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Instructor's Solutions ManualInstructor's Solutions Manual

Instructor's Solutions Manual, 6th Edition

Ben Streetman , University of Texas, Austin Sanjay Banerjee, Publisher: Prentice Hall Copyright: 2006 Format: On-line Supplement; 300 pp ISBN-10: 0131497278 ISBN-13: 9780131497276 Published: 28 Aug 2005

Solid State Electronic Devices: International Edition, 6th Edition

Ben Streetman , University of Texas, Austin Sanjay Banerjee, University of Texas, Austin Publisher: Pearson Higher Education Copyright: 2006 Format: Paper; 608 pp ISBN-10: 0132454793 ISBN-13: 9780132454797

This Solutions Manual accompanies

University of Texas, Austin

Prob. 1.

Label planes.

x y z 2 3 4 1/2 1/3 1/ 6 4 3

x y z 2 4 2 1/2 1/4 1/ 2 1 2

Prob. 1. Calculate densities of Si and GaAs. The atomic weights of Si, Ga, and As are 28.1, 69.7, and 74.9, respectively.

Si: a = 5.43-10"^8 cm, 8 atoms/cell

8 atoms (5.43-10"^8 cm)

3 5 - 1 0^

2 2 ^

3 cm

density =

5-10^22 -L - 2 8. 1 ^

6.02-10^23 mol^1

GaAs: a = 5.65 -10"^8 cm, 4 each Ga, As atoms/cell

(5.65-10'scm)

-=2.22-10^22 ^

density =

2.22-10^22 -V(69.7 + 74.9)

cnr V / mol 6.02-10^23 mol^1

= 5.33 A

Prob. 1.

For InSb, find lattice constant, primitive cell volume, (110) atomic density.

^ = 1. 4 4 A + 1.36A = 2.8A

a=6.47A

a^3 FCC unit cell has 4 lattice points /.volume of primitive cell = — = 61.lK?

area of (110) plane = V2a^2

4.1+2.1 ^/ density of In atoms = — %= —- = - ^ - = 3.37-101 4^ ^ V2a^2 a^2

same number of Sb atoms = 3.37-10^14 cm^1

Prob. 1.

Find density ofsc unit cell.

nearest atom separation = 2 • 2.5A = 5A

number of atoms per cube = 8 • | = 1 atom

5 42 -^- mass of one atom = : — ^ — = 9-10~^24 -§- 6.02-10 ^ f density = * * ™ ' ^ i & r (^) = QQ (5A)^3

Prob. 1.

Draw <110> direction of diamond lattice.

This view is tilted slightly from (110) to show the Y alignment of atoms. The open channels are 1 hexagonal along this direction.

Prob. 1.

Find Na CI density. Na : atomic weight 23g/mol, radius CI': atomic weight 35.5g/mol, radius 1.

unit cell with a = 2.8A by hard sphere approximation

1 atoms , Qg g | 1 atoms. g g g g Vi Na and lA CI atoms per unit cell = - 2 ^ ^ ^ '-^ = 4.86 -10"^23 _-_ 6.02-10^23 ^gf ^

density = 4.86-102 3^ ^ (^) =22^_ ( 2. 8 - 1 0 - ^ ) 3 ^

The hard sphere approximation is comparable with the measured 2.17-^ density.

Prob. 1.

Find packing fraction, B atoms per unit volume, and A atoms per unit area.

AhJ V rt^ A D^ A rt^ ) Note: The atoms are the same size and touch each other by the hard sphere approximation.

radii of A and B atoms are then lA

number of A atoms per unit cell = 8 • __ = 1

4 A number of B atoms per unit cell = 1

volume of atoms per unit cell = l-^-(lA)^3 + l-^-(lA)^3 = f^A^3

volume of unit cell = (4A)^3 = 64A^3

MA (^3) 71 packing fraction = 3. , = — = 0.13 = 13% 64A^3

B atoms volume density = ^- = 1.56-10^22 ~ 64A^3

number of A atoms on (100) plane = 4-1 = 1

A atoms (100) aerial density = 1 a^ ^ ° m^ = 6.25-10^14 -½ (4A)^2

Prob. 1.

Find atoms/cell and nearest neighbor distance for sc, bcc, andfee lattices.

sc: atoms/cell = 8-j = 1

nearest neighbor distance = a

bcc: atoms/cell = 8 - 1 + 1 = 2

nearest neighbor distance = L->/

fee: atoms/cell = 8 - | + 6 - f = 4

nearest neighbor distance = i-V

Prob. 1.

Draw cubes showing four {111} planes andfour {110} planes.

{111} planes

{110} planes

Prob. 1. Calculate densities ofGe andlnP. The atomic weights of Ge, In, and P are 72.6,114.8, and 31, respectively.

Ge: a = 5.66-10"^8 cm, 8 atoms/cell

8 atoms

a (5.66-l(r^8 cm)

= 4.41-10^22 ^ r

density =

4.41-102 2^ ^-72.6-!mol 6.02- (^23) J mol

= 5.32-^

GaAs: a = 5.87 -10~^8 cm, 4 each In, P atoms/cell

(5.87-10"^8 cm)

-=1.98-10^22 ^ r

1.98-10^^-(114.8+31)-^

density = , "* \ „, J-^- = 4. 7 9 ^ 6-02-102 3^ ^

Prob. 1.

Sketch diamond lattice showing four atoms of interpenetratingfee in unit cell..

Full Interpenetrating Lattice Four Interpenetrating Atoms in Unit Cell

Prob. 1.

FindAlSbxAsi-x to lattice match InP and give band gap.

Lattice constants of AlSb, AlAs, and InP are 6.14k, 5.66k, and 5.87A, respectively from Appendix III. Using Vegard's Law,

6.14A-x + 5.66A-(l-x) = 5.87A -> x = 0.

AlSbo.44Aso.56 lattice matches InP and has Eg=1.9eV from Figure 1-13.

Find InxGai-x P to lattice match GaAs and give band gap.

Lattice constant of InP, GaP, and GaAs are 5.87A, 5.45A, and 5.65A, respectively from Appendix IE. Using Vegard's Law,

5.87A-x + 5.45A-(l-x) = 5.65A -> x = 0.

Ino.48Gao.52P lattice matches GaAs and has Eg=2.0eV from Figure 1-13.

Prob. 1.

Find weight of As (kd=0.3) added to lkg Si in Czochralski growth for 10^15 cm'^3 doping.

atomic weight of As = 7 4. 9 ^

1015 -J-

C =kd - ( ^ = 1 0 1 5^ ^ - > C (^) L = ^- = 3.33-10^15 -½

assume As may be neglected for overall melt weight and volume

i ° ° 2 « ® - 429.2cm' Si 2.33A cm 3.33-10^15 -^ • 429.2cm^3 = 1.43-10^18 As atoms cm

1.43-10^18 atoms • 74.9-^mol _ 1 0 1 A - 4 „ A „ - 1 0 1 A-

6. 0 2 - 1 0 2 3^ ^ = 1.8-10^g As = 1.8-10"'kg As

Prob. 2.

(a) Find generic equation for Lyman, Balmer, and Paschen series.

AE:^ Ac ^ mq T 32712 ^^2 ¾^2 ¾^2 " 32n^2 e 02 n 22 h^2 hc mq^4 (n 2 2 - n i^2 ) _ mq^4 (n 2 2 - n i^2 )

Vn 2

2

2

2 2 „ 2 „ 2 T. 2

326^11/112^2 ¾^2 ^

mq

2 7,

1 =

te 0 xv* 8 e 0 z^ n > 2 Az^ -he^ 8s^2 A^3 c tru^n.,^2 -^^2 ) mq^4 8-(8.85 • 10"^12 1)^2 - (6.63 • 1(T^34 J-s)^3 - 2.998 • 10^8

2 2 n > 2 2 2 n 2 - ^

9.1 M0"^31 kg- (1.60- 10"^19 C)^4

2? 2 2 n 2 -n, A = 9.11-10^8 m-

rij n 2 2 n 2 -nx

^ r = 9. 1 l A -

2 2 2 2 n 2 - n (^) : nj=l for Lyman, 2 for Balmer, and 3 for Paschen

(b) Plot wavelength versus nfor Lyman, Balmer, and Paschen series.

LYMAN SERIES n 2 3 4 5

nA 2 4 9 16 25

nA2- 3 8 15 24

nA2/(nA2-1)

911*nA2/(nA2-1) 1215 1025 972 949 LYMAN LIMIT 911A BALMER SERIES n 3 4 5 6 , 7

nA 2

16 25 36 49

nA2- 5 12 21 32 45

4nA2/(nA2^)

9114nA2/(nA2-4: 6559 4859 4338 4100 3968

PASCHEN SERIES n 4 5 6 7 8 9

10

nA 2 16 25 36 49 64 81 100

nA2- 7 16 27 40 55 72 91

9*nA2/(nA2-9)

9119nA2/(nA2-9) 18741 12811 10932 10044 9541 9224 9010 PASCHEN LIMIT 8199A

BALMER LIMIT 3644A

Prob. 2.

Show equation 2-17 corresponds to equation 2-3. That is show c-R = m-q 2-K^2 -h^2 -h

From 2-17 and solution to 2.3,

c »21 =

2.998-10* f

8

m-^

n 2 V

= 3.29-10^15 Hz-' V n^ i n 2 J

From 2-3,

u. c-R^ v n^ i (^) n

= 2.998-10^8 ^. 1.097-10^7 J- •

2 J V^ n^ i^ n

= 3.29-10^15 Hz- 2 J

f

V n^ i n (^) 2 J

Prob. 2. (a) Find Apxfor Ax= Apx-Ax = A _* (^) A p x = J = 6. 6 3 - 1 0 ^ J - s (^) = 5 0 3 4 ( ). (^) 2 5 ^ F x (^) An x (^) 4;r-Ax 471-10-1¾

(b)FindAtforAE=leV. AT. * * A A 4.14-10"^15 eV-s 0 _ (^) 1 A - i (^6) AE-At = — -> At= = =3.30-10 16 s An An-AE 4jr-leV Prob. 2.

Find wavelength of WOeVand 12keVelectrons. Comment on electron microscopes compared to visible light microscopes.

E = i m v 2 -> v = J — V m

X - * = ± = _ * = 6.63-lQ-J.s .E-* = E ^ 4. 9 1. 1 0 - ¥. m p mv V2-E-m ^/2-9.11-10-^ ¾

For lOOeV, X = E_i-4.91-10"^19 P-m = (1 OOeV-1.602-10-19^)^4 -4.91-10"^19 P-m=l.23-10-^10 m =1.

For 12keV, X = E^-4.91-10"^19 Ji-m = (1.2-10^4 eV-1.602-10'^19 ^)"i-4.91-10-^19 P-m=1.12-10"^11 m = 0.112A

The resolution on a visible microscope is dependent on the wavelength of the light which is around 5000A; so, the much smaller electron wavelengths provide much better resolution.

Prob. 2. Show that T is the average lifetime in exponential radioactive decay.

The probability of finding an atom in the stable state at time t is N(t)=N 0 • e"^7. This is analogous to the probability of finding a particle at position x for finding the average.

t) = ^-

Jte'Mt (^). T X Je'dt

This may also be found by mimicking the diffusion length calculation (Equations 4-37 to 4-39).

Prob. 2.

Find the uncertainty in position (Ax) and momentum (Ap).

¥(x,t) = j l. s i n f — __ (^) Q - ^ h and ( f - f d x = l

x) = J V - x - ^ F d x ^ j x o L (^0) L ~ L (x^2 ) = J V - x - ^ c b ^ - j x

0

dx = 0.5L (from problem note)

2 • 2 •sin

7IX dx = 0.28L^2 (from problem note)

Ax = ^(x^2 ) - (x)^2 = 70.28L^2 - (0.5L)^2 = 0.17L

A p > - ^ = 0. 4 7 - ^ 4TI-AX L

Prob. 2. Calculate the first three energy levels for a 10A quantum well with infinite walls.

„ _ n^2 -7i^2 ^^2 _ (6.63-10-^34 )^2 ,-20 „ 2 2-m-L^2 8-9.11-10-31^ -(10-9)^2 Ej = 6.03-10"^20 J = 0.377eV E 2 =4-0.377eV=1.508eV E 3 = 9-0.377eV = 3.393eV

•n^2 = 6.03-10" -n

Prob. 2.

Show schematic of atom with Is 2s 2p and atomic weight 21. Comment on its reactivity.

nucleus with 8 protons and This atom is chemically reactive because 13 neutrons the outer 2p shell is not full. It will tend 2 electrons in 1s to try to add two electrons to that outer shell.

2 electrons in 2s

4 electrons in 2p © = proton = neturon = electron

Chapter 3 Solutions

Prob. 3. Calculate the approximate donor binding energy for GaAs ( e (^) r = 13.2, m* = 0.067 m 0 ).

From Equation 3-8 and Appendix II,

T3 _ m^ I-q^4 _ 0.067-(9.11-10-^31 kg)-(1.6-10"^19 C)^4 8-(e 0 er)^2 -h^2 8-(8.85-10-12^ -13.2)^2 -(6.63-10¾)^2

8.34-10'^22 J = 5.2 meV

Prob. 3.

Plot Fermi function for Ep—leV and show the probability of an occupied state AE above Ep is equal to the probability of an empty state AE below Ep so f(EF + AE) = 1 -f(EF -AE).

use f(E) = 1 1+e

E-EF^ and^ kT=0.0259eV

E(eV)

(E-EF)/kT

-9. -3. -1. -0.

f(E)

occupation probability above EF = f(EF+AE) = — l + e w 1 empty probability below EF = 1 - f(EF- AE) - 1 - 1+e^1 ^

l-f(EF-AE) = l -

-AE » k T (^) 1 1 -AE l + e k T

-AE l + e k T

AE AE = f ( E (^) F + A E ) e k T^ + l l + e k T This shows that the probability of an occupied state AE above Ep is equal to the probability of an empty state AE below EF.

Prob. 3.

Discuss m*for GaAs and GaP. What happens if a Y valley electron moves to the L valley?

From Figure 3.10, the curvature of the V valley is much greater than L or X. Thus T valley electrons have much smaller mass. The light mass V electrons in GaAs (/Xn=8500) have higher mobility than the heavy mass X electrons (/Xn=300) in GaP since /½ is inversely proportional to m*. If light mass electrons in T were transferred to the heavier mass L valley at constant energy, they would slow down. The conductivity would decrease (see discussion in Section 10.3).

Prob. 3.

Find Eg for Si from Figure 3-17.

Inn

T

fornjj andn^ on graph

n;i =3-10^14 — =2-10'^3 ^

ni2 = 10s^ : 4 - 1 0 ^

This result is approximate because the temperature dependences of Nc , N (^) v , and E are neglected.

n; =JN3Ve"2 k T^ -• E =-2kT-ln .n;

In —— =\n.-a.{l- lnn^ n (^) i 2

/

v 2kT, + InV^N;

for Si (see above) -» E = 2k • n„

V T^2 T^ l /

-»• Inn,- +

V A f

V 2 k^ T 2

2kT

lnVNcNv

InV^N,

E. ( 1 P

= 2-8.62-10^14 •

In

J 3-10^14 108

2k

4-10-3^ -2-10^ r -^^^3 l K

T T

= 1.3eV

Prob. 3.

(a) FindNd for Si with 10 cm' boron atoms and a certain number of donors so EF-Et=0.36eV.

n =n,e

kT

n 0 =Nd - N. -> Nd =n 0 +Na = n (^) i e « ' +N£

0.36eV = 1.5-10I0-^T-e00259eV^ + 10^16 -½ = 2.63-10^16 ^ r

(b) Si with 10 cm' In and a certain number of donors has EF-EY=0. 2 6eV. How many In atoms are unionized (i.e.: neutral)?

fraction of Ea states filled = f(Ea) =

E (^) a - E (^) F 1 + e kT^

0.26eV-0.36eV 1 + e - 0259eV

unionizedIn = [l-f(Ea) ]-Nt o = 0. 0 2 1 - 1 0 1 6^ ^ =2.1-10^14 i^ 3 cm

Prob. 3. Show that Equation 3-25 results from Equation 3-15 and Equation 3-19. Find the position of the Fermi level relative to Ei at 300Kfor no=10 cm.

E,

0.347eV

0.55eV

Et

(Ec -Ep) Equation 3-15 -> n 0 =Nc -e kT

E (^) r - E (^) B Ep-E; Ep-E; EE -E: n 0 = N (^) c - e kT^ = N (^) c - e kT^ -e kT^ = n j - e k T^ using 3-21 yields Equation 3-25 a Ep-Ey Equation 3-19 -> p 0 =^ N (^) v • e Ep-Ev

kT 0 •'•''V E (^) f - E (^) v E(-EF E,--E« P o = N^ v - e^ kT^

: (^) N„-e kT , (^) P kT (^) = (^) n d -e^ kT (^) using 3-21 yields Equation 3-25b

16 for Fermi level relative to Ei at 300K for no=10 cm 1 5-10^10 EF-R =0.0259eV-ln -— = 0.347eV 101