Solutions Manual to accompany PRINCIPLES OF ELECTRONIC MATERIALS AND DEVICES

Solutions_Manual_to_accompany__PRINCIPLES_OF_ELECTRONIC_MATERIALS_AND_DEVICES__SECOND_EDITION/Ch1SM.pdfSolutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001) Chapter 1

1.1

Second Edition ( 2001 McGraw-Hill)

Chapter 1

1.1 The covalent bondConsider the H2 molecule in a simple way as two touching H atoms as depicted in Figure 1Q1-1. Doesthis arrangement have a lower energy than two separated H atoms? Suppose that electrons totallycorrelate their motions so that they move to avoid each other as in the snapshot in Figure 1Q1-1. Theradius ro of the hydrogen atom is 0.0529 nm. The electrostatic potential energy PE of two charges Q1and Q2 separated by a distance r is given by Q1Q2/(4or).Using the Virial Theorem stated in Example 1.1 (in textbook) consider the following:

a. Calculate the total electrostatic potential energy (PE) of all the charges when they are arranged asshown in Figure 1Q1-1. In evaluating the PE of the whole collection of charges you must considerall pairs of charges and, at the same time, avoid double counting of interactions between the same pairof charges. The total PE is the sum of the following: electron 1 interacting with the proton at adistance ro on the left, proton at ro on the right, and electron 2 at a distance 2ro + electron 2 interactingwith a proton at ro and another proton at 3ro + two protons, separated by 2ro, interacting with eachother. Is this configuration energetically favorable?

b. Given that in the isolated H-atom the PE is 2 (13.6 eV), calculate the change in PE with respect totwo isolated H-atoms. Using the Virial theorem, find the change in the total energy and hence thecovalent bond energy. How does this compare with the experimental value of 4.51 eV?

Solution

Nucleus

Hydrogen

ro

eNucleus

Hydrogen

roe

2

1

Figure 1Q1-1 A simplified view of the covalent bond in H2: a snap shot at one instant in time. Theelectrons correlate their motions and avoid each other as much as possible.

a Consider the PE of the whole arrangement of charges shown in the figure. In evaluating the PE ofall the charges, we must avoid double counting of interactions between the same pair of charges. The totalPE is the sum of the following:

Electron 1 interacting with the proton at a distance ro on the left, with the proton at ro on theright and with electron 2 at a distance 2ro

+ Electron 2 on the far left interacting with a proton at ro and another proton at 3ro+ Two protons, separated by 2ro, interacting with each other

Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001) Chapter 1

1.2

PEe

r

e

r

e

r

e

r

e

r

e

r

o o o o o o

o o o o

o o

=

+

+

2 2 2

2 2

2

4 4 4 2

4 4 3

4 2

( )

substituting and calculating we find

PE = 1.0176 10-17 J or -63.52 eVThe negative PE for this particular arrangement indicates that this arrangement of charges is indeed

energetically favorable compared with all the charges infinitely separated (PE is then zero).

b The potential energy of an isolated H-atom is 2 13.6 eV or 27.2 eV. The difference between thePE of the H2 molecule and two isolated H-atoms is,

PE = (63.52 eV) 2(27.2) eV = - 9.12 eVWe can write the last expression above as changes in the total energy as

E = = = 12 12 9 12PE ( . )eV 4.56 eV This is the change in the total energy which is negative. The H2 molecule has lower energy than

two H-atoms by 4.56 eV which is the bonding energy. This is very close to the experimental value of 4.51eV. (Note: We used an ro value from quantum mechanics - so the calculation was not totally classical).

1.2 Ionic bonding and NaClThe interaction energy between Na+ and Cl- ions in the NaCl crystal can be written as

E rr r

( ). .= +

4 03 10 697 1028 96

8

where the energy is given in joules per ion pair, and the interionic separation r is in meters. Calculate thebinding energy and the equilibrium ionic separation in the crystal; include the energy involved in electrontransfer from Cl- to Na+. Also estimate the elastic modulus Y of NaCl given that

Yr

d E

dro r ro

=

1

6

2

2

SolutionThe PE curve for NaCl is given by

E rr r

( ). .= +

4 03 10 697 1028 968

where E is the potential energy and r represents interionic separation.

We can differentiate this and set it to zero to find the minimum PE, and consequently the minimum(equilibrium) separation (ro).


1.3

dE r

dro

o

( ) = 0

+ =55 76 110

4 031

10096 9 28 2. .r ro o

isolating ro: ro = 2.81 10-10 m or 2.81

The apparent ionic binding energy (Eb) for the ions alone, in eV, is:

EE r

q r r qbo

o o

= = +

( ) . .4 03 10 6 97 10 128 96

8

Eb =

( ) +

( )

4 03 10

2 81 10

6 97 10

2 81 10

11 602 10

28

10

96

10 8 19

.

.

.

. . m m J/eV

E b = 7.83 eVNote however that this is the energy required to separate the Na+ and Cl- ions in the crystal and

then to take the ions to infinity, that is to break up the crystal into its ions. The actual bond energyinvolves taking the NaCl crystal into its constituent neutral Na and Cl atoms. We have to transfer theelectron from Cl- to Na+. The energy for this transfer, according to Figure 1Q2-1, is -1.5 eV (negativerepresents energy release). Thus the bond energy is 7.83 - 1.5 = 6.33 eV as in Figure 1Q2-1.

Cl Na+

ro = 0.28 nm

6

6

0

6.3

0.28 nm

Potential energy E(r), eV/(ion-pair)

Separation, r

Coh

esiv

e E

nerg

y

1.5 eV

r=

Cl Nar=

Na+Cl

Figure 1Q2-1 Sketch of the potential energy per ion pair in solid NaCl. Zero energy

corresponds to neutral Na and Cl atoms infinitely separated.

If r is defined as a variable representing interionic separation, then Youngs modulus is given by:

Yr

d

dr

dE r

dr=

16

( )

Y r rr

= +

16

8 061

10501 84

11028 3 96 10

. .

Yr r

= 8 364 10 1 1 3433 10 195 1128

4. .


1.4

Substituting the value for equilibrium separation (ro) into this equation (2.81 10-10 m),

Y = 7.54 1010 Pa = 75 GPaThis value is somewhat larger than about 40 GPa in Table 1.2 (in the textbook), but not too far

out.

*1.3 van der Waals bondingBelow 24.5 K, Ne is a crystalline solid with an FCC structure. The interatomic interaction energy peratom can be written as

E rr r

( ) . .=

2 14 45 12 13

6 12

(eV/atom)

where and are constants that depend on the polarizability, the mean dipole moment, and the extent ofoverlap of core electrons. For crystalline Ne, = 3.121 10-3 eV and = 0.274 nm.a. Show that the equilibrium separation between the atoms in an inert gas crystal is given by ro =

(1.090). What is the equilibrium interatomic separation in the Ne crystal?b. Find the bonding energy per atom in solid Ne.

c. Calculate the density of solid Ne (atomic mass = 20.18 g/mol).

Solutiona Let E = potential energy and x = distance variable. The energy E is given by

E xx x

( ) . .=

2 14 45 12 13

6 12

The force F on each atom is given by

F xdE x

dxxx

xx

( )( )

. .= =

2 145 56 86 7

11

2

5

2

F xx x

( ) . .=

2 145 56 86 712

13

6

7

When the atoms are in equilibrium, this net force must be zero. Using ro to denote equilibriumseparation,

F ro( ) = 0

2 145 56 86 7 012

13

6

7 . .

r ro o

=

145 56 86 712

13

6

7. .

r ro o

=


1.5

rro

o

13

7

12

6

145 5686 7

=

.

.

ro = 1.090

For the Ne crystal, = 2.74 10-10 m and = 0.003121 eV. Therefore,

ro = 1.090(2.74 10-10 m) = 2.99 10-10 m for Ne.

b Calculate energy per atom at equilibrium:

E rr ro o o

( ) . .=

2 14 45 12 136 12

E ro( ) . . .

.

= ( ) ( )

2 0 003121 1 602 10

14 45

12 13

19

6

12eV J/eV

2.74 10 m 2.99 10 m

2.74 10 m 2.99 10 m

-10

-10

-10

-10

E (ro) = -4.30 10-21 J or -0.0269 eV

Therefore the bonding energy in solid Ne is 0.027 eV per atom.

c To calculate the density, remember that the unit cell is FCC, and density = (mass of atoms in theunit cell) / (volume of unit cell). There are 4 atoms per FCC unit cell, and the atomic mass of Ne is 20.18g/mol.

a

a

a2R

FCC Unit Cell(a) (b) (c)

a

Figure 1Q3-1

(a) The crystal structure of copper which is face-centered-cubic (FCC). The atoms are positionedat well-defined sites arranged periodically, and there is a long-range order in the crystal.

(b) An FCC unit cell with close-packed spheres.

(c) Reduced-sphere representation of the FCC unit cell.

Examples: Ag, Al, Au, Ca, Cu, -Fe (>912 C), Ni, Pd, Pt, Rh.

Since it is an FCC crystal structure, let a = lattice parameter (side of cubic cell) and R = radius ofatom. The shortest interatomic separation is ro = 2R (atoms in contact means nucleus to nucleus separationis 2R (see Figure 1Q3-1).

R = ro/2

and 2a2 = (4R)2


1.6

a R ro= = = ( )

2 2 2 22

2 2 99 10 10. m

a = 4.228 10-10 mTherefore, the volume (V) of the unit cell is:

V = a3 = (4.228 10-10 m)3 = 7.558 10-29 m3

The mass (m) of 1 Ne atom in grams is the atomic mass (Mat) divided by NA, because NA numberof atoms have a mass of Mat.

m = Mat / NA

m = ( )( )

= 20 18 0 001. . g/mol kg/g

6.022 10 mol3.351 10 kg23 -1

26

There are 4 atoms per unit cell in the FCC cell. The density () can then be found by:

= (4m) / V = [4 (3.351 10-26 kg)] / (7.558 10-29 m3)

= 1774 kg/m3

In g/cm3 this density is:

=( )

( ) =1774100

10003

kg/mcm/m

g/kg3

1.77 g/cm3

The density of solid Ne is 1.77 g cm-3.

1.4 Kinetic Molecular TheoryCalculate the effective (rms) speeds of the He and Ne atoms in the He-Ne gas laser tube at roomtemperature (300 K).

Solution

Gas atoms

Figure 1Q4-1 The gas molecules in the container are in random motion.


1.7

Current regulated HV power supply

Flat mirror (Reflectivity = 0.999) Concave mirror(Reflectivity = 0.985)

He-Ne gas mixtureLaser beam

Very thin tube

Figure 1Q4-2 The He-Ne gas laser.

To find the root mean square velocity (vrms) of He atoms at T = 300 K:

The atomic mass of He is (from Periodic Table) Mat = 4.0 g/mol. Remember that 1 mole has amass of Mat grams. Then one He atom has a mass (m) in kg given by:

mM

Nat

A

= =

( ) = 4 0

6 022 100 001 6 642 1023 1

27. .

. . g/mol

molkg/g kg

From kinetic theory:

12

32

2m v kTrms( ) =

v kTmrms

= =( )( )

( )3 33001.381 10 J K K

6.642 10 kg

-23 -1

-27

vrms = 1368 m/sThe root mean square velocity (vrms) of Ne atoms at T = 300 K can be found using the same

method as above, changing the atomic mass to that of Ne, Mat = 20.18 g/mol. After calculations, the massof one Ne atom is found to be 3.351 10-26 kg, and the root mean square velocity (vrms) of Ne is found tobe vrms = 609 m/s.

*1.5 Vacuum depositionConsider air as composed of nitrogen molecules N2.

a. What is the concentration n (number of molecules per unit volume) of N2 molecules at 1 atm and 27C?

b. Estimate the mean separation between the N2 molecules.

c. Assume each molecule has a finite size that can be represented by a sphere of radius r. Also assumethat l is the mean free path, defined as the mean distance a molecule travels before colliding withanother molecule, as illustrated in Figure 1Q5-1a. If we consider the motion of one N2 molecule,with all the others stationary, it is apparent that if the path of the traveling molecule crosses the cross-sectional area S =(2r)2, there will be a collision. Since l is the mean distance between collisions,there must be at least one stationary molecule within the volume Sl, as shown in Figure 1Q5-1a.Since n is the concentration, we must have n(Sl) = 1 or l =1/(4r2n). However, this must becorrected for the fact that all the molecules are in motion, which only introduces a numerical factor,so that


1.8

l = 1

2 41 2 2/ r nAssuming a radius r of 0.1 nm, calculate the mean free path of N2 molecules between collisions at 27C and 1 atm.

d. Assume that an Au film is to be deposited onto the surface of an Si chip to form metallicinterconnections between various devices. The deposition process is generally carried out in avacuum chamber and involves the condensation of Au atoms from the vapor phase onto the chipsurface. In one procedure, a gold wire is wrapped around a tungsten filament, which is heated bypassing a large current through the filament (analogous to the heating of the filament in a light bulb)as depicted in Figure 1Q5-1b. The Au wire melts and wets the filament, but as the temperature of thefilament increases, the gold evaporates to form a vapor. Au atoms from this vapor then condenseonto the chip surface, to solidify and form the metallic connections. Suppose that the source(filament)-to-substrate (chip) distance L is 10 cm. Unless the mean free path of air molecules is muchlonger than L, collisions between the metal atoms and air molecules will prevent the deposition of theAu onto the chip surface. Taking the mean free path l to be 100L, what should be the pressure insidethe vacuum system? (Assume the same r for Au atoms).

v

S = (2r)2

Any molecule withcenter in S gets hit.

Molecule

Molecule

(a)

Vacuumpump

Vacuum Hotfilament

Evaporatedmetal atoms

Semiconductor

Metal film

(b)

Figure 1Q5-1

(a) A molecule moving with a velocity v travels a mean distance l between collisions. Since thecollision cross-sectional area is S, in the volume Sl there must be at least one molecule.Consequently, n(Sl) = 1.

(b) Vacuum deposition of metal electrodes by thermal evaporation.

SolutionAssume T = 300 K throughout. The radius of the nitrogen molecule (given approximately) is r =

0.1 10-9 m. Also, we know that pressure P = 1 atm = 1.013 105 Pa.Let N = total number of molecules , V = volume and k = Boltzmanns constant. Then:

PV = NkT

(Note: this equation can be derived from the more familiar form of PV = RT, where is the totalnumber of moles, which is equal to N/NA, and R is the gas constant, which is equal to k NA.)

The concentration n (number of molecules per unit volume) is defined as:

n = N/V

Substituting this into the previous equation, the following equation is obtained:

P = nkT

a What is n at 1 atm and T = 27 C + 273 = 300 K?


1.9

n = P/(kT)

n = ( )( )1 013 10

300

5.

Pa

1.381 10 J K K-23 -1

n = 2.45 1025 molecules per m3

b What is the mean separation between the molecules?

Consider a crystal of the material which is a cube of unit volume, each side of unit length as shownin Figure 1Q5-2. To each atom we can attribute a portion of the whole volume which for simplicity is acube of side d. Thus, each atom is considered to occupy a volume of d3.

Length = 1

Length = 1

Length = 1

Each atom has this portionof the whole volume. Thisis a cube of side d. .d

d

d

Interatomic separation = d

Volume of crystal = 1

Figure 1Q5-2 Relationship between interatomic separation

and the number of atoms per unit volume.

The actual or true volume of the atom does not matter. All we need to know is how much volumean atom has around it given all the atoms are identical and that adding all the atomic volumes must give thewhole volume of the crystal.

Suppose that there are n atoms in this crystal. Then n is the atomic concentration, number of atomsper unit volume. Clearly, n atoms make up the crystal so that

n d 3 = Crystal volume = 1

Remember that this is only an approximation. The separation between any two atoms is d. Thus,

d

n

= =( )

1 1

2 45 101

325 33 . m

d = 3.44 10-9 m or 3.4 nm

c Assuming a radius, r, of 0.1 nm, what is the mean free path, l, between collisions?

l =

= ( ) ( )

12 4

1

2 4 0 1 10 2 45 102 9 2 25 3 r n . . m m

l = 2.30 10-7 m or 230 nm

d We need the new mean free path, l = 100L, or 0.1 m 100 (L is the source-to-substrate distance)

l = 10 m


1.10

The new l corresponds to a new concentration n of nitrogen molecules.

=

l1

2 4 2r n

=

=( ) ( )

nr

18

2 18

2

0 1 10 102 9 2 l . m m

= 5.627 1017 m-3

This new concentration of nitrogen molecules requires a new pressure, P:

P = nkT = (5.627 1017 m-3)(1.381 10-23 J K-1)(300 K) = 0.00233 PaIn atmospheres this is:

=

= P 0 002331 013 105

. .

PaPa/atm

2.30 10 atm8

In units of torr this is:

= ( )( ) = P 2 30 10 7608. atm torr/atm 1.75 10 torr5There is an important assumption made, namely that the cross sectional area of the Au atom is

about the same as that of N2 so that the expression for the mean free path need not be modified to accountfor different sizes of Au atoms and N2 molecules. The calculation gives a magnitude that is quite close tothose used in practice , e.g. a pressure of 10-5 torr.

1.6 Heat capacitya. Calculate the heat capacity per mole and per gram of N2 gas, neglecting the vibrations of the

molecule. How does this compare with the experimental value of 0.743 J g-1 K-1?

b. Calculate the heat capacity per mole and per gram of CO2 gas, neglecting the vibrations of themolecule. How does this compare with the experimental value of 0.648 J K-1 g-1? Assume that CO2molecule is linear (O-C-O), so that it has two rotational degrees of freedom.

c. Based on the Dulong-Petit rule, calculate the heat capacity per mole and per gram of solid silver.How does this compare with the experimental value of 0.235 J K-1 g-1?

d. Based on the Dulong-Petit rule, calculate the heat capacity per mole and per gram of the siliconcrystal. How does this compare with the experimental value of 0.71 J K-1 g-1?

Solution

a N2 has 5 degrees of freedom: 3 translational and 2 rotational. Its molar mass is Mat = 2 14.01g/mol = 28.02 g/mol.

Let Cm = heat capacity per mole, Cs = specific heat capacity (heat capacity per gram), and R = gasconstant, then:

Cm = = ( ) = 5252

R 8.315 J K mol-1 -1 20.8 J K mol1 1

Cs = Cm/ Mat = (20.8 J K-1 mol-1)/(28.02 g/mol) = 0.742 J K-1 g-1

This is close to the experimental value.

b CO2 has the linear structure O=C=O. Rotations about the molecular axis have negligible rotationalenergy as the moment of inertia about this axis is negligible. There are therefore 2 rotational degrees of


1.11

freedom. In total there are 5 degrees of freedom: 3 translational and 2 rotational. Its molar mass is Mat = 12.01 + 2 16 = 44.01 g/mol.

Cm = = ( ) = 5252

8 315R . J K mol1 1 20.8 J K mol1 1


This is smaller than the experimental value 0.648 J K-1 g-1, because vibrational energy wasneglected in the 5 degrees of freedom assigned to the CO2 molecule.

c For solid silver, there are 6 degrees of freedom: 3 vibrational KE and 3 elastic PE terms. Its molarmass isMat = 107.87 g/mol.

Cm = = ( ) = 6262

8 315R . J K mol1 1 24.9 J K mol1 1


This is very close to the experimental value.

d For a solid, heat capacity per mole is 3R. The molar mass of Si is Mat = 28.09 g/mol.

Cm = = ( ) = 6262

8 315R . J K mol1 1 24.9 J K mol1 1


The experimental value is substantially less and is due to the failure of classical physics. One hasto consider the quantum nature of the atomic vibrations and also the distribution of vibrational energyamong the atoms. The student is referred to modern physics texts (under heat capacity in the Einsteinmodel and the Debye model of lattice vibrations).

1.7 Thermal expansion

a. If is the thermal expansion coefficient, show that the thermal expansion coefficient for an area is2. Consider an aluminum square sheet of area 1 cm2. If the thermal expansion coefficient of Al atroom temperature (25 C) is about 24 10-6 K-1, at what temperature is the percentage change in thearea +1%?

b. The density of silicon at 25 C is 2.329 g cm-3. Estimate the density of silicon at 1000 C given thatthe thermal expansion coefficient for Si over this temperature range has a mean value of about 3.5 10-6 K-1.

c. The thermal expansion coefficient of Si depends on temperature as,

= ( )[ ]{ } + 3 725 10 1 0 00588 124 5 548 106 10. exp . .T Twhere T is in Kelvins. The change in the density due to a change T in temperature is given by

= = 0 03V T T

By integrating this equation , calculate the density of Si at 1000 C and compare with the result in b.

Solutiona Consider an rectangular area with sides xo and yo. Then at temperature T0,


1.12

A x y0 0 0=

and at temperature T,

A x T y T x y T= +( )[ ] +( )[ ] = +( )0 0 0 0 21 1 1 that is

A x y T To= + + ( )[ ]0 21 2 .We can now use that A x y0 0 0= and neglect the term T( )2 because it is very small in comparison withthe linear term T ( Eg (Chapter 5 in the textbook).

4.4 Compound III-V semiconductorsIndium as an element is a metal. It has a valency of III. Sb as an element is a metal and has a valencyof V. InSb is a semiconductor, with each atom bonding to four neighbors, just like in silicon. Explainhow this is possible and why InSb is a semiconductor and not a metal alloy. (Consider the electronicstructure and sp3 hybridization for each atom.)

Solution

The one s and three p orbitals hybridize to form 4 hyb orbitals. In Sb there are 5 valenceelectrons. One hyb has two paired electrons and 3 hyb have 1 electron each as shown in Figure 4Q4-1.In In there are 3 electrons so one hyb is empty. This empty hyb of In can overlap the full hyb of Sb.The overlapped orbital, the bonding orbital, then has two paired electrons. This is a bond between Inand Sb even though the electrons come from Sb (this type of bonding is called dative bonding). It is abond because the electrons in the overlapped orbital are shared by both Sb and In. The other 3 hyb ofSb can overlap 3 hyb of neighboring In to form "normal bonds". Repeating this in three dimensionsgenerates the InSb crystal where each atom bonds to four neighboring atoms as shown. As all thebonding orbitals are full, the valence band formed from these orbitals is also full. The crystal structure isreminiscent of that of Si, as all the valence electrons are in bonds. Since it is similar to Si, InSb is asemiconductor.

Sb In

In atom (Valency III)Sb atom (Valency V)

Sb In

hyb orbitals

Sb ion core (+5e)

Valenceelectron

hyb orbitals

In ion core (+3e)

Valenceelectron

Sb

Sb Sb

Sb Sb

Sb Sb

In

In

In In

In In

In

Figure 4Q4-1 Bonding structure of InSb.


4.6

4.5 Compound II-VI semiconductorsCdTe is a semiconductor, with each atom bonding to four neighbors, just like in silicon. In terms ofcovalent bonding and the positions of Cd and Te in the Periodic Table, explain how this is possible.Would you expect the bonding in CdTe to have more ionic character than that in III-V semiconductors?

Solution

Te CdCd Te

Te CdTeCd

Te CdCd Te

Te CdTeCd

hyb orbitals

Te ion core (+6e)

Valenceelectron

hyb orbitals

Cd ion core (+2e)

Valenceelectron

Dative bonding

Te Cd

Cd atom (Valency II)Te atom (Valency VI)

Figure 4Q5-1 Bonding structure of CdTe.

In CdTe one would expect a mixture of covalent and ionic bonding. Transferring 2 Cd electronsto Te would generate Te2 and Cd2+ which then bond ionically. In covalent bonding we expecthybridization of s and p orbitals. The one s and three p orbitals hybridize to form 4 hyb orbitals. In Tethere are 6 valence electrons. Two hyb have two paired electrons each and two hyb have 1 electron eachas shown. In Cd there are 2 electrons so two hyb are empty. An empty hyb of Cd can overlap a fullhyb of Te. The overlapped orbital, the bonding orbital, then has two paired electrons. This is a bondbetween Cd and Te even though the electrons come from Te (this type of bonding is called dativebonding). It is a bond because the electrons in the overlapped orbital are shared by both Te and Cd. Theother full hyb of Te can similarly overlap another empty hyb of a different neighboring Cd to formanother dative bond. The half occupied orbitals (two on Te and two on Cd) overlap and form "normalbonds". Repeating two dative bonds and two normal bonds in three dimensions generates the CdTecrystal where each atom bonds to four neighboring atoms as shown. Since all the bonding orbitals arefull, the valence band formed from these orbitals is also full. Strictly the bonding is neither fullycovalent nor fully ionic but a mixture.

*4.6 Density of states for a two-dimensional electron gasConsider a two-dimensional electron gas in which the electrons are restricted to move freely within asquare area a2 in the xy plane. Following the procedure in Section 4.5 (in the textbook), show that thedensity of states g(E) is constant (independent of energy).


4.7

SolutionFor a two dimensional electron gas confined within a square region of sides a we have:

Eh

m an n

e

= +( )2

2 12

22

8

Only positive n1 and n2 are allowed. Each n1 and n2 combination is an orbital state. Define a newvariable n as:

n2 = n12 + n2

2

substitute: Eh

m an

e

=2

22

8

Let us consider how many states there are with energies less than E. E corresponds to n n.

= E hm a

ne

2

22

8

= n a m Eh

e82

2

n1

n2

n1

n2

n12 + n

22 = n'2

n1 = 1

n2 = 3

0

1

2

3

4

5

1 2 3 4 5 6

n1 = 2, n

2 = 2

Figure 4Q6-1 Each state, or electron wavefunction in the crystal, can be

represented by a box at n1, n2.

Consider Figure 4Q6-1. All states within the quarter arc defined by n have E < E. The area ofthis quarter arc is the total number of orbital states. The total number of states, S, including spin is twiceas many,

S na m E

he=

=

2

14

214

822

2

2

S a m Eh

e= 42

2

The density of states g is defined as the number of states per unit area per unit energy.Therefore,


4.8

g =

= =1 1 4 42 2

2

2 2a

dS

dE a

a m

h

m

he e

Thus, for a two dimensional gas, the density of states is constant.

4.7 Fermi energy of CuThe Fermi energy of electrons in copper at room temperature is 7.0 eV. The electron drift mobility incopper, from Hall effect measurements, is 33 cm2 V1 s1.

a. What is the speed vF of conduction electrons with energies around EF in copper? By how manytimes is this larger than the average thermal speed vthermal of electrons, if they behaved like an idealgas (Maxwell-Boltzmann statistics)? Why is vF much larger than vthermal?

b. What is the De Broglie wavelength of these electrons? Will the electrons get diffracted by the latticeplanes in copper, given that interplanar separation in Cu = 2.09 ? (Solution guide: Diffraction ofwaves occurs when 2dsin = , which is the Bragg condition. Find the relationship between andd that results in sin > 1 and hence no diffraction.)

c. Calculate the mean free path of electrons at EF and comment.

Solutiona The Fermi speed vF is given by:

E m vF e F=12

2

v EmF

F

e

= =( ) ( )

( )

2 27 1 602 10

9 109 10

19

31

.

.

eV J/eV

kg

vF = 1.57 106 m/s

Maxwell - Boltzmann statistics predicts an effective velocity (rms velocity) which is calledthermal velocity given by (assume room temperature T = 20 C = 293 K):

12

32

2m v kTe thermal =

v Tkme

thermal

K J/K

kg= =

( ) ( )( )

3 3 293 1 381 10

9 109 10

23

31

.

.

v thermal = 1.15 105 m/s

Comparing the two values:

Ratio = vF/vthermal = 13.7

vF is about 14 times greater than vthermal. This is because vthermal assumes that electrons do notinteract and obey Maxwell-Boltzmann statistics (Eav = 3/2kT). However, in a metal there are manyconduction electrons. They interact with the metal ions and obey the Pauli exclusion principle, i.e.Fermi-Dirac statistics. They extend to higher energies to avoid each other and thereby fulfill the Pauliexclusion principle.

b The De Broglie wavelength is = h/p where p = mevF is the momentum of the electrons.


4.9

= = ( ) ( )

h

m ve F

6 626 10

9 109 10 1 57 10

34

31 6

.

. .

J s

kg m/s

= 4.63 10-10 m or 4.63 The interplanar separation, d, is given as 2.09 . The diffraction condition is:

= 2dsin

sin.

. = =( )( ) =

12

12

4 631 11

d

2.09

Since this is greater than 1, and sin cannot be greater than 1, the electrons will not be diffracted.

c The drift mobility is related to the mean scattering time by:

= eme

= =( ) ( )

m

ee

33 10 9 109 104 2 1 1 31 .

m V s kg

1.602 10 C-19

= 1.876 10-14 sThe mean free path, lF, of electrons with speed, vF is:

lF = vF = (1.57 106 m/s)(1.876 10-14 s) = 2.95 10-8 m or 295

The mean free path of those electrons with effective speeds ve (close to mean speed) can be foundas follows (EF has little change with temperature, therefore EF EFO):

12

35

35

2m v E Ee e FO F= =

v Eme

F

e

= =( ) ( )

( ) =

65

65

7 0 1 602 10

9 109 101 215 10

19

196

. .

. .

eV J/eV

kgm/s

le = ve = (1.215 106 m/s)(1.876 10-14 s) = 2.28 10-8 m or 228

4.8 Fermi energy and electron concentrationConsider the metals in Table 4Q8-1 from groups I, II and III in the Periodic Table. Calculate the Fermienergies at absolute zero, and compare the values with the experimental values. What is yourconclusion?

Table Q4.8-1Metal Group Mat

(g/mol)Density (g cm-3) EF (eV)

[Calculated]EF (eV)

[Experiment]Cu I 63.55 8.96 - 6.5Zn II 65.38 7.14 - 11.0Al III 27 2.70 - 11.8


4.10

SolutionSince Cu is in group I, its valency (G) is also 1. The electron concentration n is then the atomic

concentration multiplied by the group number, or:

n GN D

MA

at

= = ( )( ) ( )

=

1

6 022 10 8 96 10

63 55 108 490 10

23 1 3 3

328 3

. .

. .

mol kg/m

kg/molm

Using Equation 4.22 in the text:

Eh

m

n

qFO e=

22

3

83 1

EFO =( )( )

( )

6 626 10

8 9 109 10

3 8 490 10 11 602 10

34 2

31

28 32

3

19

.

.

.

.

J s

kg

m

J/eV

EFO = 7.04 eVComparing with the experimental value:

% difference 8.31%= =7 04 6 56 5

100. .

. %

eV eVeV

EFO can be calculated for Zn and Al in the same way (remember to take into account the differentvalencies). The values are summarized in the following table and it can be seen that calculated values areclose to experimental values:

Table Q4.8-2 Summarized values for Fermi energy at absolute zero temperature.

Metal n (m-3) ( 1028) EFO (eV)(calculated)

EFO (eV)

(experimental)

% Difference

Cu 8.490 7 . 0 4 6.5 8 . 3 1Zn 13.15 9 . 4 3 11.0 1 4 . 3Al 18.07 1 1 . 7 11.8 0 .847

4.9 Temperature dependence of the Fermi energya . Given that the Fermi energy for Cu is 7.0 eV at absolute zero, calculate the EF at 300 K. What is

the percentage change in EF and what is your conclusion?

b. Given the Fermi energy for Cu at absolute zero, calculate the average energy and mean speed perconduction electron at absolute zero and 300 K, and comment.

Solutiona The Fermi energy in eV at 0 K is given as 7.0 eV. The temperature dependence of EF is given byEquation 4.23 in the textbook. Remember that EFO is given in eV.

E EkT

EF FO FO=

1 12

2 2


4.11

EF = ( ) ( )( )

( ) ( )

7 0 1 12

1 381 10 300

7 0 1 602 10

2 23

19

2

. .

. . eV

J/K K

eV J/eV

= 6.999921 eV

% difference 0.00129%= =6 999921 7 07 0

100. .

. %

eV eVeV

This is a very small change. The Fermi energy appears to be almost unaffected by temperature.

b The average energy per electron at 0 K is:

Eav(0 K) = 3/5 (EFO) = 4.2 eV

The average energy at 300 K can be calculated from Equation 4.26 (in the textbook):

E T EkT

E qav FO FO( ) = +

35

1512

2 2

Eav ( ) . .

. . 300

35

7 0 1512

1 381 10 300

7 0 1 602 10

2 23

19

2

K eVJ/K K

eV J/eV= ( ) +

( )( )( ) ( )

Eav(300 K) = 4.200236 eVThis is a very small change.

Assume that the mean speed will be close to the effective speed ve. Effective speed at absolutezero is denoted as veo, and is given by:

E q m vav e eo( )012

2K =

veo = =( )( )

( )

20

21 602 10 4 219qE

mav

e

( ) . .

K J/eV eV

9.109 10 kg-31 = 1215446 m/s

At 300 K, the effective speed is ve:

ve = =( )( )

( )

2300

21 602 10 4 20023619qE

mav

e

( ) . .

K J/eV eV

9.109 10 kg-31=1215480 m/s

Comparing the values:

% difference 0.002797%= =1215480 12154461215446

100

%

m/s m/sm/s

The mean speed has increased by a negligible amount (0.003%) from 0 K to 300 K.

Note: For thermal conduction this tiny increase in the velocity is sufficient to transport energyfrom hot regions to cold regions. This very small increase in the velocity also allows the electrons todiffuse from hot to cold regions giving rise to the Seebeck effect.

4.10 X-ray emission spectrum from sodiumStructure of Na atom is [Ne]3s1. Figure 4Q10-1a shows formation of the 3s and 3p energy bands in Naas a function of internuclear separation. Figure 4Q10-1b shows the x-ray emission spectrum (called theL-band) from crystalline sodium in the soft x-ray range as explained in Example 4.6 (in the textbook).


4.12

a. From Figure 4Q10-1, estimate the nearest neighbor equilibrium separation between Na atoms in thecrystal if some electrons in the 3s band spill over into the states in the 3p band.

b. Explain the origin of X-ray emission band in Figure 4Q10-1b and the reason for calling it the L-band.

c. What is the Fermi energy of the electrons in Na from Figure 4Q10-1b ?

d. Taking the valency of Na to be I, what is the expected Fermi energy and how does it compare withthat in (c)?

Solution

25 26 27 28 29 30 31

Photon energy in eVIn

tens

ity o

f em

itted

rad

iatio

n1.510.50

0

5

10

Internuclear distance (nm)

3s

3p

Ene

rgy

(eV

)

(a) (b)

Figure 4Q10-1

(a) Energy band formation in sodium.

(b) L-emission band of X-rays from sodium.

SOURCE: (b) Data extracted from W. M. Cadt and D. H. Tomboulian, Phys. Rev., 59, 1941,p.381).

a As represented in Figure 4Q10-1a, the estimated nearest interatomic separation is about 0.36 0.37 nm.

b When an electron, for some reason, is leaving the closed inner L-shell of an atom, an empty stateis created there. An electron from the energy band of the metal drops into the L-shell to fill the vacancyand emits a soft X-ray photon in this process. The spectrum of this X-ray emission from metal involvesa range of energies, corresponding to transitions from the bottom of the band and from the Fermi level tothe L-shell. So all the X-ray photons emitted from the electrons during their transitions to the L-shellwill have energies laying in that range (band) and because all of them are emitted due to a transition to theL-shell, this X-ray band is called L-band.

c As explained in b, the width of the X-ray band corresponds to the distance from the bottom ofthe energy band to the Fermi level. As shown in Figure 4Q10-1b, the position of the Fermi level withrespect to the bottom of the energy band is approximately 3.2 eV.

d Theoretically, the position of the Fermi level is given by Equation 4.23 (in the textbook). Sincethe temperature dependence of the Fermi level is really very weak, we can neglect it. Then EF(T) = EF0and using Equation 4.22 (in the textbook), we receive

Eh

m

nF

e

=

22

3

83

The electron concentration in Na can be calculated, assuming that each Na atom donates exactlyone electron to the crystal. Taking from Appendix B (in the textbook) the density of Na d (0.97 g cm-3)and its atomic mass Mat (22.99 g mol

-1), we receive


4.13

nd N

MA

at

= =( ) ( )

( )

0 97 6 022 10

22 99

3 23 1

1

. .

.

g cm mol

g mol = 2.54 1022 cm-3 = 2.54 1028 m-3

Thus for the Fermi level is

Eh

m

nF

e

= =

( )( )

( )

22

334 2

31

28 32

3

83 6 626 10

8 9 1 10

3 2 54 10

.

.

. J s

kg

m = 5.05 10-19 J

Or EF = 3.15 eV,

which is very close to the value obtained from the X-ray spectrum.

4.11 Thermoelectric effects and EFConsider a thermocouple pair that consists of gold and aluminum. One junction is at 100 C and theother is at 0 C. A voltmeter (with a very large input resistance) is inserted into the aluminum wire.Use the properties of Au and Al in Table 4Q11-1 below to estimate the emf registered by the voltmeterand identify the positive end.

Table 4Q11-1Au Al

Atomic Mass, Mat, g/mol 197 27.0Density, g cm-3 19.3 2.7Conduction electrons peratom

1 3

x (Mott-Jones index) -1.48 2.78

SolutionAu

Al

100 C 0 C ColdHot

AlV

0

Figure 4Q11-1 The Al-Au thermocouple. The cold end is maintained at 0 C which is the referencetemperature. The other junction is used to sense the temperature. In this example it is heated to 100 C.

We essentially have the arrangement shown above. For each metal there will be a voltage acrossit given by integrating the Seebeck coefficient. From the Mott-Jones equation:

V SdT x k TeE

dTx k

eET T

FOT

T

T

T

FO= = =

2 2 2 2 202

3 600

( )

The emf (VAB) available is the difference in V for the two metals labeled A (= Al) and B (= Au)so that

V V VAB A B=


4.14

where in this example, T = 373 K and T0 = 273 K. We can calculate EFAO and EFBO for each metal as inExample 4.9 (in the textbook) by using

Eh

m

nFO

e

=

2 2 3

83

/

where n is the electron concentration:

n = atomic concentration (nat) number of conduction electrons per atomFrom the density d and atomic mass Mat, the atomic concentration of Al is:

nN d

MA

atAl

-- = =

6. mol kg/m

. kg/mol= . m

022 10 2700

0 0276 022 10

23 1 328 3

( )( )( )

so that n = 3nAl = 1.807 1029 m-3

which leads to: Eh

m

nFAO

e

= =

( )( )

( )

2 2 3 342

31

29 2 3

83 6 10

8 9 109 10

3 1 807 10

//

.

.6. 26

EFAO = 1.867 10-18 J or 11.66 eV

Similarly for Au, we find EFBO = 5.527 eV.

Substituting x and EF values for A (Al) and B (Au) we find,

V k xeE

T TAA

FAO

= ( )2 2

202

6= -188.4 V

and V k xeE

T TBB

FBO

= ( )2 2

202

6= 211.3 V

so that the magnitude of the voltage difference is

|VAB | = |-188.3 V - 211.3 V| = 399.7 V

Al

Au

IHot Cold

Meter

157 V

188 V

Figure 4Q11-2

To find which end is positive, we put in the resistance of the voltmeter and replace each metal byits emf and determine the direction of current flow as in the figure. For the particular circuit shown, thecold connected side of the voltmeter is positive.


4.15

4.12 The thermocouple equation

Although inputting the measured emf for V in the thermocouple equation V = aT + b(T)2 leads to aquadratic equation, which in principle can be solved for T, in general T is related to the measuredemf via

T = a1V + a2V2 + a3V

3 + ...

with the coefficients a1, a2 etc., determined for each pair of TCs. By carrying out a Taylor's expansionof TC equation, find the first two coefficients a1 and a2. Using an emf table for the K-typethermocouple or Figure 4Q12-1, evaluate a1 and a2.

0

10

20

30

40

50

60

70

80

0 200 400 600 800 1000

Temperature (C)

E-Type

J-Type

S-TypeT-Type

K-Type

EMF (mV)

Figure 4Q12-1 Output emf versus temperature (C) for various thermocouples between 0 to 1000 C.

SolutionFrom Example 4.11 (in the textbook), the emf voltage (V) can be expressed:

Vk

e E ET T

FAO FBOo=

( )2 2

2 2

41 1

Since we know that T = T - To, and therefore T = T + To, we can make the followingsubstitution:

Vk

e E ET T T

FAO FBOo o=

+[ ] ( )2 2

2 2

41 1

expanding, Vk T T

eE

k T T

eE

k T

eE

k T

eEo

FAO

o

FBO FAO FBO

= + ( ) ( ) 2 2 2 2 2 2 2 2 2 2

2 2 4 4

factoring, Vk T

e E ET

k

e E ETo

FAO FBO FAO FBO

=

+

( )

2 2 2 22

21 1

41 1


4.16

Upon inspection, it can be seen that this equation is in the form of the thermocouple equation, V= aT + bT2, and therefore we know that the coefficients a and b are equal to:

ak T

e E Eo

FAO FBO

=

2 2

21 1

bk

e E EFAO FBO=

2 2

41 1

Continuing with the thermocouple equation, we can rearrange it as follows to obtain a quadraticequation:

-b(T)2 - aT + V = 0

This is a quadratic equation in T. The solution is

T a a bVb

a

b

a bV a

b= +

= + +

2 242 2

1 42

/

T ab

a

bbV a= + +( )

2 21 4 2

1

2/

T ab

a

b

bV

ann

n

= + +

=

2 2 141 2

20

/

Taylor expansion: T ab

a

b

bV

a

bV

a

bV

a= + +

+

( ) ( )

+

( ) ( ) ( )

+

2 2

112

42

43

42

12

12

2

2 12

12

32

2

3

! !...

T Va

bV

a

b V

a= + +

2

3

2 3

52 ...

The positive root is not used because V = 0 must give T = 0, and with the positive root the a/2bterms will not cancel out. This equation is of the form given in the question,

T = a1V + a2V2 +a3V

3 + ...

such that aa

ab

a1 2 31= = and

aa k T

e E E

eE E

k T E Eo

FAO FBO

FAO FBO

o FBO FAO1 2 2 2 2

1 1

21 1

2= =

=( )

and ab

a

k

e E E

k T

e E E

e E E

k T E EFAO FBO

o

FAO FBO

FAO FBO

o FBO FAO

2 3

2 2

2 23

2 2 2

4 4 3 2

41 1

21 1

2= =

= ( )

From Figure 4Q12-1, at T = 200 C, V = 8 mV, and at T = 800 C, V = 33 mV. Using thesevalues and neglecting the effect of a3, we have two simultaneous equations we can solve:

T = a1V + a2V2

T = 200 C: 200 C = a1(8 mV) + a2(8 mV)2

a1 = (-(64 mV2)a2 + 200 C) / (8 mV)


4.17

T = 800 C: 800 C = a1(33 mV) + a2(33 mV)2

substitute for a1: 800 C = [(-(64 mV2)a2 + 200 C) / (8 mV)](33 mV) + a2(33 mV)

2

isolate a2: a2 = -0.03030 C/mV2

200 C = a1(8 mV) + (-0.03030 C/mV2)(8 mV)2

a1 = 25.24 C/mV

We can check our calculation by calculating T when V = 20 mV and comparing the value weobtain from examining Figure 4Q12-1.

T = (25.24 C/mV)(20 mV) + (-0.03030 C/mV2)(20 mV)2 = 492.7 CThis is close to that shown in Figure 4Q12-1. Obviously to get a more accurate determination,

we need a3 as well but then we must solve a cubic equation to find a1, a2 and a3 or use numericaltechniques.

4.13 Thermionic emission

A vacuum tube is required to have a cathode operating at 800 C and providing an emission(saturation) current of 10 A. What should be the surface area of the cathode for the two materials inTable 4Q13-1? What should be the operating temperature for the Th on W cathode, if it is to have thesame surface area as the oxide-coated cathode?

Table 4Q13-1

Be (A m2 K2) (eV)

Th on W 3 104 2.6Oxide coating 100 1

SolutionOperating temperature T is given as 800 C = 1073 K and emission current I is given as 10 A.

The temperature and current of the tube are related to its area by Equation 4.44 (in the textbook):

JI

AB T

kTes= =

( )

2 exp E

AI

B TkTe

s

= ( )

2 exp E

Assuming there is no assisting field emission, the area needed for Th on W is:

A =( )( ) ( ) ( )

( )( )

10

3 10 10732 6 1 602 10

1 381 10 10734 2 2 2

19

23 1

exp. .

.

A

A m K KeV J/eV

J K K

A = 467 m2 (large tube)For the oxide coating:


4.18

A =

( )( ) ( ) ( )( )( )

10

100 10731 1 602 10

1 381 10 10732 2 2

19

23 1

exp .

.

A

A m K KeV J/eV

J K K

A = 0.00431 m2 (small and practical tube)To find the temperature that the Th on W cathode would have to work at to have the same surface

area as the oxide coated cathode, the area of the oxide cathode can be used in the current equation and thetemperature can be solved for. It is a more difficult equation, but can be solved through graphicalmethods.

I AB TkTe

=

2 exp

The plot of thermionic emission current I versus temperature T is shown below, with A =0.00431 m2, Be = 3 10

-4 A m2 K2, and = (2.6 eV)(1.602 10-19 J/eV).

5

10

15

I (A)

1.6x103 1.7x103 1.8x103

T (K)Figure 4Q13-1 Behavior of current versus temperature for the Th on W cathode.

From the graph, it appears that at 10 A of current the cathode will be operating at a temperature ofT = 1725 K or 1452 C.

4.14 Field-assisted emission in MOS deviceMetal-oxide-semiconductor (MOS) transistors in microelectronics have metal gate on SiO2 insulatinglayer on the surface of doped Si crystal. Consider this as a parallel plate capacitor. Suppose the gate isan Al electrode of area 50 m 50 m and has a voltage of 10 V with respect of the Si crystal.Consider two thicknesses for the SiO2, (a) 100 and (b) 40 , where (1 = 10

-10 m). Theworkfunction of Al is 4.2 eV, but this refers to electron emission in vacuum, whereas in this case, theelectron is emitted into the oxide. Given that the potential energy barrier B between Al and SiO2 isabout 3.1 eV, and the field emission current density in Equation 4.47 (in the textbook) in full is


4.19

Je

hE

E

EE

m

ehfield emission B

cc

e B =

=

( )3 2 3 1 28

8 2

3

exp ;

/

calculate the field emission current for the two cases. For simplicity take me to be the electron mass infree space. What is your conclusion?

SolutionWe can begin the calculation of field emission current finding the values of the field independent

constants Ec, Be

h B=

3

8 and the area A of the Al electrode.

Thus

Em

ehce B=

( )=

( ) ( )[ ]( ) ( )

8 2

3

8 2 9 1 10 3 1 1 602 10

3 1 602 10 6 626 10

3 1 2 31 193

1

2

19 34

/ . . .

. .

kg J

C J s = 3.726 1010 V m-1

Be

h B= =

( )( ) ( )

3 193

34 198

1 602 10

8 6 626 10 3 1 1 602 10 .

. . .

C

J s J = 4.971 10-7 A V- 2

A = ( ) ( ) 50 10 50 106 6 m m = 2.5 10-9 m2When the thickness of SiO2 layer d is 100 , the field in the MOS device is

E = 10

100 10 10 V

m = 1 109 V m-1

and the field emission current is

I AJ ABEE

Efield emissiomc= = =

2 exp

= ( ) ( ) ( ) ( )( )

2 5 10 4 971 10 1 103 726 10

1 109 2 7 2 9 1 2

10 1

9 1. . exp.

m A V V m

V m

V m

= 8.18 10-14 A.In the second case the SiO2 layer is 2.5 times thinner ( 40 ) and the field in the device is 2.5

times stronger.

E = 10

40 10 10

V

m = 2.5 109 V m-1

The current in this case is

I AJ ABEE

Efield emissiomc= = =

2 exp

= ( ) ( ) ( ) ( )( )

2 5 10 4 971 10 2 5 103 726 10

2 5 109 2 7 2 9 1 2

10 1

9 1. . . exp.

. m A V V m

V m

V m

= 2.62 10-3 A.So as predicted by equation 4.47 (in the textbook), the field-assisted emission current is a very

strong function of the electric field.


4.20

4.15 Lattice waves and heat capacity

a. Consider an aluminum sample. The nearest separation 2R (2 atomic radius) between the Al-Alatoms in the crystal is 0.286 nm. Taking a to be 2R, and given the sound velocity in Al as 5100 ms-1, calculate the force constant in Equation 4.66 (in the textbook). Use the group velocity g fromthe actual dispersion relation, Equation 4.55 (in the textbook), to calculate the sound velocity atwavelengths of = 1 mm, 1 m and 1 nm. What is your conclusion?

b. Aluminum has a Debye temperature of 394 K. Calculate its specific heat capacity in the summer at25 C and in winter at -10 C.

c. Calculate the specific heat capacity of a Ge crystal at 25 C and compare it with the experimentalvalue.

Solutiona The group velocity of lattice waves is given by Equation 4.55 (in the textbook). For sufficientlysmall K, or long wavelengths, such that 1/2Ka 0, the expression for the group velocity can besimplified like in Equation 4.66 (in the textbook) to

g a M=

From here we cam calculate the force constant

=

Ma

g2

The mass of one Al atom is

MM

Nat

A

=

and finally for the force constant we receive

=

=( )

( )

M

N aat

A

g2 3 1

23 1

1

9

227 10

6 022 10

51000 286 10

.

.

kg mol

mol

m sm

= 14.26 kg s- 2

Now considering the dispersion relation K = 2

and Equation 4.55 (in the textbook) we receive

g Aat

aN

M

a

( ) =

1

2

cos

Performing the calculations for the given wavelengths, we receive the following results:

g m103( ) = 5100 m s-1

g m106( ) = 5099.998 m s-1

g m109( ) = 3176.22 m s-1


4.21

It is evident that for the first two wavelengths, 1/2Ka 0 and we can use the approximation inEquation 4.66 (in the textbook). For the third wavelength, this is not true and we have to use the exactdispersion relation when calculating the group velocity.

b In summer, the temperature is given to be T = 25 C = 298 K and T/TD is 298/394 = 0.756.From Figure 4Q15-1, the molar heat capacity of Al during the summer is

Cm = 0.92 (3R) = 22.95 J K-1 mol-1

The corresponding specific heat capacity is

cC

Msm

at

= =( )

( )

22 95

27

1 1

1

.

J K mol

g mol = 0.85 J K-1 g-1

In similar way during the winter we have

Cm = 0.898 (3R) = 22.40 J K-1 mol-1 and c

C

Msm

at

= =( )

( )

22 40

27

1 1

1

.

J K mol

g mol = 0.83 J K-1 g-1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0

5

10

15

20

25

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Cm = 3R

CmJ mole-1Cm/(3R)

T / TD

Al -

win

ter

Al -

sum

mer

Figure 4Q15-1 Heat capacity of Al at -10 C and at 25 C.

c We can find the heat capacity of Ge in the way described in b. Alternatively, we can find Cmperforming the integration in Equation 4.64 (in the textbook) numerically

C RT

T

x e

edx R

x e

edx Rm

D

x

x

x

x

T

TD

=

( )

= ( )

= ( )9

13 3

298360 1

3 0 9313 4

2

3 4

2

0

360

298

0

. = 23.22 J K-1 mol-1

Thus the specific heat capacity is:

cC

Msm

at

= =( )

( )

23 22

72 59 10

1 1

3 1

.

.

J K mol

kg mol = 319.9 J K-1 kg-1

The difference between the calculated value and the experimental one is less than 1%.


4.22

4.16 Thermal conductivitya. Given that silicon has Youngs modulus of about 110 GPa density of 2.3 g cm-3, calculate the mean

free path of phonons in Si at room temperature.

b. Diamond has the same crystal structure as Si but has a very large thermal conductivity, about 1000W m-1 K-1 at room temperature. Given that diamond has a specific heat capacity cs of 0.50 J K

-1 g-1,Youngs modulus of 830 GPa, and density of 0.35 g cm-3, calculate the mean free path ofphonons in diamond.

c. GaAs has a thermal conductivity of 200 W m-1 K-1 at 100 K and 80 W m-1 K-1 at 200 K. Calculateits thermal conductivity at 25 C and compare with the experimental value of 44 W m-1 K-1. (Hint:Consider Figure 4Q16-1.)

1

10

100

1000

10000

100000

1 10 100 1000

Temperature (K)

MgO

Sapphire

(W

m-1

K-1

)

Figure 4Q16-1 Thermal conductivity of sapphire and MgO as a function of temperature.

Solution

a Assume room temperature of 25 C (298 K). For this temperature from Table 4.5 (in thetextbook), we can find the thermal conductivity ( = 148 W m-1 K-1) for silicon and its specific heatcapacity Cs (Cs = 0.703 J K

-1 g-1). We can calculate the phonon mean free path at this temperature fromEquation 4.68 (in the textbook):

lph

V phC= 3

where CV is the heat capacity per unit volume. CV can be found from the specific heat capacity

CV = Cs and the phonon velocity can be obtained from Equation 4.67 (in the textbook) - phY .

Thus the phonon mean free path in Si at 25 C is

lph

s sC Y C Y= = =3 3

=( )

( ) ( ) ( )

3 148

0 703 10 2 3 10 110 10

1 1

3 1 1 3 3 9

. .

W m K

J K kg kg m Pa = 3.971 10-8 m


4.23

b The mean free path of phonons in diamond is

lphsC Y

= =( )

( ) ( ) ( )

3 3 1000

0 5 10 3 5 10 830 10

1 1

3 1 1 3 3 9

. .

W m K

J K kg kg m Pa = 1.113 10-7 m

c The temperatures at which the thermal conductivity is given can be considered as relativelyhigh. For this temperature range, we can assume that CV is almost constant and since the phononvelocity is approximately independent from temperature according to Equation 4.68 (in the textbook) thethermal conductivity is proportional to the mean free path of phonons lph . Since the phonon

concentration increases with temperature, nph T, the mean free path decreases as lph T

1 . Thus,

decreases in the same manner with temperature as in Figure 4Q16-1.

We can assume that the temperature dependence of the thermal conductivity is given by:

( )T AT

B= +

Then we have two equations and two unknowns

200100

80200

= +

= +

AB

BB

and for the coefficients A and B we receive: A = 2.4 104 W m-1 and B = -40 W m-1 K-1

The thermal conductivity at 25 C (298 K) is

=

2 4 10298

404 1

1 1.

W mK

W m K = 40.5 W m-1 K-1

which is close to the experimental value.

*4.17 Overlapping bandsConsider Cu and Ni with their density of states as schematically sketched in Figure 4Q17-1. Bothhave overlapping 3d and 4s bands, but the 3d band is very narrow compared to the 4s band. In thecase of Cu the band is full, whereas in Ni, it is only partially filled.

a. In Cu, do the electrons in the 3d band contribute to electrical conduction? Explain.

b. In Ni, do electrons in both bands contribute to conduction? Explain.

c. Do electrons have the same effective mass in the two bands? Explain.

d. Can an electron in the 4s band with energy around EF become scattered into the 3d band as a resultof a scattering process? Consider both metals.

e. Scattering of electrons from the 4s band to the 3d band and vice versa can be viewed as anadditional scattering process. How would you expect the resistivity of Ni to compare with that ofCu, even though Ni has 2 valence electrons and nearly the same density as Cu? In which casewould you expect a stronger temperature dependence for the resistivity?


4.24

4s

3d

E

g(E)

EF

Cu

4s

E

g(E)

Ni

EF

3d

Figure 4Q17-1 Density of states and electron filling in Cu and Ni.

Solutiona In Cu the 3d band is full so the electrons in this band do not contribute to conduction.

b In Ni both the 3d and 4s bands are partially filled so electrons in both bands can gain energy fromthe field and move to higher energy levels. Thus both contribute to electrical conductivity.

c No, because the effective mass depends on how easily the electron can gain energy from the fieldand accelerate or move to higher energy levels. The energy distributions in the two bands are different.In the 4s band, the concentration of states is increasing with energy whereas in the 3d band, it isdecreasing with energy. One would therefore expect different inertial resistances to acceleration,different effective mass and hence different drift mobility for electrons in these bands.

d Not in copper because the 3d band is full and cannot take electrons. In Ni the electrons canindeed be scattered from one band to the other, e.g. an electron in the 4s band can be scattered into the 3dband. Its mobility will then change. Electrons in the 3d band are very sluggish (low drift mobility) andcontribute less to the conductivity.

e Ni should be more resistive because of the additional scattering mechanism from the 4s to the 3dband (Matthiessen's rule). This scattering is called s-d scattering. One may at first think that this s-dscattering de-emphasizes the importance of scattering from lattice vibrations and hence, overall, theresistivity should be less temperature dependent. In reality, electrons in Ni also get scattered by magneticinteractions with Ni ion magnetic moments (Nickel is ferromagnetic; Ch. 8 in the textbook) which has astronger temperature dependence than T.

"After a year's research, one realises that it could have been done in a week."

Sir William Henry Bragg

Solutions_Manual_to_accompany__PRINCIPLES_OF_ELECTRONIC_MATERIALS_AND_DEVICES__SECOND_EDITION/Ch5SM.pdfSolutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001) Chapter 5

5.1

Second Edition ( 2001 McGraw-Hill)

Chapter 5

5.1 Bandgap and photodetectiona. Determine the maximum value of the energy gap that a semiconductor, used as a photoconductor,

can have if it is to be sensitive to yellow light (600 nm).

b. A photodetector whose area is 5 10-2 cm2 is irradiated with yellow light whose intensity is 2 mWcm2. Assuming that each photon generates one electron-hole pair, calculate the number of pairsgenerated per second.

c. From the known energy gap of the semiconductor GaAs (Eg = 1.42 eV), calculate the primarywavelength of photons emitted from this crystal as a result of electron-hole recombination.

d. Is the above wavelength visible?

e. Will a silicon photodetector be sensitive to the radiation from a GaAs laser? Why?

Solution

a We are given the wavelength = 600 nm, therefore we need Eph = h = Eg so that,

Eg = hc/ = (6.626 10-34 J s)(3.0 108 m s-1) / (600 10-9 m)

E g = 3.31 10-19 J or 2.07 eV

b Area A = 5 10-2 cm2 and light intensity Ilight = 2 10-3 W/cm2. The received power is:

P = AIlight = (5 10-2 cm2)(2 10-3 W/cm2) = 1.0 10-4 W

Nph = number of photons arriving per second = P/Eph

Nph = (1.0 10-4 W) / (3.31 10-19 J)

Nph = 3.02 1014 Photons s-1

Since the each photon contributes one electron-hole pair (EHP), the number of EHPs is then:

N EHP = 3.02 1014 EHP s-1

c For GaAs, Eg = 1.42 eV and the corresponding wavelength is

= hc/Eg = (6.626 10-34 J s)(3.0 108 m s-1) / (1.42 eV 1.602 10-19 J/eV)

= 8.74 10-7 m or 874 nmThe wavelength of emitted radiation due to electron-hole pair (EHP) recombination is therefore

874 nm.

d It is not in the visible region (it is in the infrared).

e From Table 5.1 (in the textbook), for Si, Eg = 1.10 eV and the corresponding cut-off wavelengthis,

g = hc/Eg = (6.626 10-34 J s)(3.0 108 m s-1) / (1.1 eV 1.602 10-19 J/eV)


5.2

g = 1.13 10-6 m or 1130 nm

Since the 874 nm wavelength of the GaAs laser is shorter than the cut-off wavelength of 1130nm, the Si photodetector can detect the 874 nm radiation (Put differently, the photon energycorresponding to 874 nm, 1.42 eV, is larger than the Eg, 1.10 eV, of Si which means that the Siphotodetector can indeed detect the 874 nm radiation).

5.2 Minimum conductivity

a. Consider the conductivity of a semiconductor, = ene + eph. Will doping always increase theconductivity?

b. Show that the minimum conductivity for Si is obtained when it is p-type doped such that the holeconcentration is

p nm ie

h

=

and the corresponding minimum conductivity (maximum resistivity) is

min = 2eni e h

c. Calculate pm and min for Si and compare with intrinsic values.

Solutiona Doping does not always increase the conductivity. Suppose that we have an intrinsic samplewith n = p but the hole drift mobility is smaller. If we dope the material very slightly with p-type then p> n. However, this would decrease the conductivity because it would create more holes with lowermobility at the expense of electrons with higher mobility. Obviously with further doping p increasessufficiently to result in the conductivity increasing with the extent of doping.

b To find the minimum conductivity, first consider the mass action law:

np = ni2

isolate n: n = ni2/p

Now substitute for n in the equation for conductivity:

= ene + eph

= +enp

epi e h2

To find the value of p that gives minimum conductivity (pm), differentiate the above equation withrespect to p and set it equal to zero:

d

dp

en

pei e h

= +2

2

+ =enp

ei e

mh

2

20

Isolate pm and simplify: p nm ie

h

=


5.3

Substituting this expression back into the equation for conductivity will give the minimumconductivity:

min

= + = +enp

epen

neni e

mh m

i e

i e hh i

e

h

2 2

min = + = +en en en eni e he

i e h i e h i e h

min = 2eni e h

c From Table 5.1, for Si: e = 1350 cm2 V-1 s-1, h = 450 cm

2 V-1 s-1 and ni = 1.45 1010 cm-3.

Substituting into the equations for pm and min:

p nm ie

h

= = ( )

1 45 10135010 3

2 1 1

2 1 1.

cm cm V s

450 cm V s

pm = 2.51 1010 cm- 3

min = 2eni e h

min . .= ( ) ( ) ( )( ) 2 1 602 10 1 45 10 1350 45019 10 3 2 1 1 2 1 1 C cm cm V s cm V s min = 3.62 10

-6 -1 cm-1

The corresponding maximum resistivity is:

max = 1 / min = 2.76 105 cm

The intrinsic value corresponding to pm is simply ni (= 1.45 1010 cm-3). Comparing it to pm:

p

nm

i

=

=

2 51 101 45 10

1 7310 3

10 3

.

. .

cmcm

The intrinsic conductivity is:

int = eni(e + h)

int = (1.602 10-19 C)(1.45 1010 cm-3)(1350 cm2 V-1 s-1 + 450 cm2 V-1 s-1)

int = 4.18 10-6 -1 cm-1

Comparing this value to the minimum conductivity:

int

min

.=

=

3 62 104 18 10

0 8666 1 1

6 1 1

. W cm. W cm- -

Sufficient p-type doping that increases the hole concentration by 73% decreases the conductivityby 15% to its minimum value.

5.3 Compensation doping in Sia. A Si wafer has been doped n-type with 1017 As atoms cm3.

1. Calculate the conductivity of the sample at 27 C.


5.4

2. Where is the Fermi level in this sample at 27 C with respect to the Fermi level (EFi) in intrinsicSi?

3. Calculate the conductivity of the sample at 127 C.

b. The above n-type Si sample is further doped with 9 1016 boron atoms (p-type dopant) percentimeter cubed.

1. Calculate the conductivity of the sample at 27 C.2. Where is the Fermi level in this sample with respect to the Fermi level in the sample in (a) at 27

C? Is this an n-type or p-type Si?

Solutiona Given temperature T = 27 C = 300 K, concentration of donors Nd = 10

17 cm-3, and drift mobilitye 800 cm

2 V-1 s-1 (from Figure 5Q3-1). At room temperature the electron concentration n = Nd >> p(hole concentration).

50

100

1000

2000

1015 1016 1017 1018 1019 1020

Dopant Concentration, cm-3

ElectronsHoles

Dri

ft M

obili

ty(c

m2

V-1

s-1 )

Figure 5Q3-1 The variation of the drift mobility with dopant concentration in Si

for electrons and holes at 300 K.

(1) The conductivity of the sample is:

= eNde (1.602 10-19 C)(1017 cm-3)(800 cm2 V-1 s-1) = 12.8 -1 cm-1

(2) In intrinsic Si, EF = EFi,

ni = Ncexp[-(Ec - EFi)/kT] (1)

In doped Si, n = Nd, EF = EFn,

n = Nd = Ncexp[-(Ec - EFn)/kT] (2)

Eqn. (2) divided by Eqn. (1) gives,

N

n

E E

kTd

i

Fn Fi= exp (3)

ln Nn

E E

kTd

i

Fn Fi

=


5.5

EF = EFn - EFi = kT ln(Nd/ni) (4)

Substituting we find (ni = 1.45 1010 cm-3 from Table 5.1 in the textbook),

EF = (8.617 10-5 eV/K)(300 K)ln[(1017 cm-3)/ (1.45 1010 cm-3)]

EF = 0.407 eV above Efi

10

100

1000

10000

50000

70 100 800

Ge

Nd =1013

Si

L T1.5

Nd =1014

Nd =1016

Nd =1017

Nd =1018

Nd =1019

T1.5

Temperature (K)

Ele

ctro

n D

rift

Mob

ility

(cm

2 V

-1s-

1 )

Figure 5Q3-2 Log-log plot for drift mobility versus temperature for n-type Geand n-type Si samples. Various donor concentrations for Si are shown, Nd are incm-3. The upper right insert is the simple theory for lattice limited mobilitywhereas the lower left inset is the simple theory for impurity scattering limitedmobility.

(3) At Ti = 127 C = 400 K, e 450 cm2 V-1 s-1 (from Figure 5Q3-2). The semiconductor is still n-

type (check that Nd >> ni at 400 K), then

= eNde (1.602 10-19 C)(1017 cm-3)(450 cm2 V-1 s-1) = 7.21 -1 cm-1

b The sample is further doped with Na = 9 1016 cm-3 = 0.9 1017 cm-3 acceptors. Due to

compensation, the net effect is still an n -type semiconductor but with an electron concentration given by,

n = Nd - Na = 1017 cm-3 - 0.9 1017 cm-3 = 1 1016 cm-3 (>> ni)

We note that the electron scattering now occurs from Na + Nd (1.9 1017 cm-3) number of

ionized centers so that e 700 cm2 V-1 s-1 (Figure 5Q3-1).

(1) = eNde (1.602 10-19 C)(1016 cm-3)(700 cm2 V-1 s-1) = 1.12 -1 cm-1

(2) Using Eqn. (3) with n = Nd - Na we have

N N

n

E E

kTd a

i

Fn Fi = exp

so that E = EFn - EFi = (0.02586 eV)ln[(1016 cm-3) / (1.45 1010 cm-3)]

E = 0.348 eV above EFiThe Fermi level from (a) and (b) has shifted down by an amount 0.059 eV.

Since the energy is still above the Fermi level, this an n-type Si.


5.6

5.4 Temperature dependence of conductivityAn n-type Si sample has been doped with 1015 phosphorus atoms cm3. The donor energy level for Pin Si is 0.045 eV below the conduction band edge energy.

a. Calculate the room temperature conductivity of the sample.

b. Estimate the temperature above which the sample behaves as if intrinsic.

c. Estimate to within 20% the lowest temperature above which all the donors are ionized.

d. Sketch schematically the dependence of the electron concentration in the conduction band on thetemperature as log(n) versus 1/T, and mark the various important regions and critical temperatures.For each region draw an energy band diagram that clearly shows from where the electrons areexcited into the conduction band.

e. Sketch schematically the dependence of the conductivity on the temperature as log() versus 1/Tand mark the various critical temperatures and other relevant information.

Solution

Si

Ge

GaAs

0C200C400C600C 27C

1018

1 1.5 2 2.5 3 3.5 41000/T (1/K)

1015

1012

103

106

109

Intr

insi

c C

once

ntra

tion

(cm

-3)

1.45 1010 cm-3

2.4 1013 cm-3

2.1 106 cm-3

Figure 5Q4-1 The temperature dependence of the intrinsic concentration.

a The conductivity at room temperature T = 300 K is (e = 1350 10-4 m2 V-1 s-1 can be found in

Table 5.1 in the textbook):

= eNde

= (1.602 10-19 C)(1 1021 m-3)(1350 10-4 m2 V-1 s-1) = 21.6 -1 m-1

b At T = Ti, the intrinsic concentration ni = Nd = 1 1015 cm-3. From Figure 5Q4-1, the graph of

ni(T) vs. 1/T, we have:

1000 / Ti = 1.9 K-1


5.7

Ti = 1000 / (1.9 K-1) = 526 K or 253 C

c The ionization region ends at T = Ts when all donors have been ionized, i.e. when n = Nd. FromExample 5.7, at T = Ts:

n N N NE

kTd c d s= =

12 2

1

2exp

T E

kN

N N

E

kN

N

s

d

c d

d

c

=

=

2 22

12

ln ln

T E

kN

N

s

c

d

=

ln2

Take Nc = 2.8 1019 cm-3 at 300 K from Table 5.1 (in the textbook), and the difference between

the donor energy level and the conduction band energy is E = 0.045 eV. Therefore our firstapproximation to Ts is:

TE

kN

N

s

c

d

=

=( ) ( )( ) ( )( )

ln

. .

. ln.

2

0 045 1 602 10

1 381 102 8 10

2 10

19

2319 3

15 3

eV J/eV

J/Kcm

cm

= 54.68 K

Find the new Nc at this temperature, Nc:

= = ( )

N NT

c cs

3002 8 10

54 68300

3

2 19 3

3

2.

.

cmK

K= 2.179 1018 cm-3

Find a better approximation for Ts by using this new Nc:

=

=( ) ( )( ) ( )( )

TE

kN

N

s

c

d

ln

. .

. ln.

2

0 045 1 602 10

1 381 102 179 10

2 10

19

2318 3

15 3

eV J/eV

J/Kcm

cm

= 74.64 K

= = ( )

N NT

c cs

3002 8 10

74 64300

3

2 19 3

3

2.

.

cmK

K= 3.475 1018 cm-3

A better approximation to Ts is:

=

=( ) ( )( ) ( )( )

TE

kN

N

s

c

d

ln

. .

. ln.

2

0 045 1 602 10

1 381 103 475 10

2 10

19

2318 3

15 3

eV J/eV

J/Kcm

cm

= 69.97 K

= = ( )

N NT

c cs

3002 8 10

69 97300

3

2 19 3

3

2.

.

cmK

K = 3.154 1018 cm-3


5.8

=

=( ) ( )( ) ( )( )

TE

kN

N

s

c

d

ln

. .

. ln.

2

0 045 1 602 10

1 381 103 154 10

2 10

19

2318 3

15 3

eV J/eV

J/Kcm

cm

= 70.89 K

We can see that the change in Ts is very small, and for all practical purposes we can consider thecalculation as converged. Therefore Ts = 70.9 K = -202.1 C.

d and e See Figures 5Q4-2 and 5Q4-3.

ln(n)

1/T

Ti

Ts

Intrinsic

Extrinsic Ionization

ni(T)

ln(Nd) slope = E/2k

slope = Eg/2k

Figure 5Q4-2 The temperature dependence of the electron concentration in an n-type semiconductor.

log(n)

LO

GA

RIT

HM

IC S

CA

LE

INTRINSIC

EXTRINSIC

IONIZATION

log( )

log( )T 3/2 T 3/2

Latticescattering

Impurityscattering

1/TLow TemperatureHigh Temperature

T

Metal

Semiconductor

Res

istiv

ity

Figure 5Q4-3 Schematic illustration of the temperature dependence of electrical

conductivity for a doped (n-type) semiconductor.

5.5 GaAsGa has a valency of III and As has V. When Ga and As atoms are brought together to form the GaAscrystal, as depicted in Figure 5Q5-1, the 3 valence electrons in each Ga and the 5 valence electrons ineach As are all shared to form four covalent bonds per atom. In the GaAs crystal with some 1023 or soequal numbers of Ga and As atoms, we have an average of four valence electrons per atom, whetherGa or As, so we would expect the bonding to be similar to that in the Si crystal: four bonds per atom.The crystal structure, however, is not that of diamond but rather that of zinc blende (Chapter 1 of thetextbook).

a. What is the average number of valence electrons per atom for a pair of Ga and As atoms and in theGaAs crystal?


5.9

b. What will happen if Se or Te, from Group VI, are substituted for an As atom in the GaAs crystal?

c. What will happen if Zn or Cd, from Group II, are substituted for a Ga atom in the GaAs crystal?

d. What will happen if Si, from Group IV, is substituted for an As atom in the GaAs crystal?

e. What will happen if Si, from Group IV, is substituted for a Ga atom in the GaAs crystal? What doyou think amphoteric dopant means?

f. Based on the above discussion ,what do you think the crystal structures of the III-V compoundsemiconductors AlAs, GaP, InAs, InP, and InSb will be?

As GaGa As

As GaAsGa

As GaGa As

As GaAsGa

AsGa

Ga atom (Valency III) As atom (Valency V)

Figure 5Q5-1 The GaAs crystal structure in two dimensions. Averagenumber of valence electrons per atom is four. Each Ga atom covalently

bonds with four neighboring As atoms and vice versa.

Solution

As GaGa As

As GaAsGa

As GaGa As

As GaAsGa

As Ga

Ga atom (Valency III)As atom (Valency V)

hyb orbitals

As ion core (+5e)

Valenceelectron

hyb orbitals

Ga ion core (+3e)

Valenceelectron

Explanation of bonding in GaAs: The one s and three p orbitals hybridize to form 4 hyborbitals. In As there are 5 valence electrons. One hyb has two paired electrons and 3 hyb have 1 electroneach as shown. In Ga there are 3 electrons so one hyb is empty. This empty hyb of Ga can overlap thefull hyb of As. The overlapped orbital, the bonding orbital, then has two paired electrons. This is abond between Ga and As even though the electrons come from As (this type of bonding is called dativebonding). It is a bond because the electrons in the overlapped orbital are shared by both As and Ga. Theother 3 hyb of As can overlap 3 hyb of neighboring Ga to form "normal bonds". Repeating this in threedimensions generates the GaAs crystal where each atom bonds to four neighboring atoms as shown.Because all the bonding orbitals are full, the valence band formed from these orbitals is also full. Thecrystal structure is reminiscent of that of Si. GaAs is a semiconductor.

a The average number of valence electrons is 4 electrons per atom.


5.10

b Se or Te replacing As will have one additional electron that cannot be involved in any of the fourbonds. Hence Se and Te will act as a donor.

c Zn or Cd replacing Ga will have one less electron than the substituted Ga atom. This creates ahole in a bond. Zn and Cd will act as acceptors.

d The Si atom has 1 less electron than the As atom and when it substitutes for an As atom in GaAsthere is a "hole" in one of the four bonds. This creates a hole, or the Si atom acts as an acceptor.

e The Si atom has 1 more electron than the Ga atom and when it substitutes for a Ga atom in GaAsthere is an additional electron that cannot enter any of the four bonds and is therefore donated into the CB(given sufficiently large temperature). Si substituting for Ga therefore acts as a donor.

f All these compounds (AlAs, GaP, InAs, InP, InSb) are compounds of III elements and Velements so they will follow the example of GaAs.

5.6 Doped GaAsConsider the GaAs crystal at 300 K.

a. Calculate the intrinsic conductivity and resistivity.

b. In a sample containing only 1015 cm3 ionized donors, where is the Fermi level? What is theconductivity of the sample?

c. In a sample containing 1015 cm3 ionized donors and 9 1014 cm3 ionized acceptors, what is thefree hole concentration?

Solutiona Given temperature, T = 300 K, and intrinsic GaAs.

From Table 5.1 (in the textbook), ni = 1.8 106 cm-3, e 8500 cm

2 V-1 s-1 and h 400 cm2 V-1

s-1. Thus, = eni(e + h)

= (1.602 10-19 C)(1.8 106 cm-3)(8500 cm2 V-1 s-1 + 400 cm2 V-1 s-1)

= 2.57 10-9 -1 cm-1

= 1/ = 3.89 108 cmb Donors are now introduced. At room temperature, n = Nd = 10

15 cm-3 >> ni >> p.

n = eNde (1.602 10-19 C)(1015 cm-3)(8500 cm2 V-1 s-1) = 1.36 -1 cm-1

n = 1/n = 0.735 cmIn the intrinsic sample, EF = EFi,

ni = Ncexp[-(Ec - EFi)/kT] (1)

In the doped sample, n = Nd, EF = EFn,

n = Nd = Ncexp[-(Ec - EFn)/kT] (2)

Eqn. (2) divided by Eqn. (1) gives,

N

n

E E

kTd

i

Fn Fi= exp (3)

EF = EFn - EFi = kT ln(Nd/ni) (4)


5.11

Substituting we find,

EF = (8.617 10-5 eV/K)(300 K)ln[(1015 cm-3)/(1.8 106 cm-3)]

EF = 0.521 eV above EFi (intrinsic Fermi level)

c The sample is further doped with Na = 9 1014 cm-3 = 0.9 1015 cm-3 acceptors. Due to

compensation, the net effect is still an n -type semiconductor but with an electron concentration given by,

n = Nd - Na = 1015 cm-3 - 0.9 1015 cm-3 = 1 1014 cm-3 (>> ni)

The sample is still n-type though there are less electrons than before due to the compensationeffect. From the mass action law, the hole concentration is:

p = ni2 / n = (1.8 106 cm-3)2 / (1 1014 cm-3) = 0.0324 cm-3

On average there are virtually no holes in 1 cm3 of sample.

We can also calculate the new conductivity. We note that electron scattering now occurs from Na+ Nd number of ionized centers though we will assume that e 8500 cm

2 V-1 s-1.

= ene (1.602 10-19 C)(1014 cm-3)(8500 cm2 V-1 s-1) = 0.136 -1 cm-1

5.7 Degenerate semiconductorConsider the general exponential expression for the concentration of electrons in the CB,

n NE E

kTcc F=

exp( )

and the mass action law, np = ni2. What happens when the doping level is such that n approaches Nc

and exceeds it? Can you still use the above expressions for n and p?

Consider an n-type Si that has been heavily doped and the electron concentration in the CB is1020 cm3. Where is the Fermi level? Can you use np = ni

2 to find the hole concentration? What is itsresistivity? How does this compare with a typical metal? What use is such a semiconductor?

SolutionConsider n = Ncexp[-(Ec - EF)/kT] (1)

and np = ni2 (2)

These expressions have been derived using the Boltzmann tail (E > EF + a few kT) to the Fermi- Dirac (FD) function f(E) as in Section 5.1.4 (in the textbook). Therefore the expressions are NOTvalid when the Fermi level is within a few kT of Ec. In these cases, we need to consider the behavior ofthe FD function f(E) rather than its tail and the expressions for n and p are complicated.

It is helpful to put the 1020 cm-3 doping level into perspective by considering the number of atomsper unit volume (atomic concentration, nSi) in the Si crystal:

nN

MatA

at

= =

( ) ( . )( . )

( . Density kg m mol

kg mol )

-3

1

2 33 10 6 022 1028 09 10

3 23 1

3

i.e. nat = 4.995 1028 m-3 or 4.995 1022 cm-3

Given that the electron concentration n = 1020 cm-3 (not necessarily the donor concentration!), wesee that


5.12

n/nat = (1020 cm-3) / (4.995 1022 cm-3) = 0.00200

which means that if all donors could be ionized we would need 1 in 500 doping or 0.2% donor doping inthe semiconductor (n is not exactly Nd for degenerate semiconductors). We cannot use Equation (1) tofind the position of EF. The Fermi level will be in the conduction band. The semiconductor isdegenerate (see Figure 5Q7-1).

(a) (b)

EFp

Ev

Ec

EFn

Ev

Ec

CB

VB

CB

E

g(E)

Impuritiesforming a band

Figure 5Q7-1

(a) Degenerate n-type semiconductor. Large number of donors form a band thatoverlaps the CB.

(b) Degenerate p-type semiconductor.

50

100

1000

2000

1015 1016 1017 1018 1019 1020

Dopant Concentration, cm-3

ElectronsHoles

Dri

ft M

obili

ty(c

m2

V-1

s-1 )

Figure 5Q7-2 The variation of the drift mobility with dopant concentration

in Si for electrons and holes at 300 K.

Take T = 300 K, and e 900 cm2 V-1 s-1 from Figure 5Q7-2. The resistivity is

= 1/(ene) = 1/[(1.602 10-19 C)(1020 cm-3)(900 cm2 V-1 s-1)]

= 6.94 10-5 cm or 694 10-7 m

Compare this with a metal allo

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Solutions Manual to accompany PRINCIPLES OF ELECTRONIC MATERIALS AND DEVICES

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