1

Solutions Manual to Accompany

SEMICONDUCTOR DEVICESPhysics and Technology

2nd Edition

S. M. SZEUMC Chair Professor

National Chiao Tung UniversityNational Nano Device Laboratories

Hsinchu, Taiwan

John Wiley and Sons, IncNew York. Chicester / Weinheim / Brisband / Singapore / Toronto

0

ContentsCh.1 Introduction--------------------------------------------------------------------- 0Ch.2 Energy Bands and Carrier Concentration -------------------------------------- 1Ch.3 Carrier Transport Phenomena -------------------------------------------------- 7Ch.4 p-n Junction -------------------------------------------------------------------- 16Ch.5 Bipolar Transistor and Related Devices---------------------------------------- 32Ch.6 MOSFET and Related Devices------------------------------------------------- 48Ch.7 MESFET and Related Devices ------------------------------------------------- 60Ch.8 Microwave Diode, Quantum-Effect and Hot-Electron Devices --------------- 68Ch.9 Photonic Devices ------------------------------------------------------------- 73Ch.10 Crystal Growth and Epitaxy--------------------------------------------------- 83Ch.11 Film Formation---------------------------------------------------------------- 92Ch.12 Lithography and Etching ------------------------------------------------------ 99Ch.13 Impurity Doping--------------------------------------------------------------- 105Ch.14 Integrated Devices------------------------------------------------------------- 113

1

CHAPTER 2

1. (a) From Fig. 11a, the atom at the center of the cube is surround by four

equidistant nearest neighbors that lie at the corners of a tetrahedron. Therefore

the distance between nearest neighbors in silicon (a = 5.43 Å) is

1/2 [(a/2)2 + ( a2 /2)2]1/2 = a3 /4 = 2.35 Å.

(b) For the (100) plane, there are two atoms (one central atom and 4 corner atoms

each contributing 1/4 of an atom for a total of two atoms as shown in Fig. 4a)

for an area of a2, therefore we have

2/ a2 = 2/ (5.43 × 10-8)2 = 6.78 × 1014 atoms / cm2

Similarly we have for (110) plane (Fig. 4a and Fig. 6)

(2 + 2 ×1/2 + 4 ×1/4) / a2 2 = 9.6 × 1015 atoms / cm2,

and for (111) plane (Fig. 4a and Fig. 6)

(3 × 1/2 + 3 × 1/6) / 1/2( a2 )( a

23 ) =

2

23

2

a

= 7.83 × 1014 atoms / cm2.

2. The heights at X, Y, and Z point are ,43

,41 and 4

3 .

3. (a) For the simple cubic, a unit cell contains 1/8 of a sphere at each of the eight

corners for a total of one sphere.

4 Maximum fraction of cell filled

= no. of sphere × volume of each sphere / unit cell volume

= 1 × 4ð(a/2)3 / a3 = 52 %

(b) For a face-centered cubic, a unit cell contains 1/8 of a sphere at each of the

eight corners for a total of one sphere. The fcc also contains half a sphere at

each of the six faces for a total of three spheres. The nearest neighbor distance

is 1/2(a 2 ). Therefore the radius of each sphere is 1/4 (a 2 ).

4 Maximum fraction of cell filled

= (1 + 3) 4ð[(a/2) / 4 ]3 / 3 / a3 = 74 %.

(c) For a diamond lattice, a unit cell contains 1/8 of a sphere at each of the eight

corners for a total of one sphere, 1/2 of a sphere at each of the six faces for a

total of three spheres, and 4 spheres inside the cell. The diagonal distance

2

between (1/2, 0, 0) and (1/4, 1/4, 1/4) shown in Fig. 9a is

D = 21

222

222

+

+

aaa = 3

4a

The radius of the sphere is D/2 = 38a

4 Maximum fraction of cell filled

= (1 + 3 + 4)3

383

4

aπ / a3 = ð 3 / 16 = 34 %.

This is a relatively low percentage compared to other lattice structures.

4. 1d = 2d = 3d = 4d = d

1d + 2d + 3d + 4d = 0

1d • ( 1d + 2d + 3d + 4d ) = 1d • 0 = 0

2

1d + 1d • 2d + 1d • 3d + 1d • 4d = 0

4d2+ d2 cosè12 + d2cosè13 + d2cosè14 = d2 +3 d2 cosè= 0

4 cosè = 31−

è= cos-1 (31−

) = 109.470 .

5. Taking the reciprocals of these intercepts we get 1/2, 1/3 and 1/4. The smallest

three integers having the same ratio are 6, 4, and 3. The plane is referred to as

(643) plane.

6. (a) The lattice constant for GaAs is 5.65 Å, and the atomic weights of Ga and As

are 69.72 and 74.92 g/mole, respectively. There are four gallium atoms and

four arsenic atoms per unit cell, therefore

4/a3 = 4/ (5.65 × 10-8)3 = 2.22 × 1022 Ga or As atoms/cm2,Density = (no. of atoms/cm3 × atomic weight) / Avogadro constant

= 2.22 × 1022(69.72 + 74.92) / 6.02 × 1023 = 5.33 g / cm3.

(b) If GaAs is doped with Sn and Sn atoms displace Ga atoms, donors are

formed, because Sn has four valence electrons while Ga has only three. The

resulting semiconductor is n-type.

7. (a) The melting temperature for Si is 1412 ºC, and for SiO 2 is 1600 ºC. Therefore,

SiO2 has higher melting temperature. It is more difficult to break the Si-O

bond than the Si-Si bond.

(b) The seed crystal is used to initiated the growth of the ingot with the correct

crystal orientation.

(c) The crystal orientation determines the semiconductor’s chemical and electrical

3

properties, such as the etch rate, trap density, breakage plane etc.

(d) The temperating of the crusible and the pull rate.

8. Eg (T) = 1.17 – 636)(

4.73x10 24

+

−

TT

for Si

∴ Eg ( 100 K) = 1.163 eV , and Eg (600 K) = 1.032 eV

Eg(T) = 1.519 –204)(

5.405x10 24

+

−

TT

for GaAs

∴Eg( 100 K) = 1.501 eV, and Eg (600 K) = 1.277 eV .

9. The density of holes in the valence band is given by integrating the product

N(E)[1-F(E)]dE from top of the valence band ( VE taken to be E = 0) to the

bottom of the valence band Ebottom:

p = ∫bottomE

0 N(E)[1 – F(E)]dE (1)

where 1 –F(E) = ( )[ ] /kT1 e/1 1 FEE −

+− = [ ] 1/)(e1−−+ kTEE F

If EF – E >> kT then1 – F(E) ~ exp ( )[ ]kTEEF −− (2)

Then from Appendix H and , Eqs. 1 and 2 we obtain

p = 4ð[2mp / h2]3/2 ∫bottomE

0 E1/2 exp [-(EF – E) / kT ]dE (3)

Let x a E / kT , and let Ebottom = ∞− , Eq. 3 becomes

p = 4ð(2mp / h2)3/2 (kT)3/2 exp [-(EF / kT)] ∫∞−

0 x1/2exdx

where the integral on the right is of the standard form and equals π / 2.4 p = 2[2ðmp kT / h2]3/2 exp [-(EF / kT)]

By referring to the top of the valence band as EV instead of E = 0 we have,

p = 2(2ðmp kT / h2)3/2 exp [-(EF – EV) / kT ]or p = NV exp [-(EF –EV) / kT ]

where NV = 2 (2ðmp kT / h2)3 .

10. From Eq. 18NV = 2(2ðmp kT / h2)3/2

The effective mass of holes in Si is

mp = (NV / 2) 2/3 ( h2 / 2ðkT )

= 3

23619

2m101066.2

×× − ( )( )( )300103812

10625623

234

−

−

××

.

.π

= 9.4 × 10-31 kg = 1.03 m0.Similarly, we have for GaAs

mp = 3.9 × 10-31 kg = 0.43 m0.

11. Using Eq. 19

4

( ) ( )CVVC NNkTEEEi ln22)( ++=

= (EC+ EV)/ 2 + (3kT / 4) ln

3

2)6)(( np mm (1)

At 77 K Ei = (1.16/2) + (3 × 1.38 × 10-23T) / (4 × 1.6 × 10-19) ln(1.0/0.62) = 0.58 + 3.29 × 10-5 T = 0.58 + 2.54 × 10-3 = 0.583 eV.

At 300 K Ei = (1.12/2) + (3.29 × 10-5)(300) = 0.56 + 0.009 = 0.569 eV.

At 373 K Ei = (1.09/2) + (3.29 × 10-5)(373) = 0.545 + 0.012 = 0.557 eV.

Because the second term on the right-hand side of the Eq.1 is much smaller

compared to the first term, over the above temperature range, it is reasonable to

assume that Ei is in the center of the forbidden gap.

12. KE = ( ) ( )

)(/)(

/

d

de

CFtop

C

Ftop

C

EExkTEEC

E

E

kTEEC

E

E C

EeEE

EEEEE

−≡−−

−−

−

−−

∫∫

= kT

∫∫

∞ −

∞ −

0

21

0

23

de

de

xx

xx

x

x

= kT

Γ

Γ

2325

= kTπ

π

50

5051

.

.. ××

= kT23

.

13. (a) p = mv = 9.109 × 10-31 ×105 = 9.109 × 10-26 kg–m/s

λ = ph

= 26

34

101099

106266−

−

××

.

.= 7.27 × 10-9 m = 72.7 Å

(b) nλ = λpm

m0 = 06301

.× 72.7 = 1154 Å .

14. From Fig. 22 when ni = 1015 cm-3, the corresponding temperature is 1000 / T = 1.8.

So that T = 1000/1.8 = 555 K or 282 .

15. From Ec – EF = kT ln [NC / (ND – NA)]which can be rewritten as ND – NA = NC exp [–(EC – EF) / kT ]

Then ND – NA = 2.86 × 1019 exp(–0.20 / 0.0259) = 1.26 × 1016 cm-3

or ND= 1.26 × 1016 + NA = 2.26 × 1016 cm-3

A compensated semiconductor can be fabricated to provide a specific Fermi

energy level.

16. From Fig. 28a we can draw the following energy-band diagrams:

5

17. (a) The ionization energy for boron in Si is 0.045 eV. At 300 K, all boronimpurities are ionized. Thus pp = NA = 1015 cm-3

np = ni2 / nA = (9.65 × 109)2 / 1015 = 9.3 × 104 cm-3.

The Fermi level measured from the top of the valence band is given by:

EF – EV = kT ln(NV/ND) = 0.0259 ln (2.66 × 1019 / 1015) = 0.26 eV

(b) The boron atoms compensate the arsenic atoms; we have

pp = NA – ND = 3 × 1016 – 2.9 × 1016 = 1015 cm-3

Since pp is the same as given in (a), the values for np and EF are the same asin (a). However, the mobilities and resistivities for these two samples are

different.

18. Since ND >> ni, we can approximate n0 = ND andp0 = ni

2 / n0 = 9.3 ×1019 / 1017 = 9.3 × 102 cm-3

From n0 = ni exp

−

kT

EE iF ,

we haveEF – Ei = kT ln (n0 / ni) = 0.0259 ln (1017 / 9.65 × 109) = 0.42 eV

The resulting flat band diagram is :

6

19. Assuming complete ionization, the Fermi level measured from the intrinsic

Fermi level is 0.35 eV for 1015 cm-3, 0.45 eV for 1017 cm-3, and 0.54 eV for 1019

cm-3.

The number of electrons that are ionized is given by

n ≅ ND[1 – F(ED)] = ND / [1 + e ( ) TkEE FD /−− ]

Using the Fermi levels given above, we obtain the number of ionized donors as

n = 1015 cm-3 for ND = 1015 cm-3

n = 0.93 × 1017 cm-3 for ND = 1017 cm-3

n = 0.27 × 1019 cm-3 for ND = 1019 cm-3

Therefore, the assumption of complete ionization is valid only for the case of1015 cm-3.

20. ND+ = kTEE FDe /)(

16

110

−−+ =

1350

16

e1

10.−+

=

14511

1

1016

.+

= 5.33 × 1015 cm-3

The neutral donor = 1016 – 5.33 ×1015 cm-3 = 4.67 × 1015 cm-3

4 The ratio of +D

D

N

N O

= 335764

.

. = 0.876 .

7

CHAPTER 3

1. (a) For intrinsic Si, µn = 1450, µp = 505, and n = p = ni = 9.65×109

We have 51031.3)(

11×=

+=

+=

pnipn qnqpqn µµµµρ Ω-cm

(b) Similarly for GaAs, µn = 9200, µp = 320, and n = p = ni = 2.25×106

We have 81092.2)(

11 ×=+

=+

=pnipn qnqpqn µµµµ

ρ Ω-cm.

2. For lattice scattering, µn ∝ T-3/2

T = 200 K, µn = 1300×2/3

2/3

300200

−

−

= 2388 cm2/V-s

T = 400 K, µn = 1300×2/3

2/3

300400

−

−

= 844 cm2/V-s.

3. Since 21

111µµµ

+=

∴ 5001

25011 +=

µ µ = 167 cm2/V-s.

4. (a) p = 5×1015 cm-3, n = ni2/p = (9.65×109)2/5×1015 = 1.86×104 cm-3

µp = 410 cm2/V-s, µn = 1300 cm2/V-s

ρ = pqpqnq ppn µµµ

1

1≈

+ = 3 Ω-cm

(b) p = NA – ND = 2×1016 – 1.5×1016 = 5×1015 cm-3, n = 1.86×104 cm-3

µp = µp (NA + ND) = µp (3.5×1016) = 290 cm2/V-s,

µn = µn (NA + ND) = 1000 cm2/V-s

ρ = pqpqnq ppn µµµ

1

1≈

+ = 4.3 Ω-cm

8

(c) p = NA (Boron) – ND + NA (Gallium) = 5×1015 cm-3, n = 1.86×104 cm-3

µp = µp (NA + ND+ NA) = µp (2.05×1017) = 150 cm2/V-s,

µn = µn (NA + ND+ NA) = 520 cm2/V-s

ρ = 8.3 Ω-cm.

5. Assume ND − NA>> ni, the conductivity is given by

σ ≈ qnµn = qµn(ND − NA)

We have that

16 = (1.6×10-19)µn(ND − 1017)

Since mobility is a function of the ionized impurity concentration, we can use

Fig. 3 along with trial and error to determine µn and ND . For example, if we

choose ND = 2×1017, then NI = ND+ + NA

- = 3×1017, so that µn ≈ 510 cm2/V-s

which gives σ = 8.16.

Further trial and error yields

ND≈ 3.5×1017 cm-3

and

µn ≈ 400 cm2/V-s

which gives

σ ≈ 16 (Ω-cm)-1.

6. )/( )( 2 nnbnqpnq ippn +=+= µµµσ

From the condition dσ/dn = 0, we obtain

bnn i / =

Therefore

9

b

b

bnq

nbbbnqì

ip

iip

i

m

2

1

)1(1

)/(

1

+=

+

+=

µρρ

.

7. At the limit when d >> s, CF = 2ln

π = 4.53. Then from Eq. 16

226.053.410501011010 4

3

3

=×××××=××= −

−

−

CFWIV

ρ Ω-cm

From Fig. 6, CF = 4.2 (d/s = 10); using the a/d = 1 curve we obtain

78.102.41050

10226.0)/(

4

3

=××

×=⋅⋅=−

−

CFWIV ρ mV.

8. Hall coefficient,

7.42605.0)101030(105.2

106.11010493

33

=×××××

×××==−−

−−

WIBAV

Rz

HH cm3/C

Since the sign of RH is positive, the carriers are holes. From Eq. 22

16

191046.1

7.426106.111 ×=

××==

−HqR

p cm-3

Assuming NA ≈ p, from Fig. 7 we obtain ρ = 1.1 Ω-cm

The mobility µp is given by Eq. 15b

3801.11046.1106.1

111619

=××××

==−ρ

µqpp cm2/V-s.

9. Since R ∝ ρ and pn qpqn µµ

ρ+

=1

, hence pn pn

Rµµ +

∝1

From Einstein relation µ∝D

50// == pnpn DDµµ

10

pAnD

nD

NN

NR.

R

µµ

µ

+

=1

1

50 1

1

We have NA = 50 ND .

10. The electric potential φ is related to electron potential energy by the charge (− q)

φ = +q1 (EF − Ei)

The electric field for the one-dimensional situation is defined as

E(x) = −dxdφ =

dxdE

qi1

n = ni exp

−

kTEE iF = ND(x)

Hence

EF − Ei = kT ln

i

D

n)x(N

dx)x(dN

)x(NqkT

(x) D

D

1

−=E .

11. (a) From Eq. 31, Jn = 0 and

aq

kTeN

eaNq

kTndx

dnDx ax

ax

n

n +=−== −

−

)(

- - )(0

0

µE

(b) E (x) = 0.0259 (104) = 259 V/cm.

12. At thermal and electric equilibria,

0)(

)( =+=dx

xdnqDxnqJ nnn Eµ

xNNLNNND

LNN

LxNNND

dxxdn

xnD

x

L

L

n

n

L

Ln

n

n

n

)(

))((1)(

)(1

)(

00

0

0

00

−+−−=

−−+

−=−=

µ

µµE

11

000

0

0

n ln)(

D-

NND

xNNLNNN

V L

n

n

L

LL

µµ−=

−+−

= ∫ .

13. 11166 10101010 =××==∆=∆ −LpGpn τ cm-3

151115 101010 ≈+=∆+=∆+= nNnnn Dno cm-3

.cm 101010

)1065.9( 3-1111

15

292

≈+×=∆+= pN

np

D

i

14. (a) 8

15715 1010210105

11 −− =

××××=≈

tthpp Nνσ

τ s

48 103109 −− ×=×== ppp DL τ cm

201010210 10167 =×××== −stssthlr NS σν cm/s

(b) The hole concentration at the surface is given by Eq. 67

.cm 10

20101032010

11010102

)10(9.65

1)0(

3-9

84

8178

16

29

≈

×+×

×−×+××=

+

−+=

−−

−−

lrpp

lrpLpnon SL

SGpp

ττ

τ

15. pn qpqn µµσ +=

Before illumination

nonnon ppnn == ,

After illumination

,Gnnnn pnonon τ+=∆+=

Gpppp pnonon τ+=∆+=

. )(

)()]()([

Gq

pqnqppqnnq

ppn

nopnonnopnon

τµµ

µµµµσ

+=

+−∆++∆+=∆

12

16. (a) diff , dxdp

qDJ pp −=

= − 1.6×10-19×12×41012

1−×

×1015exp(-x/12)

= 1.6exp(-x/12) A/cm2

(b) diff ,drift , ptotaln JJJ −=

= 4.8 − 1.6exp(-x/12) A/cm2

(c) Enn qnJ µ=drift , Q

∴∴ 4.8 − 1.6exp(-x/12) = 1.6×10-19×1016×1000×E

E = 3 − exp(-x/12) V/cm.

17. For E = 0 we have

02

2

=∂

∂+−−=∂∂

xp

Dpp

tp n

pp

non

τ

at steady state, the boundary conditions are pn (x = 0) = pn (0) and pn (x = W) =

pno.

Therefore

[ ]

−

−+=

p

p

nonnon

LW

LxW

pppxp

sinh

sinh

)0()(

[ ]

−=

∂∂

−=== pp

pnon

x

npp L

WL

Dppq

xp

qDxJ coth)0()0(0

[ ]

−=

∂∂

−===

p

p

pnon

Wx

npp

LWL

Dppq

xp

qDWxJ

sinh

1)0()( .

18. The portion of injection current that reaches the opposite surface by diffusion is

13

given by

)/cosh(1

)0(

)(0

pp

p

LWJ

WJ==α

26 105105050 −− ×=××=≡ ppp DL τ cm

98.0)105/10cosh(

1

220 =×

=∴−−

α

Therefore, 98% of the injected current can reach the opposite surface.

19. In steady state, the recombination rate at the surface and in the bulk is equal

surface ,

surface ,

bulk ,

bulk ,

p

n

p

n pp

ττ

∆=

∆

so that the excess minority carrier concentration at the surface

∆pn, surface = 1014⋅6

7

10

10−

−

=1013 cm-3

The generation rate can be determined from the steady-state conditions in thebulk

G = 6

14

10

10−

= 1020 cm-3s-1

From Eq. 62, we can write

02

2

=∆

−+∂

∆∂

pp

pG

xp

Dτ

The boundary conditions are ∆p(x = ∞ ) = 1014 cm-3 and ∆p(x = 0) = 1013 cm-3

Hence ∆p(x) = 1014( pLxe /9.01 −− )

where Lp = 61010 −⋅ = 31.6 µm.20. The potential barrier height

χφφ −= mB = 4.2 − 4.0 = 0.2 volts.21. The number of electrons occupying the energy level between E and E+dE is

dn = N(E)F(E)dEwhere N(E) is the density-of-state function, and F(E) is Fermi-Dirac distributionfunction. Since only electrons with an energy greater than mF qE φ+ and havinga velocity component normal to the surface can escape the solid, the thermioniccurrent density is

dEeEvhm

qvJ kTEExqEx

F

mF

)(21

3

23

)2(4 −−∞

+∫ ∫==φ

π

where xv is the component of velocity normal to the surface of the metal. Sincethe energy-momentum relationship

)(21

2222

2

zyx pppmm

PE ++==

14

Differentiation leads to m

PdPdE =

By changing the momentum component to rectangular coordinates,

zyx dpdpdpdPP =24π

Hence

zmkTp

ymkTp

xxmkTmEp

p

zyxp

mkTmEpppx

dpedpedppemh

q

dpdpdpepmh

qJ

zyfx

x

x

fzyx

22

2)2(

3

p

p

2/)2(

3

222

0

0 y

222

2

2

−∞

∞−

−∞

∞−

−−∞

∞ ∞

−∞=

∞

−∞=

−++−

∫∫∫

∫ ∫ ∫=

=

where ).(220 mFx qEmp φ+=

Since 21

2

=−∞

∞−∫ adxe ax π

, the last two integrals yield (2ðmkT) 21 .

The first integral is evaluated by setting umkT

mEp Fx =−2

22

.

Therefore we have mkT

dppdu xx=

The lower limit of the first integral can be written as

kTq

mkTmEqEm mFmF φφ =−+

22)(2

so that the first integral becomes kTqu

ktqm

m

emkTduemkT φ

φ

−−∞=∫

/

Hence

−== −

kTq

TAeThqmk

J mkTq mφπ φ exp

4 2*2

3

2

.

22. Equation 79 is the tunneling probability

110234

1931

20 m1017.2

)10054.1()106.1)(220)(1011.9(2)(2 −

−

−−

×=×

×−×=−=h

EqVmnβ

[ ] 6

12100

1019.3)220(24

1031017.2sinh(201 −

−−

×=

−××××××+=T .

23. Equation 79 is the tunneling probability

[ ]

403.0)2.26(2.24

)101099.9sinh(61)10(

1210910 =

−×××××+=

−−

−T

( )[ ]

( )9

12999 108.7

2.262.24101099.9sinh6

1)10( −

−−

− ×=

−×××××+=T .

19234

1931

20 m1099.9

)10054.1()106.1)(2.26)(1011.9(2)(2 −

−

−−

×=×

×−×=−=h

EqVmnβ

15

24. From Fig. 22As E = 103 V/s

νd ≈ 1.3×106 cm/s (Si) and νd ≈ 8.7×106 cm/s (GaAs)

t ≈ 77 ps (Si) and t ≈ 11.5 ps (GaAs)

As E = 5×104 V/s

νd ≈ 107 cm/s (Si) and νd ≈ 8.2×106 cm/s (GaAs)

t ≈ 10 ps (Si) and t ≈ 12.2 ps (GaAs).

cm/s105.9m/s109.5

101.9300101.382

22 velocity Thermal 25.

64

31

23-

00

×=×=×

×××=

==

−

mkT

mE

v thth

For electric field of 100 v/cm, drift velocity

thnd vv <<×=×== cm/s1035.11001350 5Eµ

For electric field of 104 V/cm.

thn v≈×=×= cm/s1035.1101350 74Eµ .

The value is comparable to the thermal velocity, the linear relationship between

drift velocity and the electric field is not valid.

16

CHAPTER 4

1. The impurity profile is,

The overall space charge neutrality 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 in the n-side, thus we can obtain the depletion layer width in the n-side region:

1414

1032

1088.0 ××=××nW

Hence, the n-side depletion layer width is:

m0671 µ.Wn =

The total depletion layer width is 1.867 µm.

We use the Poisson’s equation for calculation of the electric field E(x).

In the n-side region,

( )

V/cm108640

)100671(103

1006710m0671

3

414

4

×−===

×−××=∴

××−=⇒==

+=⇒=

−

−

.)x(

.xq

x

.Nq

K).x(

KxNq

)x(Nq

dxd

nmax

sn

Ds

n

Ds

nDs

EE

E

E

EE

ε

εµ

εε

In the p-side region, the electrical field is:

x (µm)

(ND-NA) (cm-3)

3×1014

a =1019 cm-4

17

( )

( ) ( )V/cm108640

10802

10802

0m80

2

3

242

24

2

×−===

×−××=∴

×××−=⇒=−=

+×=⇒=

−

−

.)x(

.xaq

x

.aq

K).x(

Kaxq

)x(Nq

dxd

pmax

sp

s

'p

'

spA

s

EE

E

E

EE

ε

εµ

εε

The built-in potential is:

( ) ( ) ( ) V 52.0

0

0

=−−=−= ∫∫ ∫ −− − −

nn

p p

x

siden

x

x x sidepbi dxxdxxdxxV EEE .

2. From ( )∫−= dxxVbi E , the potential distribution can be obtained

With zero potential in the neutral p-region as a reference, the potential in the p-side

depletion region is

( ) ( ) ( )[ ] ( ) (

( ) ( )

×−×−××−=

×−×−−=×−××−=−=

−−

−−−∫ ∫3424311

4243

0

0

242

108.032

108.031

10596.7

108.032

108.031

2 108.0

2

xx

xxqa

dxxaq

dxxxVs

x x

sp εε

E

With the condition Vp(0)=Vn(0), the potential in the n-region is

( )

×−×−××−=

×+×−××−=

−−

−−

73

427

73

4214

1098.0

10067.121

1056.4

1098.0

10067.121

103

xx

xxq

xVs

n ε

The potential distribution isDistance p-region n-region

-0.8 0.000

-0.7 0.006

-0.6 0.022

-0.5 0.048

-0.4 0.081

-0.3 0.120

-0.2 0.164

-0.1 0.211

0 0.259 0.25941

3333

18

0.1 0.305788533

0.2 0.3476

03733

0.3 0.384858933

0.4 0.41755

4133

0.5 0.4456

89333

0.6 0.4692

64533

0.7 0.4882

79733

0.8 0.50273

4933

0.9 0.5126

30133

1 0.51796

5333

1.06

7

0.51898

8825

Potential Distribution

-0.100

0.000

0.100

0.200

0.300

0.400

0.500

0.600

-1 -0.5 0 0.5 1 1.5

D istance (um)

Pot

enti

al

(V)

19

3. The intrinsic carriers density in Si at different temperatures can be obtained by using Fig.22 in

Chapter 2 :

Temperature (K) Intrinsic carrier density (ni)

250 1.50×108

300 9.65×109

350 2.00×1011

400 8.50×1012

450 9.00×1013

500 2.20×1014

The Vbi can be obtained by using Eq. 12, and the results are listed in the following table.

T ni Vbi (V)

250 1.500E+0

8

0.777

300 9.65E+9 0.717

350 2.00E+1

1

0.653

400 8.50E+1

2

0.488

450 9.00E+13 0.366

500 2.20E+1

4

0.329

Thus, the built-in potential is decreased as the temperature is increased.

The depletion layer width and the maximum field at 300 K are

V/cm. 10476.11085.89.11

10715.910106.1

m 9715.010106.1

717.01085.89.1122

4

14

51519

max

1519

14

×=××

××××==

=××

××××==

−

−−

−

−

s

D

D

bis

WqN

qNV

W

ε

µε

E

20

18

16

2/1

18

18

14

195

2/1

max

101

10755.1

1010

1085.89.1130106.12

1042

.4

D

D

D

D

DA

DA

s

R

NN

NN

NNNNqV

+=×⇒

+××

×××=×⇒

+

≈ −

−

ε.E

We can select n-type doping concentration of ND = 1.755×1016 cm-3 for the junction.

5. From Eq. 12 and Eq. 35, we can obtain the 1/C2 versus V relationship for doping concentration of

1015, 1016, or 1017 cm-3, respectively.

For ND=1015 cm-3,

( ) ( ) ( )VV

NqåVV

C Bs

bi

j

−×=×××××

−×=−

=−−

837.010187.110108.8511.9101.6

0.837221 16

1514192

For ND=1016 cm-3,

( ) ( ) ( )VV

NqåVV

C Bs

bi

j

−×=×××××

−×=−

=−−

896.010187.110108.8511.9101.6

0.896221 15

1614192

For ND =1017 cm-3,

( ) ( ) ( )VV

NqåVV

C s

bi

j

−×=×××××

−×=−

=−−

956.010187.110108.8511.9101.6

0.956221 14

171419B

2

When the reversed bias is applied, we summarize a table of 2jC1/ vs V for various ND values as

following,

21

V ND=1E15 ND=1E16 ND=1E17

-4 5.741E+

16

5.812E+

15

5.883E+

14

-3.5 5.148E+

16

5.218E+

15

5.289E+1

4

-3 4.555E+1

6

4.625E+

15

4.696E+

14

-2.5 3.961E+

16

4.031E+

15

4.102E+

14

-2 3.368E

+16

3.438E+

15

3.509E+1

4

-1.5 2.774E

+16

2.844E+

15

2.915E+1

4

-1 2.181E

+16

2.251E+

15

2.322E+

14

-0.5 1.587E+

16

1.657E+

15

1.728E+

14

0 9.935E+1

5

1.064E+

15

1.134E+

14

Hence, we obtain a series of curves of 1/C2 versus V as following,

The slopes of the curves is positive proportional to the values of the doping concentration.

1/C2 vs V

0

1E+16

2E+16

3E+16

4E+16

5E+16

6E+16

7E+16

-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0

Applied Voltage

1/C

^2

22

The interceptions give the built-in potential of the p-n junctions.

23

6. The built-in potential is

( )V 5686.0

1065.9106.18

0259.01085.89.111010ln0259.0

32

8ln

32

3919

142020

32

2

=

××××

×××××××=

=

−

−

i

sbi nq

kTaqkT

Vε

From Eq. 38, the junction capacitance can be obtained

( )( )

( )

3/12142019-3/12

5686.0121085.89.1110101.6

12

−×××××=

−==

−

RRbi

ssj VVV

qa

WC

εε

At reverse bias of 4V, the junction capacitance is 6.866×10-9 F/cm2.

7. From Eq. 35, we can obtain

( ) ( ) 2

2

221j

s

RbiD

Bs

bi

j

Cq

VVN

NqVV

C εε−

=⇒−

=

We can select the n-type doping concentration of 3.43×1015cm-3.

8. From Eq. 56,

169

1515

1571515

1089310659

02590020

exp1002590

020exp10

10101010

expexp

×=××

−

+

×××=

−

+

−

=−=

−−

−−

..

..

..

n

kTEE

kTEE

NUG i

tip

itn

tthnp

σσ

υσσ

and

( )

315

281419

2

cm10433

)10850(108589111061

422

−

−−−

×=⇒

××××××

×=≅⇒>>

.N

....

CqV

NVV

d

js

RDbiR ε

Q

24

( )m 266.1cm1066.12

10106.1)5.0717.0(1085.89.1122 5

1519

14

µε

=×=××

+××××=+

= −−

−

A

bis

qNVV

W

Thus

2751619 A/cm10879.71066.121089.3106.1 −− ×=×××××== qGWJ gen .

9. From Eq. 49, and D

ino N

np

2

=

We can obtain the hole concentration at the edge of the space charge region,

( ) 3170259.0

8.0

16

29)0259.0

8.0(

2

cm1042.210

1065.9 −

×=×== eeN

np

D

in .

10. ( ) ( ) ( )1/ −=−+= kTqVspnnp eJxJxJJ

V. 017.0

195.0

1

0259.0

0259.0

=⇒

−=⇒

−=⇒

V

e

eJJ

V

V

s

11. The parameters are

ni= 9.65×109 cm-3 Dn= 21 cm2/sec

Dp=10 cm2/sec τp0=τn0=5×10-7 sec

From Eq. 52 and Eq. 54

( ) ( )

( )

315

0259.0

7.029

7

19

2/

cm102.5

11065.9

10510

106.17

11

−

−−

×=

⇒

−×××

×××=

⇒

−××=−=

D

D

kT

qV

D

i

po

pkTqV

p

nopnp

N

eN

eNnD

qeL

pqDxJ

a

τ

25

( ) ( )

( )

316

0259.0

7.029

7

19

2/

cm10278.5

11065.9

10521

106.125

11

−

−−

×=

⇒

−×××

×××=

⇒

−××=−=−

A

A

kT

qV

A

i

no

nkTqV

n

ponpn

N

eN

eNnD

qeL

nqDxJ

a

τ

We can select a p-n diode with the conditions of NA= 5.278×1016cm-3 and ND =

5.4×1015cm-3.

12. Assumeôg =ôp =ôn = 10-6 s, Dn= 21 cm2/sec, and Dp = 10 cm2/sec

(a) The saturation current calculation.From Eq. 55a and ppp DL τ= , we can obtain

+=+=

00

200 11

n

n

Ap

p

Di

n

pn

p

nps

DN

D

Nqn

L

nqD

L

pqDJ

ττ

( )

212

616618

2919

A/cm1087.6

1021

101

1010

101

1065.9106.1

−

−−−

×=

+×××=

And from the cross-sectional area A = 1.2×10-5 cm2, we obtain

A10244.81087.6102.1 17125 −−− ×=×××=×= ss JAI .

(b) The total current density is

−= 1kt

qV

s eJJ

Thus

A.10244.8110244.8

A1051.41047.510244.8110244.8

170259.0

7.017

7.0

511170259.0

7.017

7.0

−−

−−

−−−

×=

−×=

×=×××=

−×=

eI

eI

V

V

26

13. From

−= 1kt

qV

s eJJ

we can obtain

V 78.0110244.8

10ln0259.01ln

0259.0 17

3

=

+

×

×=⇒

+

=

−

−

VJJV

s

.

14. From Eq. 59, and assume Dp= 10 cm2/sec, we can obtain

( ) ( )

1519

14

6

919

15

29

619

2

101061

108589112

10

106591061

10

10659

10

101061

××+×××××××+××=

+≅

−

−

−

−

−−

.

VV......

WqnNnD

qJ

Rbi

g

i

D

i

p

pR ττ

( ) V 834.01065.9

1010ln0259.0

29

1519

=×

×=biV

Thus

RR VJ +×+×= −− 834.010872.11026.5 711

27

VR Js

0 1.713E-

07

0.1 1.755E-

07

0.2 1.941E-

07

0.3 2,129E-

07

0.4 2.316E-

07

0.5 2.503E-

07

0.6 2.691E-

07

0.7 2.878E-

07

0.8 3.065E-

07

0.9 3.252E-

07

1 3.439E-

07

28

When ND=1017 cm-3, we obtain

( ) V 953.01065.9

1010ln0259.0

29

1719

=×

×=biV

15. From Eq. 39,

( )∫∞

−=

x nonpn

dxppqQ

Current Density

0.E+00

5.E-08

1.E-07

2.E-07

2.E-07

3.E-07

3.E-07

0 0.2 0.4 0.6 0.8 1 1.2

Applied Voltage

Cur

rent

Den

sity

RR VJ +×+×= −− 956.010872.11026.5 813

Current Density

0.00E+00

5.00E-09

1.00E-08

1.50E-08

2.00E-08

2.50E-08

3.00E-08

0 0.2 0.4 0.6 0.8 1 1.2

Applied Voltage

Cur

rent

Den

sity

29

( ) ( )∫∞ −−−=

x

LxxkTqVno

n

pn dxeepq

// 1

The hole diffusion length is larger than the length of neutral region.

( )∫ −='

n

n

x

x nonp dxppqQ

( ) ( )

.C/cm10784.8

1)105(10

)1065.9(106.1

1)(

1

23

5

0

5

1

0259.0

14

16

2919

x

//

'

'

n

−

−−−−

−−

−−

−−

×=

−

−×−×××=

−

−−=

−= ∫

eee

eeeLqp

dxeepq

p

nn

p

nn

n pn

L

xx

L

xx

kT

qV

pno

x LxxkTqVno

16. From Fig. 26, the critical field at breakdown for a Si one-sided abrupt

junction is about 2.8 ×105 V/cm. Then from Eq. 85, we obtain

( ) ( ) 12

22voltagebreakdown −== B

cscB N

qW

VEE ε

( ) ( ) 115

19

2514

10106.12

108.21085.89.11 −

−

−

××××××=

258= V

( )£gm4318cm108431

101061

2581085891122 31519

14

...

..qN

VVW

B

bis =×=××

××××≅−

= −−

−ε

When the n-region is reduced to 5µm, the punch-through will take place first.

From Eq. 87, we can obtain

( )/2Winsert 29 Fig.in area shaded

mcE=

B

B

V'V

−

=

mm WW

WW

2

12143.18

52

43.185

2582' =

−

×=

−

=

mmBB W

WWW

VV V

Compared to Fig. 29, the calculated result is the same as the value under the

30

conditions of W = 5 µm and NB = 101 5 cm-3.

17. We can use following equations to determine the parameters of the diode.

kTqV

D

i

p

pkTqV

r

ikTqV

D

i

p

pF e

N

nDqe

qWne

N

nDqJ /

22//

2

2 τττ≅+=

( ) 12

22−== D

cscB N

qW

VEE ε

⇒

( ) 302590

7029

7

192

102210659

101061 −

−− ×=

××××⇒= .e

N.D

.AeNnD

AqAJ .

.

D

pkT/qV

D

i

p

pF τ

( ) ( ) 1

19

2141

2

10612

10858911130

22−

−

−−

××××

=⇒== Dc

Dcsc

B N.

..N

qW

VEEE ε

Let Ec= 4×105 V/cm, we can obtain ND = 4.05×1015 cm-3.

The mobility of minority carrier hole is about 500 at ND = 4.05×1015

∴Dp= 0.0259×500=12.95 cm2/s

Thus, the cross-sectional area A is 8.6×10-5 cm2.

18. As the temperature increases, the total reverse current also increases. That is,

the total electron current increases. The impact ionization takes place when the

electron gains enough energy from the electrical field to create an electron-

hole pair. When the temperature increases, total number of electron increases

resulting in easy to lose their energy by collision with other electron before

breaking the lattice bonds. This need higher breakdown voltage.

19. (a) The i-layer is easy to deplete, and assume the field in the depletion region is

31

constant. From Eq. 84, we can obtain.

V/cm 1087.5)10(1041101054

101104

10 56153

64

0

6

5

4 ×=××=⇒=×

×⇒=

×−∫ critical

Wdx E

EE

4 587101087.5 35 =××= −BV V

(b) From Fig. 26, the critical field is 5 × 105 V/cm.

( ) ( ) 12

22voltagebreakdown −== B

cscB N

qW

VEE ε

( ) ( ) 116

19

2514

102106.12

1051085.84.12 −

−

−

×××

××××=

8.42= V.

20. 422

4

18

cm10102

102 −−

=××=a

( )

( )238

2122

21

19

1423

212123

10844

101061

1085891123

4

2

3

4

32

/c

///

c

//

s/

cB

.

...

aq

WV

E

E

EE

−

−

−

−

−

×=

×

×

×××=

==

ε

The breakdown voltage can be determined by a selected Ec.

21. To calculate the results with applied voltage of V 5.0=V , we can use a similar calculation

in Example 10 with (1.6-0.5) replacing 1.6 for the voltage. The obtained electrostatic

potentials are V 1.1 and V 103.4 4−× , respectively. The depletion widths are

cm 103.821 5−× and cm 101.274 8−× , respectively.

Also, by substituting V 5−=V to Eqs. 90 and 91, the electrostatic potentials are V 6.6

and V 1020.3 4−× , and the depletion widths are cm 109.359 5−× and cm 103.12 8−× ,

respectively.

The total depletion width will be reduced when the heterojunction is forward-biased from

the thermal equilibrium condition. On the other hand, when the heterojunction is reverse-

32

biased, the total depletion width will be increased.

22. Eg(0.3) = 1.424 + 1.247 × 0.3 = 1.789 eV

biV = ( ) −−−∆− q/EEqE

q

EVF

Cg22

2 ( ) q/EE FC 11 −

= 1.789 – 0.21–15

17

105

1074ln

××.

qkT

–15

18

105

107ln

××

qkT

= 1.273 V

1x = ( )2

1

21

212

+ ADD

biA

NNqNvNεε

εε= ( )

21

1519

14

46114121051061

27311085846114122

+××××××××

−

−

.......

= 4.1 510 −× cm.

Since 21 xNxN AD = 21 xx =∴

m0.82 cm10282 51 µ . x W =×==∴ − .

33

CHAPTER 5

1. (a) The common-base and common-emitter current gains is given by

. 199

995.01995.0

1

995.0998.0997.0

0

00

0

=−

=−

=

=×==

αα

β

γαα T

(b) Since 0=BI and A 1010 9−×=CpI , then CBOI is A 1010 9−× . The emitter current is

( )( )

. A 102

10101991

1

6

9

0

−

−

×=

×⋅+=

+= CBOCEO II β

2. For an ideal transistor,

. 9991

999.0

0

00

0

=−

=

==

αα

β

γα

CBOI is known and equals to A 1010 6−× . Therefore,

( )

. mA 10

1010)9991(

16

0

=×⋅+=

+=−

CBOCEO II β

3. (a) The emitter-base junction is forward biased. From Chapter 3 we obtain

( ). V 956.0

1065.9

102105ln0259.0ln

29

1718

2=

×

×⋅×=

=

i

DAbi

nNN

qkT

V

The depletion-layer width in the base is

34

( )

( )

. m 105.364cm 105.364

5.0956.0102105

1102105

106.11005.12

12

junction) base-emitter theof hlayer widt-depletion (Total

2-6-

171817

18

19

12

1

µ×=×=

−

×+×

××

××⋅=

−

+

=

+

=

−

−

VVNNN

Nq

NNN

W

biDAD

As

DA

A

ε

Similarly we obtain for the base-collector function

( ). V 0.795

1065.9

10102ln0259.0

29

1617

=

×

⋅×=biV

and

( )

. m 10.2544cm 104.254

5795.010210

1102

10106.1

1005.12

2-6-

171617

16

19

12

2

µ×=×=

+

×+

××

×⋅=−

−

W

Therefore the neutral base width is

. m 904.010.254410364.51 -2221B µ=×−×−=−−= −WWWW

(b) Using Eq. 13a

( ). cm 10543.2

1021065.9

)0( 3-110259.05.017

292

×=××=== ee

Nn

epp kTqV

D

ikTqVnon

EBEB

4. In the emitter region

( ). 18.625

1051065.9

cm 10721.01052 scm 52

18

29

38

=××=

×=⋅== −−

EO

EE

n

LD

In the base region

35

( ). 465.613

102

1065.9

mc 1021040 scm 40

17

292

37

=××==

×=⋅=== −−

D

ino

pppp

Nn

p

DLD τ

In the collector region

( ). 109.312

101065.9

cm 10724.1010115 scm 115

316

29

36

×=×=

×=⋅== −−

CO

CC

n

LD

The current components are given by Eqs. 20, 21, 22, and 23:

( )

. 0

A 103.19610724.10

10312.9115102.0106.1

A 101.041110721.0

625.1852102.0106.1

A 101.596

A 101.59610904.0

613.46540102.0106.1

14-

3

3219

7-0259.05.0

3

219

5-

5-0259.05.0

4

219

=−=

×=×

×⋅⋅×⋅×=

×=−×

⋅⋅×⋅×=

×=≅

×=×

⋅⋅×⋅×=

−

−−

−

−−

−

−−

CpEpBB

Cn

En

EpCp

Ep

III

I

eI

II

eI

5. (a) The emitter, collector, and base currents are given by

. A 101.041

A 101.596

A 101.606

7-

5-

-5

×=−+=

×=+=

×=+=

CnBBEnB

CnCpC

EnEpE

IIII

III

III

(b) We can obtain the emitter efficiency and the base transport factor:

9938.010606.110596.1

5

5

=××==

−

−

E

Ep

I

Iγ

. 110596.110596.1

5

5

=××== −

−

Ep

CpT I

Iα

Hence, the common-base and common-emitter current gains are

. 3.1601

9938.0

0

00

0

=−

=

==

ααβ

γαα T

36

(c) To improve γ , the emitter has to be doped much heavier than the base.

To improve Tα , we can make the base width narrower.

6. We can sketch (0))( nn pxp curves by using a computer program:

0 . 0 0 . 2 0 . 4 0 . 6 0 . 8 1 . 0

0 . 0

0 . 2

0 . 4

0 . 6

0 . 8

1 . 0

20

1

2

5

10

W / L p<0.1

DISTANCE x

p n(x)

/pn(

0)

In the figure, we can see when 1.0<pLW ( 05.0=pLW in this case), the

minority carrier distribution approaches a straight line and can be simplified to

Eq. 15.

7. Using Eq.14, EpI is given by

37

( )

( ) .

cosh

1 1 coth

sinh

1

sinh

cosh

)1 (

sinh

cosh1

sinh

cosh1

)1 (

0

0

+−

=

+

−=

−

+

−−

−−=

−=

=

=

p

kTqV

pp

nop

pp

pkTqV

p

nop

xp

ppno

p

ppkTqVnop

x

npEp

LW

eLW

L

pDqA

LW

LW

LW

eL

pDqA

LW

Lx

Lp

LW

LxW

LepqDA

dxdp

qDAI

EB

EB

EB

Similarly, we can obtain CpI :

( )

( ) . cosh 1

sinh

1

sinh

cosh

sinh

1)1 (

sinh

cosh1

sinh

cosh1

)1 (

+−

=

+

−=

−

+

−−

−−=

−=

=

=

p

kTqV

p

p

nop

p

p

p

kTqV

p

nop

Wxp

ppno

p

ppkTqVnop

Wx

npCp

LW

e

LWL

pDqA

LW

LW

LW

eL

pDqA

LW

Lx

Lp

LW

LxW

LepqDA

dxdp

qDAI

EB

EB

EB

8. The total excess minority carrier charge can be expressed by

38

[ ]

. 2

)0(2

)2

(

)1(

)(

0

2

0

0

n

kTqVno

W

kTqVno

WkTqV

no

W

nonB

qAWp

eqAWp

Wx

xeqAp

dxWx

epqA

dxpxpqAQ

EB

EB

EB

=

=

−=

−=

−=

∫

∫

From Fig. 6, the triangular area in the base region is 2

)0(nWp . By multiplying this

value by q and the cross-sectional area A, we can obtain the same expression as BQ .

In Problem 3,

. C 103.6782

10543.210904.0102.0106.1

15-

114219

×=

×⋅×⋅×⋅×=−−−

BQ

9. In Eq. 27,

( )

. 2

2

)0(2

)0(

1

2

2

2221

Bp

np

np

kTqVC

QW

D

qAQp

W

DW

pqAD

aeaI EB

=

=

≅

+−=

Therefore, the collector current is directly proportional to the minority carrier

charge stored in the base.

10. The base transport factor is

39

( )

( )

.

cosh

1 1 coth

cosh 1

sinh

1

+−

+−

=≅

p

kTqV

p

p

kTqV

p

Ep

CpT

LW

eLW

LW

e

LW

I

I

EB

EB

α

For 1<<pLW , ( ) 1cosh ≅pLW . Thus,

( ) . 21

21

1

sech

cothsinh

1

22

2

p

p

p

pp

T

LW

LW

LW

LW

LW

−=

−=

=

⋅

=α

11. The common-emitter current gain is given by

. 11 0

00

T

T

γαγα

ααβ

−=

−≡

Since 1≅γ ,

( )( )[ ]

( ) . 12

211

21

1

22

22

22

0

−=

−−

−=

−≅

WL

LW

LW

p

p

p

T

T

ααβ

If 1<<pLW , then 220 2 WLp≅β .

12. m 54.77mc 10477.5103100 37 µ=×=×⋅== −−ppp DL τ

Therefore, the common-emitter current gain is

40

( )( )

. 1500102

1077.5422

24

2422

0

=×

×=≅−

−

WLpβ

13. In the emitter region,

6871031037401

407354

1817.

..pE =

×⋅×++=

−µ

. ..DE scm 2.2668702590 =⋅=

In the base region,

6311861021069801

125288

1617.

.n =

×⋅×++=

−µ

. ...Dp scm733063118602590 =⋅=

In the collector region,

824531051037401

407354

1517.

..pC =

×⋅×++=

−µ

. ...DC scm75118245302590 =⋅=

14. In the emitter region,

cm 10506.110269.2 36 −− ×=⋅== EEE DL τ

( ). cm 04.31

103

1065.9 3-18

292

=×

×==E

iEO N

np

In the base region,

cm 10544.510734.30 36 −− ×=⋅=nL

( ). cm 13.4656

1021065.9 3-

16

29

=××=pon

In the collector region,

cm 10428.310754.11 36 −− ×=⋅=CL

41

( ). cm 5.18624

1051065.9 3-

15

29

=××=COp

The emitter current components are given by

A 1083.526105.0

13.4656734.3010106.1 60259.06.04

419−

−

−−

×=×

⋅⋅⋅×= eI En

( ) .A 10609.8110506.1

04.31269.210106.1 90259.06.0

3

419−

−

−−

×=−×

⋅⋅⋅×= eIEp

Hence, the emitter current is

.A 10839.526 6−×=+= EpEnE III

And the collector current components are given by

A 1083.526105.0

13.4656734.3010106.1 60259.06.0

4

419−

−

−−

×=×

⋅⋅⋅×= eICn

.A 10022.110428.3

5.18624754.1110106.1 14

3

419−

−

−−

×=×

⋅⋅⋅×=CpI

Therefore, the collector current is obtained by

.A 10268.5 4−×=+= CpCnC III

15. The emitter efficiency can be obtained by

. 99998.010839.526

1083.5266

6

=×

×==−

−

E

En

IIγ

The base transport factor is

. 11083.526

1083.5266

6

=××==

−

−

En

CnT I

Iα

Therefore, the common-base current gain is obtained by

. 99998.099998.010 =×== Tγαα

The value is very close to unity.

The common-emitter current gain is

42

. 5000099998.01

99998.01 0

00 ≅

−=

−=

αα

β

16. (a) The total number of impurities in the neutral base region is

( )( ) . cm 10583.51103102

1

2-13103108518

0

55

×=−×⋅×=

−==−− ××−−

−−∫e

elNdxeNQ lWAO

W

lxAOG

(b) Average impurity concentration is

. cm 10979.6108

10583.5

3-17

5

13

×=×

×= −WQG

17. For -317 cm 10979.6 ×=AN , scm 77.7 3=nD , and

cm 10787.21077.7 36 −− ×=⋅== nnn DL τ

( )( ) 999588.0

10787.22

1081

21 23

25

2

2

=×

×−=−≅

−

−

nT

L

Wα

. 99287.0

101010583.5

77.71

1

1

1

1

419

13=

⋅×⋅+

=+

=

−EE

G

n

E

LNQ

DD

γ

Therefore,

99246.00 == Tγαα

. 6.1311 0

00 =

−=

αα

β

18. The mobility of an average impurity concentration of -317 cm 10979.6 × is about

300 Vscm2 . The average base resistivity Bρ is given by

( ) . cm- 0299.01 Ω==

WQq GnB µ

ρ

43

Therefore,

( ) ( ) . 869.11080299.0105105 533 Ω=×⋅×=×= −−− WR BB ρ

For a voltage drop of qkT ,

.A 0139.0==B

B qRkT

I

Therefore,

.A 83.10139.06.131 0 =⋅== BC II β

19. From Fig. 10b and Eq. 35, we obtain

( )AµBI ( )mACIB

C

I

I

∆∆

=0β

0 0.20 ---

5 0.95 150

10 2.00 210

15 3.10 220

20 4.00 180

25 4.70 140

0β is not a constant. At low BI , because of generation-recombination current, 0β

increases with increasing BI . At high BI , EBV increases with BI , this in turn causes

a reduction of BCV since V 5==+ ECBCEB VVV . The reduction of BCV causes a

widening of the neutral base region, therefore 0β decreases.

The following chart shows 0β as function of BI . It is obvious that 0β is not a

constant.

44

0 5 10 15 20 25 301 0 0

1 5 0

2 0 0

2 5 0

IB (µA)

β 0

20. Comparing the equations with Eq. 32 gives

11 aIFO = , 12 aIROR =α

21 aIFOF =α , and . 22aI RO =

Hence,

no

EO

p

E

E

F

pn

DD

LWa

a

⋅⋅+==

1

1

11

21α

. 1

1

22

12

no

CO

p

C

C

R

pn

DD

LWa

a

⋅⋅+==α

21. In the collector region,

cm 101.414 102 36 −− ×=⋅== CCC DL τ

( ) . cm 101.863 1051065.9 3-415292 ×=××== CiCO Nnn

From Problem 20, we have

45

99995.01031.9

31.9101

10105.0

1

1

1

1

23

4

=×

⋅⋅×+=

⋅⋅+=

−−

−

no

EO

p

E

E

F

pn

DD

LW

α

876.01031.910863.1

102

10414.1105.0

1

1

1

1

2

4

3

4

=××⋅⋅

××+

=⋅⋅+

=

−−

−

no

CO

p

C

C

R

pn

DD

LW

α

A 1049.1

1031.91

105.01031.910

105106.1

14

44

2419

11

−

−−−−

×=

⋅+×

×⋅⋅×⋅×=

+==

E

EOEnopFO L

nDW

pDqAaI

. A 107.1

10414.1

10863.12

105.0

1031.910105106.1

14

3

4

4

2419

22

−

−−−−

×=

×

×⋅+

××⋅

⋅×⋅×=

+==

C

COCnopRO L

nDW

pDqAaI

The emitter and collector currents are

( )A 10715.1

1 4−×=

+−= RORkTqV

FOE IeII EB α

( ). A 10715.1

1 4−×=

+−= ROkTqV

FOFC IeII EBα

Note that these currents are almost the same (no base current) for 1<<pLW .

22. Referring Eq. 11, the field-free steady-state continuity equation in the collector region is

( ) ( ). 0

2C

2

=−

−

C

poCC

n'xn

'dx'xnd

Dτ

The solution is given by ( CCC DL τ= )

( ) . 21CC L'xL'x

C eCeC'xn −+=

Applying the boundary condition at ∞='x yields

46

. 021 =+ ∞−∞ CC LL eCeC

Hence 01 =C . In addition, for the boundary condition at 0='x ,

( )020

2 CL nCeC C ==−

( ) ( ) . 1.0 −= kTqVCOC

CBenn

The solution is

( ) ( ) . 1 CCB L'xkTqVCOC eenxn −−=

The collector current can be expressed as

( ) ( )( ) ( ) . 1 1

1 1

2221

0

−−−=

−

+−−=

−+

−=

==

kTqVkTqV

kTqV

C

COCnopkTqVnop

'x

CC

Wx

npC

CBEB

CBEB

eaea

eLnD

W

pDqAe

W

pDqA

'dxdn

qDAdx

dpqDAI

23. Using Eq. 44, the base transit time is given by

( ).

. DW pB s 101.25

1021050

2 10-

242 ×=

××==

−

τ

We can obtain the cutoff frequency :

. GHz 1.27 21 =≅ BTf πτ

From Eq. 41, the common-base cutoff frequency is given by:

. GHz 1.275 998.0

1027.1

9

0 =×=≅ αα Tff

The common-emitter cutoff frequency is

( ) ( ) . .. f f MHz 2.55102751998011 90 =××−=−= αβ α

Note that βf can be expressed by

( ) ( ) TT ffff0

000

1 1 1 β

ααα αβ =×−=−= .

47

24. Neglect the time delays of emitter and collector, the base transit time is given by

s1083.311052

12

1 12

9

−×=××

==ππ

τT

B f.

From Eq. 44, W can be expressed by

BpDW τ2= .

Therefore,

. m0.252

cm102.52

1083.311025-

12

µ=×=

×××= −W

The neutral base width should be 0.252 µm.

25. gE∆ =9.8% ×1.12 ≅ 110 meV.

oβ ~exp

∆kT

E g

( )( )C0

C100

°°

∴

o

o

ββ

= exp

−

k

k

273meV110

373meV110

= 0.29.

26.

−=

−=

0259.0

424.1)(expexp

)BJT()HBT(

0

0xE

kT

EE gEgBgE

ββ

where

. 145.0 ,143.0125.09.1

45.0 ,247.1424.1)(

≤<++=

≤+=

xxx

xxxEgE

The plot of )BJT()HBT( 00 ββ is shown in the following graph.

48

Note that )HBT(0β increases exponentially when x increases.

27. The impurity concentration of the n1 region is -314 cm 10 . The avalanche

breakdown voltage (for mWW > ) is larger than 1500 V ( m 100 µ>mW ). For a

reverse block voltage of 120 V, we can choose a width such that punch-through

occurs, i.e.,

. 2

2

s

DPT

WqNV

ε=

Thus,

. cm 1096.32 3−×=

ε=

D

PTs

qNV

W

When switching occurs,

. 121 ≅+ αα

That is,

6.04.01

ln5.00

1

=−=

=

JJ

W

Lpα

. 51.1

106.391025

5.06.0

ln4

4

0

=×

×=

−

−

JJ

49

Therefore,

250 cmA 1025.25.4 −×=≅ JJ

. cm 4.441025.2

101Area 2

5

3

=×

×==−

−

JI s

28. In the 2np2n1 −− transistor, the base drive current required to maintain current conductionis

KII )1( 21 α−= .In addition, the base drive current available to the 2np2n1 −− transistor with areverse gate current is

gA III −= 12 α .

Therefore, when use a reverse gate current, the condition to obtain turn-off of the thyristor

is given by

12 II <or KgA III )1( 21 αα −<− .Using Kirchhoff’s law, we have

gAK III −= .Thus, the condition for turn-on of the thyristor is

Ag II2

21 1

ααα −+

> .

Note that if we define the ratio of IA to Ig as turn-off gain, then the maximum turn-offgain maxβ is

121

2max −+

=ααα

β .

50

CHAPTER 6

1.

ECEF

Ei

EV

VG = VT

ψS

= 2ψB

METAL OXIDE n-TYPE SEMICONDUCTOR

51

2.

3.

ECEi

EF

EV

n+ POLYSILICON OXIDE p-TYPE SEMICONDUCTOR

n+ POLYSILICON OXIDE p-TYPE SEMICONDUCTOR

φms

VG = VFB = φφms

EFECEi

EV

EC

EiEF

EV

52

4.

0 W x

-dqND

QM

QP

ρS(x)

-d 0 W

x

E (x)

-d 0 W

x

ψ (x)

V Voψs

(a)

(b)

(c)

53

5. A

i

As

mNq

nN

kT

W2

ln

2

=ε

= 21619

9

1614

1051061

10659

105ln026010858911

×××

×

××××

−

−

.

....

= 1.5 ×10-5 cm = 0.15 µ m.

6. msox

ox

WdC

)/(min εεε

+=

µ150.Wm = m From Prob. 5

8

57

14

100361051

91193

108

1085893 −

−−

−

×=××+×

××=∴ .

...

..Cmin F/cm2 .

7. V42010659

10ln0260ln

9

17

..

.nN

qkT

i

AB =

×==ψ

s

As

WqNε

=E at intrinsic Bs ψψ =

51719

14

1074.010106.1

42.01085.89.1122 −−

−

×=××

××××==A

ss

qNW

ψεcm

5

14

51719

1011110858911

10740101061×=

××××××

=−

−−

...

..sE V/cm

Eo = Esoxε

ε s = 1.11 ×105×93911

.. = 3.38 ×105 V/cm

V = Vo+ sψ = Eod + sψ = ( )75 10510383 −×××. +0.42. = 0.59 V.

8. At the onset of strong inversion, sψ =2 Bψ ⇒ VG=VT

thus, VG =o

mA

CWqN

+2 Bψ

From Prob. 5, Wm = 0.15£gm, Bψ = 0.026ln

××

9

16

10659

105

.= 0.4 V

Co =6

14

10

1085893−

−×× .. = 3.45×10-7 F/cm2

GV∴ =7

51619

10453

10511051061−

−−

××××××

...

+0.8 = 0.35 + 0.8 = 1.15V.

54

9. ∫−

−−

−

×××==610

0

2617

6

1917 ])10(10

21

[10

106.1)10(

1dyyq

dQot

= 8×10-9 C/cm2

2

7

9

1032.21045.3

108 −−

−

×=×

×==∆o

otFB C

QV V.

10. ∫=d

otot dyyyd

Q

0 )(

1ρ

)(105 11 xqot δρ ××= , where

×≠×=∞

=−

−

7

7

105,0

105,)(

y

yxδ

19711

6106.1105105

101 −−

−××××××=∴ otQ

= 4×10-8 C/cm2

12.01045.3

1047

8

=×

×==∆∴−

−

o

otFB C

QV V

11. ∫−

×××=610

0

23 )105(1

dyyqyd

Qot

= 2.67×10-8 C/cm2

2

7

8

1074.71045.31067.2 −

−

−

×=××==∆∴

o

otFB C

QV V.

12. moo

otFB N

dd

Cq

CQ

V ××==∆ , where Nm is the area density of Qm.

11

19

7

1047.6106.1

1045.33.0 ×=×

××=×∆=⇒−

−

qCV

N oFBm cm-2 .

13. Since )( TGD VVV −<< , the first term in Eq. 33 can be approximated as

DBGon V)V(CLZ ψµ 2− .

Performing Taylor’s expansion on the 2nd term in Eq. 33, we obtain

D/

B/

BD/

B/

B/

B/

BD V)()(V)()()()V( 212321232323 223

2223

222 ψψψψψψ =−+≅−+

Equation 33 can now be re-written as

.)10(10531

106.110

1 362319

6

−−−

×××××=

55

DTGon

Do

BAsBGon

DBo

AsDBGonD

V)VV(C)LZ

(

VC

)(qNVC

LZ

VC

qNV)V(C

LZ

I

−≅

+−≅

×−−≅

µ

ψεψµ

ψε

ψµ

222

2232

32

2

where o

BAsBT C

)(qNV

ψεψ

222 += .

14. When the drain and gate are connected together, DG VV = and the MOSFET is

operated in saturation )( DsatD VV > . DI can be obtained by substituting

DsatD VV = in Eq. 33

[ ]

−+

−

−==

2/32/3 )(2)2(2

32

22 BBDsat

o

AsDsatB

DsatonVVD V

C

qNV

VC

LZ

IGD

ψψε

ψµ

where DsatV is given by Eq. 38. Inserting the condition 0)( == LyQn into Eq.27 yields

02221 =−++= GBBDsatAs

oDsat V)V(qN

CV ψψε

For BDsatV ϕ2<< , the above equation reduces to

TBo

BAsDG V

C

)(qNVV =+== ψ

ψε2

22 .

Therefore a linear extrapolation from the low current region to 0=DI will yieldthe threshold voltage value.

15. 400109.65

1050.026lnln

9

16

.)nN

(q

kT

i

A =

××==

−ψ V

7

161914

10453

105106110858911−

−−−

×××××××=≡

....

C

qNK

o

Asε

= 0.27

+−+−≅∴

2

2 2112

KV

KVV GBGDsat ψ

= ])27.0(

1011[)27.0(8.05 2

2 +−+−

= 3.42 V

27

)7.05(12

1045.380010)(

2−

××××=−=

−

TGon

Dsat VVL

CZI

µ

= 2.55×10-2 A.

56

16. The device is operated in linear region, since 1.0=DV V < 5.0)( =− TG VV V

Therefore, )(. TGonD

Dd VVC

LZ

V

Ig

constGV−=

∂∂

==

µ

= 5.01045.350025.05 7 ×××× −

= 1.72×10-3 S.

17. DonG

Dm VC

LZ

VI

gconstDV

µ=∂∂

== .

= 1.01045.350025.05 7 ×××× −

=3.45×10-4 S.

18. o

BAsBFBT C

qNVV

ψεψ

22 ++= , )

1065.9

10ln(026.0

9

17

×=Bψ

7

1019

1045.3105106.1

2 −

−

××××−−−=−= B

g

o

fFB

E

C

QV

msψφ

= 102.042.056.0 −=−−− V

7

191714

1045.3106.142.0101085.89.112

84.01−

−−

×××××××++−=∴ TV

= 49.084.01 ++−= 0.33 V

( msφ can also be obtained from Fig. 8 to be – 0.98 V).

19. 71045.3

33.07.0−×

+= BqF

11

19

7

108106.1

1045.337.0 ×=×

××=−

−

BF cm-2 .

20. 14.042.056.02

−=+−=+−= Bg

ms

Eψφ V

7

171914

1045.342.010106.11085.89.112

84.002.014.0

22

−

−−

×××××××−−−−=

−−−=o

BDsB

o

fmsT C

qN

C

QV

ψεψφ

= 49.1− V.

57

21. 71045.3

49.17.0−×

+−=− BqF

12

19

7

107.1106.1

1045.379.0 ×=×

××=−

−

BF cm-2 .

22. The bandgap in degenerately doped Si is around 1eV due to bandgap-

narrowing effect. Therefore,

86.0114.0 =+−=msφ V49.049.084.002.086.0 −=−−−=∴ TV V.

58

23. 42.01065.9

10ln026.0

9

17

=

×=Bψ V

o

oo

o

BAsB

o

fmsT

C.

.

C....

.C

..

C

qN

C

qQV

8

1917141119

10215140

106142010108589112840

101061980

22

−

−−−

×+−=

××××××++××−−=

++−=ψε

ψφ

ddCo

1314 1045.31085.89.3 −− ×=××=

14.201045.3

102.1520

13

8

>×

××⇒>−

−dVT

51057.4 −×>∴ d cm = 0.457 µm.

24. A£g 1.0at V 5.0 == dT IV

Subthreshold swing =

1

0

0

loglog−

==

−

−

T

VDVVD

V

IIGTG

0log7

5.01.0

=−−=

GVDI

12log 0 −==GVDI A 101 120

−= ×=∴

GVDI .

25.

( ) 22

2 BBSB

o

AsT V

C

NqV ψψ

ε−+=∆

V 42.0)1065.9

10ln(026.0

9

17

=×

=Bψ

7

7

14

109.6105

1085.89.3 −−

−

××

××=oC F/cm2

V 1.0=∆ TV if we want to reduce ID at VG = 0 by one order of magnitude,

since the subthreshold swing is 100 mV/decade.

( )V. 83.0

84.084.0109.6

101085.89.4106.121.0

7

171419

=∴

−+×

××××××=−

−−

BS

BS

V

V

26. Scaling factor κ =10

Switching energy = 2)(21

VAC ⋅

59

κ

κ

κε

VV

AA

Cd

C ox

=′

=′

==′

2

∴scaling factor for switching energy : 1000

1111322 ==⋅⋅

κκκκ

A reduction of one thousand times.

27. From Fig. 24 we have

−+−=∆−=∆−=+

∆−=

++−=∆

=−∆+∆∴

−+=∆+

12

1112

222

'

2'

2

022

)()(2

222

j

mj

jmjj

jmj

mmjj

rW

L

r

LLL

LLL

LL

rWrr

rWr

WWrr

From Eq. 17 we have

.12

1)2

'(

)regionr rectangula thein charge space region al trapezaid thein charge space(

−+−=−

+=

−=∆

jo

jmA

o

mA

o

mA

oT

rW

LC

rWqN

CWqN

LLL

CWqN

CV

28. Pros:1. Higher operation speed.2. High device density

Cons:1. More complicated fabrication flow.2. High manufacturing cost.

29. The maximum width of the surface depletion region for bulk MOS

nm 49

cm 109.4

105106.1)1065.9/105ln(026.01085.89.11

2

)/(ln2

6

1719

91714

2

=×=

×××××××××=

=

−

−

−

A

iAsm Nq

nNkTW

ε

For FD-SOI, nm 49=≤ msi Wd .

60

30. o

siABFBT C

dqNVV ++= ψ2

V. 18.0

28.01.0 1063.8

103105106.192.002.1

F/cm 1063.8104

1085.89.3

V 92.0)(ln22

V 02.11065.9

105ln026.0

212.1

)(ln2

7

61719

27

7

14

9

17

=+−=

××××××++−=∴

×=×

××=

=×=

−=

×

×−−=−−==

−

−−

−−

−

T

o

i

AB

i

AgmsFB

V

C

nN

qkT

nN

qkTE

V

ψ

φ

31. oC

sidqNTV

A ∆=∆

mV 461063.8

105105106.17

71719

=×

×××××=−

−−

Thus, the range of VT is from (0.18-0.046)= 0.134 V to (0.18+0.046) = 0.226 V.

32. The planar capacitor

C = 15

6

1424

1045.3101

1086.89.3)101( −−

−−

×=×

×××=dA oxε F

For the trench capacitor

A = 4× 7 µm2+ 1 µm2 = 29 µm2

C = 29 ×3.45 ×10-15 F = 100 × 10-15 F.

33. 11

3

14 101.31045.2

105 −−

− ×=×

××==d

dVCI A

34. )( Toipo VVCLZ

g −= µ

4 × 10-5 = A ( TV−− 5 ) VVT 7=∴ 1 × 10-5 = A( 25 +− ) 9)2(7 =−−=∆V V.

35. 0

2C

qNVV BAs

BFBT

ψεψ ++=

61

V 0.980.420.56

022

V 42.01065.9

10ln026.0

F/cm1045.310

10854.89.3

9

17

B

28

5

14

0

−=−−=

−−−=−=

=

×

=

×=××= −−

−

Bgf

msFB

EQV

C

ψφ

ψ

V 4.34

4.90.420.98 1045.3

106.142.0101085.89.11242.098.0 8

191714

=++−=

×××××××++−=∴ −

−−

TV

V. 2.02

32.234.4

V 2.32

1045.3106.11058

1911

0

=−=

−=×

×××−=−=∆−

−

T

fFB

V

C

QV

62

CHAPTER 7

1. From Eq.1, the theoretical barrier height is

eV 54.001.455.4 =−=−= χφφ mBn

We can calculate Vn as

V 188.0)102

1086.2ln(0259.0ln

16

19

=×

×==D

Cn N

Nq

kTV

Therefore, the built-in potential is

V 352.0188.054.0 =−=−= nBnbi VV φ .

2. (a) From Eq.11

/V/F)(cm 103.202

10)6.42.6()/1( 2214142

×−=−−

×−=dV

Cd

3-16

2 cm 107.4/)/1(

12×=

−=

dVCdqN

sD ε

V 06.0107.4

107.4ln0259.0ln

16

17

=

××==

D

Cn N

Nq

kTV

From Fig.6, the intercept of the GaAs contact is the built-in potential Vbi,

which is equal to 0.7 V. Then, the barrier height is

V 76.0=+= nbiBn VVφ

(b) 217 A/cm 105×=sJ

8* =A A/K2-cm2 for n– type GaAs

kTqs

BneTAJ /2* φ−=

eV 72.0105

)300(8ln0259.0)ln( 7

22*

=

×

×== −

sBn J

TAq

kTφ

The barrier height from capacitance is 0.04 V or 5% larger.

63

(c) For V = –1 V

m 0.222 cm 1022.2

107.4106.1)17.0(1009.12)(2

5

1619

12

µ

ε

=×=

×××+××=

+=

−

−

−

D

Rbis

qNVV

W

V/cm 1043.1 5×==s

Dm

WqNε

E

28 F/cm 1022.5 −×==W

C sε.

3. The barrier height is

V 64.001.465.4 =−=−= χφφ mBn

V 177.0103

1086.2ln0259.0ln

16

19

=

×××==

D

Cn N

Nq

kTV

The built-in potential is

V 463.0177.064.0 =−=−= nBnbi VV φ

The depletion width is

ìm 142.0463.0103106.11085.89.112)(2

1619

14

=×××××××=

−= −

−

D

bis

qNVV

Wε

The maximum electric field is

V/cm. 1054.6)1085.8(9.11

)1042.1()103()106.1(0)( 4

14

51619

×=××

×××××====−

−

WqN

xs

D

εEE m

4. The unit of C needs to be changed from µF to F/cm2, so

1/C2 = 1.74×1015 – 2.12×1015 Va (cm2/F)2

Therefore, we obtain the built-in potential at 1/C2 = 0

V 74.01012.21057.1

15

15

=××=biV

64

From the given relationship between C and Va, we obtain

152

1012.2

1

×−=

adVC

d (cm2/F)2/V

From Eq.11

×××××=

−=

−− 151419

2

1012.21

1085.89.11106.12

/)/1(12

dVCdqN

sD ε

-315 cm 106.5 ×=

V 161.0106.5

1086.2ln0259.0ln

16

19

=

××==

D

Cn N

Nq

kTV

We can obtain the barrier height

V. 901.0161.074.0 =+=+= nbiBn VVφ

5. The built-in potential is

V 605.0

195.08.0

105.11086.2

ln0259.08.0

ln

16

19

=−=

××−=

−=D

CBnbi N

Nq

kTV φ

Then, the work function is

V. 81.4

01.48.0

=+=+= χφφ Bnm

6. The saturation current density is

65

27

2

2*

A/cm 1081.3

0259.08.0

exp)300(110

exp

−×=

−××=

−

=kTq

TAJ Bns

φ

The injected hole current density is

211163

29192

A/cm 1019.1105.1101

)1065.9(12106.1 −−

−

×=×××

××××==Dp

ippo NL

nqDJ

)1(

)1(

currentElectron current Hole

/

/

−

−=

kTqVs

kTqVpo

eJ

eJ

.1031081.31018.1 5

3

11−

−

−

×=××==

s

po

J

J

7. The difference between the conduction band and the Fermi level is given by

V. 04.0101

107.4ln0259.0

17

17

=

××=nV

The built-in potential barrier is then

V 86.004.09.0 =−=biV

For a depletion mode operation, VT is negative. Therefore, From Eq.38a

m.ì 0.108 cm 1008.1

86.01019.2106.1

86.01085.84.12210106.1

2

086.0

5

2

12

2

14

172192

=×>

>××

>×××××

==

<−=

−

−

−

−

−

a

a

aNqaV

VV

s

DP

PT

ε

66

8. From Eq.33 we obtain

DbiG

P

p

Pm V

VVV

VI

g 2 2 +

=

= Dbi

Psn VVV

aLZ

εµ

62.41085.84.122

)103(107106.12 14

2516192

=×××

×××××==−

−−

s

DP

aqNV

ε V

mg ∴ = 184.062.4

105.1103.01085.84.124500105

44

144

××××

×××××−−

−−

= 1.28×10-3 S = 1.28 mS.

9. (a) The built-in voltage is

×−=−= 17

17

10107.4

ln025.09.0nBnbi VV φ = 0.86 V

At zero bias, the width of the depletion layer is

2

D

bis

qNV

Wε

= = 1719

12

10106.186.01009.12

×××××

−

−

= 1.07 × 10-5 cm

= 0.107 µm

Since W is smaller than 0.2 µm, it is a depletion-mode device.

(b) The pinch-off voltage is

14

517192

1085.84.122)102(10106.1

2 −

−−

××××××==

s

DP

aqNV

ε = 2.92 V

and the threshold voltage is

VT = Vbi – VP = 0.86 – 2.92 = –2.06 V.

10. From Eq.31b, the pinch-off voltage is

67

14

517192

1085.84.122)102(10106.1

2 −

−−

××××××==

s

DP

aqNV

ε= 0.364 V

The threshold voltage is

VT = Vbi – Vp = 0.8 – 0.364 = 0.436 V

and the saturation current is given by Eq. 39

( )2

2 TGns

Dsat VVaL

ZI −=

µε

= 42

44

144

107.4)436.00()101()105.0(245001085.84.121050 −

−−

−−

×=−××××

×××××A.

11. 0.85 – 0.0259ln 2

14

1917

1085.84.122106.1107.4

aN

ND

D−

−

×××××

−

× = 0

For ND = 4.7 × 1016 cm-3

a = DD NN

)107.3(107.4ln0259.085.0

321

17 ×

×−

= 1.52 ×10-5 cm = 0.152 µm

For ND = 4.7 × 1017 cm-3

a = 0.496 × 10-5 cm = 0.0496 µm.

12. From Eq.48 the pinch-off voltage is

TC

BnP VqE

V −∆

−= φ

= 0.89 – 0.23 – (–0.5)

= 0.62 V

and then,

14

21

181921

1085.83.122103106.1

2 −

−

×××××××== ddqN

Vs

DP ε

= 0.62 V

68

d1 = 1.68 × 10-6 cm

d1= 16.8 nm

Therefore, this thickness of the doped AlGaAs layer is 16.8 nm.

13. The pinch-off voltage is

4

7181921

1085.83.122)1050(10106.1

2 −

−−

×××××××==

s

DP

dqNV

ε = 1.84 V

The threshold voltage is

PC

BnT VqE

V −∆−= φ

= 0.89 – 0.23 – 1.84

= –1.18 V

When ns = 1.25× 1012 cm-2, we obtain

[ ] 12

70

19

14

1025.1)18.1(010)850(106.1

1085.83.12 ×=−−××++××

××=−−

−

dns

and then

d0 +58.5 = 64.3

d0 = 5.8 nm

The thickness of the undoped spacer is 5.8 nm.

14. The pinch-off voltage is

14

27171921

1085.83.122)1050(105106.1

2 −

−−

××××××××==

s

DP

dqNV

ε= 1.84 V

The barrier height is

79.084.125.03.1 =++−=+∆+= PC

TBn VqE

Vφ V

69

The 2DEG concentration is

[ ] 12

719

14

1029.1)3.1(010)81050(106.1

1085.83.12 ×=−−××++××

××=−−

−

sn cm-2.

15. The pinch-off voltage is

4

21

181921

1085.83.122101106.1

5.12 −

−

×××××××=== ddqN

Vs

DP ε

The thickness of the doped AlGaAs is

1819

14

1 10106.11085.83.1225.1

××××××= −

−

d = 4.45×10-8 cm = 44.5 nm

93.05.123.08.0 −=−−=−∆

−= PC

BnT VqE

V φ V.

16. The pinch-off voltage is

212

dqN

Vs

DP ε

=

= ( )27

14

1819

10351085.83.122103106.1 −

−

−

×××××××

= 2.7 V

the threshold voltage is

pC

BnT VqE

V −∆−= φ

= 0.89 – 0.24 – 2.7

= –2.05 V

Therefore, the two-dimensional electron gas is

[ ] 12

719

14

102.3)05.2(010)835(106.1

1085.83.12 ×=−−××+××

××=−−

−

sn cm-2.

70

CHAPTER 8

1. Z0 =CL

nH 25.1175102 21220 =××=⋅= −ZCL .

2. 222

2

+

+

=

dp

bn

amc

f r

( ) ( )

. GHz 616.11162

m/s103

40102

8

22

=××=

++= c

3. pngbi VVqEV ++≅ )/( = 1.42+0.03+0.03 = 1.48 V

( )

( )

. F/cm 1013.61089.11016.1

nm 18.9 cm1089.1

25.048.110101010

106.11016.12

2

276

12

6

1919

1919

19

12

−−

−

−

−

−

×=××==

=×=

−

×+

×××=

+

+=

WC

VVNNNN

qW

S

biDA

DAS

ε

ε

4. exp1exp 0

+

−

=

kTqV

IVV

VV

IIPP

p

From Fig. 4, We note that the largest negative differential resistance occurs between Vp<V<VV.

The corresponding voltage can be obtained from the condition d2I/d2V = 0. By neglecting the

second term in Eq. 5, we obtain

71

( )

. 2.27

0367.0 1.02.0

1exp 1.0

2.0101.0

10

V2.01.022

01 exp 2

1 exp

1

2.0

2

22

2.0

322

2

2

Ω−=

=

−=

−

×−=

=×==∴

=

−

+−=

−

−≅

−

−−

V

V

P

PP

P

P

P

PP

P

P

P

dVdI

R

dVdI

VV

VV

VVI

VI

dVId

VV

VVI

VI

dVdI

5. (a) RSC = SS

W

SS

W

vAW

dxvA

IxI

dxI εε 2

11 2

0

0 ==∆ ∫∫ E

( ) Ω=×××

×=−−

−

137101005.1 )105(2

1012

7124

24

(b) The breakdown voltage for ND = 1015 cm-3 and W = 12 µm is 250 V (Refer to Chapter 4).

The voltage due to RSC is

( ) =×××= − 13710510 43SCIR 68.5 V

The total applied voltage is then 250 + 68.5 = 318.5 V.

6. (a) The dc input power is 100V(10-1A) = 10W. For 25% efficiency, the power dissipated as

heat is 10W(1-25%) = 7.5 W.

C75)C/W10(W5.7 ooT =×=∆

(b) =×=∆ C75)C/mV60( ooBV 4.5 V

The breakdown voltage at room temperature is (100-4.5) = 95.5V .

7. (a) For a uniform breakdown in the avalanche region, the maximum electric field is

Em = 4.4 ×105 V/cm. The total voltage at breakdown across the diode is

72

( )

( ) ( )

V 8.742.576.17

100.4-3 1009.1

105.1106.1104.4104.0104.4

4

12

1219545

=+=

×

×

×××−×+××=

−

−+=

−−

−−

AS

mAmB xWqQ

xVε

EE

(b) The average field in the drift region is

( ) V/cm102.2104.03

2.57 5

4×=

×− −

This field is high enough to maintain velocity saturation in the drift region.

(c) ( ) ( ) . GHz 19104.032

10

2 4

7

=−

=−

=−

A

S

xW

vf

8. (a) In the p layer

E1(x) = s

m

xqNε

1−E 0 bx ≤≤ = 3 µm

E2(x) =s

m

bqNε

1−E b Wx ≤≤ = 12 µm

E2(x) should be larger than 105 V/cm for velocity saturation

51 10≥−∴s

m

bqNε

E

or Em 11515 1066.41010 −×+=+≥s

bqNε

N1.

This equation coupled the plot of Em versus N in Chapter 3 gives

N1 = 7 × 1015 cm-3 for Em = 4.2 × 105 V/cm

2

EE bV m

B

)( 2−=∴ +E2W = 45

45

109102

10310)12.4( −−

××+×××−

73

= 138 V

(b) Transit time t = 7

4

1010)312(

−×−=−

svbW

= 9×10-15 = 90 ps .

9. (a) For transit-time mode, we require n0L ≥ 1012 cm-2.

316412120 cm10101/10/10 −− =×=≈ Ln

(b) t = L/v = 10-4 / 107 = 10-11 s = 10 ps

(c) The threshold field for InP is 10.5 kV/cm; the corresponding applied voltage is

( ) V 525.01012

105.10 43

=×

×= −V

The current is

( ) ( ) 86.3101025.5104600106.1 4316190 =××××××=== −−AnqJAI n Eµ A

The power dissipated in the device is then

W 02.2== IVP .

10. (a) Referring to Chapter 2, we have

319172

3

0

017

23

23

2

cm1033.371107.407.02.1

107.4

22

−×=××=

×=

=

=

mm

m

mN

kTmN

nL

nUCL

nUCU

h

π

(b) For Te = 300 K

( )4104.4

97.11 exp710259.0

eV31.0exp71)/exp(

−×=

−×=

−×=∆− e

CL

CU kTENN

(c) For Te = 1500 K

74

( ) ( )

5.6

394.2 exp71300/1500 0259.0

eV31.0exp71)/exp(

=

−×=

−×=∆− e

CL

CU kTENN

Therefore, at Te = 300 K most electrons are in the lower valley. However, at Te = 1500 K ,

87%, i.e., 6.5/(6.5+1), of the electrons are in the upper valley.

11. The energy En for infinitely deep quantum well is

2

2*

2

8n

Lm

hEn =

( )( )LE

Lmnh

LEn 13

*

22 2 2

8=−=

∆∆ −

LLE

En ∆=∆ 12

meV 31 =∆∴ E , =∆ 2E 11 meV .

12. From Fig. 14 we find that the first excited energy is at 280 meV and the width is 0.8 meV. For

same energy but a width of 8 meV, we use the same well thickness of 6.78 nm for GaAs, but

the barrier thickness must be reduced to 1.25 nm for AlAs.

The resonant-tunneling current is related to the integrated flux of electrons whose energy is in

the range where the transmission coefficient is large. Therefore, the current is proportional to

the width nE∆ , and sufficiently thin barriers are required to achieve a high current density.

75

CHAPTER 9

1. Eq.9)(From eV 2.071.24/0.6)m6.0( ==µhv

α (0.6 µm) = 3×104 cm-1

The net incident power on the sample is the total incident power minus the reflected

power, or 10 mW.

( ) 31032 1051104 −×−− ×=− We

W = 0.231 µm.

The portion of each photon’s energy that is converted to heat is

%4.3107.2

42.107.2 =−=−ν

ν

h

Eh g

The amount of thermal energy dissipated to the lattice per second is

31.4%×5 = 1.57 mW.

2. For λ = 0.898 µm, the corresponding photon energy is

eV38.124.1 ==λ

E

From Fig. 18, we obtain 2n (Al0.3Ga0.7As) = 3.38

( )( )

( )[ ]( )

. %15.30315.0

38.31

2117cos138.314cos14

.21172958.038.31

sin

2221

21

2

1

==

+′−××=

+−

=

′=⇒===

oc

occ

nn

nnEfficiency

nn

θ

θθ

3. From Eq. 15

29601393

13932

..

.R =

+

−=

(a) The mirror loss

76

58.40296.01

ln10300

11ln

14

=

×=

−RL cm-1.

(b) The threshold current reduction is

( ) ( )( )

%.6.3658.4010

90.0296.01

ln103002

158.40

1ln

1

1ln

211

ln1

296.09.0,296.0296.0

4

21

21

=+

⋅×⋅−

=

+

+−

+

≈

===−=

−

RL

RRLRL

RJRRJRJ

th

thth

α

αα

4. From Eq. 10

cc nnnn

θθ sinsin 212

1 ⋅=⇒=

From Eq. 14

( ) ( )( )45 101sin16.3108exp1exp1 −×⋅−⋅×−−=∆−−≅Γ cdnC θ

For θc = 84o n1 = 3.58 Γ1 = 0.794

θc = 78o n2 = 3.52 Γ2 = 0.998 .

5. From Eq. 16 we have

Lnm 2 =λ .

Differentiating the above equation with respect to λ, we obtain

λλλ

dnd

Lmddm

2=+ .

77

Substituting λLn2 for m and letting dm/dλ = - ∆m/∆λ, yield

−

∆=∆∴

=+

∆∆−

λλ

λλ

λλλλ

dnd

nLn

m

dnd

LLnm

12

22

2

and( )

( ) ( ).nm 97.0m107.9

5.258.389.0

130058.32

189.0 42

=µ×=

−

×=∆ −λ

6. From Eq. 22

+

Γ=

RLg

1ln

11α and Eq. 15

2

1

1

+

−=n

nR

R1 = 0.317, R2 = 0.311

( )

( ) 217311.01

ln10100

1100

998.01

78

270317.01

ln10100

1100

794.01

84

4

4

=

×+=

=

×+=

−

−

o

o

g

g

×=

⋅′ − RRL1

ln10100

199.0

1ln

21

4

For R1 = 0.317

50317.01

ln10100

199.0317.0

1ln

21

4 =′⇒

×=

⋅′ −L

L µm

For R2 = 0.311

43.50311.01

ln10100

199.0311.0

1ln

21

242

=′⇒

×=

⋅′ −L

L µm.

7. From Eq. 23a

78

For R = 0.317

2

4cmA 1000

99.0317.01

ln101002

110010 −

−=

×××+×=thJ

and so mA 5105101001000 44 =××××= −−thI

For R = 0.311

β ∝Γ ⇒ β2 = 0.1×0.998/0.764 = 0.13

2

4cmA 766

99.0311.01

ln101002

110066.7 −

−=

×××+×=thJ

and so mA 83.310510100766 44 =××××= −−thI .

8. From the equation, we have for m = 0:

0442 =± oBB LnLn λλλ m (25a)

which can be solved as

+±−±=

2

16164 22o

B

LnLnLn λλ (25b)

There are several variations of ± in this solutions. Take the solution which is the only

practical one, i.e., λB ≈ λo, gives λB = 1.3296 or 1.3304 µm.

m196.04.32

33.1 µ=×

≅d .

9. The threshold current in Fig. 26b is given by

Ith = I0 exp (T/110).

Therefore

( ) 1C0091.0

11011 −

==≡ oth

th dTdI

Iξ .

If T0 = 50 oC, the temperature coefficient becomes

79

( ) 1C02.0

501 −== oξ .

which is larger than that for T0 = 110 oC. Therefore the laser with T0 = 50 oC is worse

for high-temperature operation.

10. (a) ∆I = q (µn + µp) ∆nEA

( )

( )( )( )313

219

3

cm1010126.01017003600106.1

1083.2

−−−

−

=××+×

×=

+∆=−=∆∴

AqI

pairsholeelectronnpn Eµµ

(b) sµτ 1206.231083.2 3

=×=−

(c) ( ) ( ) 9

4

313 105.2

102.110

exp10exp ×=

×

−=−∆=∆ −

−

τtntn cm-3 .

11. From Eq. 33

A55.21055.2

101050001063000

310

85.0

6

4

106

µ=×=

×

⋅×⋅⋅

×

⋅=

−

−

−−

qqI p

and from Eq. 35

Gain 91010

500010630004

10

=×

⋅×⋅==−

−

Ln Eτµ

.

12. From Eq. 36

⋅=

⋅

=

⋅

=

−

qh

Rq

hP

I

h

P

q

I

opt

poptp ννν

η1

The wavelength λ of light is related to its frequency ν by ν = c/λ, where c is the velocity of

light in vacuum. Therefore νh /q = hc / λq and h = 6.625×10-34 J-s, c = 3.0×1010 cm/s, q =

1.6×10–19 coul, 1 eV = 1.6×10–19 J.

80

Therefore, h v/q = 1.24 / λ (µm)

Thus, η = (R×1.24) / λ and R = (ηλ) / 1.24.

13. The electric field in the p-layer is given by

( )s

m

xqNx

ε1

1 −= EE bx ≤≤0 ,

where Em is the maximum field. In the p layer, the field is essentially a constant given by

( ) WxbbqN

xs

m ≤<−= 1

εEE 2

The electric field required to maintain velocity saturation of holes is ~ 105 V/cm.

Therefore

51 10≥−s

m

bqNε

E

or

111515 1066.41010 N

bqN

sm

−×+=+≥ε

E .

From the plot of the critical field versus doping, the corresponding Em are obtained :

N1 = 7×1015 cm-3

Em = 4.2×105 V/cm

The biasing voltage is given by

( ) ( ) ( )

V.138

109102

103102.32

4545

22

=

×+×××=+−

= −−

Wb

V mB E

E E

The transit time is

( )ps. 90109

10

109 117

4

=×=×=−≈ −−

s

bWt

ν

81

14. (a) For a photodiode, only a narrow wavelength range centered at the optical signal

wavelength is important; whereas for a solar cell, high spectral response over a broad

solar wavelength range are required.

(b) Photodiode are small to minimize junction capacitance, while solar cells are large-area

device.

(c) An important figure of merit for photodiodes is the quantum efficiency (number of

electron-hole pairs generated by incident photon), whereas the main concern for solar

cells is the power conversion efficiency (power delivered to the load per incident solar

energy).

15. kTE

p

p

Dn

n

AVCs

geD

ND

NNAqNI −

+=

ττ11

)a(

( )( )( )

( )

( )( )1

A1028.2

1067.51043.2

1055.2

1051

103.9

107.11

1066.21086.2106.12

12

2.431420

12.1

719516

191919

−−=

−=

×=

××=

×

××

+×

××××=

−

−−

−−−

−

kTqVsL

LkTqV

s

kTeV

eIII

IeII

e

e

V 0 0.1 0.2 0.3 0.4 0.5 0.6 0.65 0.7 (V)

)1( −kTqV

s eI0 1.1×10-

75×10-6 2.5×10-

41.1×10-

20.55 26.8 179 1520 (mA)

IL 95 95 95 95 95 94.5 68.2 -84 -1425 (mA)

(b) V 68.01028.2

1095ln0259.0ln

12

3

=

××=

=

−

−

s

LOC I

Iq

kTV

(c) ( ) VIeVIP LkTqV

s −−= 1

82

( )( ) L

kTqVs

LkTqV

skTqV

s

IVeI

IekT

qVIeI

dV

dP

−+≈

−+−==

1

10

( )

V.V

VI

Ie

m

s

LkTqV

640

1

=

+=∴

( )[ ]

mW52

0259.07.241ln0259.068.01095

1ln

3

=

−+−×=

−

+−==

−

qkT

kTqV

qkT

VIVIP nOCLmmm

0

50

100

0 0.2 0.4 0.6 0.8

V (V)

I (m

A)

16. From Eq. 38 and 39

( ) LkT/qV

s IeII −−= 1 and

≅

+=

s

L

s

Loc I

Iq

kTII

qkT

V ln1ln

A1049323 1002585060 −−− ×=⋅== .eeII ..kTqVLs

02585.01010493.23 VeI ⋅×−= − and P = I V

83

V I P0 3.00 0.00

0.1 3.00 0.300.2 3.00 0.600.3 3.00 0.900.4 3.00 1.200.5 2.94 1.47

0.51 2.91 1.480.52 2.86 1.490.53 2.80 1.480.54 2.71 1.460.55 2.57 1.41

0.6 0.00 0.00

0.00

0.50

1.00

1.50

2.00

0 0.5 1

V (volts)P

(wat

ts)

∴Maximum power output = 1.49 W Fill Factor

.83.0

36.049.1

=×

==≡ocL

m

ocL

mm

VIP

VIVI

FF

17. From Fig. 40

The output powers for Rs = 0 and Rs = 5 Ω can be obtained from the area

P1 (Rs = 0) = 95 mA×0.375V = 35.6 mW,

P2 (Rs = 5 Ω) = 50 mA×0.18V = 9.0 mW

∴For Rs = 0 P1/ P1 = 100%

For Rs = 5 Ω P2/ P1 = 9/35.6 = 25.3%.

18. The efficiencies are 14.2% (1 sun), 16.2% (10-sun), 17.8% (100-sun), and 18.5%

(1000-sun).

Solar cells needed under 1-sun condition

84

( ) ( )( ) ( )

sun.-1000forcells13001%2.1410%5.18

sun-100forcells1251%2.1410%8.17

sun-10for cells4.111%2.1410%2.16

-1-1

ionconcentrationconcentrat

=×

×=

=×

×=

=×

×=

××

=sunPsun

P

in

in

ηη

85

CHAPTER 10

1. C0 = 1017 cm-3

k0(As in Si) = 0.3

CS= k0C0(1 - M/M0)k0-1

= 0.3×1017(1- x)-0.7 = 3×1016/(1 - l/50)0.7

x 0 0.2 0.4 0.6 0.8 0.9

l (cm) 0 10 20 30 40 45

CS (cm-3) 3×1016 3.5×1016 4.28×1016 5.68×1016 1.07×1017 1.5×1017

0

2

4

6

8

10

12

14

16

0 10 20 30 40 50

l (cm)

N D (1016 cm-3)

2. (a) The radius of a silicon atom can be expressed as

Å175.143.583

so

83

=×=

=

r

ar

(b) The numbers of Si atom in its diamond structure are 8.

So the density of silicon atoms is

322

33atoms/cm 100.5

)Å43.5(88 ×===

an

(c) The density of Si is

86

3

23

2223

cm / g 1002.6

10509.28

/1

1002.6/

×

××=

×=

n

Mρ = 2.33 g / cm3.

3. k0 = 0.8 for boron in silicon

M / M0 = 0.5

The density of Si is 2.33 g / cm3.

The acceptor concentration for ρ = 0.01 Ω–cm is 9×1018 cm-3.

The doping concentration CS is given by

1

0

000)1( −−= k

s MM

CkC

Therefore

318

2.0

18

1

0

0

0

cm108.9

)5.01(8.0109

)1( 0

−

−−

×=

−×=

−=

k

s

MM

k

CC

The amount of boron required for a 10 kg charge is

2218 102.4108.9338.2000,10 ×=×× boron atoms

So that

boron g75.0atoms/mole1002.6

atoms102.4g/mole8.10

23

22

=×

×× .

4. (a) The molecular weight of boron is 10.81.

The boron concentration can be given as

318

2

233

atoms/cm1078.9 1.014.30.10

1002.6g81.10/g1041.5

fersilicon wa of volumeatomsboron ofnumber

×=××

×××=

=

−

bn

(b) The average occupied volume of everyone boron atoms in the wafer is

87

3

18cm

1078.911×

==bn

V

We assume the volume is a sphere, so the radius of the sphere ( r ) is the

average distance between two boron atoms. Then

cm109.243 7−×==πV

r .

5. The cross-sectional area of the seed is

22

cm 24.0255.0 =

π

The maximum weight that can be supported by the seed equals the product of the

critical yield strength and the seed’s cross-sectional area:

kg 480g108.424.0)102( 56 =×=××

The corresponding weight of a 200-mm-diameter ingot with length l is

m. 56.6cm 656

g 4800002

0.20)g/cm33.2(

2

3

==∴

=

l

lπ

6. We have1

000

0

1/−

−=

k

s MM

kCC

Fractional 0 0.2 0.4 0.6 0.8 1.0solidified

0/CsC 0.05 0.06 0.08 0.12 0.23 ∞

88

7. The segregation coefficient of boron in silicon is 0.72. It is smaller than unity, so the solubility

of B in Si under solid phase is smaller than that of the melt. Therefore, the excess B atoms will

be thrown-off into the melt, then the concentration of B in the melt will be increased. The tail-

end of the crystal is the last to solidify. Therefore, the concentration of B in the tail-end of

grown crystal will be higher than that of seed-end.

8. The reason is that the solubility in the melt is proportional to the temperature, and the

temperature is higher in the center part than at the perimeter. Therefore, the solubility is higher

in the center part, causing a higher impurity concentration there.

9. The segregation coefficient of Ga in Si is 8 ×10-3

From Eq. 18

Lkxs ekCC /

0 )1(1/ −−−=

We have

cm. 24

ln(1.102) 250

105/10511081

ln 1082

/11

ln

1615

3

3-

0

==

××−

×−×

=

−

−=

−

CCk

kL

xs

10. We have from Eq.18

])/exp()1(1[0 LxekekCsC −−−=

So the ratio )]/exp()1(1[0/ LxekekCsC −−−== 1/at 52.0)13.0exp()3.01(1 ==×−•−− Lx= 38.0 at x/L = 2.

0.01

0.11

0.21

0 0.2 0.4 0.6 0.8 1

Fraction Solidified

Cs/

Co

89

11. For the conventionally-doped silicon, the resistivity varies from 120 Ω-cm to 155 Ω-cm. The

corresponding doping concentration varies from 2.5×1013 to 4×1013 cm-3. Therefore the range of

breakdown voltages of p+ - n junctions is given by

V 11600 to7250/109.2)(106.12

)103(1005.1

)(2

171

19

2512

12

=×=××

×××=

≅

−−

−

−

BB

Bcs

B

NN

Nq

VEε

V 4350725011600 =−=∆ BV

%307250/2

±=

∆

∴ BV

For the neutron irradiated silicon, ρ = 148 ± 1.5 Ω-cm. The doping concentration is

3×1013 (±1%). The range of breakdown voltage is

. V 9762 to9570

%)1(103/109.2/103.1 131717

=

±××=×= BB NV

V 19295709762 =−=∆ BV

%19570/2

±=

∆

∴ BV.

12. We have

ls

CCCC

MM

ms

lm

l

s =−−

==b

b

Tat liquid ofweight Tat GaAs ofweight

Therefore, the fraction of liquid remained f can be obtained as following

65.03016

30 =+

≈+

=+

=ls

lMM

Mf

ls

l .

13. From the Fig.11, 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 is increased. Thus the

composition of liquid GaAs always becomes gallium rich.

90

14.

−×=×=−=)300/(

8.88exp105)/eV 3.2exp(105)/exp( 2222

TkTkTENn ss

= K 300 C27at 0cm1023.1 0316 =≈× −−

= K 1173 C900at cm107.6 0312 =× −

= K 1473 C1200at cm107.6 0314 =× − .

15. )2/exp(` kTENNn ff −=

= )300//(7.94242/1.1/8.32722 1007.7101105 TkTeVkTeV eee −−− ××=××××

= 1027.5 17−× at 27oC = 300 K

=2.14×1014 at 900oC = 1173 K.

16. 37 × 4 = 148 chips

In terms of litho-stepper considerations, there are 500 µm space tolerance between the

mask boundary of two dice. We divide the wafer into four symmetrical parts for

convenient dicing, and discard the perimeter parts of the wafer. Usually the quality of the

perimeter parts is the worst due to the edge effects.

91

17. MkT

dvf

dvvf

v

v

av πν

8

0

0 ==∫∫

∞

∞

Where

−

=

kTM

kTM

f2

exp2

4 22

2/3ν

νπ

ν

M: Molecular mass

k: Boltzmann constant = 1.38×10-23 J/k

T: The absolute temperature

ν: Speed of molecular

So that

cm/sec1068.4m/sec 4681067.129

3001038.122 4

27

23

×==××

×××=−

−

πνav .

18. cm)Pain (

66.0P

=λ

92

Pa 104.4150

66.066.0 3−×===∴λ

P .

19. For close-packing arrange, there are 3 pie shaped sections in the equilateral triangle.

Each section corresponds to 1/6 of an atom. Therefore

dd

N s

23

21

61

3

triangle theof area trianglein the contained atoms ofnumber

×

×==

=282 )1068.4(3

2

3

2−×

=d

= 214 atoms/cm 1027.5 × .

20. (a) The pressure at 970°C (=1243K) is 2.9×10-1 Pa for Ga and 13 Pa for As2. The

arrival rate is given by the product of the impringement rate and A/πL2 :

Arrival rate = 2.64×1020

2L

A

MT

Pπ

= 2.64×1020

×

×

× −

2

1

12

5

124372.69

109.2

π = 2.9×1015 Ga molecules/cm2 –s The growth rate is determined by the Ga arrival rate and is given by

(2.9×1015)×2.8/(6×1014) = 13.5 Å/s = 810 Å/min .

(b) The pressure at 700ºC for tin is 2.66×10-6 Pa. The molecular weight is 118.69.

Therefore the arrival rate is

s⋅×=

×

×

××

−210

2

620 cmmolecular/ 1028.2

125

97369.118

1066.21064.2

π

If Sn atoms are fully incorporated and active in the Ga sublattice of GaAs, we have an

dd

93

electron concentration of

. cm 1074.12

1042.4109.21028.2 3-17

22

15

10

×=

×

××

21. The x value is about 0.25, which is obtained from Fig. 26.

22. The lattice constants for InAs, GaAs, Si and Ge are 6.05, 5.65,5.43, and 5.65 Å,

respectively (Appendix F). Therefore, the f value for InAs-GaAs system is

066.005.6)05.665.5( −=−=f

And for Ge-Si system is

. 39.065.5)65.543.5( −=−=f

94

CHAPTER 11

1. From Eq. 11 (with ô=0)

x2+Ax = Bt

From Figs. 6 and 7, we obtain B/A =1.5 µm /hr, B=0.47 µm 2/hr, therefore A=

0.31 µm. The time required to grow 0.45µm oxide is

min 44hr 0.720.45)0.31(0.450.47

1 2 ==×+=+= Ax)(xB1

t 2 .

2. After a window is opened in the oxide for a second oxidation, the rate constants are

B = 0.01 µm 2/hr, A= 0.116 µm (B/A = 6 ×10-2 µm /hr).

If the initial oxide thickness is 20 nm = 0.02 µm for dry oxidation, the value

ofôcan be obtained as followed:

(0.02)2 + 0.166(0.02) = 0.01 (0 +ô)

or

ô= 0.372 hr.

For an oxidation time of 20 min (=1/3 hr), the oxide thickness in the window

area is

x2+ 0.166x = 0.01(0.333+0.372) = 0.007

or

x = 0.0350 µm = 35 nm (gate oxide).

For the field oxide with an original thickness 0.45 µm, the effectiveôis given by

ô= .hr 72.27)45.0166.045.0(01.01

)(1 22 =×+=+ AxxB

x2+ 0.166x = 0.01(0.333+27.72) = 0.28053

or x = 0.4530 µm (an increase of 0.003µm only for the field oxide).

3. x2 + Ax = B )( τ+t

)(4

)2

(2

2 τ+=−+ tBAA

x

95

++=+ )(

4 )

2(

22 τt

BA

BA

x

when t >> τ , t >> B

A4

2

,

then, x2 = Bt

similarly,

when t >> τ , t >> B

A4

2

,

then, x = )( τ+tAB

4. At 980 (=1253K) and 1 atm, B = 8.5×10-3 µm 2/hr, B/A = 4×10-2 µm /hr (from Figs. 6

and 7). Since A a2D/k , B/A = kC0/C1, C0 = 5.2×1016 molecules/cm3 and C1 =

2.2×1022 cm-3 , the diffusion coefficient is given by

.s/cm104.79

hr/m101.79

hr/m 102.5

102.22105.8

222

29-

23

2

16

223

0

1

0

1

×=

×=×××

=

=

⋅==

−

µ

µ

CCB

CC

ABAAk

D

5. (a) For SiNxHy

2.11

NSi ==

x

4 x = 0.83

atomic % 2083.01

100H =

++=

yy

4 y = 0.46

The empirical formula is SiN 0.83H0.46.

(b) ñ= 5× 1028e-33.3×1.2 = 2× 1011 Ù-cm

As the Si/N ratio increases, the resistivity decreases exponentially.

96

tA

3C 01

OTa 52

εε=

AAA 020302ONO

tttC

1εεεεεε

++=

( )tA

32

032ONO 2

Cεε

εεε+

=

( )tA

tA

32

2032101

23 εεεεεεε

+=

6. Set Ta2O5 thickness = 3t, ε1 = 25

SiO2 thickness = t, ε2 = 3.9 Si3N4 thickness = t, ε3 = 7.6, area = A then

( ) ( )37.5

6.79.336.729.325

3

2

C

C

32

321

ONO

OTa 52 =××

×+=+

=εε

εεε.

7. Set BST thickness = 3t, ε1 = 500, area = A1

SiO2 thickness = t, ε2 = 3.9, area = A2

Si3N4 thickness = t, ε3 = 7.6, area = A2

then

.0093.0A

A

2

1 =

8. Let

Ta2O5 thickness = 3t, ε1 = 25 SiO2 thickness = t, ε2 = 3.9 Si3N4 thickness = t, ε3 = 7.6 area = A then

dA

tA 0201

3εεεε =

.468.03

1

2 tt

d ==εε

97

9. The deposition rate can be expressed as

r = r0 exp (-Ea/kT)

where Ea = 0.6 eV for silane-oxygen reaction. Therefore for T1 = 698 K

−== 11

6.0 exp2)()(

211

2

kTkTTrTr

ln 2 =

− 300

698300

0259.0

6.0

2T

4 T2 =1030 K= 757 .

10. We can use energy-enhanced CVD methods such as using a focused energy

source or UV lamp. Another method is to use boron doped P-glass which will

reflow at temperatures less than 900 .

11. Moderately low temperatures are usually used for polysilicon deposition, and

silane decomposition occurs at lower temperatures than that for chloride

reactions. In addition, silane is used for better coverage over amorphous

materials such SiO 2.

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 reactions occur at lower temperatures, resulting in smoother and better

adhering films. Another reason is that the polysilicon will have small grains. The

finer grains are easier to mask and etch to give smooth and uniform edges.

However, for temperatures less than 575 ºC the deposition rate is too low.

13. The flat-band voltage shift is

FBV∆ = 0.5 V ~ 0C

Qot

28

8

14

0 F/cm109.610500

1085.89.3 −−−

−

×=×

××==d

C oxε.

4 Number of fixed oxide charge is

98

2-11

19

80 cm101.2

106.1109.65.05.0

×=×

××=−

−

q

C

To remove these charges, a 450 heat treatment in hydrogen for about 30

minutes is required.

14. 20/0.25 = 80 sqs.

Therefore, the resistance of the metal line is

5×50 = 400 Ω .

15. For TiSi2

30 × 2.37 = 71.1nm For CoSi2

30 × 3.56 = 106.8nm.

16. For TiSi2 :

Advantage: low resistivity

It can reduce native-oxide layers

TiSi2 on the gate electrode is more resistant to high-field-

induced hot-electron degradation.

Disadvantage: bridging effect occurs.

Larger Si consumption during formation of TiSi2

Less thermal stability

For CoSi2 :

Advantage: low resistivity

High temperature stability

No bridging effect

A selective chemical etch exits

Low shear forces

Disadvantage: not a good candidate for polycides

99

Ω×=×××

××==−−

− 3

44

6 102.3103.01028.0

11067.2

AL

R ρ

F..

...

S

TL

d

AC 13

4

64414

109210360

1010110301085893 −−

−−−

×=×

×××××××===

εε

ns 42.0101.2102 133 =×××= −RC

Ω×=×××

××== − 3

44

6 102.3103.01028.0

11067.2

AL

R ρ

F107.81036.0

31103.01085.89.3 13

4

414−

−−

×=×

××××××===STL

dA

Cεε

.

17. (a)

ns 93.0109.2103.2 155 =×××= −RC

(b) Ω×=×××

××==−−

− 3

44

6 102103.01028.0

1107.1

LA

R ρ

F101.21036.0

1103.01085.88.2dA 13

4

414−

−

−−

×=×

×××××===STL

Cεε

(c) We can decrease the RC delay by 55%. Ratio = 093

42.0 = 0.45.

18. (a)

RC = 3.2 ×103 ×8.7 × 10-13 = 2.8 ns.

(b) Ω×=×××

××== − 3

44

6 102103.01028.0

1107.1

AL

R ρ

F103.61036.0

31103.01085.88.2d

13

4

414−

−−

×=×

××××××===STLA

C εε

ns. 5.2107.8102.3

ns 5.2107.8102133

133

=×××=

=×××=−

−

RC

RC

19. (a) The aluminum runner can be considered as two segments connected in series:

20% (or 0.4 mm) of the length is half thickness (0.5 µm) and the remaining

1.6 mm is full thickness (1µm). The total resistance is

××

+×

×=

+= −−−−

−

)105.0(10

04.0

1010

16.0103 4444

6

2

2

1

1

AAR

llρ

= 72 Ù.

100

The limiting current I is given by the maximum allowed current density times

cross-sectional area of the thinner conductor sections:

I = 5×105 A/cm2× (10-4×0.5×10-4) = 2.5×10-3 A = 2.5 mA.

The voltage drop across the whole conductor is then

A 105.272 3−××Ω== RIV = 0.18V.

20.

=

h: height , W : width , t : thickness, assume that the resistivities of the cladding

layer and TiN are much larger than CuA and ρρ l

5.0)1.05.0(7.2

×−=

××= ll

WhR AlAl ρ

)25.0()25.0(7.1

ttWhR CuCu −×−

=×

×= llρ

When CuAl RR =

Then 2)25.0(

7.15.04.0

7.2t−

=×

⇒ t = 0.073 µm = 73 nm .

40 n m

60 n m

AlCu

0.5 µm

0.5 µm

101

CHAPTER 12

1. With reference to Fig. 2 for class 100 clean room we have a total of 3500 particles/m3 withparticle sizes ≥ 0.5 µm

350010021 × = 735 particles/m2 with particle sizes ≥ 1.0 µm

3500100

5.4 × = 157 particles/m2 with particle sizes ≥ 2.0 µm

Therefore, (a) 3500-735 = 2765 particles/m3 between 0.5 and 1 µm(b) 735-157 = 578 particles/m3 between 1 and 2 µm(c) 157 particles/m3 above 2 µm.

2. AD

neY 1

9

1

−

=Π=

A = 50 mm2 = 0.5 cm2

%1.302.1)5.01(1)5.025.0(4)5.01.0(4 ==××= −×−×−×− eeeeY .

3. The available exposure energy in an hour is 0.3 mW2/cm2 × 3600 s =1080 mJ/cm2

For positive resist, the throughput is

wafers/hr71401080 =

For negative resist, the throughput is

r wafers/h0129

1080 = .

4. (a) The resolution of a projection system is given by

178.065.0

m£g193.06.01 =×==

NAklm

λ µm

== 222 )65.0(

m193.05.0

)(µλ

NAkDOF = 0.228 µm

(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 that provide lower k1 and higher k2 for better

resolution and depth of focus.

(c) PSM technique changes k1 to improve resolution.

5. (a) Using resists with high γ value can result in a more vertical profile but throughput

decreases.

(b) Conventional resists can not be used in deep UV lithography process because these resists

have high absorption and require high dose to be exposed in deep UV. This raises the

102

concern of damage to stepper lens, lower exposure speed and reduced throughput.

6. (a) A shaped beam system enables the size and shape of the beam to be varied, thereby

minimizing the number of flashes required for exposing a given area to be patterned.

Therefore, a shaped beam can save time and increase throughput compared to a Gaussian

beam.

(b) We can make alignment marks on wafers using e-beam and etch the exposed marks. We can

then use them to do alignment with e-beam radiation and obtain the signal from these marks

for wafer alignment.

X-ray lithography is a proximity printing lithography. Its accuracy requirement is very high,

therefore alignment is difficult.

(c) X-ray lithography using synchrotron radiation has a high exposure flux so X-ray has better

throughput than e-beam.

7. (a) To avoid the mask damage problem associated with shadow printing, projection printing

exposure tools have been developed to project an image from the mask. With a 1:1

projection printing system is much more difficult to produce defect-free masks than it is

with a 5:1 reduction step-and-repeat system.

(b) It is not possible. The main reason is 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 X-ray lithography

system.

8. As shown in the figure, the profile for each case is a segment of a circle with origin at the

initial mask-film edge. As overetching proceeds the radius of curvature increases so that

the profile tends to a vertical line.

9. (a) 20 sec

0.6 × 20/60 = 0.2 µm…..(100) plane

0.6/16 × 20/60 = 0.0125 µm……..(110) plane

0.6/100 × 20/60 = 0.002 µm…….(111) plane

22.12.025.120 =×−=−= lWWb µm

(b) 40 sec

0.6 × 40/60 = 0.4 µm….(100)plane

103

0.6/16 × 40/60 = 0.025 µm…. (110) plane

0.6/100 × 40/60 = 0.004 µm…..(111) plane

4.025.120 ×−=−= lWWb = 0.93 µm

(c) 60 sec

0.6 ×1 = 0.6 µm….(100)plane

0.6/16 ×1 = 0.0375 µm…. (110) plane

0.6/100 ×1= 0.006 µm…..(111) plane

=×−=−= 6.025.120 lWWb 0.65 µm.

10. Using the data in Prob. 9, the etched pattern profiles on <100>-Si are shown in below.

(a) 20 sec l = 0.012 µm, 5.10 == bWW µm

(b) 40 sec l = 0.025 µm, 5.10 == bWW µm

(c) 60 sec l = 0.0375 µm 5.10 == bWW µm.

11. If we protect the IC chip areas (e.g. with Si3N4 layer) and etch the wafer from the top, thewidth of the bottom surface is

18846252100021 =×+=+= lWW µm

The fraction of surface area that is lost is

221

2 /)( WWW − × 100%=(18842-10002) /18842× 100% = 71.8 %

In terms of the wafer area, we have lost

71.8 % × 2)2/15(π =127 cm2

Another method is to define masking areas on the backside and etch from the back. The widthof each square mask centered with respect of IC chip is given by

6252100021 ×−=−= lWW = 116 µm

Using this method, the fraction of the top surface area that is lost can be negligibly small.

12. 1 Pa = 7.52 m Torr

104

PV = nRT

7.52 /760 × 10-3 = n/V ×0.082 × 273

n/V = 4.42 × 10-7 mole/liter = 4.42 × 10-7 × 6.02 × 1023/1000 =2.7 ×1014 cm-3

mean–free–path

P/105 3−×=λ cm = 5× 10-3 ×1000/ 7.52 = 0.6649 cm = 6649 µm

150Pa = 1128 m Torr

PV = nRT

105

1128/ 760 × 10-3 = n/V × 0.082 × 273

n/V = 6.63 × 10-5 mole/liter = 6.63 ×10-5×6.02×1023/1000 = 4 × 1016 cm-3

mean-free-path

P/105 3−×=λ cm = 5× 10-3 ×1000/1128 = 0.0044 cm = 44 µm.

13. Si Etch Rate (nm/min) = 2.86 × 10-13 × RTE

F

a

eTn−

×× 21

= 2.86 × 10-13 ×3×1015× 298987.1

1048.2

21

3

)298( ××−

× e

= 224.7 nm/min.

14. SiO2 Etch Rate (nm/min) = 0.614× 10-13 ×3×1015× 298987.1

1076.3

21

3

)298( ××−

× e = 5.6 nm/min

Etch selectivity of SiO 2 over Si = 025.07.224

6.5 =

Or etch rate (SiO 2)/etch rate (Si) = 025.086.2614.0 298987.1

)48.276.3(

=× ×+−

e .

15. A three–step process is required for polysilicon gate etching. Step 1 is a nonselective etch

process that is used to remove any native oxide on the polysilicon surface. Step 2 is a high

polysilicon etch rate process which etches polysilicon with an anisotropic etch profile. Step 3 is

a highly selective polysilicon to oxide process which usually has a low polysilicon etch rate.

16. If the etch rate can be controlled to within 10 %, the polysilicon may be etched 10 % longer or

for an equivalent thickness of 40 nm. The selectivity is therefore

40 nm/1 nm = 40.

17. Assuming a 30% overetching, and that the selectivity of Al over the photoresist maintains 3.

The minimum photoresist thickness required is

106

(1+ 30%) × 1 µm/3 = 0.433 µm = 433.3 nm.

18. e

e mqB=ω

31

199

101.9106.1

1045.22−

−

×××=×× B

π

B = 8.75 × 10-2(tesla)

= 875 (gauss).

19. Traditional RIE generates low-density plasma (109 cm-3) with high ion energy. ECR and ICP

generate high-density plasma (1011 to 1012 cm-3) with low ion energy. Advantages of ECR and

ICP are low etch damage, low microloading, low aspect-ratio dependent etching effect, and

simple chemistry. However, ECR and ICP systems are more complicated than traditional RIE

systems.

20. The corrosion reaction requires the presence of moisture to proceed. Therefore, the first line of

defense in controlling corrosion is controlling humidity. Low humidity is essential,. especially

if copper containing alloys are being etched. Second is to remove as much chlorine as possible

from the wafers before the wafers are exposed to air. Finally, gases such as CF4 and SF6 can be

used for fluorine/chlorine exchange reactions and polymeric encapsulation. Thus, Al-Cl bonds

are replaced by Al-F bonds. Whereas Al-Cl bonds will react with ambient moisture and start

the corrosion process , Al-F bonds are very stable and do not react. Furthermore, fluorine will

not catalyze any corrosion reactions.

107

CHAPTER 13

1. Ea(boron) = 3.46 eV, D0 = 0.76 cm2/sec

From Eq. 6, /scm10142.4122310614.8

46.3exp76.0)exp( 215

50−

−×=

××−=−=

kTE

DD a

cm 1073.2180010142.4 615 −− ×=××== DtL

From Eq. 9,

××==

−6

20

1046.5 erfc108.1)

2(erfc)(

xLx

CxC s

If 320 /cmatoms 108.1)0(,0 ×== Cx ; x = 0.05 ×10-4, C(5× 10-6) = 3.6 × 1019

atoms/cm3; x = 0.075 ×10-4 , C(7.5×10-6) = 9.4 ×1018 atoms/cm3; x = 0.1×10-4,

C(10-5) = 1.8 × 1018 atoms/cm3;

x = 0.15× 10-4, C(1.5×10-5) = 1.8× 1016 atoms/cm3.

The m15.0)(erfc 2 1- µ==s

subj C

CDtx

Total amount of dopant introduced = Q(t)

= 141054.52 ×=LCsπ

atoms/cm2.

2. /scm1096.4132310614.8

46.3exp76.0exp 214

50−

−×=

××−=

−

=kT

EDD a

From Eq. 15, 319atoms/cm10342.2),0( ×===Dt

StCCS

π

××=

=

−5

19

10673.2erfc10342.2

2erfc)(

xLx

CxC S

If x = 0, C(0) = 2.342 × 1019 atoms/cm3; x = 0.1×10-4, C(10-5) = 1.41×1019 atoms/cm3;

x = 0.2×10-4, C(2×10-5) = 6.79×1018 atoms/cm3; x = 0.3×10-4, C(3×10-5) = 2.65×1018

108

atoms/cm3;

x = 0.4×10-4, C(4×10-5) = 9.37×1017 atoms/cm3; x = 0.5×10-4, C(5×10-5) = 1.87×1017

atoms/cm3;

x = 0.6×10-4, C(6×10-5) = 3.51×1016 atoms/cm3; x = 0.7×10-4, C(7×10-5) = 7.03×1015

atoms/cm3;

x = 0.8×10-4, C(8×10-5) = 5.62×1014 atoms/cm3.

The m 72.0ln4 µπ

==DtC

SDtx

B

j .

3.

××

×=× −

−

t13

81815

103.2410

exp101101

t = 1573 s = 26 min

For the constant-total-dopant diffusion case, Eq. 15 gives Dt

SCS

π=

2131318 atoms/cm104.31573103.2101 ×=××××= −πS .

4. The process is called the ramping of a diffusion furnace. For the ramp-down situation, the

furnace temperature T is given by

T = T0 - rt

where T0 is the initial temperature and r is the linear ramp rate. The effective Dt product

during a ramp-down time of t1 is given by

dttDDtt

eff )()(1

0 ∫=

In a typical diffusion process, ramping is carried out until the diffusivity is

negligibly small. Thus the upper limit t1 can be taken as infinity:

...)1(111

000

++≈−

=T

rt

TrtTT

109

and

2

0

02

000

0000 exp)(...))(exp(exp...)1(expexp

kT

trETD

kT

trEkT

ED

Trt

kTE

DkTEDD aaaaa

−≈

−−=

++

−=

−=

where D(T0) is the diffusion coefficient at T0. Substituting the above equation into the

expression for the effective Dt product gives

∫∞

=−

≈

0

2

002

0

0 )(exp)( )(a

aeff rE

kTTDdt

kT

trETDDt

Thus the ramp-down process results in an effective additional time equal to kT02/rEa at

the initial diffusion temperature T0.

For phosphorus diffusion in silicon at 1000°C, we have from Fig. 4:

D(T0) = D (1273 K) = 2× 10-14 cm2/s

sKr /417.06020

7731273=

×

−=

Ea = 3.66 eV

Therefore, the effective diffusion time for the ramp-down process is

min5.191)106.166.3(417.0

)1273(1038.119

2232

0 ≈=××

×=

−

−

srE

kT

a

.

5. For low-concentration drive-in diffusion, the diffusion is given by Gaussian distribution.

The surface concentration is then

==

kTE

tDS

DtS

tC a

2exp),0(

0ππ

tCt

kTE

D

SdtdC a ×−=

−

= 5.0

22exp

2/3

0π

110

or t

dt

C

dC×−= 5.0

which means 1% change in diffusion time will induce 0.5% change in surface

concentration.

220 222

expkT

EC

kT

E

kT

E

tD

SdTdC aaa −=

−

=

π

or T

dT

T

dT

T

dT

kT

E

C

dCa ×−=×

×××

××−=×

−=

−

−

9.1612731038.12

106.16.3

2 23

19

which means 1% change in diffusion temperature will cause 16.9% change in

surface concentration.

6. At 1100°C, ni = 6×1018 cm-3. Therefore, the doping profile for a surface concentration

of 4 × 1018 cm-3 is given by the “intrinsic” diffusion process:

=Dt

xCtxC s

2erfc),(

where Cs = 4× 1018 cm-3, t = 3 hr = 10800 s, and D = 5x10-14 cm2/s. The

diffusion length is then

m232.0cm1032.2 5 µ=×= −Dt

The distribution of arsenic is

××=

−5

18

1064.4erfc104)(

xxC

The junction depth can be obtained as follows

×

×= −5

1815

1064.4erfc10410 jx

xj = 1.2× 10-4 cm = 1.2 µm.

7. At 900°C, ni = 2× 1018 cm-3. For a surface concentration of 4×1018 cm-3, given by

111

the “extrinsic” diffusion process

/scm1077.3102104

8.45 216

18

1811731038.1

106.105.4

0

23

19

−××

××−−

×=×××=×= −

−

enn

eDDi

kT

Ea

nm 3.32cm1023.3108001077.36.16.1 616 =×=××== −−Dtx j .

8. Intrinsic diffusion is for dopant concentration lower than the intrinsic carrier

concentration ni at the diffusion temperature. Extrinsic diffusion is for dopant

concentration higher than ni.

9. For impurity in the oxidation process of silicon,

=t coefficein nsegregatio 2SiO inimpurity of ionconcentrat mequilibriu

silicon inimpurity of ionconcentrat mequilibriu .

10. . 006.0500

3105103

13

11

==××=κ

11. 0.5 = 1.1 + qFB/Ci

7

6

14

1045.310

1085.89.3 −−

−

×=××==d

C sox

ε

212

19

7

cm103.16.0106.11045.3 −

−

−

×=×××=BF

1912

2

5

106.1103.1)16.10(

10 −−

×××=tπ

the implant time t = 6.7 s.

12. The ion dose per unit area is

212

2

19

6

ions/cm 1038.2)

210

(

106.1

6051010

×=×

××××

==−

−

πAqIt

AN

From Eq. 25 and Example 3, the peak ion concentration is at x = Rp. Figure. 17

indicates the σp is 20 nm.

112

Therefore, the ion concentration is

317

7

12

cm1074.421020

1038.2

2−

−×=

××=

ππσ p

S.

13. From Fig. 17, the Rp = 230 nm, and σp = 62 nm.

The peak concentration is

320

7

15

cm1029.121062

102

2−

−×=

××=

ππσp

S

From Eq. 25,

−−×=

2

2

2015

2

)(exp1029.110

p

pj Rx

σ

xj = 0.53 µm.

14. Dose per unit area = 211

198

140 cm106.8

106.11025011085.89.3 −−−

−

×=×××

×××=∆

=q

VC

qQ T

From Fig. 17 and Example 3, the peak concentration occurs at 140 nm from the

surface. Also, it is at (140-25) = 115 nm from the Si-SiO 2 interface.

15. The total implanted dose is integrated from Eq. 25

∫∞

×=−=

−+=

−−=

0 2

2

9989.12

)]3.2(erfc2[2

)2

(erfc1122

)(exp

2

SSRSdx

RxSQ

p

p

p

p

p

Tσσπσ

The total dose in silicon is as follows (d = 25 nm):

8991.12

)]87.1(erfc2[2

)2

(erfc1122

)(exp

2

2

2

∫∞

×=−=

−−+=

−−=

dp

p

p

p

p

Si

SSdRSdx

RxSQ

σσπσ

the ratio of dose in the silicon = QSi/QT = 99.6%.

113

16. The projected range is 150 nm (see Fig. 17).

The average nuclear energy loss over the range is 60 eV/nm (Fig. 16).

60× 0.25 = 15 eV (energy loss of boron ion per each lattice plane)

the damage volume = VD = π (2.5 nm)2(150 nm) = 3× 10-18 cm3

total damage layer = 150/0.25 = 600

displaced atom for one layer = 15/15 = 1

damage density = 600/VD = 2×1020 cm-3

2×1020/5.02×1022 = 0.4%.

17. The higher the temperature, the faster defects anneal out. Also, the solubility of

electrically active dopant atoms increases with temperature.

18. ox

t CQ

V 1V 1 ==∆

where Q1 is the additional charge added just below the oxide-semiconductorsurface by ion implantation. COX is a parallel-plate capacitance per unit area

given by d

C sox

ε=

(d is the oxide thickness, sε is the permittivity of the semiconductor)

cm104.0

F/cm1085.89.31V6

14

1 −

−

××××=∆= oxtCVQ = 8.63

2

7

cmC

10−×

19

7

106.11063.8

−

−

××

= 5.4 ×1012 ions/cm2

Total implant dose = %45104.5 12×

= 1.2 × 1013 ions/cm2.

19. The discussion should mention much of Section 13.6. Diffusion from a surface film

avoids problems of channeling. Tilted beams cannot be used because of

shadowing problems. If low energy implantation is used, perhaps with

preamorphization by silicon, then to keep the junctions shallow, RTA is also

114

necessary.

20. From Eq.35

84.022.0

6.04.0erfc

21 =

−=S

Sd

The effectiveness of the photoresist mask is only 16%.

023.022.0

6.01erfc

21 =

−=S

Sd

The effectiveness of the photoresist mask is 97.7%.

21. 510 2

1 −==u

eT

2-u

π

02.3=∴u

d = Rp + 4.27 pσ = 0.53 + 4.27 × 0.093 = 0.927 µm.

115

CHAPTER 14

1. Each U-shape section (refer to the figure) has an area of 2500 µm × 8 µm = 2 ×

104 µm2. Therefore, there are (2500)2/2 ×104 = 312.5 U-shaped section. Each

section contains 2 long lines with 1248 squares each, 4 corner squares, 1 bottom

square, and 2 half squares at the top. Therefore the resistance for each section is

1 kΩ /¡ (1248×2 + 4×0.65 +2) = 2500.6 kΩ

The maximum resistance is then

312.5×2500.6 = 7.81 × 108 Ω = 781 MΩ

2. The area required on the chip is

25

14

127

cm 1035.41085.89.3

)105)(1030( −−

−−

×=××××==

ox

dC

Aε

= 4.35 × 103 µm2 = 66 × 66 µm

116

Refer to Fig.4a and using negative photoresist of all levels

(a) Ion implantation mask (for p+ implantation and gate oxide)

(b) Contact windows (2×10 µm)

(c) Metallization mask (using Al to form ohmic contact in the contact window

and form the MOS capacitor).

Because of the registration errors, an additional 2 µm is incorporated in all critical

dimensions.

117

3. If the space between lines is 2 µm, then there is 4 µm for each turn (i.e., 2×n, for

one turn). Assume there are n turns, from Eq.6, L ≈ µ0n2r ≈ 1.2 × 10-6n2r, where r

can be replaced by 2 × n. Then, we can obtain that n is 13.

118

4. (a) Metal 1, (b) contact hole, (c) Metal 2.

(a) Metal 1,

(b) contact hole,

(c) Metal 2.

119

5. The circuit diagram and device cross-section of a clamped transistor are shown in

(a) and (b), respectively.

6. (a) The undoped polysilicon is used for isolation.

(b) The polysilicon 1 is used as a solid-phase diffusion source to form the

extrinsic base region and the base electrode.

(c) The polysilicon 2 is used as a solid-phase diffusion source to form the emitter region

and the emitter electrode.

7. (a) For 30 keV boron, Rp = 100 nm and ∆Rp = 34 nm. Assuming that Rp and ∆Rp

for boron are the same in Si and SiO 2 the peak concentration is given by

316

7

11

cm 104.9)1034(2

108

2−

−×=

××=

∆ ππ pR

S

120

The amount of boron ions in the silicon is

( )

211

11

2

2

cm1088.7

3402

750erfc2

2108

2erfc2

2

2exp

2

−

∞

×=

×

−×=

∆

−−=

∆

−−

∆= ∫

p

p

dp

p

p

R

dRS

dxR

Rx

R

SqQ

π

Assume that the implanted boron ions form a negative sheet charge near the Si-

SiO2 interface, then

( ) 91.01025/1085.89.3

)1088.7(106.1/

714

1119

=×××

×××=

=∆−−

−

oxT CqQ

qV V

(b) For 80 keV arsenic implantation, Rp = 49 nm and ∆ Rp = 18 nm. The peak arsenic

concentration is 321

7

16

cm 1021.2)1018(

10

2−

−×=

××=

∆ ππ pR

S.

121

8. (a) Because (100)-oriented silicon has lower (~ one tenth) interface-trapped charge and

a lower fixed oxide charge.

(b) If the field oxide is too thin, it may not provide a large enough threshold

voltage for adequate isolation between neighboring MOSFETs.

(c) The typical sheet resistance of heavily doped polysilicon gate is 20 to

30 Ω /¡, which is adequate for MOSFETs with gate lengths larger than 3 µm. For

shorter gates, the sheet resistance of polysilicon is too high and will cause large

RC delays. We can use refractory metals (e.g., Mo) or silicides as the gate

material to reduce the sheet resistance to about 1 Ω /¡.

122

(d) A self-aligned gate can be obtained by first defining the MOS gate structure, then

using the gate electrode as a mask for the source/drain implantation. The self-

aligned gate can minimize parasitic capacitance caused by the source/drain

regions extending underneath the gate electrode (due to diffusion or

misalignment).

(e) P-glass can be used for insulation between conducting layers, for diffusion and

ion implantation masks, and for passivation to protect devices from impurities,

moisture, and scratches.

9. The lower insulator has a dielectric constant ε1/ε0 = 4 and a thickness d1= 10 nm The

upper insulator has a dielectric constant ε2/ε0 = 10 and a thickness d2 = 100 nm. Upon

application of a positive voltage VG to the external gate, electric field E1 and E2 are

established in the d1 and d2 respectively. We have, from Gauss’ law, that ε1E1 = ε2E2

+Q and VG = E1d1 + E2d2

where Q is the stored charge on the floating gate. From these above two equations, we

obtain

( ) ( )212121211 // dd

QddVG

εεεε ++

+=E

QQ

J 5

14

77

1 1026.22.01085.8

10010

104104

10010

101010 ×−=

××

+

+

+

×==

−

−Eσ

(a) If the stored charge does not reduce E1 by a significant amount (i.e., 0.2 >> 2.26×105

Q, we can write

( )∫ −− ×=××=∆≈=

0

861 1051025.02.02.0'

tCtdtQ Eσ

( ) ( ) 565.010100/1085.810

105714

8

2

=×××

×==∆−−

−

CQ

VT V

123

(b) when 0, →∞→ Jt we have ≅×→ 51026.2/2.0Q 8.84×10-7 C.

Then ( ) 98.910/1085.810

1084.8514

7

2

=×××==∆

−−

−

CQ

VT V.

10.

124

11. The oxide capacitance per unit area is given by

7105.32 −×==d

CSiO

ox

εF/cm2

and the maximum current supplied by the device is

( ) ( ) 5105.35.0

521

21 272 ≈−×=−≈ −

TGTGoxDS VVm

mVVC

LW

Iµ

µµ mA

125

and the maximum allowable wire resistance is 0.1 V/5 mA, or 20Ω. Then, the

length of the wire must be

074.0cm107.2

cm10208

28

=−Ω×

×Ω=×≤−

−

ρAreaR

L cm

or 740 µm. This is a long distance compared to most device spacing. When

driving signals between widely spaced logic blocks however, minimum feature

sized lines would not be appropriate.

12.

126

127

13. To solve the short-channel effect of devices.

14. The device performance will be degraded from the boron penetration. There are

methods to reduce this effect: (1) using rapid thermal annealing to reduce the

time at high temperatures, consequently reduces the diffusion of boron, (2) using

nitrided oxide to suppress the boron penetration, since boron can easily combine

with nitrogen and becomes less mobile, (3) making a multi-layer of polysilicon

to trap the boron atoms at the interface of each layer.

15. Total capacitance of the stacked gate structure is :

C =

+×=

+×

1025

5.07

1025

5.07

2

2

1

1

2

2

1

1

ddddεεεε

= 2.12

d9.3

= 2.12

12.29.3=∴d =1.84 nm.

16. Disadvantages of LOCOS: (1) high temperature and long oxidation time cause VT

shift, (2) bird’s beak, (3) not a planar surface, (4) exhibits oxide thinning effect.

Advantages of shallow trench isolation: (1) planar surface, (2) no high

temperature processing and long oxidation time, (3) no oxide thinning effect, (4)

no bird’s beak.

17. For isolation between the metal and the substrate.

18. GaAs lacks of high-quality insulating film.

128

19. (a)

( )

( ) ns. 38.11038.11003.692000

105.01011

1085.89.3105.01

110

914

4

414

8

5

=×=××=

××××××

×××=

=

−−

−

−−

−−

s

dA

AL

RC oxερ

(b) For a polysilicon runner

( )ns 207

s 1007.21003.6910

130 714

4

=

×=×

=

=

−−−

dA

WL

RRC oxsquare ε

Therefore the polysilicon runner’s RC time constant is 150 times larger than the

aluminum runner.

20. When we combine the logic circuits and memory on the chip, we need multiple

supply voltages. For reliability issue, different oxide thicknesses are needed for

different supply voltages.

21. (a) nitrideOTatotal

11152

CCC +=

hence 3.17710

2575

9.3 =+=EOT Å

(b) EOT = 16.7 Å.