132654164 Solution Manual for Semiconductor Devices Physics and Technology Sze S M Solution
Transcript of 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
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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
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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
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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 .
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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 .
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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 :
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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)' '^o'
p,dp
rdp
=
,r2o
ff,,
"to'-'^Et)l2^tr
p,dp,!-_
"-oll'^o'dp,
f*
"^o1/'^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 - '
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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
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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:
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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-')
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[[ -
!!-
)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_!{
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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
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+.
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,
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[_ 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.
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The
nterceptions ive
the built-in
potentialof
hep-n
junctions.
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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
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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
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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.
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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
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(
._-[*-__-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!
!^
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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.
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=
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
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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
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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-
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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'
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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
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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
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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)
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(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
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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
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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
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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,
))
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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
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/ ^ \ ,
,.^
-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
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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
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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.
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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
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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
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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
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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.
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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
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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
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1 .
CHAPTER
6
\
...-.................-.-
-
METAL
OXIDE
n-TYPE
SEMICONDUCTOR
Ec
Ep
Ei
Ev
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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
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4.
(a)
(b)
E
(x)
v6)
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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 .
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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
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-
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) '
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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
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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.
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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
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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.
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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
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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
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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.
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(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
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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
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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.
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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
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, ,
_ 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
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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
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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
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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
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#=(+-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
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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
' - "
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: 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 .
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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.
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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
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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^
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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.
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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
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;.
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
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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
.
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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
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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.
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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
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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
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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
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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 .
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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
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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
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.
(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
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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.
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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
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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
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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.
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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
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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
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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
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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
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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
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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%.
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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
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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.
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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
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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.
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( 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.
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4.
(a)
Metal 1,
(b)
contacthole,
c)
Metal2.
(a)
Metal l,
(b)
contacthole,
I
(c)
Metal2.
I
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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-')
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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- ' )
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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.
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(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)
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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
+
+
+
+
+ +
+
+
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+
{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
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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.
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s i
3N4
x x
I
l,*f
v
-
EPITAXY
FIELDOXIDE
( b l
( o )
( d )
( e )
i
tap*
RESIST
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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 .
7/21/2019 132654164 Solution Manual for Semiconductor Devices Physics and Technology Sze S M Solution
http://slidepdf.com/reader/full/132654164-solution-manual-for-semiconductor-devices-physics-and-technology 130/130
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"