cross sectional view of fet

ECE 584, Summer 2002 Brad NobleChapter 3 Slides

1

Cross Sectional View of FET


2

FET I-V Characteristic


3

Saturation Voltage• Vpinchoff = VDS,sat = VGS – VTH

– Separates resistive from saturation region• The drain current is given by

• Solving for VDS,sat:

PNoxnL

WNTHTN

TNGSNDS

SKCVVVVI

and NFETfor ,2

21

N

DSTNGSsatDS

IVVV

2,


4

Early Voltage Function of Length


5

Early Voltage in MOSFETs• Due to channel length modulation:• Good to solve for quiescent voltage-current.

E

DSTHGSPNDS

THGSPN

DSTHGSoxnDS

VVVVSKI

LLVVSK

VVVL

WCI

121

121

121

2

2

2


6

Ex: Find VDS,sat for an NFET


7

Body Effect


8

Variations in VTH Across Channel• We assume VTH is constant across channel

THIS IS NOT TRUE!• Depletion region is thick at S and thin at D.

Fox

depmsTH

CQV 2 Cox

Cdep

inversionlayer

Gate oxidecapacitance

Depletion cap,function of x

5.11factor slope ldsubthresho2

2

ox

dep

TNGSN

DS

CCn

VVn

I


9

Small Signal Equivalent Ckt


10

Parasitic Capacitance


11

Capacitance Equivalent Circuit


12

Variation in Capacitance


13

Notes on PFETs• PFETs typically have a shape factor 3 or 4

times larger than NFETs• Body effect can be eliminated in PFETs by

tying the n-well to VDD

– Need 6m spacing between n-wells to isolate.– Dr. Engel always does this on input devices,

not always elsewhere.


14

Subthreshold Conduction


15

Weak Inversion• What really happens if VGS < VTN?

• In digital design, IDS = 0.• We call it “weak inversion” or W.I.• IDS is primarily due to Idrift in strong inversion

and Idiffusion in weak inversion.

V1V950

TH

GS

V.V


16

Modes of Inversion• IDS = Idrift + Idiffusion

• If VGS > VTN the channel has been inverted.• To be more precise, we can say the channel

has been “strongly inverted” (S.I.) due to an abundance of carriers in the channel.

• Inversion is independent of whether the FET is in the linear or saturation region.


17

Weak Inversion Idiffusion

• Drain is more reverse biased than source:

• To find Idiff, compute gradient• Because no carriers are lost as they travel

from S to D, current is the same for all x and gradient is not a function of x.

• Note: This is not really true due to recombination, but its close!

kT

VVqNN SGOS

exp0

kTVVqNN SGO

D

exp0

dxdN


18

W.I. Surface Potential

oxd

d

oxd

oxs

CjCj

CjCC

C

11

1

oxC

dC

GV

potentialsurface

S

factor slopeldsubthresho ,5.11

ox

d

CCn

device law lExponentiaU

exp1U

expnU

exp0

T

DS

T

S

T

GDDS SII v-v-v


19

Deriving Weak Inversion IDS

T

S

T

D

T

GSSD

LNN

LNN

dxdN

Uexp

Uexp

Uexp0 v-v-v

iprelationshEinstein thea.k.a. ,

coeff.diffusion is where,

:unit widthper current The

qkTD

DdxdNqD

WI

nn

nnDS

T

D

T

S

T

GDDS I

LWI

Uexp

Uexp

nUexp

0

v-v-v


20

W.I. FET As Exp. Law Dev.• S must be big for device to be useful.• If VDS = 100mV, can be neglected.

• For W.I. vDS,Sat 100mV• Looks like a BJT

T

DS

Uexp v-

mV100for ,U

expnU

exp0

DS

T

S

T

GDDS VSII v-v

T

BESC II

Uexpv

BEv

Ci


21

Inversion Coefficient• Let

• Shape factor as a function of :

Lets you chose shape to match inversion mode.

2U2 t coefficieninversion

T

DS

ni o

< 0.1 Weakly Inverted (W.I.) > 10 Strongly Inverted (S.I.)

0.1 < < 10 Moderately Inverted (M.I.)

2U2 TPN

DS

KniS

o


22

Ex. Using Inversion Coeff.

W.I. 049.0(100))(26mV)2(1.5)(100

1uA ,100 uA,1 2

SiO

DS

M.I. 9.4(100))(26mV)2(1.5)(100

100uA ,100 uA,100 2

SiO

DS

S.I. 49(100))(26mV)2(1.5)(100

1mA ,100 mA,1 2

SiO

DS


23

Small Signal Analysis

GSV

gsGS vvBias DCChange SmallVoltage Total

O

GSv O

BSv

O

DSv

O

DSi

GSO V GSvVoltageQuiescent


24

Ex: Quiescent PointV75.00TV

V81EV

59.02V

A50 PNK

mmS

10μμ100

O

GSv O

BSv

O

DSv

O

DSi

EFFPN

VSK

OOOO DSSBGSDS

vvvvi T0 12221 2

mA34375.1815175.0350

10100

21 2

2

VA

mmO

DSi

Question: How many digits are significant?

V3O

GSvV5O

DSvV0O

BSv


25

Small Signal Model Limits• Suppose the previous circuit is the input

device of an amplifier.

• Small-signal model holds as long as the deviations are small qkT


26

Taylor Series Expansion• Taking a Taylor expansion of one variable:

SBSB

SBDS

DS

DSGS

GS

DSDSDS

iiiii vv

vv

vv

0000

mg dsg mbg

gsvv-vv GSGSGS O

dsvvDS sbvvSB

DSDS iii ODS

202

100 ))(())(()()( xxxfxxxfxfxf

Approx.Linear


27

Small Signal Model Params

dsE

DS

DS

DSds

rViig

O 10

v

1typically ,22

where0

O

SBFm

SB

SBmb gig

vv

OOO

DSmEDSEGS

DSm igVi

Vig 2 , largefor ,12

0

DSvv

dsmrgGain


28

Example: Small Signal Analysis

μA57.30)26)(0166.0()26)(1592.1(

2

:analysis signal small UsingμA65.31 mA 3754.1

:) work!of (lotsequation full UsingV026.5 V,0 ,)by (up V026.3Let

250 V,81 mA,34375.1

mVmSmVmS

2V

μA

dsE

DSgsdsdsgsm

DSBSGS

E

Viiggi

ii

qkT

Vi

OO

O

vvvv

vvv

DS

DS

DS

DSDS


29

Small Signal Low-Freq Model

gsmg v dsmbg v dsr

dsi

gsv

small

signals

S.I.

Sat

1.5factor slope ldsubthresho where,2 n

nig

ODS

m

OO

ds

E

ds

Eds

iLV

iVr


30

Ex: Find gm and rO

μm10μm4μA10

LWiO

DS

M10μA10

μm10μmV10

μS235.1

uA10μA/V1001042

so, If saturated? and S.I.it Is2

ODS

ENds

m

iLVr

g


31

Transconductance: W.I. & M.I.• What is gm for a weakly inverted FET?

• What is gm for a moderately inverted FET?

T

DS

GS

DSm

niig

O

U

0

v

exp1 where,

U

0

T

DS

GS

DSm

niig

O

v

modes allfor E

DSds

Vig

O

Not

in te

xtbo

oks!

cross sectional view of fet

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