132654164 solution manual for semiconductor devices physics and technology sze s m solution

7/21/2019 132654164 Solution Manual for Semiconductor Devices Physics and Technology Sze S M Solution

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ch.l

ch.2

ch.3

ch.4

ch.5

ch.6

ch.7

ch.8

ch.9

ch.10

ch. r l

ch.t2

ch.13

ch.r4

Contents

Introductiorr-----

---

0

Energy

Bands

and

Carrier

Concentration

I

CarrierTransport

henomena

-----

----

7

p-n

Junction

----

---

16

BipolarTransistorndRelated evices- --------- 32

MOSFET

nd

Related evices---------

----------

48

MESFET

andRelated

Devices-------

--

60

Microwave

Diode,

Quantum-Effect

ndHot-Electron

Devices

-----

68

Photonic

evices

----------

-----------

73

Crystal

Growth

and Epitaxy------

Film

Formation------

-

92

Lithography

ndEtching

----

99

Impurity

Doping--

-

105

Integrated

evices---------

--

l 13



CHAPTER

2

1.

(a)

From

Fig. llq

the atom

at the center

of the cube is

surround

by four

equidistant

earest

eighbors hat lie

at the corners

of a tetrahedron.

Therefore

the distance etween earest eighborsn silicon a: 5.43A) is

l/2

[(a/2)'

*

(Jzo

/2127t/'

J-zo

4

:

235 A.

(b)

For the

(100)

plane,

here

are wo

atoms

one

central

atom and4 corner

atoms

eachcontributing

l4

of an atom

for a total of two atoms

as shown n

Fig. 4a)

for an

area f d,

therefore

we have

2/

&:2/

(5.43

.

l0-8)z

618

*

10la

toms

crt

Similarlywe

have or

(110)

plane Fig.

4a

andFig. 6)

(2+2x l l2+4x l l4 ) / JTo2

:9 .6 , .

10rs tomsc r1 . ,

and or

(111)

plane Fig.

4aandFig.

6)

(3

/2+

x

r/6)rlz|mlf

,ffi"

I

:

:

7.83 10la

toms

crrt.

(9.

2. The

heights t X, Y,

and

point

are

/0, %,^O

%.

3.

(a)

For the

simple

cubic,a unit cell

contains1/8of a

sphere

t

eachof the eight

corners or a total of one sphere.

-

Ma><imum

raction

of cell filled

:

no.

of sphere

volume

of

each

sphere unit cell volume

:1x

4ng /2 )3

a3

:52o/o

(b)

For

a face-centered

ubic,

a unit cell

contains

1/8

of a sphereat each

of the

eight

corners or a total

of one

sphere. The fcc also contains

half a sphere

t

each

of the

six faces or

a total of three

spheres.

The

nearest eighbordistance

is

l/2(a

J;

).

Thereforehe

radiusof eachsphere

s

l/4

1a

Jz

).

-

Maximum

fraction

of cell filled

-

(1

+

3)

{4[ [(a/2)

4It

I 3] / a3

:74o/o.

(c)

For

a diamond

affice,a unit

cell contains1/8

of

a

sphereat eachof the

eight

corners

or

a total of

one sphere, /2

of

a

sphere t eachof the

six faces or

a

total

of three

spheres, nd 4

spheres nside he cell. The

diagonaldistance



between

112,0,

) and

114,

l4, Il4)

shown n

Fig.

9a

s

The

radius

f the sphere

sDl}:

1Jj

8

- Maximum fractionof cell filled

:

(t

+

3

+

4)

yErtj l '

'o,

:nJT

t16

34

%.

1 3 \ 8

) )

This s

a relatively

ow

percentage

omparedo

other

attice

structures.

4.

la,l la,l

la,l: aol

a

4 *4+4+4 :o

4 .

( 4

4

+ 4 4 ) : 4 .

o o

la, l ' *4

4

*4.4

+

4.L:

o

--d2+ d2coflrz

+

dcoiln

I dcoflr+ !

dz

+3

d2coil!

0

-

coil:

+

[:

cos-r

+

[ 109.4/

5. Taking

the reciprocals

f these ntercepts

we

get

ll2, ll3

and

l/4.

The smallest

three integers

having

the

same atio

are 6, 4, and

3. The

plane

is referred

o as

(643)

plane.

6.

(a)

The

latticeconstant

or

GaAs s

5.65A, and he

atomic

weights

of Ga and

As

are 69.72 and7492 glmole,respectively. Thereare four gallium atomsand

four

arsenic

atoms

per

unit

cell,

therefore

4/a3

4/

(5.65

x

10-8)3

4.22

x

lTn

Ga

or As atoms/cr*,

Density:

(no.of

atoms/crrf

x

atomic

weight)

Avogadro

constant

:

2.22

1022(69.72

74.92)

6.02* 1023

5.33

g

I

cni.

(b)

If

GaAs is

doped with

Sn and

Sn atomsdisplace

Ga atoms,

donors

are

formed,

because

n has our valence

electronswhile

Ga has

onlv three.

The

resulting

emiconductors

n-type.

7. (a)Themelting emperatureor Si is l4l2 oC,and or SiOz s 1600oC.Therefore,

SiOz has

higher

melting temperature.

t is

more diflicult

to break

he

Si-O

bond han

he

Si-Si bond.

(b)

The

seedcrystal

s used

o initiated

the

growth

of the

ingot

with

the correct

crystal

orientation.

(c)

The

crystal

orientation

etermines

he semiconductor's

hemical

and

electrical

; ) '

. ( ; ) '

: 1

2



properties,

uchas he

etch rate,

rap

density,

breakage

lane

etc.

(d)

The

emperating

f

the

crusible

and he

pull

rate.

4.73x10u

'

ErQ): l . l7

for

Si

.'. Es

100K)

=

1.163

V,

andEs(600

K): 1.032

V

E,(D=

.5

n

-t'o-l!j_"t!1,!'

or

GaAs

(T

+ 204)

,.Er(

100K)

:

1.501

V,andEs

600

K)

:

1.277

Y .

9. The

density of holes

n the

valence

band

s

given

by integrating

he

product

N(E)tl-F(DldE

from top

of

the valeri

:e band

En

taken o

be

E

:

0)

to the

boffom of the valence

band

Rottoml

p:

ytu' n^

N(qtl

_

F(DldE

where -F(E): I -

{ t

r [ t * "( t- Ee)*t

] :

[ t

*"(E-Ei lr t ' r l t

If Er-

E

>>

kT then

|

-

F(E)

exp

(n,

-

r)lwl

e)

Then

rom

Appendix

H and, Eqs.

and2we

obtain

p

:

4Df2mp

h2f3D

I:""^

EtD

exp

-@r

-

E)

/ kT

ldE

(3)

Letxt+

E

lkT, and

etEbooo*:

@,

Eq.3

becomes

p

:

4\-2mo

/ rtflz

(kTlttz

exp

[-(Ep

kl)i

I

xtDe*dx

where he

integral

on the right is

of the

standardorm

and

equats

G

tZ.

-

p

:2l2Dmo

kT / h213D

xp

[-(Ep

kI)j

By

refening o

the op

of the valence

and

as ETinstead

f E:0

we have

or

p:2Qo';f:"i1ff;Trrlf;,

, krl

where

Nv:2

(Nmo

kT / rtf

.

10. From

Eq. 18

Nv

:2QDmo

kT

I h2f

D

The effective

mass

of holes n

Si is

mp

-

(Nvt

21ztt

rt

tzDkT)

( 1 )

1.38

10-23

3oo)

:

9.4 10-3

kg

1.03mo.

Similarly,

we have

or

GaAs

f f ip:3.9

x

10-31

g: 0.43

mo.

UsingEq.

19

1 .



E

=

8,

+ il'

.

(%)^

(N,

w,)

=

(Ec*

Ey)l

2

+

(*T

I

4)

ln

At77

K

E,:

(1.t6/2)

+

(3

x

1.3g

t}-,tT)

(4

x

1.6

x

10-,r)

n(l.0/0.62)

:0.58

+

3.29

x

10-5

=

0.5g

+

2.54

x

10-3

0.5g3 v.

At 300K

Ei:

(1.12/2)

(3.29

10-sX300;

0.56

+

0.009

0.569 V.

At

373

K

Ei:

(1.0912)

(3.2g

l0-sx3731

0.545

0.012:0.557

ey.

Because

he

second

erm

on

the right-hand

side

of the

Eq.l

is much

smaller

compared

o

the first

term,

over the

above

emperafure

ange,

t

is reasonable

o

assume

hat

Ei is

in the

center

of the forbidden

gap.

"-@-r

Yw

6B

l'=rr-rr,

lr*,1*,)(o%f

(1 )

I::@_EC

T2.KE

:

1 3 .

f:

JE -Er"-@-tP)/*r6P

1 . 5 x 0 . 5 " G

0.sJ;

?

=

;or.

(a)

p:

ftw:9.109

x

10-3r

105

9.109

x

10-26

g-mA

1

:

h

-

6 '626x10-14-

:7.27

x

r0-e

m:72.7

A

p 9.109 l0- ' "

m ^ ^

I

(b)

1"

'L

:

*x

72.7: I154

A

.

m

p

0.063

From

Fig.22when

nr:

l0t5

cd3,

the

corresponding

emperature

s

1000

T: l.B.

So hat

:

1000/1.8:555

K

or 282

[

From

E"

-

Er:

kT

lnlNc

/

(No

-

N,q)]

which

can

be rewritten

as

No

-

N.e: l/c

exp

IaE,

-

Er)

kf

I

Then

No-N.a:2.86

*

10re

xp(-0.20

0.0259):

1

26

x

1016

rn3

or No:1 .26x1016+. t :2 .26 1016c rn3

A compensated

emiconductor

an

be fabricated

o

provide

a

specific

Fermi

energy

evel.

16. From

Fig.28a

we

candraw

he

following

energy-band

iagrams:

14 .

1 5 .



AT 77K

Ec(0.ssevi'

EF(0.s3)

E;(o)

Ev(-0.59)

Ec(0.56 v)

EF(0.38)

Ei

(o)

Ev(-0.56)

E9(0.50eV)

AT 3OOK

ri

AT 6OOK

0.s0)

17.

(a)

The ionization

energy

for

boron in

Si

is

0.045

eV.

At

300 K,

all boron

impurities

are onized.

Thus

pp:

N.a:

l0ls crn3

np: t?i2

n.a:

(g.6i

"

K;9f / l} ts

:9.3

*

lOa

crn3.

The

Fermi

level

measured

rom the

top of the

valence

band

is

given

by:

Ep-

Ev:

kTln(N/ND):0.0259ln(2.66

x

10re

l0r5;

0.26

eY

(b)

The

boron

atoms

compensate

he

arsenic atoms;

we

have

p p :

N e _ N n : 3

x

1016_

. 9

x

l 0 1 6 : l O l s c r n 3

Sincepo

is

the same

as

given

in

(a),

the values for

no and

Ep

are he

same

as

in

(a).

However,

the

mobilities

and resistivities

for these

two samples

are

different.

18.

Since Np

>>

ni, wa

can approximate

e

:

Nt and

po:

n? no:9.3

x l } te

I l0 l7 9.3

x

ld

crn3

( p

- r ' \

From

fio: txiexP

"t

"'

|

\ . k T ) '

we

have

Ep

Ei:

kT ln

(no

n)

:

0.0259 n

(1017

9.65 10e)

0.42

eY

The

resulting

lat

banddiagram

s :



E c

EF

A.rr2eY

- E -

F .

t - l

19. Assuming

complete

onization,

he Fermi

level

measured

rom

the intrinsic

Fermi

evel s

0.35

eV for

10rscm-3,0.45

V for 1017

rn3,

and

0.54eV

for

10le

crn3.

The

number

of electrons

hat

are onized

s

given

by

n = Npfl

-

F(En)l:

Np /

fl

+

"-(ro-rr)r*r

Using he

Fermi

evels

given

above,

we obtain

he

number

of ionized

donors

as

l

.12

V

l

20.

No*

Therefore,

he

assumption

f complete

onization

s

valid

onty for the

case

of

10ls

rn3.

10 tu

1016

: _

l +

e - {Eo -E r ) / k r

1

+e -0 . 13s

n: !0 t5crn3

n: 0.93 1017rn3

n :0 .27

*

lO le

rn3

:5 .33

t

10 ls

rn3

for

Na

:

1015

rn3

forNo: 1017 rn3

forNo:

10le

rn3

_

l0 t u

. l

1 . t45

Theneutral

onor:

1016-

.33

.l0ls

crr3

:

4.67

x

1015rn3

-The

ratio

of

N; - 4'76

0.g76

N;

s.33



CHAPTER

I .

(a)

For ntrinsic i,

4,

1450,

+

:505,

and

n:

p

:

lti:9.65x lOe

We have

p

-

-

3.31x

0'

C)-cm

qntth

+

{lpltp

en,(lt,

+

pr)

(b)

Similarly

or

GaAs,

h:

9200,

b

320,

and :

p

:

ni:2.25x106

We have

p

-

=

2.92x

lOt

O-crn

qnltn

+

llPIIp

For

laffice scatterinE,

-h

n

73/2

en,(ltn

+

po)

2.

3. Since

1 1 1

- = - + -

p 250 500 Fr 167 cr#N-s.

4.

(a)

p:5x l01s

cd3, n: n / /p :

(9.65x

0e12l5x10rs

1.86x104

m-3

14:4lo

cm2lv-s,

Lh:1300

cm2lv-s

r3oox

#:2388

cm2lv-s

T

:

4ooK,

1^4,:

3oox

%

:

844

m'lv-s.

300-rt

2

r

I

:3

C)-cm

T

: 200

K,

Lh:

l l l

- = - + -

P l-t' l-t"

p :

qpon

+

qlrpp

qppp

( b ) p

: N t - N o

: 2 x 1 0 1 6 -

1 . 5 x 1 0 1 6 : 5 x 1 0 l t

* ' , n :

L 8 6 x 1 0 a

m - 3

!-b: Ib

(M

+

No)

:

I+

(3.5xl0tu)

:290

cm2A/-s,

fh: th(Nt

+

Np):

1000 cm2ny'-s

=

|

:4.3

C)-cm

qLthn

+

q[Lpp

qppp



6.

(c)

p:N,q

@oron)

Nn+

N,q(Gal l ium):5x10lsd3, n: L86xl0a

cm-3

I+: I$

(M

r

Np* Ne):

14

Q.05x10tt)

150

cm2A/-s,

l-h: th

(Ne

*

Np* N,q) 520 crr?A/-s

p:8.3 C)-cm.

5. AssumeNo- N1>>n;,theconductivity

s

given

by

ox

qn[h:

elh(No

-

Nd)

We have hat

16

(1.6x

Oae)1^6Qtp-0tt)

Since

mobility

is a function

of

the ionized mpurity concentration,

we can use

Fig. 3 along with trial

and

error o

determine

7^6,

nd No. For example,

f we

choose o

:

2xl0r7,then

Nr

:

ND*

+

Nd- 3x

Qtt,

so hat

Lh

x

510 cm2lV-s

which

gives

o: 8.16.

Further rial and effor

yields

Nn=3.5x1017 rr t.3

and

lh

x

400 cm2lV-s

which

gives

6x

16

(O-cm)-t

o-

q(lt

n

+

Fo

p)

=

ello(bn

+ ni t n1

From he

conditiondddn: 0. we

obtain

f t : n i I

{b

Therefore



i . At the limit when d >> s, CF:

:4.53. Then rom Eq.16

0.226 10-'

-

10.78

mV.

50x10-o

4 .2

R u =

V,A 1 0 x 1 0 - 3

1 . 6 x l 0 - 3

-

426.7

cnf

tC

IB ,W 2.5x l0-3

(30x10-n

l0o )x0 .05

Since he sign

of R'7 s

positive,

he carriers re

holes.From Eq.22

V l0

x

l0-3

p = i x W x C F

-

x 5 0 x 1 0 - 0 x 4 . 5 3 - 0 . 2 2 6

) - c m

FromFig.

6, CF:4.2

(d/s:

10);

using hea/d: 1

curve

we obtain

p *

_ q i

o ( b n ,

4 b + ^ , l b n , )

_ b + l

P,

2JE

QFpni(D

+ l)

=

380 cm2lv-s.

Wp

1.6x10- ' n

1 .46x10 '6

1 .1

7C

lnz

V = p . I / ( W . C F ) _

8. Hall coefficient,

l l 1

'

eR,

l , .6xl0- 'n

426.7

Assuming ex p, fromFig.7 we obtainp: 1.1 )-cm

The mobllity

pg

is

given

by Eq.

15b

1

I

: - =

,

9. SinceR

n

pand

p-

qnph

+

wl_rp

FromEinstein elation D n lt

,

hence R

o.

1

.

nl+ +

pLrp

H l l r o = D n l D o - 5 9



R , _

0.5Rr

We have

N.e:50

Nn

.

N

olt^

N o l t , + N e q p

10. The electric potential

@is

elated

o electron potential

energy

by the charge (-

q)

I

Q:*=(Er_

n i )

q

The

electric

field for

the

one-dimensional

situation

is

defined

as

e(x)

-!!-:l

dEi

d x

q d x

(

n-

-

n \

ft:

niexplT):

No(x)

Hence

l l .

E p -

E ; : k T h (

* r@)

l . )

E

&)

-

-(

tt\

t dN

o(x)

\ q ) N o ( * ) d x

(a)

From

Eq.

31,

Jn:0 and

r

(x)

-

-

D,

d/d*

-

-

kT No!:a)9-

-

*kT

o

H n q N o " n ' q

(b)

E

(r):

0.0259

100)

259

V/cm.

At thermal

and

electric

equilibria,

J

,

=

qgn(x)e

+

qD,

4:!')

-,

dx

12.

Dn

I

dn(x)

D,

N,

- ff .

L

E

( x ) =

F,

- - D n

p,

n(x)

dx

Lt,

N,

-No

N o * ( N r - N ' ) ( * l

L )

L N o

+ ( N ,

- N o ) t

l0



t ,

[ ' -

D"

J o p

I

\a)

Tp

=

oor,rry

=

N,

-No

- -o-nNL

LNo +

(N,

-

N,

)r

p"

N,

13. Nt

=

Lp

-

ToG,

=

10

x

10-6 l0 tu

=

10tt

crn3

f l : f t no + Ln : No +

Ln : 10 I s +10 t t

- 19 t s

c rn3

n?

e.65

on

'

+ lor ,

=

o, , cm-,

-

\ + 4 p =

1 0 "

t4 .

5 x 0 - t 5 1 0 7 2 x l 0 t s

fln

=

tlno,

Pn

=

Pn o

=

10- t

s

10-8 20

3 x 1 0 - a

1 0 - 8

2 0

-

3

x 1 0 {

cm

Sr,

:

v,r,o,N,,, 107

x

2x 10-t6

10to

20

cm/s

(b)

The hole concentration

at the surface s

given

by

F,q.67

p

^(o)

pno

oG

lt

-,

"11'-

I

\

u p * T o S , , )

_

(9.g5

1 9_') ' 10-, t0,,( _

2

xl}tu

(.

=

lOe cm-' .

6=

QnLIt, wqy

Before illumination

1 5 .

After illumination

f l n = f r n o * L r t - f t r o * T r G ,

p n = p n o * L p - p n o * t r G

Ao

-

tqtt,(nno

*

Ln)

+

qlrp(p,"

+

4p)]

-

(qlt

n,"

*

QFo

," )

-

q(l+

+

Fp)r

oG

.

l l

9

x

10-8

oTo



ta)

r .ain

-nD

r#

-

-

l.6x

10-lex Zx

. -

l-

x

l0rsexp -x/12)

L2xl0-"

:

1.6exp

-x/12)

Alcrrl

(b)

/ " ,d r i f t

=

J,o,o,-Jp,a in

:

4.8

-

1.6exp(-x/12)

Alcni.

(c)

't

Jn.a,in

-

qnl+E

-'-

4.8

l.6exp(-xl l2):

1.6x

0-1ex

016x

000x8

E

:3-exp(-x/t2)

V/cm.

1 7 .F o r E : 0 w e h a v e

a _ _ P n _ P n o

D , I *

= g

0t

tp

o

Ax'

at

steady

state, he

boundary

onditions

are

p"

@

-

0)

:

p" (0)

and

pn (,

:

lV)

:

Pno.

Therefore

'*[fj

'*F)

J

(x-

)

-

QDpH,.-

q[.p,(o)

o,"l+-4+)

J

o(x

w)

=

-

QD.H,=,

-

ql.p

18.

The

portion

of injection

current

hat

reaches

T2

p,(x)=

pno

+[1t , (0)-

p^.

n

,(o)-

^,1?4.

'o

,inhf

w

I

lL, )

the

opposite

urface

by

diffirsion

IS



given

by

a $ =

J

o(W)

JoQ) cosh(WLe)

Lo= ^ t ' r%= '& tso* to* -5x lo -2cm

; .

d o

- -

0.98

cosh(10-25x 1O- ' )

Therefore,9SYof

the njected urrent

an

each he

opposite urface.

19. In steady

tate,

he recombination

ateatthe surface nd n the bulk is equal

APr,ou,u

-

LPn,"urf^u

so hat heexcess inorirv *i:;:*..J#'ff at hesurface

;,,surrace

lora.

g

:1013

cm-3

1 0 - o

The

generation

ate can be determinedrom the steady-stateonditions n the

bulk

G:

loto

:

lom crn3s- l

l 0 - 6

From Eq. 62, we can write

D ^ a ' L ! G

& : o

"

Axt

To

The boundary

onditions re

4(.

=

-

):

l01acrn3and

4(*

-

0):

1013

rn3

Hence

where

4( i:

lo la(1 o.ge-'

o

)

Lp:. f io- to-u

31.6

m.

20. The

potential

barrierheight

Qa

=

Q^

X:

4.2 4.0:0.2 volts.

The number

of electrons ccupying

he energy

evel betweenE and

E+dE is

dn: N(DF(DdE

whereN(E^)s the density-of-stateunction,and

F(E) is Fermi-Diracdistribution

function. Sinceonly electronswith an energy

greater

han E, +

eQ^

and having

a velocity component

ormal

o

the

surface

can escapehe solid, the

thermionic

currentdensity s

- | r- +Az.T)%

v,E%e-G-rr)ln g

--

JQt,

=

Jrr*qq^

ht

'

where v, is the component f

velocity

normal o the surfaceof the metal. Since

the

energy-momentumelationship

n

P 2 I

. ) ) ? r

E - -

2m 2m

2 t .

l 3



Differentiation

leads

o dE

-

PdP

m

By

changing

the momentum

component to rectangular

coordinates,

47iP2

P

-

dp,dp

,dp

"

f

-

Z?

, S f

[-

p

,u-'ol

*

p2,+]-z^t1)

zmkr

dp

,dp

dp

,

Hence

=Ht',

^t'r'-,;t, '^rl')u,,

,oo,ll-

,-01r,^0,

p,

f-

,-r)/z^kr

p,

mht

J

p, o

where

p',o

=

Zm(E, +

qQ).

l E

t " l l

2

Since

L

e-o"dx

-{

ll

,

the ast wo

integrals

ield (2dmkT)v'

\ a )

The

first integral

s evaluated

y setting

oi:'9'

-

u

Thereforewe

have

du

-

P'dP'

2mkT

mkT

The lower limit of the first integralcan bewritten as

2m(E,

+qQ)-2mE,

_qQ*

2mkT

kT

so hat he

first integral

ecomes

mkT

fr^,0,

e-"du

-

mkT

e-qL-l

r

Hence

-4tqmkz

72o-a0^lk,

A*7,

"*(-fg^\

h '

\ k T )

22. Equation

79 is

the unneling

probability

r

_ { t *

[ 2 0

s i n h (

. 1 7 x

0 o

g x t o - ' o ] ' ]

: 3 . r 9 x

l 0 * .

L

4 x 2 x ( 2 0 - 2 )

)

23. Equation

79 is the

unneling

probability

p -

r '0-,0)

-

{,

*

[6x

sinh(

.99 10n 10- 'o)f

]- '

_

0.403

L

4 x 2 . 2 x ( 6 - 2 . 2 )

)

Tr0-\

_

{r

*

[o

rinh

q.qq

:.

o'

*lo-'l '

]- '

=

7.8 r0-,.

L^

'

4

x2.2"(e

z.z)

l

t4



Differentiation eads

o dE

=

PdP

m

By changing the

momentum

component

to rectangular

coordinates,

47d,

dp

:

dp,dp

,dp

"

Hence

t

=

+[,

[=-

[ l

=*o,r-rr2'+pi+p2,-znr1)' 'ô'

p,dp

rdp

=

,r2o

ff,,

"to'-'Êt)l2^tr

p,dp,!-_

"-oll'ô'dp,

f*

"ô1/'^0,

p,

rvherep' o

=

2m(E,

+

eQ).

r : l l

2

r' *' -

=[

Z1

,

the ast

wo

ntegrals

ield

2dmk|)v,

rnce

J__

e-"^dx

\" )

The first integral s evaluated y setting

pi;29'

:,

.2mkT

Therefore

we have 4y

-

P'dP'

mkT

The lower imit

of the first integral

can

be written

as

2m(E,+qQ)-2mEo

=qQ.

2mkT

kT

so hat he

irst ntegral

ecomes 0,

fr^,o

e-"du: mlsT

-tQ./

r

Hence =4tq*k' 72"-t0^lw A'7, "*"(

-

uh).

-

h 3

-

k r ) '

Equation79 is the tunnelingprobability

o

_

l2* , (qVo

E)

_

P_tl - - - - - - - - - - ' i -_

2(9. l Ix

0_3'x20

2)(1.6x

10- 'e)

= 2 . 1 7

1 O t o m - t

(1.054

10- 'o ) t

) - '

= r . ,n , ,0 *

- lr- [20

x

sint( .17

100

3

x

10-'o] '

r

- \ r

|

4

x2 x (20

2 )

9 is the unneling robability

B =

r'0-,0)

=

{r*

[6x

sintr(

.99

10'

x10-'o)I

]

=

o.oo,

| .

4

x2 .2x (6

2 .2)

)

r '0 -n l

=

[

*

[o* r inh

q.qq I

o '

* lo - ' l '

I '

=

7 .8x0-e .

l ' ' 4 x 2 . 2 " ( a - z . z )J - '



24.

FromFig. 22

A s s : 1 0 3 V / s

ua

=

l.3xl06

cm/s

Si)

and ua

x

8.7x106

m/s

GaAs)

t

x

77

ps Si)

and

x

11.5

s GaAs)

As E

:5x104

V/s

va

x

107

m/s

Si)

and,

a

x

8.2x106

m/s

GaAs)

t

x

l0

ps (Si)

and

=

l2.2ps

(GaAs).

25. Thermal velocitv

=

9.5

x

10nm/s

=

9.5

x

10u mls

For

electric field

of 100 vlcm, drift

velocity

v

a

=

l_4,E

1350

100

=

1.35

105 m/s < v,,

For electric ield

of 104V/cm.

FoE

=

1350 104 1.35

107 m/s

y,r,

The

value is comparable

o the thermal

velocity,

the

drift velocity and he

electric ield

is not

valid.

linear

relationship

between

2

x|.38

x

10-2'

300



CHAPTER

4

The mpurity

profile

s,

(Np-N,a) cm")

3x Ora

x (pm)

The overall

space

charge

neutra

c :101e

cm-a

of the semiconductor

requires

that

the total negative

space

charge

per

unit

area in

the

p-side must equal

the total

positive

space

charge

per

unit area n the n-side, thus we can obtain the depletion layer width in the n-side region:

0 . 8 x 8 x 1 0 ' o

Wn

3 x

10 'o

Hence,

he

n-side

depletion

ayer width

is:

W,

=1.067

m

The total

depletion

ayer

width

is 1.867

pm.

We

use he

Poisson's

quation

or calculation

f the electric

field n(x).

In

the

n-side

egion,

L = 3 - * r + r ( x ^ ) = L N o x + K

d x t " " E s

E

x ,=1 .$67

m ) = O + K

= - + N ,

x l . 0 6 7 x

0 - a

€.,

" ' E

t , \ = L * 3 x l o r a

(x_ l ' 067

l o r )

E

E

o,

=E

(

x,

=

0

)=

-4.

86

x

103V/cm

In the

p-side region,

he

electrical

ield is:



9=L*^

=e

(x

)

=

=n

rax2

K '

dx t r

P '

2€,

E

x

p

=

_0.81m)

0

+ K'

=-t*

o*(0.S"

0-o) '

. . ,G") :a *o, l ' - (o.s, , ro- ' ) ' l

\

P '

2 t ,

L

'

'

)

E,o,

=E

(ro

=0) =

-4.86

x

103V/cm

The built-in

potential

s:

v,,

=

-

I

_:,,

&W

=

-

l'0,,,

r)d*l

o

,,0"

;",

(')4

n

s

de

0.s2

.

From Vo,

=

-

J

r

(*V,

,

the

potential

distribution

canbe obtained

With

zero

potential

n the

neutral

p-region

as a reference,

he

potential

in the p-side

depletion egion s

v,(,)

=

li e)a* ft *o l'- (o.r10''f* =

ftLir -

(o.sr0-o)'

-l{o

x10

=

-7.5e6x

0"

[1"'

-

(0.r,

o-') '

-?r(0.t,.

o-') ']

With the condition Vp(0):V"(0),

he

potential

n the

n-region s

v^&)

t":'

ro''[]" '

t.o67xro-ax.UF,.ro-')

=

-4.56xr0'

()*

-1.067x

o-ox

T,.

,o-')



[[ -

!!-

)D---

lr-

l[_ l l . - , .

!_-.[.-_

m_

l!_.-

n!

!m

l[

n_

_tr- ,

[!-

l !

,

f l _

-____t

,_

[[^

l!-

-^-[-

m_

l![[--

l--

!!t

[ [ [ .

. ,

__fl-

l[l-_J_

t--

---l[-

t![.

f_

___ r

["Jr][+n_Jo_!{



The intrinsic carriers

density n

Si at different

temperatures

an

be obtained

by using

Fig22 in

Chapter2:

Temperature

K)

lntrinsic

carrier

density

n,)

250

1.50+108

300

9.65+10'

350

2.oo+10"

400

8.50]l10''

450

9.00+10'3

500

2.20+10t4

The

Vu canbe obtained

y using Eq.12,

and he results

are isted

n the following

table.

J!

--lmIml

![ -

__m

m

nmt

m-

__tr

.

n[[![_

m , n

__tr

l![[[-

m_

ml

mlm!- Iu-

MD

, n , n [ ! *

!!-J

Thus, he built-in

potential

s decreased

s he

temperature

s increased.

The depletion ayer

width and

he maximum

ield at

300 K are

I T :

_ m u

=qN

pf t r

_

1.6 10- 'e

10 '5

9.715 10-5

q

1 1 . 9

8 . 8 5 l O - ' a

=0.9715

lnt

=1.476

104

V/cm.

2

xll.9

x

8.85 10-'o

0.717

1 . 6 x 1 0 - t e x l 0 t 5

No



+.

E- - .

l z rn (

y^*+ - l " '=4xr0s- f2x t .6x l0

'x30 [

'9 ] ' t ,

) l ' "

u

L

e

[ N r * N r ) )

[ l t . e x 8 . 8 5 x l 0 - ' o

1 0 ' ' +

N " ) ]

> 1 .755x

0 tu

N

D

r + #

We can select

n-typedoping

concentration

f Nr:

1.755x10t6

rnt

for

the

unction.

Eq. 12 and

Eq. 35, we can obtain the l/C2 versus Zrelationship for doping concentration of

l0't, 10t6,

or lOtt crn3,

respectively.

ForNr:16r5 "*-:,

+=

'V: ,

_ ' )

-

zx(o .s t t

v )

-=1.187x

0,6

o. tz t

v )

C , '

Qd "Nu

1 .6x

10 - ' n

l l . 9 x 8 . 8 5 x

10 - ' n

l 0 '

For

No:19r6 *-:,

c , '

e d " N u

l . 6 x 1 0 - r e

1 1 . 9 x g . g 5 x

0 - , 0 ; 1 0 "

= 1 ' 1 8 7 x

o " ( o ' s l o

r )

For

Nlr:1017

crn3,

I

_

zVu,

v)

_

zx(o.gsa-v)

C , ' 4 d , N u l . 6 x l 0 - l 9 ' t t f f i = L . | 8 7 x t o ' o ( o . e s o - r , )

When

he reversed

ias s

applied,

we summarize

atable

of I /Ct,

vs V for various

Np

values

as

following,



[_ l!.-._lI [[_-*JI

'-l

l!

, -

fl-l

Lt__-[[

x. fr!

L-

] [ , . f lt

._n

l ! . f [n-

3r

nl![[!_

rr__Jm

.

f-l

n n m

[-JD

TIT]

4U/4U

iL__.__.![

.-n

n . n n n

[ "

__[-_.n

n-

__-!-_ul

.-n

D[!!D[-

[.

nD

-n.^---_l

n.

_J____-[!

--!

_

nn.

n[[_

! *

.--8.-.-D

[-

J_n._nl

._n

n n n

1!J1_!U

mm

._-[!__^-l!

-_rJJil

--[

-i--m

!

][l [![-

I

Jln__-lil

.-n

[-__n[

Hence,

we obtain a series

of curvesof I/A

versusZas following,

. | I I -

!

t r s L

tk<{mth4}g

The slopes

of the curves

s

positive proportional

o the values

of

the

doping

concentration.



The

nterceptions ive

the built-in

potentialof

hep-n

junctions.



6. The builrin

potential

s

2 k T , ( a ' e " k r \

?

(

Vr,

=- ln l

- - i ;

l= i*0.0259x

hl

'

3 q \ \ q ' n i 3 t

=

0.5686

From Eq.

38, thejunction

capacitance

an

8x 1.6 0-'n (l.os to' I

e)

-)

0 'o x1020

I 1 .9

8.85x

10- 'a

0.025

] " '

be

obtained

x

o'o

x

( t

.e

8.85xo"o) '

7.

[ . 6x

10- ' '

rz(o.s6s6

vR)

At reverse

iasof 4V,

the

unction

capacitance

s 6.866x

0-e

F/crf

.

( -

- 4 - l

q o t , '

, l ' " = [

,=T=lq;4j

=L

FromEq.

35,we

canobtain

+=ry#+N,=&*r|

' . 'vR>>vu,+

No

=2(v^)

'

QE"

r

2 x 4

x (0.85

10-')2

1 . 6

l 0 - t e

x

1 1 . 9 x 8 . 8 5 1 0 - ' a

+No

=3.43x10"cm- '

We can select he

n-type

doping

concentration

of 3.43x

015crn3

8. FromEq.56,

,-, rr f

"

ocnD,rN, l

= - L : l

l n ,

Lo,"*ol-

f

)+

o,e*n[-;-J]

[

. r - l 5 v l n 7 y l n l 5

I

_ l

10 - t5

l 0 - t 5

107

l 0 r t

=l

. lxe .65x10e:3.8exr0 '6

and



2

xI l .9

x

8.85

10- 'o

(0.717

0.5)

1 . 6 x 1 0 - ' e l 0 t t

=12.66

l0-5cm=1.266

m

9.

Thus

J

g" ,=

qGWr

=

1.6x

Otn

3.89x10 'u

12.66x l0- t

=

7.879x l0- t

A /cm' .

F romE q .49 ,and - ^= i l ' '

no

ND

we can obtain he

hole

concentration

t

the edge

of the space

harge egion,

n . ,

(

o r

)

/ - \ r / n c \

,^ =#"ii; i '

-g{#lel; '- 'J

=2.42x0,, cm-,.

l v D

J

=

J

o(r , )

+ J, (-

*

o)

=

J,G" ' r '

_ t )

I v

)

"

- e o . o 2 5 e

l

J"

v

3 0 . 9 5

e o o 2 s e

l

=V

=

0.017 .

The

parameters

re

ni:

9.65x10e m-3

Dn: 21

cm2lsec

Do:|0 cm2lsec

eo:T.o:5xI0-7

ec

From

Eq. 52 andBq.54

J, ( r , \

=n ' :o-

(suvr*

'

L p

10.

l l .

-,)=nF*,

f

( q v " \

1

! i -" lel t rr

J

l

I

ND

L ]

I r--qrt I

x l

e \ o o 2 s e l

l

I

t l

L J

"

(q.os,.

on

'

-

7

= 1 . 6 x 1 0 - t t

l 0

5

><10-

N o = 5 . 2 x 1 0 1 5 c m - '

ND



2 5

= 1 . 6 x

1 0 - t t

::>

N

t

=

5.278 1016

m

3

We can

selecta

p-n

diode

with the

conditions

of Nn:

5.279xl016crn3

nd

Np

:

5.4x 0lscrn3.

12. Assume[

s

4p {,r:

1O6

,Dn:21

cr*/sec,

andDo:

10

c#/sec

(a)

The

saturation

urrentcalculation.

From

Eq.

55aand to

=,[Drrr,

we can

obtain

J^ ( - * r )

=no: ' *

(4v r t r

- l )=q

Ln

=

1 -_QDoP no

Q D , n p o

on : (LE _*

t

'

L e

L n

" | . N D I ? p o

N A

=

.6

o-' e

e.os',0') ' [*.,8.

+.8]

[ lo ' '

l l to-u

to 'u

l to- "

)

=

6.87

xl}-t'Ncmt

And

from the

cross-sectional

rea

A: 1.2x10-5

m2,

we

obtain

I , = AxJ" =1 .2x10-5 6 .87 l0 - t ' = 8 .244x l0 - t7A .

(b)

The total

current

density s

( q v

\

J

=

J"l '

- t l

\ /

Thus

[;"t"1't*t-'],

F;)

! ^ )

I o rn

I

_o.r ,

( o r \

= 8 . 2 4 4

l 0 - r ? [ e o o r t ,

r | | = t .

( u ,

\

=8.244

10- r? [eo .o"n

t

r= f

244

x

l0-t7

x

5.47

x

10"

=

4.51

x

l0-5A

.244

10-"A.



13.

From

( q v

\

J

= J " l r n

- l l

\ /

we can

obtain

v

=

[f+]*

r =

v

0.025e*

[l ,

to- '

,=*

, l

-

0.78

.

0.02se

L\/"

I

L\t.z++"

0-' '

I

14. From

Eq. 59, and

assume

o: l0

cnflsec,

we

can

obtain

J-=af i " : *Q ' 'w

'1 r, No t,

=

r.6 o.n /- lL

(q'os

ton)'

*

lJ

o*

lo '5

l . 6 x

lO - t e

9 . 6 5

x10e

l0-u

vo,

=0.0259,r,

lo ' '

x

lo'5-

=

0.834

(9.65

l0 ' ) '

Thus

J

n

:

5.26x

0- ' l+

1.872x

0- '

J0.s34

V^

2x l l .9

x8.85

10-ra

(Vo,

V^

1 . 6 x 1 0 - t e x 1 0 ' 5



(

._-[*-__-m

l^

Il._

--l--lm

[^

[l.-

._ff_-_m

[^

m_

J. "nm

! ^

m

mI

_lm__nI

[ ^

DI

-n__1._nt

[^

![^

-J--

.![

!^

m_

_ff__!m

n ^

!m

__n-._.[_-!l

[^

__n m!

!^



ll-fi_

ro0ri

m -

m

Jm ^

_!00^

Jm ^

o:B:_.

olo0t0

WhenNo:1017

n

3,

we

obtain

v^,

=0.025g,

-10' '

x

l0 ' t -

=0.g53

(9.65 l

o') '

J

n

=

5.26x

0-'' 1.872*

o-'.ft.lso

u*

llJr]-

trri.a.O

From

Eq. 39,

Q o = 4 f , b " - p ^ " \

*

-

E

J

u-L

uL

u1_

R<dIIl ll"llrtl

15.



=

n[l,o*(ftvrn

-t)

e-G-',\/ '04*

The hole diffusion

ength

s larger han

he length

of neutral

egion.

Qo=ef ; {0 , -p , , )d *

=

a

ln

*Q{

rtcr

l) e-G-.,),t"

*

(

qv

\ f

' ; - t

- t - t

)

=epnoct)1"'

l j le

Lp

e

"

)

: r .6xr0- 'e" i#t l (-5xro .,[" - '1"*-" ;)

=8.784

10-3

lcm'

16. From Fig.

26, the

critical field

at breakdown

or

a Si one-sided

abrupt

junction

s about2.8

+105

V/cm.

Then rom

Eq. 85,

we

obtain

zr(breakdown

ottugq=E{

=

t:'"'

(t,

)'

2 2 q

_

1 1 . 9 x 8 . 8 5 x 1 0 - ' o( 2 . 8 x 1 0 ' ) '

. - \ ,

2xl .6xro-, ,

-z-(10' ' /

=2ss

:1.843

x

10-3

m= 18.43fem

when the

n-region is reduced

o

5pm, the

punch-through

will take

place

first.

From Eq.

87, we can

obtain

Zr'

_

shadedar_eain_is.ginsert

=(!_)(r_y_)

vB

G.w' z \w^)\"

w^

( w \ (

w \

(

s

\ / s \

Vu '=Vu l

; l l

2 -

- l=258x1

l l z - - - : - - =1 2 l

\w

^

)\

It/^

)

\

18.43l\

t8.43

Compared

o Fig.

29, thecalculated

esult

s the

same s

he value

under he

2xll.9x

8.85 0-'a

x

258

1 . 6 x 1 0 - r e x l 0 t 5



condit ions

fW:5

pm

andNa:

1015cnL3.

17. we can use ollowing

equations

o determine

he

parameters

f the

diode.

J ,

=e ,p -n , '

,u ,0 , *Qwt l ,

ecv t2k r

"E tLesv tk r

' \ " ,

N ,

2 T ,

- ' \ t , N o -

vr=t"!

=t,!" '

*).

2 2 q

=

AJ,

=

n"

E:-eqv

,kt

+ Axr.6x

0-re

E

b

os lot

'

. , 1 l T o N o , r , + A x | . 6 x l 0 _ , " x i , ; ? e a o 2 5 9 _ 2 . 2 x 1 o - 3

v,

E+

ff{* ")-r

130

aH*#5

(ru,)-'

LetE.:4x105 Y/cm,we

can

obtain

No:4.05x101s

rn3.

The

mobilityof minority

carrier

hole s

about500

at No:4.05x

l0r5

.' Dp:0.0259x500:12.95

nf

s

Thus, hecross-sectional

reaA is

8.6x10-5

m2.

18. As

the temperature

ncreases,

he total

reverse

current also increases.

hat

is.

the total electron

current

ncreases.

he impact

onization akes

place

when

the

electron

gains

enough

energy

rom

the electrical

ield

to createan electron-

hole

pair.

When the temperature

ncreases,

otal number

of electron

ncreases

resulting n easy o lose their energyby collision with other electronbefore

breaking he

lattice

bonds.This

needhigher

breakdown

oltage.

19.

(a)

The i.layer

is easy

o deplete,

and assume

he field in

the depletion

region

is



constant.From Eq.

84, we canobtain.

" w . ( r \ u

. ( e \ o t /

,

I

l 0 ' l = - -

|

d * = l = 1 0 0 1 . - -

|

x l 0 - '

= l l E " , n i , o t : 4 x

1 0 5

( 1 0 ) l o

5 . 8 7

1 0 5

V / c m

J o

\ 4 x 1 0 ' / [ a x l 0 5 /

4V B= 5 .87 10 ' x 1 0 - 3 587V

(b)

From

Fig.26, the critical

field is 5

x

105V/cm.

Z"(breakdownottu)

{

=''i;'

(w

r)'

-

12.4

8.8510-ta

ab

x

105I

Q

*

t ' ,uf '

2

x l . 6x l0

=

42.8

V.

I - t n l 8

2 0 , e - o n ' u

= 1 0 2 2 c m - o

2

xl } -o

y,

=Y

-

+

" '

l2e

1" '

1o1'" '

^

r

3

L q l

-

4E

r"'

|z

*

n,g

l,s,Ls-.:to'

1"'

*

1ror,1-,,,

3

L

1.6 0- ' '

, j

=

4.84

xl}-,E"3

/t

The breakdown voltage

can be determined

by a selected ".

21. To calculate he results

with appliedvoltage

of V

=

0.5V

,

we can use a

similar calculation

in Example

10 with

(1.6-0.5)

replacing 1.6 for

the voltage. The obtainedelectrostatic

potentials

are l.lV

and 3.4x l0-o V, respectively.

The depletion

widths are

3.821

l

0

-s

cm and

1.27

x

10

8

cm,

respectively.

Also,

bysubstituting V

=-5

V

to Eqs.90 and

91,

the

electrostatic

otentials

re

6.6V

and 20.3 l0-4 V

,

and

the depletionwidths

ue 9.359 l0-5 cm and 3.12

x

l0-8 cm,

respectively.

The total depletion

width will

be

reduced

when he heterojunctions forward-biased

rom

the thermal

equilibrium condition.

On the other hand,

when

the

heterojunction s reverse-



biased, he total

depletion

width will

be increased.

22. Es(0.3)

=

1.424

+

1.247

x

0.3

=

1.789

eV

r/

-

Esz

A-E

/

b i

_

_ _ ' c

_ ( E o r _ E r r ) /

q _

( E r r _

E r r ) /

q q

=

.tle o.^-gh

o

I \l=orr'

4r

1

.

l:,,

=

r.273

q

5x10"

e

5x l0

r ^ , ,

t V r

|

2N

E$.v^, l" I

^ , - , - , - ,

'

L q N o ( e , N r + e r N n ) ) t

2x12.4xLI.46

8.85

l}- ta

x1.273

1.6 to- 'ex 5x1or5tz .++11.46)

: 4 .1x10-5cm.

Since Nrr,

=

N

,txz

.-.

r

=

xz

. ' .

W

:

2x ,

:9 .2x

10-5 m

=0 .82

p tm.

1'



CHAPTER

5

l.

(a)

The common-base

nd common-emitter

urrent

gains

s

given

by

do

1&r

=

0.997

0.998 0.995

o

ao

0.995

l - d ^

l - 0 . 9 9 5

= 1 9 9

(6)

Since 1a

=0

and lgr=l}xl0-eA,then

lruo is

toxlO-eA.Theemittercurrentis

Icno

=(t+

p)truo

= ( t + t r ) . to x o-e

= 2 x 1 0 { A .

2. For an ideal ransistor,

d ' o = f = Q ' ! ) )

P' :&:ses

Iruo is known

and equals o

t0

x

t0-5 A

. Therefore,

Iceo

=

(*

gotruo

:

( l

+

999) .10 10-6

= 1 0 r n A .

3.

(a)

The emitter-base

unction

is

forward

biased.From

Chapter3

we obtain

vo,

kr

1n(N

t!

l=

o.orrnrf

l o''

z

-t-9''

l=

o.nru

.

q

\

ni-

)

t

p.65

lo"f

J

Thedepletion-layer

idth

n the

bases



4

=(

,,

N

n=,

I

gotAa"ptetion

layer

widthof the

emitter base

unction)

(N ,

+No

)

-

=

lz,(yolf

_+lr,,u,

r,)

1 l

q

[ l r o ) \ ' , n - ' , o 1

_

/z

r.os

ro_,,

f:::g:)f

, \.

=

I

'-

,-''

t;

,

rorj(.;,.

oq;Ar.,l(0'es6

0's)

=5.364

10-6m

=5.364

l0-2

trrm

Similarly we obtain

for the

base-collector

function

v^,

o.o25e"[?'

o' ' '

o:

-l

=

o.rn, .

LP.oslo'Jr

l

4.254x706

m:4.254x10-2

m .

Therefore

he neutral

basewidth is

Vl

=Wn

Wr-W,

=l-5.364x10-2

4.254x10'2

0.904

m

(b)

UsingEq.l3a

. | ^ \ 2

p,(0)

=

pnoe4vralkr

l- i '

,rrolor

-

(9'65

x

l0-'f

eosloo2ss

2.543x

l0r cm-3

N D -

2 x 1 0 t 1

4. [n the emitter egion

D,

=52

cmls L,

=62,10-t

n

Eo

(q

gs

lo- ' ) '

=rg.625

5

x l 0 ' 8

In the

base

egion

= 0 . 7 2 1 x 1 0 ' c m



Do=40cmls

o=f f i

=J+o '10-7

2x10-3

m

^

-

n,'-=(q'os

ro-nY

=465.613

ro

=

Vtr-

=

-,

"10--

=

+o).1

In the collector egion

Dc

=ll5

cm/s Z.

=fis

to*

=10.724x.10

cm

,rn

=futrt=9.3r2x

lo3

l o ' u

Thecurrent omponents

re

given

by Eqs.20,2I,

22, and23:.

Iuo

Iro

Iun

1,,

Iuu

l .6x 0- 'e 0.2xI0-240.465.613

eos loozse

1.596x

or

A

0.904

10-o

7 I

uo

1 .596x

0- ' A

_l .6x l0 ,n

.

0 .Zx l0- .2

52.

19.625

(ror t ror rn_

)= t .04 lx

I 0_ i

A

0.721x10-3

v

d , = f U r = 0 . 9 9 3 8

Fo

=

:ao

:160.3

l - d o

5.

(a)

1.6

10- ' ' 0 .2x10-2

l5

.9.3

2

xl03

=

3.196x O- to

10.724x10- '

= I ro - I ro

=0

The emitter,collector,

and

base urrents

are

given

by

I,

=

I

to

* Irn

=

l'606x10-5 A

Ic

=

Icp *

Icn

=1.596

xl0

5

A

I

u

=

I

rn

* I

u ,

-

I c^

= l ' 041x10-7

A '

We can

obtain the emitter

efficiency

and the base ransport

factor:

1, ,

1.596 lo-5

Y-

" '

=

=0.9938

'

I" 1.606

l0- '

Iro

1.596 l0-5

* r

-

I t u

l . 596x lo -5

Hence, the common-base

and common-emitter

current

gains

are

(b)



(c)

To improve

y,the

emitterhas

o be doped

much heavier

han he

base.

To improve

a,

we can make he

basewidth narrower.

6. We cansketch

p^(x)l p,(0)

curves

y usinga

computer

rogram:

0 . 4

0 . 6

DISTANCE

In the figure, we

can seewhen WlLo.O.L (WlLe=0.05

in this

case), he

minority carrier distribution

approaches

straight ine and

can be simplified

to

Eq. 15.

7 . UsingEq.l4, 1r, is

given

by

o

c

o-

x

v c

0 . 4



l

t .ornEl

Le lL, )

-a"".n[l]

Lp

lL, )

t

, ( w

* \

L p

I z o )

,t"hf4l

\1 , )

'*g)

rn o

-r )

t l

t l

tl

ll

'ulkr -

t-rrf

L

rfr

lkr

_

. l

?)l

' a V

e '

'alkt

W

L,

\e

74Yt

I

'hl

\

n o

e '

'th

|

,,[---L1:{All

qoTI*'^'-L,*[fJ]'ry';]i

=

AT;E

lQ,,

,

,

,).

".{fJ]

I ,o:A(-ro,H,^)

Similarly, we can

obtain lro:

,r,=u(-no,H,=,)

=u(nilf

I

=

nn.'=O"f

o l

=uDrf

"o

l_

I ' n o

,*r4)

\1, )

=

-

,rln

.

"'

'

ttl

8. The totalexcessminority carrierchargecanbe expressedy



g,:nn([r,e)-

p,"fdx

, w l -

- 1

=qAl

I

p,o"o"u,o,1 l-

| - ; lax

J o

L - " "

W

)

" t w

:qApnoeorol*

O-+)l

2prh

otr14tr.."evrolkI

2

_

qAWp"(0)

4

L

From

Fig.

6, the

triangular

area

n the

base egion

,

wpo(O).

By

multiplying

this

z

valueby q

andthe

cross-sectional

rea

A, we

can

obtain he

same

expression

s

en.

In Problem

,

1.6

x

l0-te

.0.2x10-2

.0.9M

x

l0{

.

2.543

xl } t l

2

= 3 . 6 7 8 x 1 0 - t 5 C .

9. lnEq.27,

I

c

=

or,

(f

avulw

-l)+

a,

_

qAD,

p,(0)

W

=2Do

qAep,(o)

w.2

2

2D

=

*?Q'

'

Therefore,

he

collector

current

s directly

proportional

o the

minority

carrier

charge

tored

n the

base.

10. Thebase ransportactor s



l{""*'*

r)*

-.

l

' l

sllillt

-

|

lL ' )

"^

d , Z J =

'

Ib

t

HV

L

Thus,

:.""nIlL)

\1 , )

t - t (w) '

2 lL '

)

=r-0t ' lzro

""'nI

For t4fLo<<1,

cosh(Vf)=1.

d r =

,,,,,i',hfzl"rh(L\

\1, ) \1, )

z )

11. The common-emitter

urrent

ain

s

given

by

R :

d o

-

W r

' o -

l - o o

l - w r '

Since

7

=

1,

8..= a'

l - d ,

_

r-frr,'lz4')

,,

t - [ -Vy' lzr , ' ) ]

=Qto' fw') - t

rf WfLp <<1,

hen

o=zLo'f

W' .

12 . Lo=r lDo t ,

=^/100.3

10-7

=5.477

10-3

m

=54.77

m

Therefore, the

common-emitter

current

gain

is

."./zll

\1,

))



P'

=2L,2

lW,

=z(s! . l l

to. : \ '

P

t

Q " l o - ' ) '

=

1500 .

13 . In the emitter

egion,

In the

base region,

In the collector egion,

In the emifter egion,

In the base egion,

Fpr

=54.3+

l + 0 . 3 7 4

l 0 - r t

3

x l 0 r 8

=87 .6

D

t

=

0.0259

87.6

2.26

ml

p n

=88+

1252

=

1186.63

I + 0.698

l0-r t

.2x l0 t6

Dp

=0.0259.1186.63

30.73

m/s

Lr.r-= 54.3

407

.= ,, = 453.82

l + 0 . 3 7 4 x 1 0 - " . 5 x 1 0 ' '

Dc

=

0.0259

453.82 I .75cm/s

I

L,

=

^tDp,

=^lZ.Z6g.l0*

1.506

l0

3

cm

n__

=

f r i

=

(9 .65 r ' ^nY

r

L,(,,

N

F

3r.l ir

=

3r.04

m-j

L,

=

J3u734.10u

5.544x10-3m

6.oston

n ^- = Y:::::LJ- = 4656 13 cm-3

2

x

0 ' o

14.

ln

the collector egion,

L.

--"111.754-10{

3.428x10-3

m



/ ^ \ ,

,.^

-19'65

xlol-

f

=

18624.5 m'3

5

x l 0 ' '

The emiffer current

components

are

given

by

,

r,

:Weo.6ro

o2se

=

526.g3xo-6

A

0.5x0+

r

_1.6xrc-t:..1_0:

.?.?69.3r.04

Qo

4o.ozse_r)=

t.OOe 0_e

.

EP

r.506

x

lo-3

\

Hence, he emitter

cunent

s

It = I tna Itu = 526.839 lOj A .

And the

collectorcurrent

components

re

given

by

, r^

_l.6xl} ' "

.10-o 30.734.4656.13

eoqo.o2ss

526.g3x

0_6

0 .5

l 0

{

I -

_1.6x101e

104 11.754.19624.5

=1.022x10rn

.

cP

-

3.428

ro-3

Therefore,

the collector

current is

obtained by

Ic

=

Ico

'1

Ico

:5-268x

l0- A

.

15. The emitter fficiency

an

be obtained

y

do

--

Tdr

=

I

x

0.99998

=

0.99998

The value s

very close o

unigz.

The common-emitter

urrent

sain

is

y -

I

tn -

526'83xo{-

= 0.99998

IE

526.839x0-

The

base ransport

factor is

I"-

526.83 l0*

f l

_

\ r t

* ' -

h

- 5 2 6 ^ 8 3 * l o - u - ' '

Therefore,

he common-base

urrent

gain

is

obtainedby



B^

=

oo

-

0'99998

= 5oooo

l -a"

l -0.99998

16.

(a)

The otalnumber

f impurities

n theneutral

ase egion

s

%

=

fr

Nnoe-,/tdx=

nol(l-

-*/,)

=

2x l0 r8

3

x

l0-5

-e- ' " 'o ' / ' . 'o- ' )=

5.583

l0t3

cm-2

(b)

Average mpurity

concentration

s

9c

_5 .583x10 ' 3

W

8 x 1 0 - 5

=6.979

1017m-3

L7. For N,

=

6.979x10tt

*-t, D,

=7.77

cm3/s,and

L,

=

,tD^r,

=

tll. l l

.

0{

=

2.787

10

3

cm

_,

w2

,

(sr lo- ' I

Ar

=l

- - - - - - - - -== l - - r

- I - -

2Ln'

zQ1u

"

to- ' f

=

0.999588

I -

=

0.99287

l +

I

.5 .583x10 ' 3

7 .77 lOte

10-4

,

Therefore,

ao=wr=0 .99246

B o

: u : 1 3 1 . 6

.

l - C / o

18. The mobility

of an average

mpurity

concentration f e glg

x

l0r7 cm-3 s

about

300 cm' Vs.

The average

base esistivity

pu

is

given

by

, ,

D t

Q o

D, N

ELE



Therefore"

Ru

-

5

x

to1

p-

u

w)

=

5

x

0'

-

o.ozes

ltx

10r

)

=

1.869 .

Fora voltage ropof kr fq

,,

=

#=

o.ol3e

Therefore.

I c

=

Fo

a

=

131 .6 .0 .0139

1 . 8 3 .

19. From

Fig. 100and Eq. 35,we

obtain

1"

r,A)

1.

mA)

e,

=#

0

5

l0

l 5

20

25

0.20

0.95

2.00

3 .10

4.00

4.70

150

210

220

180

140

Bo

is not a constant.

At low 1,

,

because

f

generation-recombination

urrent,

Fo

increases ith

increasing1, .

At high I

B,

vEB

ncreases ith

l,

,

this in turn

causes

a reduction f lzr.

since Vru+Vur=Ytc=sV.

The reduction

f Vu, causes

widening

of

the

neutralbase egion,

herefore

po

decreases.

The following chart shows

Bo

as function

of I

B.

lt

is

obvious hat

Fo

is not a

constant.



20.

Comparing he equations

ith

Eq. 32

gives

5 1 0 1 5 2 0 2 5 3 0

ls (trA)

I r o = a r t ,

a ^ I o o = a ,

d

rI

ro

=

ar,

artd I

no

=

dzz

o

Hence,

21. In

the collector

egion,

,1

/ \

/ \

/ t

/ \

/ \ .

d -

:

az t

* ' -

o r -

t Z . % . " *

LE De p,o

cr r .

I

- : -

- - n

e r ,

, , W

D c f l c o

r J _ . - . _

Lc Dp pno

Lc

^[Dc%

=

Jr.tou

=1.414

o

.3

cm

ftco

f t ,2

lN,

=(e.Os

rct l

f

sx 0r5

1.g63

l0a

cm- i

From

Problem20,

we

have



I

"

.

14/ D, npn

l + - . - . :

LE D,

P,O

=

0.99995

, * 0 . 5 x 1 0 4 . 1 .

9 . 3 1

lo-3

lo

9.31 lo-2

d R - 1

, ,

W Dc

ftco

0.5

x

l0{

2 1.863

lOa

' -

+

, ,

, -

t - * * " -

m 9 3 r . r o -

=

0.876

rro=ctr,=n4+.ry:)

. ( r c . 9 . 3 1 x 1 0 2

1 . 9 . 3 1 )

1 . 6 x 1 0 - t e . 5 x l 0 - a . l

. * - l

|

0.5x 0* t0*

)

: I .49

x

10- to

rno=ozz=r4+.T)

=

1.6

l0-re

5

xl0+

.(!2' t t"to '

2' l '863x

104

')

\

0 . 5 " 1 0 - t

-

l A A " l l - t

)

=1.7

x10-r4

.

The emitter

and collector

currents

are

I . . , . - \

I,

=

I

ro\ea,ett*t

-

1)+

u^l

^o

=1 .715x10 r

L . , . - \

I

c

=

dr

I

rope'utr '

-

l )+ I

*o

= 1 . 7 1 5 x 1 0 {

.

Note thatthesecurrentsarealmost hesame nobasecurrent) for WfLo <kl .

22. Refening

Eq. 11,

he field-free

steady-state

ontinuity

equation

n the collector

egion s

D-ld'",y)l_n,(,')-,0"

o

L

d t z

)

t ,

The solution s givenby ( L, = ^lD , )

"r(r\

=

gr"''/r'

*Cre-''lLc

.

Applying

the

boundary

ondition

at x'="o

yields



Cre4r,+Cre-41.

=0

.

Hence Cr

=

0

.

In

addition,

or the

boundary ondition

at x'= 0

,

Cre-oltt :C, =nc(o)

n.(o)

=

nroQ*'"ln

-|]1

The solution s

n,

&)

=

nro

(eav"

l

w

-

tp-'l

t"

The

collectorcurrentcan

be expressed s

,,

=

n(- ,*|.=.J.,

[-

n

,#1.

=,)

=

q

AL#(*il

m

*r

t-

r^(2"t-.

"f)

Qtrc,r

r

-

r)

ar,

(dr*

lkr

-

l)

-

ou

(f+vcnltr

1)

23. Using E,q.44, he

base

ransit

ime is

given

by

tu

=w'/2D,$ji#

=t.25xro-ro

We can obtain he cutoff frequency

f,

=lf2m

u:l'27

Grrz

From Eq. 41,

the common-baseutoff

frequencys

given

by:

.f"

=

f,

ao

=#

=1.27

GHz

The common-emitter utoff

frequencys

,f

p

=(t-

a)f"

=(t

-

o.f8)"

1.275

10e

2.55

MHz

Note that

"f

p

can be expressed

by

f

B

=

(t

-

uo)f"

=(t

-

aoY o

*

.f,

=*

f,

.

Po



24. Neglecthe ime

delays

f emitter ndcollector,

he

baseransit

ime s

given

by

t , = - J -

= 3 1 . 8 3 x

0 - " s .

24, 2nx5

xl}"

FromEq. 44, /'can

beexpressed

y

w

=

^lr4r,

.

Therefore,

w=ffi

=2.52x

10-5cm

=

0.2521"rn

The neutralbasewidth

shouldbe

0.252

pm.

25 L,E

,

:9.8Yo

l

.I2

=

1

10

npV.

(

tn-\

D,-.wl

i l

\

^ ,

. ,

. . .

, ( t ,oo"?)

,

( t tomev

l l0mev)

:0 .2g .

7F-q-exp[

373r

mk

)

26

ffi="*(!o#" J="*L

where

Eru(x)

1.424

0.02s9

E

rr@)

=1.424

1.247

, x

<

0.45

=

1.9

0.125x 0.143x,

.45 x

<

l .

The

plot

of

B,(HBT)/F,

(Bni

is shown n

the following

graph.



t.e.o.tesx*d.r/t3*

x

Note

hat

B'(IST)

increases xponentially

when x

increases.

27. The impurity concenhation

of the

nl region is t0racm-3 The

avalanche

breakdown oltage

for

w rW^)

is larger han

1500V

(ry^

>l00pm).

For

a

reverseblock vottage

of 120V, we can choose

a width such hat

punch-through

occurs,.e.,

Thus,

Whenswitching

occurs,

That

is,

v

-

qNoW'

,

D T

-

-

-

28,

*-(#l=r .nuxro-3cm.

a , r + a , r = l

o,=osp'(*)

=

. t J l

u r l

-

|

t r I

\ " 0 , /

= 0.6- 0 . 4

25x70-a

39.6 l0-u

0.6

= -

0.5

=

1 .51



Therefore.

J

=

4.5J0

2.25

0-5A/cm2

/ lx l0 - l

A rea :

- ' = " " "

-

=44 .4cmt

J

2.25

lo-5

28. In the nl

-p2

-

n2 trarsistor,

he base

drive current

equired

o maintain

current

conduction

is

1,'

=

(l

-

u")I* .

In addition,

he basedrive current

available

o the nl

-

p2

-

n2

transistor

with

a

reverse

gate

currert is

I ,

=

q I

n

-

I

r .

Therefore,

when use a reverse

gate

current,

he condition

to obtain

tum-off

of the thyristor

is

given

by

Ir 1I ,

o r q I n - I s < G - u r ) I * .

Using Kirchhoffs

law, we have

I * = I u - I s .

Thus,

the condition for hrm-on

of the thyristor

is

,

q + 4 - 1

,

t s

/ - t A

'

- o q

Note that

if we define the

ratio of llto

1" as tum-off

gain,

then the

maximum

tum-off

gatn

h*

is

F * =

%

, .

oc,

+

a"

_l



1 .

CHAPTER

6

\

...-.................-.-

-

METAL

OXIDE

n-TYPE

SEMICONDUCTOR

Ec

Ep

Ei

Ev



2.

EeEc

Ei

Ev

Ec

Ei

Er

Ev

3 .

n- POLYSILICON

OXIDE p-TYPE

SEMICONDUCTOR

Ec

Ei

Er

Ev

-l

a",l

v

n- POLYSILICON

OXIDE

p-TYPE

SEMICONDUCTOR



4.

(a)

(b)

E

(x)

v6)



5. W,

=2

C-,"

J "

n *885x ro ' ooo26h f

3+ l

_

. , 1

\

9 . 6 5

0 ' ,

- ' | J

l i x t o r v " * r ; "

:

1.5 0-s

m

0.15

m.

tot

6.

Vn

.

d

+(e,,

/e")W^

W^

=o.l5p

m From

Prob.

i 'C^ i n

:

3.9

x

8. 5

x

0

-ra

=

6.03 l0{

F/cm2

= 0 . 7 4 x

10 -5cm

8 x l 0 - 7

3 ' 9

* 1 . 5 x 1 0 - 5

I 1 . 9

=t!n!o:o.o26h

to"

=

=0.42y

q

ni

9.65

x

l0'

, "=4at in t r ins ic

v "=vn

e"

E J

E o : E r € " : 1 . 1 1

€o *

V:

Vo+yr,: Eod

+

y

":0.59

V.

1 . 6

1 0 - t e

5 x

1 0 1 6

l . 5 x l 0 - 5

_

l . 6 x l 0 - r e

l 0 r 7

0 . 7 4 x 1 0 - 5

1 1 . 9 x

. 8 5 " 1 0 F =

l ' 1 1

l o '

v / c m

. 1 t

x l 0 5 x

" ' '

: 3 . 3 8

x l 0 5

V / c m

3.9

:(r

rt

x

lo5

x

5

x

o-?

)+0.+2.

8 . At

the

onsetof

strong nversion,

V

":2V

n

) V6:V7

thus,

6=$*r,

u

co

From

rob.

,

W*

:0. t5fgm,

u:0.0261n[- t , ! to ' l r ] :

.O

[9.65

10'

L o

:--------:-

:

3.45x 0-7F lcm2

10*

. r /

-

. . v c -

3.45 l0-?

+ { . 8 : 0 . 3 5 + 0 . 8 : 1 . 1 5 V .



s.

Q.,

:

[:,-

QO'

coA,V..=

yq(rl,,

0,

ryfft)*

rc',

:8x10-e

C/cr*

-

8x lo -e

-

=2.3zx lo - ' .3 .45 0- '

x (10-u

' l

/

=

5xl0-7

y

*

5xl0-7

l r a

10.

Q*

=i

lo

p",Q)dy

po,=e x5x10t t {x) ,

where

(_r)

. ' . O ^ , -

I

" 5 x 1 0 " x 5 x 1 0 - ' x l

10 '

:4x10-8

C/cnf

: ' L V ' o - Q o '

4 x 1 0 - 8 -

= 0 ' 1 2

C , 3 . 4 5 x 1 0 - '

(

l @ '

= 1

10,

. 6

x

1Orn

V

:+

x

1 .6 O- ' n

l "

S "10"110* ; , .

10 - " 3

I r l o {

I

1.

Q", : ;

l ,

y(qx5 xLo23

y)dy

:2.67x10'8

Clcn?

: '

LV"

-Qo '

-2 '67x10-B

7 '74x10- '

'

C

3 .45 10- '

12. LVr,

=% =

+"4x

N

^,where

N, is he

area ensity

f 'e*.

D

C C d

M )

- l t { ,

- ! ' vou*c" -

0 '3 -xJ '+ }10

?

=6.47x10 ' ,

rn2 .

q

1 . 6 x 1 0

13.

Since Z, <<

(Vo

Vr)

,

the irst

erm n

Eq. 33 can

beapproximated

s

7

I

lt,C

(vo

-ht/

u

)Vo

Performing

aylor's

expansion

n the 2nd erm n

Eq. 33, we

obtain

(vo

+2tyu

t , ,

- (2Vu)t , ,

=(2Vu)t , ,

+){zwu)r , 'vo

-(2Vu)r , ,

=}f rw, l , rvo

Equation33 can

now be re-written

as



-

z

n

'{'"

l*"

E

'n\'*'1I'^

T l ' t ' t ' l -

L

Lo

r l

=(|)p,c"(vG

-vr)vD

where,

=ayu.JO'tTy

14. When he drain

and

gate

are

connected

ogether,

Vo

=Vo

and he

MOSFET

s

operatedn

saturation

V

o

> V

o"*)

. I

o

canbe

obtained

by substituting

V, =V^* in Eq.33

I

ol,o*o

1

n,

"{W

2v,u)v^",

{ry)[,""^

+2vru)'''

where

Vo,o,s

given

by Eq.

38. Inserting

he condition

Q,(y

=

L)=0

into Eq.

27

yields

VD"o,=

.

^l2e"Own(Vo,*

2Vr)

+Attu

Vc

=0

L ,

For Vo,o, <2eu, theabove quatioh

educeso

vo

=vo

J'4@J

+2tyu

v,

.

c o

Thereforea

linearextrapolation

rom the

low current

egion o

I

o

=0

will

yield

the

threshold

oltagevalue.

o.o26rn

s to'u

--l=

.oo

19.65 0- 'J

-rfv)'''l

k T . . N ,

| ) .

V

=- ln ( - )

=

q n i

/-------:-;-

Y: ' ' l t 'QNn

-

C

3.45

l0-7

:s-0.8

+(0.27)' Ir-

: 3 .42V

r

D"o,

zt!:l '

vo

-vr)

2 L

\ U

l 1 + -

- -

I

,l^

(o.zt),'

_

1 0 x 8 0 0 x 3 . 4 5 x 1 0 - ?

2x l

:2.55x10-2

A.

(s

0.7) '



17.

a l t T

Therefore,

,

:

frl

n=*^,

f

F,C

(vo

: 5 * 5 0 0 x 3 . 4 5 x 1 0 - '

0.25

:

1.72x10-3

.

oln

I

z

g.

=

frlvpuo,"t

=

Tlt

c,vo

-

5

x 5 0 0 x 3 . 4 5 x 1 0 - ' x 0 . l

0.25

:3.45x10-4

S.

1 8 .

V,

=Vru

+ Ay

u

l o l ?

,Vn=0 .026 t { - - : i - - )

9 .65

10"-

E"

- 7 - u n -

-0 .56

0 .42-0 .02 :

- l

V

2 . l l l . 9x

8 .85x

0 - 'o

10"

x

0 .42x1 .6x10- 'n

16. The

device

s operated

n

linear

egion,

since

Vo

=0.1

Y

<

(Vo

Vr )=

0 .5

V

- v r )

x 0.5

1 . 6 x 1 O - t e5 x 1 O t o

3.45

x10-7

o"

V , o = 6

- : ! - =

c"

=

" 'V r =

-1+

0 .84

3.45

l0 -?

: -

1+

0.84

0.49

: 0 . 3 3

V

(Q.,

can

lso

be

obtained

rom

Fig.

8 to

be

0.9S

).

n F

= 0 . 3 3 +

a ' B

3.45

0-?

_

0.37

x

3.45

x

10-?

1.6

10- 'e

= 8 x l O t t c r n 2

0.7

9.

FB

V

0.

f . ,

=

-t i*U/a

=-0.56

0.42=

0.14

v,

=

e.,

?

2rru

2.[€JN

"W

(-

h?;, : . l I l .9

8.85x

10*'a l .6x

10-' t xl0t7

x0.42

= -0.14 0.02 0.84

:

- 1 . 4 9

V

3.45xl0-7



21.

-0J

:

-1.4g

nFu

=

3.45

l0-'

, ,

-o '79x3 '45x lo r

= l . 7x lo ' '

c rn2 .

l . 6 x 0 - ' '

22. The

bandgapn degenerately

oped

Si is around

eV due

o bandgap-

narrowingeffect.

Therefore,

Q^"

=-0.14+

1= 0.86

V

. ' .V,

=

0.86-0.02-0.84

0.49

=

-0.49

V.



z ) .

v.

:o .o26r [

to ' '

- ] :0.0,

u

[9 .65

l0 '

/

Vr=Q^"

?+2ryu+'ry

=-0 .9g

l ' 6 x10 - te

l 0 t t

* 0 . * O

co

=

4.14*

l5 '2

x

0-8

co

3.9

8.85

0- 'o

3.45

10- '3

- d - d

v r>20=

d . l l t l t 9 , '

>20 .14

3.45 10- ' '

.'.d > 4.57

x

l0-5

cm

0.457prn

24.

Vr

=0.5Y

at

o

=

0.lfg{

9.1 --- iJ-

-7

-logI

olr.=o

bg 1rl

vo=o

-12

. ' .

olro*

=

I

x

10-' '

A.

25 .

^vr=a*utp;-lv;)

c o

Vn

=o'026n(

= - t -o ' ' -. )=0.42Y9.65x 10"

a

-

3'9

x

8'85

-l0-ro

6.9

x

ro-?Frcrfi

5 x 1 0 ' '

AY,

=

0.

V if we

want o

reduce

p at

V6: 0

by one

orderof magnitude,

since he

subthreshold

wins

is 100

mV/decade.

0 . ,_

/2x l . 6x l0

'

x4 .9x8 .85x r0 - ' o

r017

( ro ra . r r_

_Jo84)

6.9 10- '

i .Vu,

=0.83V.

26.

Scaling actor

r

:10

Switching

nergy

*,a

.

A)v'

,v

co



C ' - t *

- t C

d

A , -

A

{

V ' = L

If

.'.

scaling

actor

or switching

energy

A reduction

of one

housand

imes.

From

Fig. 24

wehave

(r,

+

L)2

(r,

+[ry^\'

-W.'

.'.A2

+ 2Lr,

-2lf/.r

.

=

0

L=-r j+\ l ; ;zw*

1 1 1 1

n . - - . - - - . - = - - - - : . = -

t<

t(

i

1000

27.

L '= L-2A,

L+-21 -2^= r -A

t -2 (

2 L 2 L L L I

From

Eq. 17

we have

28.

Pros:

Cons:

no.

_

(space

charge

n

tre toapezaid

l regio!

-

spacecharge

n

tre

rectangula

region

/ r

-

' c o

_QN

nW^

L

+ L',

QN

W,

qN

nW,r,

f

W

,')

- - : -

r , r T - - r ! .

C o 2 L '

C o

C , L

[ 1

r j

I

Higher

operation

peed.

High device

ensity

More

complicated

abrication

low.

High

manufacturing

ost.

l .

2.

l .

2.

29.

The maximum

width

of the

surface

depletion

region

for

bulk

MOS

nl _" le , kT ln(Nn/n , )

, m

- L 1 l -

Y

Q, N n

=r .

1/

1.6;m

=

4.9x l0*

cm

=

49

rnn

ForFD-SOI,

, i

1W^ =49rwr.



30.

v-

=v-^

+2tu- *4Nnd"'

rD

t

D

C.

v,u

e^"

+

-+^(+)

=

f

-o

026tnt##)

,

o,

2uru

z*!!n(L)

=0.92Y

3 1 .

At ,

qNnLd

r i

l\l/

.7

=

' C o

1 . 6 x 0 - t n

5 x l 0 t t

x 5 x

1 0 - 7

For

he rench apacitor

A:4x 7

prfr+ pm2

29

pn?

C

:29

x3.45x10-1s

:

100

x

l0-1s .

3 3 . I = C d V

= 5 x 1 0 - t a

' ' t

- = 3 . 1 x 1 0 - " A

d 4 x 1 0 - '

q n i

3.9x .85x 0 - 'a

C o =

4xl}-'

".V,

=

-1.02+0.92+

=

-0 .1+0.28

= 0.18 .

7

34.

f "

= ; ! - tpc i (V ,

-Vr)

4 x l 0 - 5 : A ( - 5 - V r )

I ' 1 0 - 5 : A ( - 5 + 2 )

LY

= 7

- ( - 2 ) = 9 Y .

=

8.63

10-t

F/cm'

l . 6x10- te

5x10 t t

3x

0-u

8.63x10- i

8 .63 10-?

=

46mV

Thus, the rangeof 27"s

from

(0.18-0.046):

0.134

V to

(0.18+0

046)

0.226

V .

32.

The

planar

capacitor

^ . F , z

_

(1x10 r ) ' 3 .9x8 .86x

10 -14

= 3 . 4 5 x

1 0 _ r ,

: A'"/a

---a.lo*

35. V,

=Vru

+2r,ltu

' V r

= 7 V



8.854x0 - 'o

Co

=3'9

x

l 0 - '

V-

=4.98

+0.42+

=3.45

10-'

Flcm2

(

r o ' i )

V"

=0.0261n1

-

= l=0.42Y

( 9 . 6 5 0 ' /

O ,

E

vou

Q^,

t=- ; -vu

-o

=-0.56

-0.42

=

-{.98

V

2

3.45

x

10-8

Q , 5 x 1 0 " x l . 6 x l O - ' e

LV--=

co

3.45 10-8

V

-2.32

Vr

=4.34

2.32

=2.02Y.

= - 0 . 9 8 + 0 . 4 2 + 4 . 9

=4 .34

Y



CHAPTER

7

1. FromEq.l, the heoretical anierheight s

Qa,

Q.

-

X=

4.55 4.01

0.54

eV

We

can calculate Z, as

v,

=!!

yr5=

g.g259

*2'86

x

1o''

,

=

0. 88

q

N D

2 x l 0 r 6

Therefore,he built-in

potential

s

Vo ,=Qr ,

Vn

=

0.54-0 .188

0 .352V.

2.

(a)

FromEq.l1

de

I C,)

_

(6.2

L6):I0',

=

_2.3x

10,0

cm,/F),/V

d v

- 2 - 0

) f _ t - l

N ^ =

-

| l = 4 . 7 x 1 0 ' u c m - '

qe"

Ld(l

C') I dV

)

v-

=t!n

lk-

=

s.6259(

+:

*

to".l=

o.*

u

q

ND

\ 4 . 7

x

l0 ' "

/

From Fig.6, he

intercept f the

GaAs contact s the

built-in

potential

V6i,

which s

equal a 0.7 V. Then,

he barrierheight s

Qu,

Vo,

Vn

=0.76V

(b)

J"

=5x10'7

Ncmt

Ar

=

8

NK2-cr* for n- type

GaAs

J,

=

A.T'e-qQBt1

tkr

eu.

'{I\,

=

0.025erf

I

(1ggJ'

I

=

o.rr.,,

q

/ "

|

5 x l 0 - '

J

Thebarrier eight

romcapacitance

s 0.04V or 5Yoarger.



(c)

For V:

-l

Y

=2.22x

10-t

cm=0.222

m

4N'w

=1.43x105

/cm

es

C

=L=5.22x10-8

F/cm2

W

3. The barrierheight s

The maximum

electric ield

is

l..l

=1.

r=

o)l

QN

y4t

q

Qu,

Q^ x=

4.65 4.01=

.64

v,

=L!

yr{s-=6.6259

vr(2.8orlo'n

=0.,r,

u

q

N D

[

3 x l o ' u

)

The built-in

potential

s

Vo ,

Qr , -V ,=0 .64-0 .177

0 .463V

The depletion

width is

=

0.142

hn

(1 .6

10 ' ' )

x (3x

101u)

(1 .42x

0 r )

=6.54

x

100V/cm.

11.9

(8.85

10- 'n

4. The

unit of

C

needs

o be changed

rom

pF

to

F/c#,

so

1rc2

-

t.74xl01s

2.r2xl}ts

4

1crfinf

Therefore,

e obtain he

built-in

potential

t IlC2

:0

,^

-

l '57

x lo t j -

=

0.74y

'

2.12x10"

2e,(Vo,

V)

4No

2x l l .9

x8.85

l0 - ' o

1 . 6 x 1 0 - t e x 3 x 1 0 t 6



From he

d(J-)

\ c ' )

dn

FromEq.

N r =

given

relationship

etween

C and Vo,we

obtain

:

-Z.l2xl}ts

1cn?ny2V

t 1

2 f

- 1

I

t

_

|

qe, lag tc ' ) /

v

l

2 _ [ r ' )

ffil.rlr.ro"j

=

5.6x10" cm-'

v-

={1n

&-

=0.025e*[z.so*

0' '

]=0.,u,

q

ND

I

S.0x l0 ' "

We can obtain the

barrier height

Qun

Vo, Vn =0.74+0.1610 .901V.

5. The built-in

potential

s

v t ,=Qan_T"

*O

=

0.8 o.o25gr[z'se

ro"

]

I

t .s><lo"

)

=

0.8

0.195

=

0.605

Then, he work function

is

Q^

=Qr^

*

X

:0 .8

+

4 .01

=

4 . 8 1 V .

6. The

saturation unent density

s



J .

=

ArT2

" *o ( -

q Qu " )

\ r r )

=

o

x

(3oo), (

*ol,

-,9

,

)

\0.025e

:3.81

x

l0*1

Alcm2

The injectedhole

current

density s

,

- Q D o n , '

- 1 . 6 x 1 0 - ' n

1 2 x ( 9 . 6 5 x 1 0 ' ) '

r , ^ . .

J

=

i:t=

ffi

=

.lex

0-"Ncm'

Hole cunent

J

,"1eqn'k'

-11

Electron

urrent

J"(etvlkr

-l)

= J

r o

_

t . l g x o - r r

= 3 x l o _ 5 .

J

"

3 . 8 1 x

0 - '

7.

The

difference

between he

conduction

andand he Fermi

level is

given

by

(

d t ' t o " \

V^

=0.0259

"l

*

l:0.04V.

\

1 x 1 0 "

The built-in

potential

barrier

s then

Vt,

=0.9

0.04

=

0.86

V

For

a depletionmode

operation,

VTis negative.

herefore,From

Eq.38a

V r

= 0 . 8 6 - V ,

< 0

q a ' N ^

l . 6 x l O - ' n a ' x l O ' t

r r= -zE:=m>0 '86

1 .6 10 -2

----------------

a"

>0.86

2.19

x l0- ' '

a

>

1.08 10-5

f i r

=

0.

108

m.



8. From

8q.33

we

obtain

5

x

10-o 4500 12.4x8.85

10-'a

o =

" 6 m

0 . 3 x 1 0 - o1 . 5 x 1 0 - a

:1 .28x10-3

S:

1.28mS.

_

Q N o a '

_

1 . 6 x 1 O - t e7 x l 0 t 6 x ( 3 x 1 0 - s ) 2

_

A A . ,

2e.

2x12.4

8.85

l0- 'o

Es2

i l *o . t

VP

9.

(a)

The built-in

voltage

s

vo,

Qun

v,

=0.9

0.025hIo

1,t0 ' ': 0.86

[

1 0 "

/

At zerobias,

he width

of the

depletion

ayer s

:1 .07

x

10-s

m

: 0 . 1 0 7p m

Since / is

smaller

han

0.2

pm,

it is a

depletion-mode

evice.

(b)

The

pinch-off

voltage

s

v "

_

e N o a '

_

l . 6 x

0 - ' n

1 0 t t ( 2 x0 r )

: 2 . 9 2 y

'

2e"

2x12 .4x8 .85x

0 - 'o

and he threshold oltage s

Vr

:

Vri-

Vp

0.86 2.92

-2.06

V.

10.

From

Eq.3lb,

he

pinch-offvoltage

s

2x1.09x10-" 0.86

[ . 6 x 1 0 - t n x 1 0 t 7



, ,

_ Q N o a '

_ l . 6 x l o

' n

x l o ' t ( 2 x l o

, - - - 7 - -

l):

o.ruo

2t ,

2x l2 -4

8 .85x0 - 'o

The

threshold oltage s

Vr

:

Vni-

Vo 0.8 0.364

0.436

Y

and he

saturation urrent s

given

by Eq.

39

I

r"o,

t44v"

-Vr)'

2 a L \ s

50 1:0r;+2'1:j':851t0:

: i

4s00

(0

0.436),4.7

x10*

A.

2 x ( 0 . 5 x 1 0) x ( l x l 0 - o )

l l ' 0 '85-

o 'ozsgh(4 '7"10"

- - lux l ' - rexN^

)

\

N ,

)

2 * 1 2 . 4 ,

J 5 l 0 - "

a -

:

0

For No

:

4.7

x

1016

rn3

(

_ , t ,

a:

|

0.85

0.0259^4'7

xl}"

)"

(3'7xr}')

-

t

N D

)

^ l N "

: 1 . 5 2

x l 0 - s

m : 0 . 1 5 2

m

For

Nl

:

4.7

x

1017rn3

a:0.496

"

10-s m:0.0496

m.

12. From

Eq.48 he

pinch-off

voltage

s

. M

v,

=

Qnn

- ' -v ,

q

:0.89

-0.23

(-{.5)

: 0 . 6 2 V

and hen,

eNnd, '

1 .6x l0 - te3x10 t t

"

y .

= - :

' a '

: 0 . 6 2 V

'

2e.

2x12.3

8.85 10- 'o



d r : 1 . 6 8

1 0 6

m

dt:

16.8

nnr

Therefore,

his

thickness

f the

doped

AlGaAs

layer s

16.8

nm.

13. The

pinch-offvoltage

s

v,

-_

QNodl

l .6x

O-rn

l0 's

x

(50x

0-7)

:

l .g4v

2e"

2x12 .3

8 .85 l0 -o

The threshold

oltage

s

AE,

r./

V ,

= Q u , - "

_ r ,

q

:0.89

-0.23

1.84

:

- 1 . 1 8

V

When

r:1.25x

1012

rn2,we

btain

12 .3

8 .85

x

lO- t o

r - i

'"

:

ffi

x

o

(-l'18)]

'25 rot2

and then

do

+58.5

64.3

do:

5.8nm

The hickness f the undoped

pacers

5.8 nm.

14.

The

pinch-offvoltage

s

r r

q N r d l

l . 6 x l 0 - ' e x 5 x l 0 ' 7

( 5 0 x 1 0 ' ) '

l - ' U | _ \ - - - ' - 1 o ' | \ . '

'

28"

2x12.3x8.85

10- 'o

Thebarrierheight s

AF

Qun

V,

+

*c

*V,

=

-1.3

+

0.25+ .84

0.79V

q



The

2DEG

concentration

s

1 2 . 3 x 8 . 8 5 x 1 0 - ' o

l . 6 x

0

r n

x ( 5 0 +

0 + 8 )x l 0 - ?

*

[o

-

1-

.:yf

=

|.zgx

t o, cm-2

r r =

1 5 . The

pinch-off

voltage

s

vr

t6.

, ,

-

QNod l

r <

l .6x lo - tn

l x lo t8

xd f

'

2e

2x12 .3x8 .85

10*

The hickness

f the doped

AlGaAs is

d r =

1.5 2

x

12.3

8.85

I n - r+

f f i :4 .45x10-8cm:44.5nm

A F

=

Q

B,

_

*_=S_

_

V

p

=0.9

_

0.23

_

1.5

_0.93

V.

The

pinch-off

voltage

is

aN^

v ^ - '

"

d :

2 e

1 . 6 x 1 0 - ' t

3 x 1 O t t

(rs

to'

)'

z.t

v

2 x 1 2 . 3 x 8 . 8 5 x 1 0 - t o

the threshold

voltage

is

Atr

V,

=

Qr,

-

*c

-V,

q

:

0.89

0.24 2.7

:

-2.05

Y

Therefore,

he wo-dimensional

lectron

as

s

1 2 . 3 x 8 . 8 5 x 1 0 * ' o

f l r =

l .6x

10-'n

(35

+

8)

x

10-7

*

[o

1-z.os)]=

.zx o crn2



l . Z o :

r

: . 2

CHAPTER

t o

=2 *10 -12

752

= 1 1 . 2 5

nH .

= C

2.

3 .

3

x

108m/s

r:==

"

x { l

1 6

= 1 . 6 1 6

t l z .

a

Vo, (E, / q) +V^*Vo : 1.42+0.03+0.031.48V

'=^W

1 /

q I N , N ,

)

= {

t - r - ' '

\ l o " x r o ' ' / '

=

1.89 10-6cm

18.9nm

c

=k-r ' l6xlo- t2

=

6.13

lo '

F/cmz

W

1.89 10- '

4 ="(t)"*?

t).,,*o(#)

From

Fig. 4, We note hat

he argest egative

differential

esistance

ccurs

between Vp<V<Vq

The conesponding oltagecan be obtainedrom the condition &ttdV: 0. By neglecting he

seconderm n Eq.

5, we obtain

:{ro)'.0;GI



#=(+-T)"*l

)

#=(+.#).*['

+)=,

. ' .V

2Vp

2x0 .1=

0 .2V

#|"'.=ifr-

#]"4'

#)

=

0

0367

^=(#1,,")- '=

27.2,,.

5

a)

":l

(

n

a*=+t:&d-=#*

| , \ 2

_

( l 2 x l 0 - , 1

=137

l

2(5

x l } -a

)

1 .05

10- ' '

x

l0 '

(b)

The

breakdown

oltage

or

Np:

101s

rn3and

W:

12

pm

is 250

V

(Refer

o Chapter

).

The

voltage due

to

{sc

is

1Rr.

=

( tO'

S

xt0 ' )x

137

68.5

V

The otal

applied

oltage

s then

250

+

68.5

318.5

V.

6.

(a)

The

dc input

power

s

l00V(10-'a1

tOW.For

25Yo

fficiency,

he

power

dissipated

s

heat s I0W(l-25%):7.5

W.

LT

- -7 .5W

x(10 'C /W)

75 'C

(b)

AVB

=(60mV/"C)x75'

=4.5

V

The

breakdown

oltage

t

room

emperature

s

(100-4.5):95.5V.

7. (a) For a uniformbreakdownn the avalanche

egion,

he

maximum

electric

ield

is

E^:4.4

xld

V/cm. The

otal

voltage

t

breakdown

cross

he

diode

s



4

=E,xu.[-

f),

w-xn)

=

4Axt

(0.+

on) (+.+.

r

O

)r

.6

l0- 'e

1.5

1d2

3-0.a)x

0*

1.09x10-"

=17.6+57.2=748V

(b)

The average

ield

in the

drift

region

s

572 =Z.2x l05V/cm

(3

0.4)x0

"

This

field is

high

enough

o maintain

velocity

saturation

n

the

drift region.

( c )

= ,

u '

, = ,

l o t .

,

= l 9 G H z .

2 l t r - xn )

2 (3 -0 .4 )10-4

8.

(a)

In the

p

layer

s r ( x ) :

^ - q y ' t

o < x <

b : 3 1 t m

E"

E2(x)+ .^ -W

b<

x< I l :

t21 tm

q

E2(-r) houldbe arger han105V/cm for velocitysaturation

" 'E^

QN'b

'1g '

ts

o r E m

>

105

qN ' b

=105

+

4 .66x10 - "N1 .

q

This

equation

oupled

he

plot

of

E, versus

N in

Chapter

3

gives

Nt

:7

t

1015

m3 or

E^:

4.2

x

lOsV/cm

. . . v *

( e .

- e r ) b

+ E 2 w : ( 4 . 2 - l ) x 1 0 5

3 x l 0 {

+ l ' s x g x l 0

"

o

2

z

' - "



: 1 3 8 V

W

- b

_ ( 1 2 - 3 ) x 1 0 -

b)

Transittime

:

;=ffi

:

Qxlg-rs

90

ps

10.

(a)

Refening

o

Chapter2,

we have

Ncu:r(ry\,

=*,,(;),

=4.7

x1g,r(

'z*o

) '

=o.rx

10r7

7l=3.33x

0, ,cm-.

\0I7m0 )

(b)

For

T":300

K

NCU

oun, . -AE t L , r , \

-

?r

. , ̂ - . ^ (

_ 0 '3

lev

) - t ,

Ncz-exp(-aE

kq):

ztxexn[ffiff

)=rtxexp

-

1r.e7)

:4 .4x104

( c )

F o r L : 1 5 0 0 K

9.

(a)

For

transit-time

mode,

we require

n6L

>

l0rz

cni2

no

xl}tz

/ L

=lotz

/1x

or

-

lorucm-'

(b)

r:

L/v: I}a

/ 107

l0-rr

s: 10

ps

(c) Thethresholdield for InP is 10.5kV/cm; thecorrespondingppliedvoltage s

v

=(to's:to'

(, ',0-')=

.52sv

[ 2 ) "

The

current

s

I

=

JA=

qnry,

)a

=(t

A"

0-te

4600x

016

5.25 10,

x

10{

:

3.g6

A

Thepowerdissipatedn thedevice s then

P = I V = 2 . 0 2 W .



Ncu

o-^(- , r t ] L.r \

- - r . ,

^- ,J

-0'3iev

)

.rr/.r-

xP(-Ar:

k

T)

=

7

"*pl

'oxlpoo

l :oo)

=

7

"

"*p

(-

n94)

=

6.5

Therefore,at T" : 300 K mostelectrons re n the lower valley. However,at T" : 1500K ,

87Yo,

.e.,6.5/(6.5+D,

f the electrons

re n

the upper

alley.

I

l. The

energy

E" for

infinitely

deep

quantum

well is

E n =

h "

n '

8m't

l*:1#e'x',)l+LE,=4*

L

.'.

LEt

=

3 meV

LE2

=

11

meV

12.

From Fig.

14 we

find that

the

first

excited

energy

s at 280

meV

and the

width

is 0.8

meV.

For

same

energy

but a width

of 8

meV,

we

use he

same

well

thickness

of

6.78

nm for

GaAs,

but

thebarrier hicknessmustbe reducedo 1.25nm for AlAs.

The

resonant-tunneling

urrent

s related

o

the integrated

lux

of

electrons

whose

energy

s in

the

range

where

he transmission

oefficient

s

large.

Therefore,

he

current

s

proportional

o

the width

A,En,

and

sufficiently

hin

baniers

are equired

o

achieve

a high

current

density.



CHAPTER 9

1. fw (0.6gtllrr) I .24 0.G 2.07eY (FromEq.9)

oc(0.6

m):3xlOa

crnr

The net incident

power

on

the

sample s the

total incident

power

minus he reflected

power,

or 10 mW

I , \

1 0 " [

_

e - 3 , 1 0 ' t { J

5

x

t O r

W:0.231 u rn

The

portion

of each

photon's

energy

hat s convertedo

heat s

hv

-

E,

_2.07

1.42

=3l.4yo

hv 2.07

The amount

of

thermal

energy

dissipatedo the attice

per

seconds

3l.4Yox5:1.57 W.

2. For ": 0.898

pm,

the conesponding

hoton

nergys

E

=

l'24

=

1.38 eV

h

FromFig. 18,we

obtain

r

1arc.fCa0.7As):3.38

sin

Q

=

-+

=

+=

0.2958

o"

=

17"12'

fiz 3.38

Effi e cy

4n'n'

r

c?:

o")-

+

x

t

x

g'l

s

[t

c9!(l

'

tz')]

@,

f i,) ' (1+

.38f

=

0.0315

3 . ISYo

/ \ 2

_ f 3 . 3 9 - l l

R = l - l

= 0 . 2 9 4

\3 .39

1 /

(a)

The mirror

loss

3. FromEq. 15



f i l l ) =

t ,

h , , f - l - ' ) = 4 0 . 5 8 c m - , .

L

\ R /

3 0 0 x 1 0 - "

0 . 2 9 6 )

(b)

The

threshold

urrent eduction

s

J,o(R

=

0.296)

J,o(R,

=

0.296,R,

=

0.9)

lo*!

,f1)l

l

o*rr[--r-)l

_L

r

\R /J

L

2r

[R ,R , / j

o*

"fr')

L \ R i

4 0 . 5 8 -

t

, h f

t

)

_

2 .300

l0 - -

\0 .296 '0 .90 /

l0 + 40.58

=36.6%o.

4. From Eq.

10

s i n @

\ - i r = i r . s i n O .

n2

From

Eq. l4

r

=

I

-

exp

-

c Li

d)= t

-

""p(-

8

x

r0'

.

3.6(1

sine"

).

1

x

1

0

*

)

F o r

4

: 8 4 o

n r :

3 . 5 8

n : 0 . 7 9 4

Q:

78onz:

3.52 ;

:

0.ggS

5. From

Eq. 16we have

m L = 2 i L .

Differentiating

he

aboveequation

with

respectto ,

we obtain

n d m ^ , d n

/ 1 e - + m - z L _ -

dL d^



Substituting

Zitll"

for

m and

etting

dmldJ,:

-

Lm/L2,yield

{-}!\*z!

=2LE

\ ^ 2 )

1

i l "

:.

a,).=

and

fttm

znrl

-(1\(

\l

|

\n ) \dL))

6. From

Eq.22

r:!I..| '(fl

andEq

5

=

=)'

Rr

:

0.3 7,

R2 0.31

g(a+"

=

#a["'.#'(#)]

=

270

g(zs'

=

*['*.

**o"'(#)]

=

zt.

t

rnf

'

)=

'

hf l)

2 L '

( R . 0 . 9 9 l

t 0 0 x l 0 - a

( R /

ForRr

0.317

*^Gi

"*l

---1-_-'(#),

L,

50

n

For

R2

0.311

#'["#-) =-*-'(#) >Li=5043 t' t

7.

From

Eq.23a



J,o

ro.

ftoo

For

R

:0.317

'

,

h(

r

)l=

,ooo

cm',

2 x l 0 0 x [ 0 '

\ 0 . 3 1 7 x 0 . 9 9 ) )

'

and

o I,n 1000x100l0-o

x5x

10+

=

5 nrA

For

R: 0.311

FnI]>

Fz

0.1x0.998/0.7

4 0.

3

J,o

=7.66,.[roo

h(

|

)'l

=

,uu

Alcm-,

L

2 x l 0 0 x l 0 - *

\ 0 . 3 1 1 x 0 . 9 9 ) )

'

andso I,a

=766

x

100 10*

x

5

x

10-o 3.83

nA

8. From the

equation,we have

for

m:0:

4

t4nL4+

4nLL

=

o

which can be solved as

There

are several

variations

of

+

in

this

solutions.

Take the

practical

ne, .e., uB

fuo,

ives

"s

:1.3296

or 1.3304

m.

d-

1'33

=0.196

rn

2 x 3 . 4

9. The threshold urrent

n Fig.

26b s

given

by

I*:

Io exp

Zl

10).

Therefore

E

=

J-dI

,o

I

=

o.oo9

(,

c)-'

-

I ,o dT

l lo

If To:50

oC,

the

emperature

oefficient

ecomes

(2sa)

(25b)

solution which

is the

only



1

t = ! = 0 . 0 2 ( , C ) ' .

5 0 \ /

which

s larger

han

hat

for To:

110

oC.

Therefore

he

aserwith

To:

50

oC

is

worse

for

high-temperature

peration.

10.

(a)

Atr:

QQt"+ 1:n\

nA

- ' -

n

=electron-

ole

oairs=,

M

.

oltt,

+

prb,a

2.83x0- '

-

= t u

c i l l

l .6x

0-' '(3600

t700{10/0.0)(zx

x

l0-' ,f

0)

?

-

2'83

lo

3

=l2o

us

23.6

(c)

nn(t)=^n

xp(-lc)=r0""*n[-#h]

=2.5x10e

m-3.

l l . From

Eq.33

r

_

=

n(

.ss.

o*

).

rooo

o

ro-''

sooo

)

' \ '

3 " q

) \

t 0 x l O - o

)

=2.55x10-u

2.55

1tA

and

rom

Eq.35

t t -E

3000.

x

l0- 'o

.5000

(raln=

r

=---

1o; r-,

=

t '

12. From

Eq.

36

,

(

L\('--l '

=fgl

P\=

o

41)

( q ) \ n v l

l P . o , ) \ q )

\ q )

Thewavelength "of light is relatedo its frequencyv by v = c/L wherec is the velocityof

light in

vacuum.

herefore

v

/q: hc 2q

andh:

6.625x10-34

-s,

c

:

3.0x

0l0 cmls,

q

:

1.6x10-re

oul.

eV

:

1.6x0-re

J.



Therefore,

v/q:1.24

/ ),fum)

Thus,

:

(Rx1.24)

LandR:

(rZU

1.24.

13. The

electric

ield in

the

p-layer

s

given

by

r , ( x ) = u ^ - T ' N t t

< x < b ,

€"

where

E,n s

the maximum

ield.

In

the

p

layer,

he

field is

essentially

constant iven

by

r, (x)= ^ -9!t!- b <x <w

ts

The

electric

ield required

o

maintain

velocity

saturation

f holes s

-

lOs

V/cm.

Therefore

qN'b

E-

-J I -J"

>

16s

€ r

or

E .

> 1 0 5

4 1 ! ' b

= 1 0 ' +

4 . 6 6 x 1 0 - , , N r .

q

From the

plot

of

the

critical

field

versus

doping,

he corresponding

m

areobtained

N1

7xl01s

cm-3

E^:4.2x105

V/cm

The

biasingvoltage

s

given

by

,,

=fui&

*E2w

":,q*-lo*)

+ o,

r

o-o)

=

138

V.

The transit ime is

,

-

(w

u)

_n1

lg*

=e

x

o-,,

=eo

s.

vs

10 7



14.

(a)

For a

photodiode,

nly a naffow

wavelength ange

centered

t the optical signal

wavelength s important;whereas

or a solarcell, high

spectral esponse

ver a broad

solar

wavelength

angeare required.

(b)

Photodiode re small o minimize

unction

capacitance, hile

solarcells are arge-area

device.

(c)

An important igure of merit for

photodiodes

s the

quantum

effrciency

number

of

electron-hole

airs generated

y

incident

photon),

whereas

he main concern

or solar

cells is the

power

conversion fficiency

(power

delivered o

the load

per

ncident

solar

energy).

15

(a)

"

AqN,N,(+^E.!^T-)"-",u

[ N ,

1 C

N o l t o

)

=zft.sx o-'n[z.soto'nfu..66x0'nx

(

== t * .8

* ,

= E) ,s - , , , " ,1 r ,

I t . z * 1 9 t e ! 1 9 - s

x r o ' e

s * t o - '

J

-

=

2.43

x

to2o

s.al *

1g+p-+t

z

=2 .28x

0 - ' ' A

I : 1 ,Q* l r ' ) -

t ,

l l l=tr-1"(etvtr'

-r1

0.3 0.4 0.5 0.6

0.65 0.7

2.5x10-

1.1x10- .55 26.8179 1520

(mA)

/ 1

95

95 94.5 68.2

-84

(b)

v^.

=u^(

!-\=

0.025e*[

t

o'.-

)=

o.u,

q

[1" ,

\ .2.28x10-"

(c)

P= I,VQevrw

t)-trrt



o

=

1"(rev/t'

t) * t,

#enrlr,

-

I,

I,etvlkr

(l

+

tt)

-

t,

dP

dV

.'. eon/o'

-

IL

_t!n(r.sL\_tr1

q

\ .

k r ) q )

1"

(1

=

0.64

r

I

=

t l vo ,

L

+ v )

V

m

P^

=

I,V^

=

9s

10-3

0.68

0.025e(t

+ z+.t)-

0.}zssl

=52 mW

1 5 0

16. From Eq.38

and

39

I

=

I,("0,0,

- t ) -

t ,

and

t r . ( r ,

. )

r r . ( r , \

y _

- _ n l

"

+ l

l = _ l r l l

-

|

q

\1"

)

q

1 .1"1

I ,

=

I t€-qv lkr

-

3. e-o6fao7ss5

2.493

x

I

=

3

-

2.493

xlO-ro

.

evloo2sss nd

P

:

I

V

1 0 - t 0 A



V

I

P

0

3 . 0 0

0 . 0 0

0 . 1

3 . 0 0 0 . 3

0 . 2

3 . 0 0

0 . 6 0

0 . 3 3 . 0 0 0 . 9 0

0 . 4

3 . 0 0 20

0 . 5 2 . 9 4

. 4 7

0 . 5 1 2 . 9 1

4 8

0 . 5 2

2 . 8 6

4 9

0 . 5 2 . 8 0

4 8

0 . 5 2 . 7

4 6

0 . 5 2 . 5 7

4 l

0 . 6

0 . 0 0 0 . 0 0

.'.Maximum

power

output

1.49

W

Fill Factor

pp:I^V^

=

P-

=

IrV*

I ,V*

0.6x3

=

0.83.

17. FromFig.40

The

output

powers

or

&:

0 and

&:

5

C)

can

be obtained

rom the

area

Pr

(&:0):95

mAx0.375V:35.6

W,

Pz(&:

5 C)) :

50

mAx0. l8v:9.0

mW

.'.For4,:6

Pr/

Pt

:

L00%o

For& :5

O Pz l

P t :9 /35 .6 :25 .3Yo.

18. The

efficiencies

re

14.2o/o

1

sun),

16.2%

1O-sun),

7.g%(100-sun),

nd

18.5%

(1000-sun).

Solar ells eeded

nder

-sun

ondition

a

I

€

B

2 . 0 0

1 . 5 0

1 . 0 0

0 . 5 0

0 . 0 0

0 . 5

V

(vo l t s )

t .49



4(concentrat

on)

x

P,,

concennat

on)

rfi

-

sun)

P,,

l

-

sun)

-16'2Yoxlo

11.4

cel ls

or

1o-sun

14.2Yo

l

=ll,Y|19

=ns

cells

or

loo-sun

14.2o/ol

-

l8'5%

x

10

=

1300

cells

for

1000

sun.

14.2Yol



CHAPTER

1

O

l . C o : 1 0 1 7

c r n 3

/.g(As

n Si):0.3

0.9

45

0.6

30

0.4

.2

l 0

1 6

1 4

o

1 0

- b 8

6

2

0

Cs ftoCo(l

-

1,11M0)*-t

:

0.3x1017(1-

)-o

:

3x

101641

il50)0.7

/(cm)

Cs(cm-3)

3 .5x

0

4.28x10

5.68x

0

2.

(a)

The

radius

of a silicon

atom

can

be expressed

s

It

f

= - g

8

t;

V J

s o r

= - x 5 . 4 3 = 1 . 1 7 5 A

8

(b)

The

numbers

f Si

atom n

its diamond

structure

re

8.

So he density

of

silicon

atoms

s

8 o

n

=---==

--+

=

5.0x 1022

tomVcm

o' (5.434)3

(c)

The

density

of Si is



M

/ 6.02x10

28.09x5

l0

22

O

=

-ff

=

-ffi

gl

cm3

2.33

ctlf

3. ltu:0.8 for

boron n silicon

M /N4o:0 .5

The

density f Si is 2.33

g

/

ctrl

The

acceptor oncentration

or

p

:

0.01

Clcm

is 9x 1018

rn3.

The

dopingconcentration

s is

given

by

Il f

C

= k ^ C ^ ( l - - '

1& - t

M ^ '

Therefore

C "

9 x 1 0 "

-o

------------7--

k"Q

!)a '

0 '8(1-0 '5)-o '

: 9.8 l0t tcm- '

The amount

of boron equired

or a 10

kg charge s

10'000

x

9.8 0''

=

4.2x|0"

boron

toms

2.338

So

that

ro.8g/moreA#ffi,=o.7sg boron

4.

(a)

The

molecularweight

of boron s

10.81.

The boronconcenhation

an

be

siven

as

number

of boron aioms

/ I ^ = -

"

volume

of siliconwa fer

_ 5.41 10-3/10.819x.02x 0 ' 3

1 0 . 0 2

3 . 1 4 x 0 . 1

=

9.78 l0r8 tomVcm

(b)

The average

ccupied olume

of

everyone

oronatoms n

the wafer

is



t t

=L=--- l -cm3

nb

9.78 0 '"

We

assumehe

volume s a

sphere,

o he radius

of the

sphere

r

)

is

the

average istance

etween

wo

boron atoms.

Then

t v

r

= ^ l - '

= 2 . 9 x I 0 - t c m .

\ 4 n

5. The cross-sectional

reaof the

seed s

/

o.ss

'

7d

- ' - -

|

=0 .24cm2

( 2 )

The maximum

weight

that can

be supported

y the

seedequals

he

product

of the

critical

yield

strength nd he

seed's

ross-sectional

rea:

(2

x

106)x

.24

4.8x105

=

480

g

The corresponding

eight

of a 2O0-mm-diameter

ngot

with length

is

(2.33!cm'

,{

y)'

/

:

48oooo

l

\ z /

I

=656cm

=6.56m.

6. We have

c,/co=^(-#)^'

0 0.2

ractional

solidified

0 .8 1 .

cslc0

0.05

0.06

0.08

0. t2

0.23



r u .

tm_

!

m - m q _ m _

lJ-Sflo"mm

The

segregationoefficient

f boron

n

silicon s

0.72.It is

smaller

hanunity,

so the solubility

of

B

in Si under

solid

phase

s

smaller han

hat

of the melt.

Therefore,

he

excess

B atoms

will

be thrown-off

into the

melt,

hen the

concentration

f B in the

melt

will be increased.

he tail-

end of the

crystal is

the last

to solidify.

Therefore,

he concentration

f B

in the

tail-end

of

grown

crystalwill

be higher

han

hat

of seed-end.

The reason s that

the

solubility

in

the melt

is

proportional

o the

temperature,

nd the

temperature

s higher n

the center

part

than

at the

perimeter.

Therefore,

he

solubility

is higher

in the

center

part,

causing

a

higher mpurity

concentration

here.

The

segregation oefficient

of

Ga

n

Si is

8

x10-3

From

Eq. 18

C" / Co

=

1- (1-

k)"-o' t

We have

=

2501n(1.102)

=24cm.

rjl

E

I

7.

9.

10. We have

rom Eq.18

Cs

=

Co[

-(l -

k") exP(krx L)]

So he

ratio

Cs C0

=[1-(1-

k")exp(4rx/Q]

:

1-

(1

0.3)

exp(-0.3

1)

=

6.52

:

0.38

at x/L:2.

a t x l L = l



11. For he

conventionally-doped

ilicon,

he resistivity

aries rom 120

C)-cmo

155 )-cm.

The

corresponding

oping

concentration

aries

rom 2.5xl0t'

to 4" 1013

rn3.Therefore

he ranse

of

breakdown oltagesof p* - n unctions s givenby

t E 2

v,

=.t(i l,)

'

-

1 . 0 5 x 1 0 - " ( 3 x 1 0 5 ) 2

( N r ) - ,

= 2 . 9 x 1 0 , ,

N B = 7 2 5 0

o l 1 6 0 0 V

2x l . 6x

10 - ' t

LVB

=

11600

7250

=

4350

V

(

nv ^ \

|

- .

l l7250

=+30%o

\ 2 )

For the neutron

rradiated

ilicon,

p:

148

+

1.5C)-cm.

The doping

concentration

s

3xl0l3

(tl%).

The range

of breakdown

oltage

s

vB

=

1.3 l0t1 N B

=

2.9xr0t7 3x 1013t l%s

= 9 5 7 0 t o 9 7 6 2 Y .

LVB

9762

-9570=192V

(

tv" \

l - ^ a

l l 9 5 7 0 = * t y o .

\ 2 )

12.

We have

M,

_

weight f

GaAsatTo

_C^-C,

_s

Mr

weight

of liquid

at To

C,

-C^

I

Therefore,

he fraction

of liquid

remained/can

be obtainedas following

f

= , M ,

-

I

=

3 o

= 0 . 6 5 .

"

M , + M ,

s + /

1 6 + 3 0

13. From the Fig.ll,

we

find the

vapor

pressure

of

As is

much higher

than

that

of

the

Ga.

Therefore,

the As

content

will

be lost

when the temperature

s

increased.

Thus the

composition

f liquid

GaAs

always

becomes

allium

rich.



14. n

":

Nexp(-n"

/ kT)

5

x1022

r,p(2.3

y

/ kz)

=

5

x

ro" "*p[-

{'8--l

L(r

3oo)j

:

1.23x

0-t6cm-3

0 at 27o

C

=

300K

: 6 . 7 x 1 0 ' ' c m - 3

a t 9 0 0 0 C

1 1 7 3 K

:

6.7

lO'ucm*'

at 12000

=1473K.

15. n,

=JNi/

exp(-y,

/zkT)

:W

*"-r . tevt2kr

=7.07

xl1 to

xe-t0.7/(r /3oo)

:5 .27

10- ' t

a t27oC:300

:2.l4xl0ta

at 90fC

:

1173

K.

16.37

x

4:

148 hips

In terms

of litho-stepper

onsiderations,

here

are 500

pm

space olerance

between he

mask

boundary

of two

dice.

We

divide the

wafer

into

four

symmetrical parts

for

convenient

icing,

and discard

he

perimeter

parts

of the

wafer.

Usually

the

quality

of the

perimeter

arts

s

the worst

due

o the

edgeeffects.



300

nrn

T o to t D i e s , 148

ll vf,dv tykr

I I r t

= j i _ -

l _

ff

f'o'

\l781/t

4 ( M \ ' ' '

"

(

M ! \

where

=Glfr)

v'

"*o[

=ro,

I

M: Molecular

mass

k Boltzmannonstant:1.38x10-23/k

T: The absolute

emperature

v:

Speed

f molecular

So hat

2

'*=

J;

"

0.66

1 6 .

L =

P( in Pa)

2 x 1 . 3 8 x 1 0 - 2 3

3 0 0

29xI.67 xl}-"

m/sec

=

4.68

xlOncm/sec



;.

P

-

o'!u

=

0'66

-

4.4xto-3

a

L

150

19.

For close-packing

affange,

here

are 3

pie

shaped

sections n

the equilateral

triangle.

Each

section corresponds

o 1/6

of an

atom. Therefore

^. _

number

of aloms

contained

intre

tiangle

'1Y."

-

areaof

fire hiansle

. l

J X _

6

1

J1 ,

c l

x - d

2 2

2

= _ -

Jia'- -Jlg.as"ro*)'

:5.27

xlOto

atomVcmt.

20. (a)Thepressuret970C(:1243K) s 2.9x10-t a or Gaand13pafor Asz- he

arrival ate

s

given

by the

product

f the

mpringement

ate

and

NnL2 :

Arrivat ate

2.64xto2o(-LY4)

l"lvt )\'t'

1

:2.9x101s

Gamolecules/c#

s

The

growth

rate s

determined

by

the

Ga arrival rate

and is

given

by

(2.9x

0rs)x .81

6xt0ta)

I 3.5

A./s 8

0 A"imin

(b)

The

pressure

t 700"C

or tin

is 2.66x

0-6Pa.The

molecular

weight s

118.69.

Therefore

he

anival rate

s

2.64xrc^(

]j54Y-ll =2.zlxt0,0

molecul

rt cm,

s

\

" / l

18.69

973

/ \

x12 '

)

If

sn atoms

are ully

incorporated

nd

active

n the Ga

sublattice

f GaAs,

we have

an



electron

oncentration

f

(z.ztx

o'o

(4.42xo" )

I t - t -

- -

l l

" ' - "

^ "

|

=1 .74x10 t t

m- '

( z . e " l o ' , J [

2

)

21. The.r

value

s about

0.25,which

s obtained

rom Fig.

26.

22.

The attice

onstants

or

InAs,

GaAs,Si

and

Ge are

6.05,

5.65,5

43,

and.5.65

,

respectively Appendix

F).

Therefore,

he

value

or InAs-GaAs

system

s

f

=

(5.65

6.05)/6.05

_0.066

And

for

Ge-Sisystem

s

f

=

(s.43

s.65)

15.65

-0.39

.



CHAPTER

1

l. From

Eq.

1

(with

:0)

x"+Ax:

Bt

From

Figs.

6 and

7, we

obtain

B/A

:r.5

pm

/hr,

8=0.47

pm2/hr,

herefore

=

0.31

pm.

The ime

required

o

grow

0.45pm

oxide

s

t=*G'

+ Ax)=

=lr=10.45t

0.31x

.45)=g.72tr :44min.

B '

0 .47

2. After

a window

is

opened

n

the

oxide for

a

second

xidation,

he rate

constants

re

B:

0.01

pmz/hr,

: 0.116 m

(B/A:6

'I0'2

pm

hr).

If

the initial

oxide thickness

s 20

nm

:

0.02

pm

for

dry

oxidation,

the

value

oflcan

be

obtained

s ollowed:

Q.0D2

+

0.166(0.02):0.0r

0

+[)

or

J:0.372tu.

For

an

oxidation

time

of

20 min (:1/3

hr),

the

oxide

thickness

n

the window

area

s

,t

+

0.166x

0.01(0.333+

0.372\:

0.007

or

x

:0.0350

pm:35

nm

(gate

xide).

For

the

field oxide

with an

original

hickness

.45

pm,

the effective[is

given

by

p:11r,

+

Ax)=-J-10.+S,

0.166

0.45)

27.72ttr.

B '

0 . 0 1

,2

+

0.166x:

0.01(0.33

+27.72):

0.2g053

or x

:0.4530

pm

(an

ncrease

f 0.003pm

nly for

the ield

oxide).

3 .

* + A x : B ( r + c )

G * ! 1 ' - A '

- B ( t + r \

' 2 ' 4



whenf

)

t t t ru

A '

.

4 8 ,

then, 2

:Bt

similarly,

w h e n /

)

r ; t u ,

A t

.

4 8 ,

then.x:

! - f t *c\

A '

4.

At980

1:1253K)

nd

atm,,B:8.5x10-t

lr^'/lo,B/A:

4xl0'2

pm

lfu

(from

Figs.

6

and7).

Since r>2D/k,

B/A:

kColCr,

Co:5.2x1016

molecules/cm3

nd

C,

:

2.2xI022

crn3

,

the diffi.rsion

oefficient

s

given

by

8.5

x

10-3 2.2x1022

=-?;ffi1tm'tttr

=1.79

xl03

1tm2

hr

=4.79

xlO-ncm2

s.

s*!t '

=tl#+tr+zl]

o=L=4(

.r ' )=g[-E-)

2 2 \ A C o ) 2 \ c , )

5.

(a)

ForSil{"It

s i I

, ^

= - = r . !

N x

x

: 0 . 8 3

atomic

/o

H=

looY

=20

l + 0 . 8 3 + y

- v : 0 . 4 6

TheempiricalormulasSiN6 3FI06.

(b)

!=

5x 1028e-33

" '

:2,

10t' -cm

As the SiA{ ratio

increases,

he

resistivity decreases

exponentially.



6. SetTh2O5

hickness

3t, e1:25

SiOz hickness

t, ez:

3.9

SLN+

hickness

t,

E3:7.6, area:

A

then

cTo r o r :

I

qqA

3t

t

=

_ + _ + _

Co*o

Er€oA

ereoA

er4A

co*o=ffi,

Ct" ro ,

-er (er+Zer)

Co*o

3Er€,

_ z s ? . g + z x t . a )

5 . 3 7 .

3x3 .9 7 .6

7. Set

8. Let

TazOs

hickness

3t"

e1

25

SiOz hickness

t, e2:

3.9

Si:Nq hickness

t,

4:7.6

area: A

then

qqA

_4€oA

d =3t ' t =0 .46g t .

q

then

BST

hickness

3t,

e1 500,

area

41

SiOz hickness

t, a2:3.9,

area:

Az

Si:Na hickness

t,

E3:7.6,

area:

Az

o'

=

o.oonr.

A2

444,

=

3t



9.

The

deposition ate

can

be expressed

s

r: ro

exp

-E^/kT)

whereE":

0.6 eV for

silane-oxygen

eaction.

Therefore or

Tt

:

698 K

r(r,)

=2=€XDlo.{_

t

) I

4 r , )

' L

( k 4

k r , ) )

, 'z :

06

[ f ry_3oo]

o.o2s9

L\6e8

r,

)

)

_

Tz:1030

:

757

We can use energy-enhancedVD methodssuch as using a focusedenergy

sourceor

UV lamp. Another

method

s to use

boron doped

P-glass

which

will

reflow at temperatures

ess

han

900 .

I 1. Moderately ow

temperatures

re

usually used for polysilicon

deposition,

and

silane decomposition

occurs at lower

temperatures

han

that

for chloride

reactions. In

addition,

silane is

used for

better coverage

over

amorphous

materials uch

SiO-,.

12. There

are two reasons.

One is to

minimize

the thermal

budget

of the wafer,

reducing

dopant

diffusion

and material

degradation.

In

addition,

fewer

gas

phase

eactions

occur

at lower

temperatures,

esulting

n smoother

and

better

adhering

ilrns. Another

eason

s that he

polysilicon

will have

small

grains.

The

finer grainsare easier o mask and etch to give smoothand uniform edges.

However,

or temperatures

ess han

575

C

the

deposition ate

s too low.

13. The flat-band

voltase

shift is

L V F B D [ l ! [ [

QO,

co

10.

c^

=to '-

3 '9x8'85

lo- 'a

=6.9

x l0*

F/cm-rI

d 5 0 0 x 1 0 - o

-

Number

of fixed oxide

charse s



0.5C,

_

0.5

6.9 l0-8

=2.1xl0r,

m-,

q

1 .6

10 - ' '

To remove hese

charges,

a 450

heat

treatment

n

hydrogen

for

about

30

minutes s required.

14.

20/0.25:

80 sqs.

Therefore,

he

resistance

f the metal

ine

is

5 x 5 0 : 4 0 0

.

15. For TiSiz

3 0

x 2 . 3 7 : 7 l . l n m

For

CoSio

30

x

3.56: 106.8nm.

16.

For

TiSiz:

Advantage:

lowresistivity

lt canreducenative-oxideayers

TiSiu

on the

gate

electrode

s more

resistant

o high-field-

induced

hot-electron

egradation.

Disadvantage:

ridging

effect

occurs.

Larger

Si consumption

uring

formation

of TiSiu

Less

hermal

stabilifv

For

CoSb:

Advantage: lowresistivity

High

temperature

tability

No

bridging

effect

A selective

hemical

etchexits

Low

shear

orces

Disadvantage:

ot

a

good

candidate

or

polycides



1 7 .

a )

O = e + = 2 . 6 7 x t 0 6 x

= 3 . 2 x 1 0 3 Q

0 . 2 8 x 1 0 - a

0 . 3 x l 0 - a

tA

eTL

d s

3 . 9

8 . 8 5

l 0 - r a

0 . 3

1 0 +

x

l x l 0 a

x

1 0 - 6

0 .36

l0 -a

RC

=3.2x105

2 .9x l } - t s

=0 .93ns

(b)

R=

+=r .7 10-6

- .=+

= 2 x I 0 3 C )

A

0.28 10-a

0.3 10r

=

2.9

x

l0

-r'

F-

€A {L

2.8

8.85 l0-'o

x

0.3 10-a 1

( _

- = - :

-

d

s

0.36 l0-o

2'

x

10- '3

RC=2 x 10 tx 2 .1x 0 - '3 0 .42ns

o 4 )

(c)

We can

decreasehe

RC delay

by 55%.Ratio

,r;

:0.45.

18.

(a)

R

=

eL=2.67x

l0*

=

3.2 l0 'C)

A

0 . 2 8 x l O o x 0 . 3 x l 0 o

C

= 4 = { L -

3 . 9 x 8 . 8 5 x

0 - ' a

0 . 3 x

0 {

x l x 3

= g . 7

x l 0 - , r F

d .S 0.36 l0o

RC 3.2

x

103

8.7

x

10-13

2.8ns.

T 1

( b )

R =

O : = l . 7 x 1 0 - 6

4 = 2 x

l 0 r C )

A

0 . 2 8 x l 0 o x 0 . 3 x l O o

a 4 { L 2 . 8 x 8 . 8 5 x 10 - ' o0 . 3 x l 0 {

x l x 3

d

s

0.36 10a

= 6'3x l0-t3F

R C

= 2 x 1 0 3

8 . 7 x 1 0 - t 3

2 . 5 n s

RC

=3.2

x

103

8.7 l0- '3

=

2.5

s.

19.

(a)

The

aluminum unner

can be

considered s

wo segments onnected

n series:

20%o

or

0.4 mm)

of the length

s half thickness

0.5 pm)

and

the

remaining

1.6mm is

full thickness

lpm).

The otal

resistances

*=JL . ! "1=g* ro -u f

' 16 , * ,

o ' oo

- l

' l A ,

A r )

[ 1 0 - * x l 0 - *

0 - o x ( 0 . 5 x 1 0 - ' ) l

: 7 2 U .



20.

The

limiting current

1 is

given

by the

maximum

allowed current

density

times

cross-sectional

reaofthe

thinner

conductor

ections:

. / : 5x10s cr* " (10-ax0.5x10-a; :5x103A:2.5 mA.

The voltage

drop across

he whole

conductor

s then

V: N

=72Qx2.5x

0- '

A :0 .18V.

40nm

60nm

h:height,

W;width, r:

thickness,

ssumethattheesistivities

f the cladding

layer

and TiN aremuch

arger han

p

n,

and

pru

R , , = D , , *

(

= 2 . 7

I

h

xW

(0.5

0.1) 0.5

R.. .= p, . . -J- =1.7

(

hxtr (0.5

2t)

x(0.5

2t)

When Rn,

=

Rru

2 . 7

1 . 7

l n e n

= -

0.4

x

0.5

(0.5

2t)'

=

t :0 .073

pm:

73nm

0.5

pm



CHAPTER

12

.

With referenceo Fig.2

for class100

clean oom

we have

a totalof 3500 particlesinfwith

particle

izes

>0.5

pm

2L

t

3596: 735

particles/r#

ith

particle

izes

1.0

pm

100

**

35OO:157

particles/#

with

particle

izes

2.0

1tm

r00

Therefore,

a)

3500-73

:

2765

particles/nf

etween .5

and

pm

(b\

735-157

578

particles/m3

efween

and2

pm

(c)

157

particles/#

above

pm.

.

y

=

fr,r-o,n

A

:50

mm2:0.5

cm2

l r

_

e4(0. rx0.s)

"4(o2sxo.s)

" - t { txo.s ;

=g- r .z

=30. lyo

.

The available

exposure nergy n

an hour s

0.3mW2/cmz

3600s:1080 mJlcr*

Forpositive esist, he throughputs

1080

_

=

7 wafers/ty

140

For negative esist,

he throughput

s

loSo

-

120

wafervtrr.

9

(a)

The resolution

f a

projection

ystems

given

by

L

9.6,1

:193fgt

=

o.r7g

pm

^

=Ktfr f .=u 'u

0.65

DoF k,-L=

o.sl

,le,3f

l:

o.r28

*

(NA)'

L

(0.65)'

I

(b)

We can increase

NA to improve

the resolution.

We

can

adopt resolution

enhancement

techniques

RET)

such as optical

proximity

correction OPC)

and

phase-shifting

Masks

(PSM).

We can also develop

new resists

hat

provide

ower

h

and

higher k2

for better

resolution

and depth

of focus.

(c)

PSM technique hanges

r to improve

esolution.

(a)

Using

resists

with high

y

value

can result in

a more vertical

profile

but

tkoughput

decreases.

(b)

Conventional

esists

can not be used

n

deep

UV

lithography

process

ecausehese esists

have

high

absorption

and require

high dose o

be exposed n

deep

[fV. This

raises he



.

(a)

.

(a)

( b )

(c)

(b)

8 .

9.

concern

of damage o

stepper

ens, ower

exposure

peed

nd reduced

hroughput.

A shapedbeam system

enables

he

size and

shape

of the beam

to

be varied,

thereby

minimizing the number

of flashes equired

for exposing

a

given

area to

be

patterned.

Therefore,a shaped

beamcan

save

ime

and increase

hroughput

compared

o a

Gaussian

beam.

We

can

makealignment

marks

on wafers

using

e-beam

nd etch he

exposed

marks.

We can

then use hem o

do alignment

with

e-beam

adiation

and

obtain he

signal

rom these

marks

for wafer alignment.

X-ray lithography

s a

proximity

printing

ithography.

ts accuracy

equirement

s

very

high,

therefore

alignment s difficult.

X-ray lithography

using synchrotron

adiation

has a high

exposure

lux

so X-ray

has better

throughput han

e-beam.

To avoid the mask

damage

problem

associated

with

shadow

printing,

projection

printing

exposure ools have

been developed

o

project

an image

from the

mask.

With

a 1:l

projectionprinting

system

s much

more

difficult

to

produce

defect-free

masks

han

it is

with a 5:l reduction

step-and-repeat

ystem.

It is not

possible.

The main

reason

s that

X-rays

cannot be

focused

by an

optical lens.

When it is

through the

reticle.

So we

can not

build a

step-and-scan

-ray

lithography

system.

As

shown n the figure,

the

profile

for

eachcase

s a

segment f

a circle with

origin at

the

initial mask-film

edge.

As overetchingproceeds

he radius

of curvature

ncreases

o that

the

profile

tends

o a vertical

ine.

(a)

20 sec

0.6

x

20/60:0.2

pm....(100)

lane

0.6/16

20/60

0.0125

m........( l

0)

plane

0.6/100 20/60:

0.002

m.......(t

1)

plane

wo

=

wo .l-zt

1.5

"li x0.z

1.22

m

40 sec

b)

0.6

x

40/60:0.4pm....(100)plane



0.6/16 40/60 0.025 m.... l

t0)

ptane

0.6/100

40/60: 0.004

m.....( l

l )

plane

wu

=wo

Jzt

=

1.5-

i

xo.a:0.93

m

60sec

0.6

x l

:0.6

pm.. . . (100)plane

0.6/16

l:

0.0375

m....

l

l0)

plane

0.6/100 : 0.006

m.....(l

l)

plane

W o : W o - J z t

=

1 . 5 -

2 x 0 . 6 : 0 . 6 5

m .

data n Prob. , theetchedattem rofiles n<100>-Si reshownn below.

0

sec

I

:0112

ltm,

Wo

=

Wu

=

1.5

m

10

ec

l:0.025

ltm,

Wo

l[t

=

1.5

pm

sec

l:0.0375

1tm

Wo

Wu

=

1.5

pm.

u'e protect

he

IC chip areas

e.g.

with

SfNa layer)and

etch he wafer from

the top,

the

idth

of the bottomsurfaces

=tTt

+

Jzt

=

1000

JT*625

=

1884

m

of surface rea

hat s lost s

-t4/:)/w2

x

100%:(1g842-10002)t8842x100%:71.8%

of the wafer area.we have

ost

71.8%

x

tdl5

2)'

:127

cr*

method s to define

maskingareas

n the

backside nd etch from

the back.

The width

square

mask

centeredwith respect

f IC chip

is

given

by

v [

=Wr-^ l -z t=1000-

J ix625 :116

pm

sing his method, he fraction

of the op

surfacearea

hat s lost

can be negligibly

small.



PV: NRT

7.52 760, 10-3

n/V

x0.082

273

n/V: 4.42

x

l0-7

mole/liter:4.42

x

l0-7

x

6.02

1023/1000

2-7

,l}ta

cm3

mean-free-path

2,=

5xI0-3 P cm:5x

l0-3

1000/

.52:0.6649

m:6649

pm

l50Pa:

ll28 m Torr

PV

nRT



ll28l

760

x

10-' n/V

x

0.082

x

273

n/V

:

6.63

x

10-s

mole/liter

6.63

x

Q-sxg.Q/x

023/1000

-4

x

1016rn3

mean-free-path

2": 5

xl}- '

/Pcm

:

5* 10-3

1000/1128

0.0044

m 44

um.

Si EtchRate

nm/min)

2.86

x

10-13

n,

x7/'

,r-tiA'

:2 .86x

l0 - r3

lx lSrsx (2

t1 / ,

r#

:224.7

nm/min.

SiOzEtchRate

nm/min):

Q.Sl{x

10-13

3"

Iytsx(2Oq% "#:

5.6nm/min

Etch

selectivity f SiOz

overSi

:

#=

0.025

or etch ate

Sio2)/etch

ate

Si)

:

':u-)!

""t-t'u*'ot/n'7x2e8

0.025

2.86

A three-step

process

s

required or

polysilicon

gate

etching.

Step 1 is

a nonselective

etch

process

hat is used o

remove any

native

oxide on

the

polysilicon

surface.

Step 2

is a high

polysilicon

etchrate

process

which etches

olysilicon

with

an anisotropic

tch

profile.

Step3

is

a

highly selective

olysilicon

o

oxide

process

which

usuallyhas

a

low

polysilicon

etch

rate.

If the etch rate

can be conholled

o within

l0 %o,he

polysilicon

may be

etched 10

o/o

longer or

for an equivalenthickness f 40 nm. The selectivity s therefore

40

nm/l nm:40.

Assuming

30Yooveretching,

nd that

he selectivity

f Al over the

photoresist

maintains

3.

The minimum

photoresist

hickness

equired

s



8 .

(l+

30%)

x

I

pm/3

0.433

pm:433.3

nm.

qB

( I ) " =-

me

2nx2.45 10e

[ . 6 x 1 0 - ' e x B

9 . l x

1 0 - 3 1

B: 8.75

x

10-2(tesla)

:875

(gauss).

Traditional

RIE

generates

ow-densityplasma

10e

crn3)

with high ion

energy.

ECR

and

ICp

generate

igh-density

plasma

10rr

to 1012

rn3)

with low

ion energy.

Advantages

f ECR

and

ICP are low etch damage,ow microloading, ow aspect-ratio ependent

tching effect,

and

simple chemistry.

However,

ECR and

ICP

systems

re more complicated

han traditional

RIE

systems.

The

corrosion eaction equires

he

presence

f moisture

o

proceed.

Therefore,

he

first

line of

defense

n controlling

corrosion

s controlling

humidity. Low

humidity

is essentiaf.

especially

if coppercontaining

alloys

are being

etched.

Second s

to removeas

much chlorine

as

possible

from the wafersbefore he wafersareexposedo air. Finally,gases uchasCF+andSF6can

be

used or fluorine/chlorine

xchange

eactions

nd

polymeric

encapsulation.

Thus,

Al-Cl

bonds

are replacedby Al-F

bonds.

Whereas

Al-Cl

bonds will

react with

ambient moisture

and start

the corrosion

process

Al-F

bonds

are very

stableand

do not react.

Furthermore,

luorine

will

not catzlyze ny

corrosion

eactions.

J.



l. E"(boron)3.46

eV,

D

:0.76

cn?/sec

From

Eq.

6,

CHAPTER

13

4.142xl0-"

x

1800

2.73x10a

m

-3 .46

8.614x

}-s

x1323

-

F

o . le r *o(

-3 '46

D = D o e x ? ( # ) =

.

t g . 6 t 4 x l o _ 5

1223

)=o' t+zx

1o-"cm' ls

L=JDt

=

From

Eq. 9,

c

(x)

=

c,ertufi)=

1.8 I

02oerfc

(17{6;)

I f

x

=0 ,C(0 )=

1.8x1020

ioms

cm3;x :0 .05

t104,C(5*

105) :3 .6

x

l0 le

atoms/cm3;

:

0.07

5

r

I 0-4,

C(7.5

x

1

0-6

: g.4

x

1

gls

atoms/c

rf

,

:0.

I

"

I 0-a,

C(10-s):

1.8

x

l0r8

atoms/cm3;

x :0 .15x

104,

C1 l .5x l0 -5 ) :

.gx

l016

toms/c#.

The

x.,

=ZJDI

(efc

gs!--

)

=o.l5Arn

Total

amount

f dopant

ntroduced

Q(0

1

:#C,L

=

5.54x

Otaatoms/c#.

4n

2 ' D=Do*(#)=oz6exn(

)=

o' 'ux1o- 'ocm'/s

From

Eq. 15,

Cs

=

C(Q,t)=

+

=2.342x

10'natomVcm

Jtilt

C(x\

=C-"rf.[a)

=2.342*to'nerrc[

"

)

\2L ) \2 .673xr0 - '

Ifx

:0,

C(0): 2.342

10le

toms/crrf

*:0.lxl0-a,

C(l0r)

:

l.4lxl0te

atoms/cm3;

x

:

0.2x104,

C12xl0't)

6.7911018

toms/c

f

;

*

:0.3

x

6-a,

C(3 o-5;

2.65

10t8



3 .

1 x 1 0 "

= t * t O " " * p [

t : 1573 :26 min

atoms/cm';

x:0.4x104,c14x10-s; :

.37x10t7

toms/cm3;x

0.5x10-4,

c15xlO-s):

.g7x1gtz

atoms/cm3;

x

:

0.6x104, (6x10-t)

3.51* 016

toms/crrt;

:0.7x10-a,

c(7x 0-5)

7.03x

0r5

atoms/cm3;

.r

0.8x104,

C(8x10-s)

5.62x101a

toms/cm3.

The

x,

=

4Dtht--

=

0.72qm.

C

uJ

tDr

l0- '

)

4x2 .3x10 ' ' t

)

For he constant-total-dopantiffusioncase,Eq. 15gives C, = -L

*Dt

s

=

1

x

lottm

=

3.4xlo*atomVcm

.

4. The

process

s

called he ramping

of a diffusion

umace.For

the ramp-down

situation,

he

furnace emperature

T is

given

by

T : T o - r t

where To s the

initial temperature

nd r is the linear

amp rate.

The effective

Dt

product

duringa ramp-down

ime of tr is

given

by

(Dt)"r

=

l^'

ogyat

In

a typical diffirsion

process,

amping

s carriedout until the

diffirsivity is

negligibly small.Thus

he upper imit t1 con

be akenas nfinity:

l l l r t

- = - *

( l + - + . . . )

T

T

- r t

T ' T

0 0 0



and

( - n

/ \

- l - r

r t

. l

- E - , .

- r E t

- r E t

D=Do

xp[

hr l=Do

"*pl

1+A*;*. . . )

l :Oo("*p

"_

Xexp-#.. . )=

D(T;"ry#

\

/ r u r /

L k I o

I o

J

k l o

k f o -

\

u / T

k T ^ ,

where

D(To) s

the

diffusion

coefficient

at

T6.

Substituting

he

above

equation

nto

the

expression

or

the effective

Dr

product

gives

(Dt)"n

[i

otr,\"*p-]#dt

D(r)+

k T o '

' " ' ' E o

Thus he ramp-downprocessesults n an effbctiveadditional ime equal o Kls2/rgu t

the

initial

diffi.rsion

emperature

6.

For

phosphorus

ifflrsion

n

silicon

at 1000"C,

we

have rom

Fig.

4:

|{7d

:

D

(1273

K)

:

Zx

10-ra

m2ls

1273

773

r = - = 0 . 4 1 7 K / s

20x60

E" :3 .66

eY

Therefore,

he effective

diffusion

ime

for the

ramp-downprocess

s

oro'

_.

l.3g

xlo-r'(1273)'

=

9ls

=

1.5min

,Eo

0.417(3.66x

.6 10- 'n)

5. For low-concentration

rive-in

diffusion,

he diffusion

s

given

by

Gaussian

istribution.

The

surface oncentration

s

then

, s

, s

( r " \

C(0.t)

+=---exDl -i

i

4d)t

Ji l ;

'\2kr)

{

=

_L.,*l,&Y-r"'

)

=

_0.5

g

dt

.|1i l ,

' \2ftr l(

2

)

t



dC

dt

= - 0 . 5 x -

C t

which means

Yo

changen

diffusion

ime

will induce

0.5Yo hange

n

surface

concentration.

d c , s

( E - \ ( - E - )

E _

- - - - : ' s . v r - [ -

dT

Jrilo,

'

\zkr )\2kT'

)

zkT'

dc

-

E

dT

-

3.6 1.6

lo- ' '

dT

dT

nr

, . . . . . . . . . . . . . ._ -

a

1L i . .

C

2 k T

T

2 x l . 3 8 x l 0 - " x 1 2 7 3 T

T

which meanso/ochangen diffirsion emperature ill cause16.9% hange

n

surface

oncentration.

6. At

I

100oC,

i

6x 1018 rn3.

Therefore,

he

doping

profile

or a

surface

oncentration

of 4

x

1018

rn3 s

given

by the

"intrinsic"

diffi.rsion

rocess:

c(x.t)= "..r"[4]

\2''lDt )

where

C,

:

{x

1018 rn3,

: 3

hr: 10800

,and

D

:

5x10-1a

m2ls.

he

diffusion length

s then

J

Dt

=

2.32x10-5

m

=

0.232

m

The

distribution

f arsenic

s

C(x)

=

4

x

1018

rf"[---:-)

[4.64

xl0-'

)

The

unction

depth

can be

obtained

as ollows

10"=4x l0" " . f r f

t '

- l

(

4.64

10- '

J

xi: I.2x 104 m 1.2pm.

7. At 900oC,

ni:2x

1018 rn3.

For a

surface

oncentration

f 4x10tt

"-',

given

by



the

"extrinsic"

diffusion

process

8 .

x

t

=1.6J

Dt

=

1.6m

=

3.23x

0*ucm

32.3 m

lntrinsic diffusion

is for dopant

concentration

ower than

the intrinsic

carrier

concentration

i at the diffusion

temperature.

xtrinsic

diffi.rsion

s for dopant

concentration

igher han n;.

9. For impurity in the oxidationprocess f silicon,

segregatio coefficein t

=

3 x 1 0 " 3

f=--- . . -=- : -=0.006.

5

x

10' ' 500

[0.5: [1.1

+

qFp/Ci

-E,

-4 .05x1.6x10- le

D

=

D oe

k r

x

"

=45 . 8e t

38 " 10

r * r l ? l

ni

r ' -€ " -

" *

-Z

, ,

-3 '45x lor

xo.6

=

1.3 lo ,cm- '

"

I .6

x

l0 - ' "

10-t

;

t

=

1 .3 10 ' t

x

l . 6 x l 0 - t n

i4r0.16)'

the mplant ime r

:6.7

s.

12. The ion dose

per

unit area s

" l#

=3.77x10- 'ucm2ls

10.

1 1 .

3 . 9 x 8 . 8 5 x 1 0 - r a

=3 .45x

10 -7

10-u

f t

1 0 x 1 0 {

x 5 x 6 0

N

=

q

=

l .6x lQ- te

=2.3gx1012

ons/cm2

A

A , I 0 . ,

n"\T)-

From

Eq. 25 andExample

3, the

peak

on concentration

indicateshe

oo is 20 nm.

is

at x

:

Ro.Figure.

7



Therefore, he ion

concentration

s

^9 2.38

10' '

=

-___-___-___-_

:.

=

4.74

x

l0'tCm

-3

.

oo42n 20x10- '42 t r

13. FromFig. 17,the

&:230

nm,and

oo =

62 nm.

The

peak

concentration

s

S

2 x 1 0 ' 5

:

----------------

=

l.29x

l0'ocm-l

oo{2 t t 62x10- '42n

FromEq. 25,

t.2ex,o"

*[3:i{l

|

260-

J

x; :0.53

pm.

14. Dose

er

unit

area

9

=CoLVr

-

3'9

x8'85

10*ta

l

q

q

250 10-s

l-6

x

10- 'n

=

8'6

10"cm-2

From Fig.

17 andExample

3, the

peak

concentration

ccurs

at 140nm from

the

surface. lso,

t is at

(140-25):

115nm

from he

Si-SiOz nterface.

15. The total implanted

dose s

integrated

rom

Eq. 25

e,= f

s-"*[-t*I{ l*=l{r*fr-

errcl

e-,- l}=lo

-erfc(z.

s

ur

-

lo

@-

|

2oo.

J

z

I

L

o,.lz

l)

z-

3)l=

-xt'ee8e

The

otaldose n

silicon s as

ollows

d:25

nm):

Q,,=I: ; fu"*|#]*=i{ ' - [ '_", f"(* i , ] }=; ' -er|c(| .87)]={

the ratio

of dose n the

silicon:

Qs/Qr:99.6%.



16. The

projected

ange

s 150nm

(see

Fig.

17).

The average

uclearenergy

oss

over he range

s 60 eV/nm

(Fig.

16).

60x 0.25: 15

eV

(energy

oss

of boron on

per

each attice

plane)

the

damageolume:

Vo: n

(2.5

nm;21tSO

m):3" 10-18

m3

total damageayer

150/0.25:

600

displaced tom or

one ayer:

l5l15

:

I

damage

ensitY

600/Vo 2"

l02ocm3

zxlo2o

5.02x102

0.4%.

17. The higher he temperature,

he faster

defectsanneal

out. Also, the

solubility

of

electricallyactive

dopantatoms

ncreases

ith temperature.

o

1 8 . L V , = l V : = '

Cor

where

p1

is the

additional

charge

added

ust

below

the oxide-semiconductor

surfaceby ion

implantation.

Cox

is

a

parallel-plate

apacitance

er

unit

area

given

by C""

=1

a

(d

is the

oxide hickness,

€r

s the

permittivity

of the semiconductor)

e,

=

LV,c".

1v

x

3'9

x

8'85 10-'4F/cm

:

g.63x

0-'

c;

0.4

x

l0-6cm cm'

8'63 1o-'

:

5.4

x1ol2

ons/c#

1.6 10-te

Total mplant

ose:

t'0,\!9"

:1.2

x

1013

ons/c#.

45%

19. The discussionhouldmentionmuchof Section13.6.Diffi.rsionrom a surfaceilm

avoids

problems

of channeling.

Tilted

beams cannot be

used because

of

shadowing

problems.

If

low energy

implantation

is used,

perhaps

with

preamorphization

y silicon,

then to

keep the

junctions

shallow,

RTA is

also



necessary.

20. FromEq.35

&

=

1

"rp"[o

-g)=

0.84

^ s 2

[ 0 . 2 J 2 )

The

effectiveness f the

photoresist

mask s

only 16%o.

s,

=

l.rr"ig)

=0.023

.s

2

\0.2J2

The effectiveness

f the

photoresist

mask s 97.7%.

2 1 . r = ] - { u

- l o - ,

24n

u

. ' . u

3 . 0 2

d:

&+

4.27

p: 0.53 4.27 0.093

0.927

m.



CIIAPTER 14

l. EachU-shape

ection

refer

o the figure) hasan area

of

2500

pm

x

8

pfft:2

x

10a

m2.

herefore,

here

re

250(|)2/2*104:312.5

-shapedection. ach

section

contains2long

lines

with

1248

squares ach,4 corner

squares, bottom

square, nd

2 halfsquaresat the top. Therefore he resistanceor each

section

s

l kO/i l

(1248x2

4x0.65+2):2599.6

kO

The maximum esistances then

3r2.5x2500.6:7.8t 108 ) : 781MO

2.5

mm

2.

The area equiredon the chip is

,

c o

( 3 0 x l o - ' X 5 x l 0 - , , )

Eo, 3.9

x

8.85 l0-tu

= 4.35x10-5 m2

4Pm

(PITCH)

:4.35

"

103

mz

66

x

66

pm



Refer o Fig.4a

and usingnegative hotoresist

f all levels

(a)

Ion implantation

mask

for p+

mplantation

and

gate

oxide)

(b)Contact

windows 2x10pm)

(c)

Metallization

mask

(using

Al to

form

ohmic contact n the

contact

window

and form the

MOS capacitor).

Because

f the registration

errors,

an additional2 pm

is incorporated

n

all critical

dimensions.



( o )

56

pm

I

I

I

{ l + 2 F m

Zli*

( b )

1

73

If the spacebetween ines s

2

p*,

then here s 4

pm

for each urn

(i.e.,

2xn,

for

one um).Assume herearen turns, rom 8q.6, [, x 1tan2r 1.2x 10-6r?r,where

can be replaced y 2

x

n. Then,we canobtain hat n is 13.



4.

(a)

Metal 1,

(b)

contacthole,

c)

Metal2.

(a)

Metal l,

(b)

contacthole,

I

(c)

Metal2.

I



5.

The circuit

diagramand device

cross-section

f a clamped

ransistor

are shown n

(a)

and

(b),

respectively.

COLLECTOR

E M I T T E R

Si02

(a)

The undoped

olysilicon

s

used or isolation.

(b)

The

polysilicon

I is used

as a solid-phase

iffusion

source o

form the

extrinsicbase egion

and he

baseelectrode.

(c)

The

polysilicon

2

is used

as a solid-phase

iffusion

source o form the emiffer

region

and he emitterelectrode.

(a)

For 30 keV

boron,

&

:

100

nm

and A,Ro:

34 nm. Assuming

hat

.Ro

ndA,Ro

for

boronare he same

n Si and

SiOz he

peak

concentration

s

given

by

( o )

( b )

6.

7.

P-SUBSTRATE

oHilrc

CONTACT

J-zntn,

J:19f-

=9.4xro,u

m-,

t l2nQa

x

l0-')



The amount

of boron

ons

n the

silicon s

f;=rffiq"*lW

s [

(

n " -d . t l

==12

_

erf ,c

_+_

l l

2L

[^/2m,.,1-1

8 x l o " [ ^

^ (

7 5 0

) l

= - t

z - E l t v t

- = : -

I I

2

L

\Jz *340 ) l

=

7.88 10"cm-'

Assume

hat the

implanted

oron

ons orm a negative heet hargenear he Si-

SiO2 nterface,

hen

LVr

=

re)/

Co,

-J:6

1

10-''

x

(7'88

10"

' l q )

3 .e *8 .8m=o 'e l v

(b)

For

80

keV

arsenic

mplantation

&:49

nm and

A

&

:

1Snm.

The

peak

arsenic

concentration, J- - -

10'u

= 2.21x102, m-,

42nA,R"

r /ax( l8x

l0- ' )



Rp

=49oi

(As)

tu;\

\

\.o"r.*.

\

\

\

\

\

I

BORON

\

(LOWER

ScaLsl

1

/ar\

Re-toooi

B)

I

ol5

o d

tooo

-zm

-tso.r-zsoi

x(i)

-

FoR

cHAls{EL

ectfi

8.

(a)

Because 100)-oriented

ilicon

has ower (- one enth) nterface-trappedhargeand

a

lower

fixed

oxide

charge.

(b)

If

the field

oxide is

too

thin, it

may

not

provide

a

large

enough

hreshold

voltage

or

adequate

solation

befween

eighboring

MOSFETs.

(c)

The

typical

sheet

esistance

f heavily

dopedpolysilicon

gate

s 20

to

30

Q fl, which

is adequate

or

MOSFETs

with

gate

engths

arger

han

3

pm.

For

shorter

gates,

he

sheet esistance

f

polysilicon

s too

high

and

will

cause

arge

RC

delays.

We can

use

refractory

metals (e.g.,

Mo)

or

silicides

as

the

gate

material

o

reduce

he

sheet

esistance

o

about

O

/t1.



(d)

A

self-aligned

ate

can

be obtained

by

first defining

he

MOS

gate

structure,

hen

using the

gate

electrode

as

a mask

for the

source/drain

mplantation.

The

self-

aligned

gate

can minimize

parasitic

capacitance

aused

by the

source/drain

regions extending

underneath

the

gate

electrode

(due

to

diffusion

or

misalignment).

(e)

P-glasscan

be used or

insulation

between

onducting

ayers,

or

diffi.rsion

and

ion implantation

masks,

and for passivation

o

protect

devices

rom impurities,

moisfure,

and

scratches.

9. The lower

insulator

has a dielectric

constant

Alq:

4

anda thickness

dr

l0 nm

The

upper

nsulator

hasa dielectric

constant

A/q

:

10

and a thickness

dz:

100 nm.

Upon

application

of a

positive

voltage

Vc

to the

extemal

gate,

electric

field E

1

and E2

arc

established

nthe dt

and dz

respectively.

We have,

rom

Gauss' aw,

that

e1E1:

e2E2

+Q andVc: Erdt *Ezdz

where

Q

is the

stored

charge

on the floating

gate.

From

hese

above wo

equations,

we

obtain

" -

v o

-

Q

-l

d,

+

d,(q

ltr)

'

g

*

er(a, ar)

a

I

r - , ^ - ,

l o x l o ?

O

'

| | 4 \ | / r ^ t ,

Iro+roolrJ

. 'tff iJ],.r.rsx

o-'o

If

the stored harge

oesnot

reduceE1

by a

significant mount

i.e.,0.2>>

2.26x10s

lQ l,

*"

can write

Q

=

foadf

x

0.2 t

=

o.z

x

(o.zsx

O-u)=

x

t0-s

C

L V - : Q

= ,

5 x l o - 8

'

c2

( lo*8.s@=o'565

v

(a)



10.

(b)

when t

-)

a,J

-+}we

have

lgl-+

0.212.26x10s

8.84x10-7

.

8.84x

0-?

Then ,Vr=t=

,

( t0x8.85

1g- t4)716-5

=9.98

V.

(o)

p-TUB

(b)

POLYSILICOT{

GATE

(c)

n-TYPE

O|FFUSION

+

+

+

+

+ +

+

+



+

{d)

p-TypE

DtFFUstoN

+

t r n

t r n

(e)

CONTACT

INDows

+

(f)

METALLTZATIOTTI

11. The

oxidecapacitance

er

unit area s

given

by

€"rn

C

^-

=

' '7

=

3.5

l0

t

F/cm2

d

and he

maximum

current

supplied

y the

device s

r

,

=

!\

rc, ,(vo

v,) '

=: :+3.5xt0r

(vo

v,) '=

5mA

2

L -

2 0 . 5 t t n



and the maximum

allowablewire

resistance

s 0.1

V/5 mA.

or 20Q.

Then. the

length

of

the

wire must

be

- Rx Area 20Ox I0- tcm'

t - _

= 0.074 m

p

2.7

l }

'Q

-cm

or 740

pm.

This

is a long

distance

compared

o most device

spacing.

When

driving

signalsbetweenwidely

spaced

ogic

blocks however,

minimum

feature

sized ines

would not

be

appropriate.

12.



s i

3N4

x x

I

l,*f

v

-

EPITAXY

FIELDOXIDE

( b l

( o )

( d )

( e )

i

tap*

RESIST



1 3 .

14.

To solve he short-channel

ffect ofdevices.

The device

performance

ill be degradedrom the

boron

penetration.

There are

methods o reduce

his effect:

(1)

using rapid thermal annealing o reduce he

time at high temperatures,

onsequentlyeduceshe

diffusion of boron,

(2)

using

nitrided oxide to

suppresshe

boron

penetration,

inceboron can easily

combine

with nitrogen nd becomes

essmobile,

3)

making

a multi-layer

of

polysilicon

to

trap the boron atomsat the

interface

of

each ayer.

Total capacitancef the stacked atestructure s :5 .

c

_q

z

l(z.e-)

d, d, l

\d,

d,

)

7 2 s / ( 7 2 5 \

=

- x - /

l - + - l : 2 . 1 2

0.s l0 l

\0 .5

to

16.

3 ' 9

= 2 . 1 2

d

. ' .

d

=

3'9

=1.84

m.

2 . 1 2

Disadvantagesf LOCOS:

1)

high temperature nd ong

oxidation ime cause

V1

shift,

(2)

bird's

beak,

3)

not a

planar

surface,

4)

exhibits oxide thinning

effect.

Advantages of

shallow trench isolation: (l)

planar

surface,

(2)

no

high

temperature

rocessing

nd ong

oxidation ime,

(3)

no

oxide

thinning

effect,

(4)

no bird'sbeak.

For isolationbetween

he

metaland he substrate.

GaAs acksof high-quality

nsulating ilm.

t7 .

1 8 .



19.

(a)

:

2000

(el.o:x

l0-'o

)=

1.38 10-'s

=

1.38 s.

(b)

For a

polysilicon

unner

(

r . \ (

l )

RC

=

|

R,ouo,"#

ll

",-

|

\

w ) \ d )

(

t \ ,

=

391

-l-

l(ol.or><

0-'o)

=2.07

xl0-7

s

\ 1 0 - " , / '

= 2 0 7n s

Therefore he

polysilicon

unner'sRC time constant s 150 imes larger han

the

aluminum unner.

20. When

we combine he logic circuits

and

memory

on the chip, we needmultiple

supplyvolrages.For reliability issue,differentoxide hicknesses

re needed or

different supply

voltages.

21'

(a)

/r^^,

=

/cr^,o,

*

/r,u,,o"

132654164 solution manual for semiconductor devices physics and technology sze s m solution

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