Distortion in elementary transistor circuits

Willy Sansen 10-05 181

Distortionin elementary transistor circuits

Willy Sansen

KULeuven, ESAT-MICASLeuven, Belgium

[email protected]


Why distortion ?

mc

f

mc

fCM3

Non-linearity :distortion

Ch1 Ch2 Ch1 Ch2

Mixing up channels !!!


Table of contents

Definitions : HD, IM, intercept point, ..Distortion in a MOST

Single-ended amplifier

Differential amplifier

Distortion in a bipolar transistorReduction of distortion by feedbackDistortion in an opampOther cases of distortion and guide lines


Linear distortion

vIN

t

vOUT

t

vOUTvIN

f

High-pass filter


Linear distortion

vIN

t

vOUT

t

vOUTvIN

f

Low-pass filter


Non-linear distortion

vIN

vOUT

t

tvIN

vOUT

QRounded or compressed

Quiescent point

Expanded


Soft and hard non-linearity

vIN

vOUT

t

tvIN

vOUT

QSoft non-lin.

vIN

vOUT

t

tvIN

vOUT

QHard non-lin.


Non-linearity : by power series

vIN

vOUT

t

tvIN

vOUT

QSoft non-lin.

vOUT = a0 + a1vIN + a2vIN

2 + a3vIN3 + ...

vIN(t) vOUT(t)


How to find a0, a1, a2, a3, ...

y = a0 + a1u + a2u2 + a3u3 + …

a0 = y a1 =dydu u = 0

a3 = d3y

du3 u = 0 61

u = 0

a2 = d2y

du2 u = 0 21


Definition of harmonic distortion HD

y = a0 + a1u + a2u2 + a3u3 + …

With u = U cos ωt 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 ωt + U2 cos 2ωt + U3 cos 3ωt

HD2 = U

4a22

a34

3

21

41a2

a1

a3a1

HD3 = U2


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-distortionregion :slopes 1 and 2

Noise


HD of a resistor

HD (dB)

Output Voltage (Vptp)

slope 1

slope 1

slope 2

slope 2

THD = HD22 + HD3

2 + …

HD2


Output spectrum

Freq. (Hz)

Output amplitude (dB)

HD2 HD3

2f 3ff


Definition of intermodulation distortion IM

y = a0 + a1u + a2u2 + a3u3 + …

with u = U (cos ω1t + cos ω2t )

y = a0 + …

IM2 = 2 HD2 = U

IM2 at ω1± ω2 IM3 at 2ω1± ω2 and ω1± 2ω2

a2a1

a3a14

3IM3 = 3 HD3 = U2


IM components

f2f1-f2 f1 f2 2f2-f19 10 11 12

2f1 f1+f2 2f2 20 21 22

3f1 2f1+f2 2f2+f1 3f230 31 32 33 MHz

0 f2-f10 1

IM3IM3IM3IM3IM2IM2 HD2 HD2 HD3 HD3

F F


IM components

f2f1-f2 f1 f2 2f2-f19 10 11 12

2f1 f1+f2 2f2 20 21 22

3f1 2f1+f2 2f2+f1 3f230 31 32 33 MHz

0 f2-f10 1

IM3IM3IM3IM3IM2IM2 HD2 HD2 HD3 HD3

a1U

2 x or 6 dB

3 x or 9.5 dB

a3U334 a3U31

4a2U2a2U2 a2U21

2

IM3 3 x larger !Next to F’s !


Output spectrum of amplifier for IM

Output amplitude (dB)

Freq. (MHz)

IM3IM3

Ref.: J.Silva-Martinez, Kluwer 1993


Amplitude IM3 versus input signal

Vin (dB)Noise

Vout (dB)

a1Vin

43 a3Vin

3

VNout

IM3

IMFDR3

DRN

IP3

1 dB -1 dBCompressionpoint

IP3 is the IM3Intercept point

SFDR3


Relation IP3 and IM3

Vin

Vout

a1Vin

43 a3Vin

3

VNout

IM3

IMFDR3

DRN

IP3

1 dB IP3 is Vinwhere IM3 = F

IP3 =a1a33

4

= VinIM3

1

= VindB - IM3dB21


Vin

Vout

a1Vin

43 a3Vin

3

VNout

IM3

IMFDR3

DRN

IP3

1 dB

Relation IMFDR3 and IP3

IMFDR3 = max DR@ VNout = 4

3 a3Vin3

IMFDR3 =

a1a33

4 1VNin

2

3

= IP3VNin

( )2/3

= (IP3dB-VNindB)32


Vin

Vout

a1Vin

43 a3Vin

3

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


Relationship exercise

Vin

Vout

a1Vin

43 a3Vin

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 µVRMS (-78 dBm)

IMFDR3 = 119 = 73 dB32

Vin1dBc = 21 dBm


Definition of crossmodulation distortion CM

y = a0 + a1u + a2u2 + a3u3 + …

with u = U cos ω1t + U (1 + mc cos ωct ) cos ω2t

CM3 = mc U2 = mc IM3 43

mc

f

mc

fCM3

a3a1

ω1 ω2 ω1 ω2


Table of contents






iDS = K (vGS - VT)2 WL

IDS + ids = K (VGS + vgs - VT)2

IDS is the DC componentiDS is the DC + ac componentids is the ac componentIds is the amplitude of the ac component

Distortion in a single-MOST amplifier

K = K’


DC and ac components

vIN

iDS

t

tvIN

Q

IDS : DC componentiDS : DC + ac componentids : ac componentIds : amplitude of the ac

component

Ids

IDSiDS

ids


IDS = K (VGS - VT)2 WL

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’


ids = 2K (VGS - VT) vgs + K vgs2

or ids = g1vgs + g2vgs2 + g3vgs

3 + …

g1 = 2K (VGS - VT)g2 = Kg3 = 0

IM2 = Vgs =g2g1

& IM3 = 0Vgs

2(VGS-VT)

Coefficients a1, a2, a3 by comparison

WL

K = K’


Normalized current swing

ids = 2K (VGS - VT) vgs + K vgs2 iDS = K (vGS - VT)2

or y = a1u + a2u2 + a3u3 + ..

y = =ids

IDS

y = = u + u2Ids

IDS

2 vgs

VGS - VT+

14

2 vgs

VGS - VT( )2

14

Vgs

(VGS - VT)/2U =

y is the relative current swing !


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 !


In generalids = gmvgs + K2gmvgs

2 + K3gmvgs3 +

govds + K2govds2 + K3govds

3 +gmbvbs + K2gmbvbs

2 + K3gmbvbs3 +

K2gm&gmbvgsvbs + K3,2gm&gmbvgs2vbs

+ K3,gm&2gmbvgsvbs2 +

………… +K3gm&gmb&govgsvdsvbs

More coefficients a1, a2, a3 ...


Distortion of a MOST diode

iDS = K (vDS - VT)2

y = =ids

IDS

y = = u + u2Ids

IDS

2 vds

VDS - VT+

14

2 vds

VDS - VT( )2

14

Vds

(VDS - VT)/2U =

Same as for a MOST transistor amplifier !


The zero HD3 point for smaller L

gm

VT

wi

vs

VGSgm’

gm’’

IDS’’’ = gm’’

gmsat = WCoxvsat

HD3 = 0 at VGS = VT ?

VGS

si


Derivatives of gm

Ref. Fager JSSC Jan. 2004, 24-33

W = 60 µmL = 0.6 µmVDS = 2 V


A differential pair is symmetrical

vOd

vIN

t

t

0

RLIB

rounded: compressed

vId

symmetrical: no 2nd order


y = =iOd

IB

vId

VGS-VT

14

vId

VGS-VT

( )2

IBVGS - VT

1 -

vId is the differential input voltageiOd is the differential output current (gmvId) or

twice the circular current gmvId /2IB is the total DC current in the pair

Note that gm = = K’ (VGS - VT)

Distortion in MOST differential pair

WL


y = = U 1 - U2IOd

IB

14

VId

VGS - VTU =

IM3 = U2332

≈ U - U318

1 - x ≈ 1 -x2

Distortion in MOST differential amplifier

y = =iOd

IB

vId

VGS-VT

14

vId

VGS-VT

( )21 -

IP3 = 4 (VGS - VT) ≈ 3.3 (VGS - VT)23

IM2 = 0U is the relative current swing


Distortion in linear region

VDS1 = RDID ≈ 0.2 V

IDS1 = β1VDS1(VGS1-VT)

gm1 = β1VDS1 is constant

Ref. Alini,JSSC, Dec.92, pp.1905-1915

Low distortion !


Table of contents






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 << 1 u2

2u3

6

Distortion in a bipolar transistor amplifier

ICE DC componentiCE DC + ac componentice ac componentIce amplitude of

the ac component


is the non-linear equation

y ≈ u + + + ... u2

2u3

6U =

Vbe

kTe/q

y is the relative current swing !

a1 = 1a2 = 1/2a3 = 1/6

IM2 = U =

IM3 = U2 = ( ) 2

a2a1

Vbe

kTe/q12

34

a3a1

Vbe

kTe/q18

Distortion in a bipolar transistor amplifier


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 mVthen yp = 0.1/0.026 ≈ 4 (must be << 1 !!) gives IM2 = ?? Too high distortion !!


y = = u + +IdID

U =

Same as for a Bipolar transistor amplifier !

Vd

kTe/q

y ≈ u + + + ... u2

2u3

6

u2

2u3

6

iD = IS exp( )vD

kTe/q

Distortion in a diode


Distortion in bipolar differential amplifier

y = = tanhiOd

IB

y = ≈ U - U3 IOd

IB13

VId

2kTe/qU =

IM3 = U214

VId

2kTe/qtanh x = e

x - e-x

≈ x - x313

ex + e-x

U is the relative current swing

IP3 = 4 kTe/qIM2 = 0


C = C0 ( 1 + a1V + a2V2 + ... )

For poly-poly caps : a1 ≈ 20 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


Non-linearity depletion capacitance

VIN0-VB

Cj

C0

Cj =C0

1 -vIN

Φ

Cj = C0B (1 + x)-1/2 = C0B (1 - 1/2 x + 3/8 x2 - 5/16 x3 + ..)

vIN = VB + vin

Cj =C0

1 +VB

Φ

1

1 +VB + Φ

vin

x


Table of contents






Distortion reduction by feedback

a1 a2 a3

d1 d2 d3

F

Σv u

v

y

y

u = v - Fy

y = a1u + a2 u2 + a3 u3

y = d1v + d2 v2 + d3 v3

elim. uelim. y coeff v2 : d2

coeff v : d1

coeff v3 : d3

-


Distortion reduction by feedback

a1 a2 a3

d1 d2 d3

F

Σv u

v

y

y

Loop gain 1+T = 1+a1F u 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

≈ 1F

-


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)2a2

a1-

V2

(1 + T)2IM3f = V2d3

d1

34

= ( )[ ]34

2

Distortion components with feedback

reduction in current swing

expansioncompression

reduction by loop gain

reduction in current swing


1

(1 + T)

a3

a1

2T

(1 + T)2a2

a1-

V2

(1 + T)2IM3f = V2d3

d1

34

=

(

)[ ]34

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

T1

a12

a3

T1

a1 a3a11 -

2 a22

(

)For large T : =


IM2f = Vin

(1 + T)212

1(1 + T)

=U2

Emitter resistor to reduce distortion IM2f

1kTe/q

Vin

(1 + T)1

kTe/qU = is the relative current swing

T = gm RE =VRE

kTe/qa2

IM2f decreases linearly with T for constant U !

a1 2=

1


IM3f =

Null for T = 0.5

Emitter resistor to reduce distortion IM3f

1 - 2T(1 + T)2

U2

8

Vin

(1 + T)1

kTe/qU = is the relative current swing

IM3f also decreases with T for constant U for large T !!

a2a1 2

=1 a3

a1 6=

1


Null in IM3 ifa3 (1 + T) = 2f a2

2

a3 (1 + T) = 2T

0.01 T

RE

IM%

1

0.1

0.01

0.1 1 10 100

1 10 100 1kΩ

a1

1

a22

T = - 1a1a3

2a22T = 0.5

IM2

IM3Same slopes !

Null in IM3 by RE (Bipolar trans. ICE = 1 mA)


Vin

T

1

kTe/qU = =

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 T3

Vin 1=)2


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)23U2

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


Iout

M2

R

M1VG

+

-

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

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= =


vinvOUT

IBVB

vOUT = vIN - vGS

K’W/L

IBvGS = VT +

vOUTF = |2ΦF| +vOUT - |2ΦF|

vIN = vOUT + VT0 + γ [ vOUTF ] +K’W/L

IB

VT = VT0 + γ [ vOUTF ]

Distortion Source follower with substrate effect

CL


VOUT

VIN

γ = 0slope 1

γ = 0.8 V 1/2

slope 1/n

0

vIN = u2 + γ u + B

u2 = vOUT + |2ΦF|B = VGS0 - |2ΦF| - γ |2ΦF|

K’W/L

IBVGS0 = VT0 +

VT0 = 0.6 V ; VGS0 = 0.9 V; 2ΦF = 0.7 V; B = -0.47 V; 1/n = 0.73a1 = 0.765; a2 = 0.02; a3 = -0.0035VINp = 1 Vp; HD2 = 1.32 %; HD3 = -0.114 %

2 3 V0

2

V

1 1

1.37

0.9 2.27

1

Distortion Source follower - Example


Increasing the IP3 by feedback

M1Vid/2 -Vid/2

2Ibias

-ioutiout

M1 M1Vid/2 -Vid/2

Ibias

-ioutiout

M1

IbiasR R 2R

IP3 ≈ 3.3 (VGS-VT)(1+gm1R)2 HD3/n2 n= 1+gm1R

HD3 = - 60 dB for Vid = 1 V requires VGS-VT = 0.38 V and gm1R = 3 !!!


Increasing the IP3 by feedback

Additional local FB

2R

More FB with opamps

2R


Distortion cancellation

Vid/2 -Vid/2

IB2IB1

iout

M1

iout

M2M1

Parameters :α = IB2 / IB1

≈ 0.25v = VGST1 / VGST2

≈ 1.6VGST = VGS-VT

IM3 ≈ 0 if v = α-1/3

then iout = gm1 Vid (1 - α2/3 )


Distortion cancellation

IM3 ≈

iDS

IB= U - U3 U =

Vid

VGS - VT

18

IM3 = U2332

Vid

VGS1- VT

332

)2( 1 - α v3

1 - α v

IM3 ≈ 0 if v00 = α -1/3

at which point iout = gm1 Vid (1 - α 2/3 )

iout = 2 (iDS1 - iDS2)


Compensation of IM3

-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

α = 0.20v00 = 1.71

α = 0.25v00 = 1.6

α = 0.33v00 = 1.44

v

1 - α v3

1 - α v


Output signal vs current ratio

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 α

1 - α 2/3

α ≈ 0.25v00 = 1.6

x 0.6


Table of contents






Miller CMOS opamp with Feedback

GBW = 10 MHz & Avc = 10 ZL = 100 kΩ//5pF

R2

+-

R1vIN vOUT

ZL

Av

Av0

f

BW = 1 kHz

GBW = 10 MHz

101 MHz


Distortion in input stage

a1 a2 a3

d1 d2 d3

F

Σv u

v y

u = v - Fy

y = B1(a1u + a2 u2 + a3 u3)

y = d1v + d2 v2 + d3 v3


coeff v : d1

coeff v3 : d3

B1y

-1+T = 1 + B1a1F


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)2a2

a1-

V2

(1 + T)2IM3f = V2d3

d1

34

= ( )[ ]34

2

Distortion in input stage

1+T = 1 + B1a1F

Same as before but with different Loop gain :


Distortion in input stage with LPF

a1 a2 a3

d1 d2 d3

F

Σv u

v y

u = v - Fy

y = B1p(a1u + a2 u2 + a3 u3)

y = d1v + d2 v2 + d3 v3


coeff v : d1

coeff v3 : d3

yB1fp

-

1+T = 1 + B1pa1F


IM2f = V

(1 + T)2a2

a1

1

(B1pa1F)2

a2

a1= V

1

(1 + T)

a3

a1

V2

(1 + T)2IM3f =

34

Distortion in input stage with LPF

diff.pair

=a3

a1

34

1

(B1pa1F)3V2

f

ffp

fp

40 dB/dec

60 dB/decIM3f =

Single trans.

a22

a12

34

2

(B1pa1F)3V2


Distortion in output stage

A1

d1 d2 d3

F

Σv u

v y

u = v - Fy

y = A1b1u+A12b2u2+A1

3b3 u3

y = d1v + d2 v2 + d3 v3


coeff v : d1

coeff v3 : d3

yb1 b2 b3

-1+T = 1 + A1b1F


IM2f = V

(1 + T)2b2

b1

A1

(A1b1F)2

b2

b1= V

Distortion in output stage

f

f

= b2

2

b12

34

2 A12

(A1b1F)3V2

IM3f =

Single trans.

2T

(1 + T)2b2

b1

V2

(1 + T)2( )

34

2

x A1

x A12


IM2f

Two-stage opamp a & b

ffp

V

T2

a2

a1 fp

f

V

T2

b2

b1a1 T = a1b1F

IM3f

ffp

V2

T3

a3

a1

V2

T3

b3

b1a1

2

34

( )2

fp

f( )

3( )

a1a3 - 2 a22

a1a3

( )b1b3 - 2 b2

2

b1b3

34


IM2f

Three-stage opamp a & b & c

ffp

V

T2

a2

a1 fp

f

V

T2

c2

c1a1b1 T = a1b1c1F

IM3f

ffp

V2

T3

a3

a1

V2

T3

c3

c1a1

2b12

34

( )2

fp

f( )

3( )

a1a3 - 2 a22

a1a3

( )c1c3 - 2 c2

2

c1c3

34


Distortion in an opamp at low frequencies

12

34

VOUT

= 1 VM1 M1

M2 M2

M4

+- Cc

M3

GBW = 10 MHzAv0 = 10.000BW = 1 kHz Avc = 10

Vin = 0.1 mV

Vm = 10 mV

ZL IDS1 = 6 µAgm1 = 60 µSIDS3 = 120 µAgm3 = 1.2 mSRL = 100 k ΩCL = 5 pFCc = 1 pF


Low-distortion amplifier

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+Cc

Cc

≈ 0.3

vin = constant

≈fnd

GBW



12

34

VOUT

= 1 VM1 M1

M2 M2

M4

+- Cc

Vnode at 100 Hz ?M3

GBW = 10 MHzAvc = 10

Vin = 0.1 mV

Vm = 10 mV

ZL

U1 = gm1Vin/IDS1 = 5 10-4

IDS1 = 6 µAIDS3 = 120 µA

U3 = gm3Vm/IDS3 = 0.1


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

12

34

VOUT

= 1 VM1 M1

M2 M2

M4

+- Cc

Vnode at 100 kHz ?M3

GBW = 10 MHzAvc = 10

Vin = 10 mV

Vm = 10 mV

ZL

U1 = gm1Vin/IDS1 = 5 10-2

IDS1 = 6 µAIDS3 = 120 µA

U3 = gm3Vm/IDS3 = 0.1


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 in an opamp at high frequencies

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

HD2 (dB)

Freq. (Hz)


1.8 V Low distortion CMOS Opamp

Ref.Hernes Kluwer 2003

GBW ≈ 3 GHz

CL = 8 pF

fP = 380 MHzat 0.38 Vpeak

SR ≈ 900 V/µs

fP =2π Vpeak

SR

Large VGS4-VT


HD2 & HD3 vs Amplitude

large distortion

Slope 1

Slope 2


HD2 & HD3 vs Frequency

HD2 HD3dBdB

Hz Hz


Table of contents






Other cases of distortion and guide lines

Distortion caused by limited SRDistortion of a switchDistortion at high frequencies :

Volterra series instead of power seriesDistortion in continuous-time filtersGuide lines


Guide lines for low distortion

Scaling such that voltage amplitudes

are limited

Scaling such that relative current swings

are limited

Feedback

All fully differential


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


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

Distortion components with Feedback (T > 5)


References

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


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Distortion in elementary transistor circuits

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