbep = yp(kTe/q)= 2.6 mVp (1.8 mVRMS)
IP3 = 8 (kTe/q) = 74 mVp or 50 mVRMS or -13 dBm
2. Vbep = 100 mV
then yp = 0.1/0.026 4 (must be
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y = = u + +Id
IDU =
Same as for a Bipolar transistor amplifier !
Vd
kTe/q
y u + + + ...u2
2
u3
6
u2
2
u3
6
iD = IS exp( )vD
kTe/q
Distortion in a diode
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Distortion in bipolar differential amplifier
y = = tanhiOd
IB
y = U - U3IOd
IB
1
3
VId
2kTe/qU =
IM3 = U21
4
VId
2kTe/q
tanh x =ex - e-x
x - x31
3
ex + e-x
U is the relative current swing
IP3 = 4 kTe/qIM2 = 0
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C = C0 ( 1 + a1V + a2V2 + ... )
For poly-poly caps : a120 ppm/V
a2 2 ppm/V2
R = R0 ( 1 + a1V + a2V2 + ... ) [JFET with large VP]
For diffused resistors : a1 5 ppm/V
a2 1 ppm/V2
Distortion in a resistor or capacitor
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Non-linearity depletion capacitance
VIN0-VB
Cj
C0
Cj =C0
1 -
vIN
Cj = C0B (1 + x)-1/2 = C0B (1 - 1/2 x + 3/8 x
2 - 5/16 x3 + ..)
vIN = VB + vin
Cj =
C0
1 +VB
1
1 +VB +
vin
x
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Table of contents
Definitions : HD, IM, intercept point, ..
Distortion in a MOST
Single-ended amplifier
Differential amplifier
Distortion in a bipolar transistor
Reduction of distortion by feedback
Distortion in an opamp
Other cases of distortion and guide lines
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Distortion reduction by feedback
a1 a2 a3
d1 d2 d3
F
v u
v
y
y
u = v - Fy
y = a1u + a2 u2 + a3 u
3
y = d1v + d2 v2 + d3 v
3
elim. uelim. y coeff v2 : d2
coeff v : d1
coeff v3 : d3
-
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Distortion reduction by feedback
a1 a2 a3
d1 d2 d3
F
v u
v
y
y
Loop gain 1+T = 1+a1Fu is (1+T) times smaller than v :
v is reduced by loop gain (1+T)
u = v - Fy
a1
1 + T
a3 (1 + T) - 2F a22
(1 + T)5
d1 =
d2 =
d3 =
a2
(1 + T)3
1
F
-
i i i f
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IM2f = V =d2 V
(1 + T)2d1
a2
a1
1
(1 + T)
a2
a1
=V
(1 + T)
1
(1 + T)
a3
a1
2T
(1 + T)2
a2
a1
-V2
(1 + T)2IM3f = V
2d3
d1
3
4= ( )[ ]
3
4
2
Distortion components with feedback
reduction in
current swing
expansioncompression
reduction by loop gain
reduction in
current swing
Di i i h f db k l
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1
(1 + T)
a3
a1
2T
(1 + T)2
a2
a1
-V2
(1 + T)2IM3f = V
2d3
d1
3
4=
(
)[ ]3
4
2
Distortion components with feedback : examples
MOST : a3 = 0 : a2 dominant
Diff. pair : a2 = 0 : a3 dominant
Bipolar : a1 = 1 a2 = 1/2 a3 = 1/6 : a2 dominant
a3 a1 - 2 a22
T
1
a1
2
a3
T
1
a1 a3a11 -
2 a22
(
)For large T : =
E itt i t t d di t ti IM
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IM2f =Vin
(1 + T)2
1
2
1
(1 + T)
=U
2
Emitter resistor to reduce distortion IM2f
1
kTe/q
Vin
(1 + T)
1
kTe/q
U = is the relative current swing
T = gm RE =VRE
kTe
/q
a2
IM2f decreases linearly with T for constant U !
a1 2=
1
E itt i t t d di t ti IM
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IM3f =
Null for T = 0.5
Emitter resistor to reduce distortion IM3f
1 - 2T
(1 + T)2
U2
8
Vin
(1 + T)
1
kTe/q
U = is the relative current swing
IM3f also decreases with T for constant U
for large T !!
a2
a1 2
=1 a3
a1 6
=1
N ll i IM b R (Bi l t I 1 A)
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Null in IM3 if
a3 (1 + T) = 2f a22
a3 (1 + T) = 2T
0.01 T
RE
IM
%
1
0.1
0.01
0.1 1 10 100
1 10 100 1k
a1
1
a22
T =
- 1a1a3
2a22
T = 0.5
IM2
IM3
Same slopes !
Null in IM3 by RE (Bipolar trans. ICE = 1 mA)
Emitter resistor R reduces distortion for large T
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Vin
T
1
kTe/q
U = =
REICE
Vin
IM2fT = =U
2T
Vin kTe/q
2 (REICE )2
IM3fT = = (
U2
4T
Emitter resistor RE reduces distortion for large T
=
kTe/q 2 T2
Vin 1
Vin2 kTe/q
4 (REICE )3kTe/q 4 T
3
Vin 1
=)2
Source resistor R to reduce distortion
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IM2f =1
(1 + T)
U
4
Source resistor RS to reduce distortion
is the relative current swing
T = gm RS =VRS
(VGS-VT)/2
Vin
(1 + T)
1
(VGS-VT)/2U =
IM3f =T
(1 + T)2
3U2
32
a2
a1 4=
1a3 = 0
(VGS-VT)/2 4 T2
Vin 1=
Vin (VGS-VT)/2
4 (RSIDS )2
(VGS-VT)
2/4 32T3
Vin2 3
=3Vin
2 (VGS-VT)/2
32 (RSIDS )3
Current source with series R
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Iout
M2
R
M1
VG
+
-
Iout
(VGS-VT) (VGS-VT)
(W/L) (W/L)
Same Iout & same VG :
Same gain !
Same output noise !
Same distortion ?
Current source with series R
IM2f
IM2= VR
VGST1
1 -
( 1 + ) 2
VR
VGST1
Source & Emitter Follower
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Source & Emitter Follower
vin vout
IBVB
RS
U =
Vin
VEnL
Vin
(VGS - VT)/2
1
gmrDS
If vBS = 0 !!
CL
vin vout
IBVB
RS
CL
U U
U =
Vin
VE
Vin
kTe/q
1
gm ro= =
Distortion Source follower with substrate effect
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vin vOUT
IBVB
vOUT = vIN - vGS
KW/L
IBvGS = VT +
vOUTF = |2F| +vOUT - |2F|
vIN = vOUT + VT0 + [ vOUTF ] +KW/L
IB
VT
= VT0
+
[v
OUTF ]
Distortion Source follower with substrate effect
CL
Distortion Source follower - Example
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VOUT
VIN
= 0
slope 1
= 0.8 V 1/2
slope 1/n
0
vIN = u2 + u + B
u2 = vOUT + |2F|
B = VGS0 - |2F| - |2F|
KW/L
IBVGS0 = VT0 +
VT0 = 0.6 V ; VGS0 = 0.9 V; 2F = 0.7 V; B = -0.47 V; 1/n = 0.73
a1 = 0.765; a2 = 0.02; a3 = -0.0035
VINp = 1 Vp; HD2 = 1.32 %; HD3 = -0.114 %
2 3 V
0
2
V
11
1.37
0.9 2.27
1
Distortion Source follower - Example
Increasing the IP3 by feedback
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Increasing the IP3 by feedback
M1Vid/2 -Vid/2
2Ibias
-ioutiout
M1 M1Vid/2 -Vid/2
Ibias
-ioutiout
M1
Ibias
R R 2R
IP3 3.3 (VGS-VT)(1+gm1R)2 HD3/n
2 n= 1+gm1R
HD3 = - 60 dB for Vid = 1 V requires VGS-VT = 0.38 V and gm1R = 3 !!!
Increasing the IP3 by feedback
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Increasing the IP3 by feedback
Additional local FB
2R
More FB with opamps
2R
Distortion cancellation
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Distortion cancellation
Vid/2 -Vid/2
IB2IB1
iout
M1
iout
M2M1
Parameters :
= IB2/ IB1
0.25
v = VGST1/ VGST21.6
VGST = VGS-VT
IM3 0 if v = -1/3
then iout = gm1 Vid (1 - 2/3 )
Distortion cancellation
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Distortion cancellation
IM3
iDS
IB= U - U3 U =
Vid
VGS - VT18 IM3 = U
23
32
Vid
VGS1- VT332 )
2( 1 - v3
1 - v
IM3 0 if v00 = -1/3
at which point iout = gm1 Vid (1 - 2/3 )
iout = 2 (iDS1 - iDS2)
Compensation of IM3
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p
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
0 0.25 0.5 0.75 1 1.25 1.5 1.75
= 0.20
v00 = 1.71
= 0.25
v00
= 1.6
= 0.33
v00 = 1.44
v
1 -
v3
1 - v
Output signal vs current ratio
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p g
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
1 -
2/3
0.25
v00 = 1.6
x 0.6
Table of contents
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Definitions : HD, IM, intercept point, ..
Distortion in a MOST
Single-ended amplifier
Differential amplifier
Distortion in a bipolar transistor
Reduction of distortion by feedback
Distortion in an opamp
Other cases of distortion and guide lines
Miller CMOS opamp with Feedback
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p p
GBW = 10 MHz & Avc = 10
ZL = 100 k//5pF
R2
+
-R1vIN
vOUT
ZL
Av
Av0
f
BW
= 1 kHz
GBW
= 10 MHz
101 MHz
Distortion in input stage
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a1 a2 a3
d1 d2 d3
F
v u
v y
u = v - Fy
y = B1(a1u + a2 u2 + a3 u
3)
y = d1v + d2 v2 + d3 v
3
elim. uelim. y coeff v2 : d2
coeff v : d1
coeff v3 : d3
B1y
-
1+T = 1 + B1a1F
Distortion in input stage
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IM2f = V =d2 V
(1 + T)2d1
a2
a1
1
(1 + T)
a2
a1
=V
(1 + T)
1
(1 + T)
a3
a1
2T
(1 + T)2
a2
a1
-V2
(1 + T)2
IM3f = V2
d3
d1
3
4
= ( )[ ]3
4
2
1+T = 1 + B1a1F
Same as before but with different Loop gain :
Distortion in input stage with LPF
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a1 a2 a3
d1 d2 d3
F
v u
v y
u = v - Fy
y = B1p(a1u + a2 u2 + a3 u3)
y = d1v + d2 v2 + d3 v
3
elim. uelim. y coeff v2 : d2
coeff v : d1
coeff v3 : d3
yB1fp
-
1+T = 1 + B1pa1F
Distortion in input stage with LPF
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IM2f =V
(1 + T)2
a2
a1
1
(B1pa1F)2
a2
a1
= V
1
(1 + T)
a3
a1
V2
(1 + T)2IM
3f
=3
4
diff.pair
=a3
a1
3
4
1
(B1pa1F)3 V
2
f
ffp
fp
40 dB/dec
60 dB/decIM3f =
Single trans.
a22
a12
3
4
2
(B1pa1F)3
V2
Distortion in output stage
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A1
d1 d2 d3
F
v u
v y
u = v - Fy
y = A1b1u+A12b2u
2+A13b3 u
3
y = d1v + d2 v2 + d3 v
3
elim. uelim. y coeff v2 : d2
coeff v : d1
coeff v3 : d3
yb1 b2 b3
-
1+T = 1 + A1b1F
Distortion in output stage
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IM2f =V
(1 + T)2
b2
b1
A1
(A1b1F)2
b2
b1
= V
f
f
=
b22
b12
3
4
2 A12
(A1b1F)3 V2
IM3f =
Single trans.
2T
(1 + T)2
b2
b1
V2
(1 + T)2
( )3
4
2
x A1
x A12
Two-stage opamp a & b
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IM2f
ffp
V
T2
a2
a1 fp
f
V
T2
b2
b1a1 T = a1b1F
IM3f
ffp
V2
T3
a3
a1
V2
T3
b3
b1a1
2
3
4
( )2
fp
f( )
3
( )a1a3 - 2 a22
a1a3
( )b1b3 - 2 b22
b1b3
3
4
Three-stage opamp a & b & c
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IM2f
ffp
V
T2
a2
a1 fp
f
V
T2
c2
c1a1b1 T = a1b1c1F
IM3f
ffp
V2
T3
a3
a1
V2
T3
c3
c1a1
2b12
3
4
( )2
fp
f( )
3
( )a1a3 - 2 a22
a1a3
( )c1c3 - 2 c22
c1c3
3
4
Distortion in an opamp at low frequencies
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12
34
VOUT= 1 V
M1 M1
M2 M2
M4
+- Cc
M3
GBW = 10 MHz
Av0 = 10.000
BW = 1 kHz
Avc = 10Vin =0.1 mV
Vm =
10 mV
ZL IDS1 = 6 A
gm1 = 60 S
IDS3 = 120 A
gm3 = 1.2 mS
RL = 100 k CL = 5 pF
Cc = 1 pF
Low-distortion amplifier
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Av
Av0
f
BW
= 1 kHz
GBW= 10 MHz
10
Av
Av0
f
vout
vm
RL(CL+Cc)
1 MHz
256 kHz
Av2
Av1
30 MHz
gm1
gm3
CL+C
cCc
0.3
vin = constant
fnd
GBW
Distortion in an opamp at low frequencies
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34
VOUT= 1 V
M1 M1
M2 M2
M4
+- Cc
Vnode at 100 Hz ?
M3
GBW = 10 MHz
Avc = 10
Vin =0.1 mV
Vm =
10 mV
ZL
U1 = gm1Vin/IDS1 = 5 10-4
IDS1 = 6 A
IDS3 = 120 A
U3 = gm3Vm/IDS3 = 0.1
Distortion in an opamp at low frequencies
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Distortion generation by nonlinear output stage :
Distortion reduction by feedback :
T = 1000 IM2f= 2.5 %/1000 = 0.0025 % Negligible !
U3 = gm3Vm/IDS3 = 0.1
IM2 = U3/4 = 0.25 0.1 = 2.5 %
Distortion in an opamp at high frequencies
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12
34
VOUT= 1 V
M1 M1
M2 M2
M4
+- Cc
Vnode at 100 kHz ?
M3
GBW = 10 MHz
Avc = 10
Vin =10 mV
Vm =
10 mV
ZL
U1 = gm1Vin/IDS1 = 5 10-2
IDS1 = 6 A
IDS3 = 120 A
U3 = gm3Vm/IDS3 = 0.1
Distortion in an opamp at high frequencies
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Distortion generation by nonlinear output stage :
Distortion reduction by feedback :T = 10 IM2f= 2.5 %/100 = 0.25 %
U3 = gm3Vm/IDS3 = 0.1
IM2 = U3/4 = 0.25 0.1 = 2.5 %
Distortion generation by nonlinear input stage :
U1 = gm1Vm/IDS1 = 0.05
IM3 = U12/10 = 0.0025/10 = 0.025 % Negligible !
Miller CMOS OTA Measured Distortion
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HD2 (dB)
Freq.
(Hz)
1.8 V Low distortion CMOS Opamp
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Ref.Hernes Kluwer 2003
GBW 3 GHz
CL = 8 pF
fP = 380 MHz
at 0.38 Vpeak
SR 900 V/s
fP =2 Vpeak
SR
Large VGS4-VT
HD2 & HD3 vs Amplitude
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large distortion
Slope 1
Slope 2
HD2 & HD3 vs Frequency
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HD2 HD3dBdB
Hz Hz
Table of contents
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Definitions : HD, IM, intercept point, ..
Distortion in a MOST
Single-ended amplifier
Differential amplifier
Distortion in a bipolar transistor
Reduction of distortion by feedback
Distortion in an opamp
Other cases of distortion and guide lines
Other cases of distortion and guide lines
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Distortion caused by limited SR
Distortion of a switch
Distortion at high frequencies :
Volterra series instead of power series
Distortion in continuous-time filters
Guide lines
Guide lines for low distortion
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Scaling such that voltage amplitudes
are limited
Scaling such that relative current swings
are limited Feedback
All fully differential
Distortion components
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Distortion comp. IM2 IM3x Up x Up
2 Up = Vref=
Bipolar 1/2 1/8 kTe/q
MOST 1/4 0 (VGS-VT)/2
Bip. diff.pair 0 1/4 2kTe/q
MOST diff.pair 0 3/32 (VGS-VT)
Vref
Vip
Distortion components with Feedback (T > 5)
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Distortion comp. IM2 -IM3x Up x Up
2 Up = Vref=
Bipolar 1/2T 1/4T kTe/q x T
MOST 1/4T 3/32T (VGS-VT)/2 x T
Bip. diff.pair 0 1/4T 2kTe/q x T
MOST diff.pair 0 3/32T (VGS-VT) x T
Vref
Vip
References
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P.Wambacq, W.Sansen : Distortion analysis of analog Integrated
Circuits, Kluwer Ac. Publ. 1998
W.Sansen : Distortion in elementary transistor circuits
IEEE Trans. CAS II Vol 46, No 3, March 1999, pp.315-324
J. Silva-Martinez, etal : High-performance CMOS continuous-time
filters, Kluwer Ac. Publ. 1993
B. Hernes, T. Saether : Design criteria for low-distortion in
feedback opamp circuits, Kluwer Ac. Publ. 2003
G. Palumbo, S. Pennisi : Feedback amplifiers, Kluwer Ac. Publ. 2002
Table of contents
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Definitions : HD, IM, intercept point, ..
Distortion in a MOST
Single-ended amplifier
Differential amplifier
Distortion in a bipolar transistor
Reduction of distortion by feedback
Distortion in an opamp
Other cases of distortion and guide lines