Chapter 18

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Willy Sansen 10-05 181

Distortionin elementary transistor circuits

Willy Sansen

KULeuven, ESAT-MICAS

Leuven, Belgium

[email protected]

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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


Distortion in a MOST

Single-ended amplifier






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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









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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









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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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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 !!!


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Additional local FB

2R

More FB with opamps

2R

Distortion cancellation

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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 )


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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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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


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


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


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


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12

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


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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

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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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Chapter 18

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Transcript of Chapter 18