Semiconductor Devices [Kanaan Kano]

APPENDICES A: UNITS FUNDAMENTAL QUANTITIES Quantity

Unit

Symbol

Capacitance (C)

Farad

F

Conductance (G)

Siemens

S

C u r r e n t (I)

Ampere

A

Electric Charge (Q, q)

Coulomb

C

E n e r g y (W, E )

Joule

J

Force (F)

Newton

Ν

Frequency (f)

Hertz

Hz

Length (L)

Meter

m

Mass (m)

Kilogram

kg

Potential (V, φ)

Volt

V

Power (P)

Watt

W

Resistance (R)

Ohm

Ω

T e m p e r a t u r e (T)

D e g r e e s Kelvin

Κ

Time (t)

Second

s

DERIVED QUANTITIES A n g u l a r Frequency (ω)

Radian/sec

rad/s

Conductivity (σ)

Siemens/meter

S/m

Dielectric constant (Relative Permittivity) Permittivity (ε, ε ) οχ

Dimensionless Farad/meter

F/m

Resistivity (p)

Ohm-m

Ω-m

Velocity (v)

Meters/second

m/s

Wavelength (λ)

Meters

m

465

466

Appendix Β: Physical Constants and Factors

B: PHYSICAL CONSTANTS AND FACTORS B.1. PHYSICAL CONSTANTS Quantity

Value

Symbol

Avogadro's N u m b e r

Ν

6.022 X 1 0 / m

B o l t z m a n n Constant

k

1.38066 X 1 0 - J / K

B o l t z m a n n Constant

k/q

8.61738 X 1 0 - e V / K

23

2 3

5

1 9

q

1.602 X 10

Electronvolt

eV

1.602 X 10~ J

E l e c t r o n R e s t Mass

m

9.109 X 1 0 " k g

Permittivity in Vacuum

ε

M a g n i t u d e of Electronic Charge

Planck's Constant Speed of Light in Vacuum Thermal Voltage

C

19

31

0

8.854 X 1 0 - F / m 1 2

0

h

6.626 X 1 0 - J - s

c

2.998 X 1 0 m / s

34

8

V =

0.02586 V ( T = 300K,27°C)

kT/q

B.2 P R E F I X E S 6

9

3

k = kilo, Μ = mega = 10 , G = giga = 10 , m = milli = 10~ , 10" 10 pico micron = 10 , η = n a n o 6

B.3 C O N V E R S I O N F A C T O R S 8

10

1A = 1 0 - c m = 1 0 - m 9

7

l n m = 10~ m = 10A = 1 0 - c m 4

6

Ιμπι = 10" cm = 10" m l e V = 1.602 X 1Q- J 19

C: THE DENSITY OF STATES N(E) By assuming the free electron m o d e l for a metal, the energy states for electrons will be the same as for electrons in a three-dimensional potential well where the poten­ tial is V = 0 inside the well and infinite elsewhere. Schrodinger's equation in rectangular coordinates inside the well for electrons having constant total energy Ε becomes 2

2

2

2

ΘΨ 3Ψ δΨ 8ττ π\Ε —τ + — τ + — j + — Γ — Ψ = 0 dx dy dz h τ

2

2

27

2

v

(C1) '

This equation is solved by the m e t h o d of separation of variables by assuming * =

g (x)g (y)g (z) x

y

z

(C.2)

Appendix C: The Density of States N(E)

467

By substituting this solution in E q . ( C . l ) we have 2

2

1 dg

2

1 dg

x

2

+

g dx

8-n mE

z

+

g ^f

x

2

1 dg _ g dz

y

2

h

z

2

(

C

3

)

Since the right-hand side of E q . (C.3) is a constant, it follows that each of the terms is a constant so that £

(

C

.

4

,

We assume for t h e sake of simplicity that each of the sides of the cubic well has length a. T h e b o u n d a r y conditions for ψ are \\> = 0atx

= y = z = 0

χ =y

(C.5)

=ζ =α

The solution to each of t h e equations in (C.4) contains sine and cosine terms. Possible solutions, wherein the cosine terms d r o p out and, subject to t h e b o u n d a r y conditions in E q . (C.5), b e c o m e g= x

'3-

Asin^—

(a)

= B s i n ^ a

(b)

g = C sin

(C.6)

(c)

z

where n , n and n are q u a n t u m n u m b e r s and A , B, and C are new constants. Using the above expressions in E q . (C.3) we have x

y

z

, , , n\ + n) + n =

8mEa

2

2

(C.7)

z

In the three coordinate system that contains η , η, and n , we assume that their m a x i m u m values are N , Ν , and N . T h e reason for this is to d e t e r m i n e the n u m b e r of states that have energies less t h a n some value Ε given by χ

x

z

z

E

^h\N

2 x

+

N

2

2

+

N) z

8 ma We are actually interested in a section of the sphere of the solid that is m a d e by joining t h e t h r e e m a x i m u m values for which the radius of t h e sphere R is V7V2 + N + N . That section is o n e for which t h e values of n , η , n are all positive. This section forms one eighth of the sphere, shown in Fig. CF.l, having volume vR /6 or 2

2

Z

x

z

3

V o l u m e = ^(N

2 X

+ Ν

2

2

+ Νψ ζ

2

(C.9)

468

Appendix C: The Density of States N(E) n, y

0

Figure C.F.I Section of sphere having n , n , and n as coordinates. x

y

z

By substituting Eq. (C.8) in E q . (C.9), we obtain ττ/8ιηα Ε', 2 \ 3 / 2 V o l u m e = — —— — 2

£

(CIO)

Ί

E a c h point inside the section of Fig. C.F.I, corresponding to integer values of n , n and n represents an energy state that contains a pair of electrons. Thus the total volume inside this section represents t h e total n u m b e r of states. Since each q u a n t u m state, because of t h e q u a n t u m spin, can a c c o m m o d a t e u p to two electrons, at the absolute zero of t e m p e r a t u r e the volume represents one half the n u m b e r of electrons N. We can then write x

y

z

(Gil) The question is, what is t h e distribution of the density of the electrons as a function of energy? We first introduce the following terms defined as N(E)

= volume density of states per unit energy at energy Ε

N(E)dE

= volume density of states with energies from Ε to Ε + dE

N = volume density of occupied states where s

1

N u m b e r of electrons

1 ./V

Volume The term Ν can be written as (C.12) where N represents the total density of states from zero energy to E. T h e r a t e of change of t h e density of states becomes s

(C.13)

Appendix D: Einstein Relation

469

Thus E q . (C.13) represents t h e increment in the density of states in an incre­ m e n t d E at energy E. This is what we defined as N(E) so that (G14) A t absolute zero, electrons occupy the lowest levels possible. We assume that all states u p to χ = y = ζ = a are occupied and for χ > a there are n o electrons at all. Thus one can d e t e r m i n e the m a x i m u m energy that electrons can have at absolute zero from E q . ( G i l ) . We label that energy t h e Fermi energy E . O n c e E is fixed, the highest q u a n t u m states at absolute zero are fixed and E is defined as the m a x i m u m energy that an electron in a certain material can have at absolute zero. We d e t e r m i n e £ f r o m E q . ( G i l ) as F

F

F

f

(C.15) A s an example, assume that the density of free electrons in silver is 5.9 X 1 0 m " = N/a .We solve for E to b e 5.51eV. 28

3

3

F

D: EINSTEIN RELATION By considering the expressions for the drift and diffusion current densities given by Eqs. (4.15) and (4.16), it seems reasonable to expect that, other t h a n t h e particle charge, there must b e some o t h e r commonality b e t w e e n these 2 equations since both refer to t h e motion of carriers in t h e same solid. This commonality exists b e t w e e n the mobility and t h e diffusion constant. To determine the relation b e t w e e n t h e mobility and the diffusion constant, we consider a slab of semiconductor that has b e e n d o p e d to a nonuniform distribution of electrons in a region, as shown in D . F l a . For χ < 0 t h e material is intrinsic, for 0 < χ =£ d t h e distribution of electrons is increasing while for χ > d the density of elec­ trons is constant. A t t h e r m a l equilibrium electrons diffuse from the region a r o u n d χ = d to that at χ = 0 leaving behind positively ionized atoms. A n electric field is thus established b e t w e e n χ = d and χ = 0 directed from χ = d to χ = 0. Electrons continue to diffuse from χ = d to χ = 0, and since at thermal equilibrium t h e electron current is zero, the electric field returns t h e same n u m b e r of electrons from χ = 0 to χ = d. In accordance with E q . (4.15) t h e electron drift current is directed from right to left and in the direction of the electric field while t h e electron diffusion current is from left to right. For the total electron current to b e zero we have qO dn a

dx

(D.l)

We show, in Fig. D . F l b the potential distribution in the slab, and in Fig. D . F l c the energy b a n d diagram. The electron density in Fig. D.F1 can be expressed as

Appendix D: Einstein Relation E

n{x) = n exp

£,-(0) - £,·(*)

- £,(x) kT

F

i

«, exp

kT

The electric field intensity and the potential energy of an electron to the potential by

«

™ E (x)=

-qtfx)

i

(b)

Since φ(0) = 0 and hence Ε (0) = 0 we can write Eq. (D.2) as ;

n(x) = n exp i

~q

V, V V (V ), V (V )

=

B J T D C voltage; emitter to base, collector to base, emitter to col­ lector D C supply voltage to BJT; collector circuit, base circuit

=

F E T D C voltage; gate to source, drain to source

V,

=

M O S F E T D C voltage; source to bulk, channel to source

EC

cc

BB

C

GS

D

DS

SB

V

cs

V

p

=

F E T D C pinchoff voltage

V

FB

=

M O S F E T D C flatband voltage

V

T

=

F E T D C threshold (turn-on) voltage

=

t h e r m a l voltage k T/q

V

t

=

F E T D C voltage at saturation

v

V

=

drift velocity

ν

=

saturation velocity

v

=

t h e r m a l velocity

W

=

depletion layer width, potential energy

SAT

d

lh

W

=

B J T base width

W ,W

=

width of Ρ region, Ν region

Ζ

=

F E T depth

R

Greek a

= BJT: D C ratio of collector to emitter current =

ot

D

=

BJT: D C ratio of collector to emitter current with collector shorted to base in active region BJT: D C ratio of collector to emitter current with emitter shorted to base in inverse active region

β

= =

BJT: D C ratio of collector to base current BJT: D C ratio of collector t o base current with collector shorted to base in active region

β

Λ

=

BJT: D C ratio of collector to base current with emitter shorted to base in inverse action region

β

0

=

BJT: A C ratio of collector to base current with collector shorted to base

Appendix Ε: Frequently Used Symbols γ

=

BJT: emitter injection efficiency

δ

=

BJT: base transport factor

ε e

r

=

permittivity or dielectric constant

=

relative permittivity

ε

ο

= permittivity of free space

ε

οχ

= permittivity of oxide

λ

=

wavelength

λ

=

M O S F E T channel length modulation factor

μ μ, μ η

ρ

μ, μ η

mobility mobility of electron, holes

=

M O S F E T effective mobility of electrons, holes

=

frequency of light

ρ

ν

= =

ρ

=

resistivity

p

=

volume charge density

=

conductivity

c

σ σ, σ η

τ

= conductivity of Ν region, of Ρ region

ρ

= carrier base transit time

B T

V « φ, φ

=

lifetime of holes in Ν region, of electrons in Ρ region =

potential, surface potential

=

potential from E. to E in bulk of M O S F E T

(£>

=

metal work function voltage

Φ

=

semiconductor work function voltage

φ

5

Ρ

m

ί

F

Φ^,

=

X

= affinity of semiconductor, E — E

s

work function difference from metal to semiconductor, Φ Q

c

- Φ

475

chapter 1 ATOMIC STRUCTURE AND QUANTUM MECHANICS

1.0 I N T R O D U C T I O N Major advances in the fields of semiconductor materials and semiconductor devices have t a k e n place within the last four decades. These advances have b e e n largely responsible for the information revolution, b o t h in the processing and transmission of intelligence. Materials used in t h e fabrication of large-scale circuits may be classified as: good conductors, insulators, and semiconductors. T h e basic difference b e t w e e n the three is their resistance to current flow, defined in terms of t h e resistivity of the material. G o o d conductors, such as copper and aluminum, have a resistivity of less t h a n 10~ ohm-cm and are used for low-resistance wiring and interconnections in electric circuits. Insulators that have resistivities greater than 10 ohm-cm are used as isolators of circuits and devices, and in the formation of capacitors. B e t w e e n these two limits of resistivities lie the semiconductor materials: the p u r e elemental semiconductors, silicon and germanium, and some II-VI, IV-VI, and III-V compounds, t h e most i m p o r t a n t of which is gallium arsenide. Their mid-range resistivity, or conductivity, is not the direct reason for the importance of semiconductors. Rather, it is the extent to which their properties are influenced by light, temperature, and m o r e importantly, by the addition of minute amounts of special impurities. Extensive changes take place in the resistivity of silicon when one part of an impurity is added to a million parts of silicon. Silicon is abundant in nature in the form of sand (silica) and clay. However, before silicon can b e used in devices, major purification of the material is required. A n application that illustrates the use of the three types of materials in the formation of integrated circuits follows. Integrated circuits are fabricated on 8 to 20cm diameter circular sections of silicon k n o w n as wafers. T h e major c o m p o n e n t s in integrated circuits are transistors, 3

5

1

2

Chapter 1

Atomic Structure and Quantum Mechanics

capacitors, and resistors. In the fabrication process, controlled a m o u n t s of particular impurities are a d d e d to the silicon in order to form the various regions of a transis­ tor. Certain sections of the wafers are oxidized to form silicon dioxide, which serves as an insulator and as an isolator of different circuit parts. After the various devices are formed, a good conductor, such as aluminum, is deposited on the surface to interconnect t h e various parts of the circuits. Finally, the complete circuit, with external leads attached, is encapsulated in a plastic or ceramic package. O u r objective in this book is to study, first, some of the properties of pure semiconductors and those semiconductors to which impurities are added. Having studied these properties, we then investigate the operation and the current-voltage characteristics of semiconductor devices, such as diodes and transistors. To do that, we n e e d to consider the internal structure of semiconductors, t h e types of current carriers, and the modes of transport of these carriers. 1.1

C R Y S T A L S A N D T H E UNIT CELL Based on t h e internal a r r a n g e m e n t of the atoms, a solid is labeled as amorphous, crystalline, or polycrystalline. In crystalline solids, the atoms are arranged in an orderly three-dimensional array that is r e p e a t e d throughout t h e structure. A m o r p h o u s solids have their atoms arranged in a very r a n d o m m a n n e r with no r e p e a t e d pattern. The atoms in polycrystalline solids are so arranged that, within certain sections, s o m e sort of a p a t t e r n of the atoms exists but the various sections are randomly arranged with respect to each other. Most semiconductors are crystalline in nature. Let us look into the internal a r r a n g e m e n t of the atoms of semiconductors in the basic building block k n o w n as t h e unit cell. T h e a r r a n g e m e n t inside the unit cells of silicon and g e r m a n i u m is k n o w n as t h e diamond lattice because this a r r a n g e m e n t is a characteristic of dia­ m o n d . D i a m o n d is a form of carbon, which is an element in Column I V of the peri­ odic table. In the diamond lattice, shown in Fig. 1.1, t h e atoms are arranged within a cube having dimension L, w h e r e L is k n o w n as the lattice constant. The unit cell for t h e d i a m o n d lattice has an a t o m in each corner of the cube (8), one at the center of each of the six faces (6), and four (4) internal to the cube located along the diagonals. F r o m this total of 18 atoms, the a t o m at each corner is shared by eight cells and each face a t o m is shared by two cells. We will use the above information in the following examples to calculate the density of atoms and the mass density of silicon.

EXAMPLE

1.1

a) Determine the number of atoms in each cell of the diamond lattice. b) Determine the density of atoms in silicon, given the lattice constant L = 5.43A. Solution a) Eight atoms are shared by eight cells, six atoms are shared by two cells, and four atoms are internal to the cell so that: the number of atoms in each cell = 8/8 + 6/2 + 4 = 8.

Section 1.2

W h a t Are W e Looking For Now?

3

b) Volume of cell = (5.43 x 1(T ) cm . 8

3

3

Density = 8/volume = 5 x 10 atoms/cm . 22

3

E X A M P L E 1.2 Determine the density of silicon given Avogadro's number is 6.023 Χ 10 atoms/mole. 23

SOLUTION From the periodic tabic of elements, the atomic weight ol silicon is 28.09. I he density ol silicon atoms was found in Example I I to be 5 Mi "cm \ . 5 Χ 10 atoms/cm Χ 28.09 g/mole 2.33 g Densitv = 7— -π ; f = 6.023 X 10" atoms/mole cm 22

3

Figure 1.1 Unit cell of the diamond lattice. Source: Electronic Materials Science by Mayer/Lau, © 1990. Reprinted by permission of PrenticeHall, Inc., Upper Saddle River,NJ.

Gallium Arsenide, a III-V compound formed from Gallium (III) and Arsenic (V), crystallizes in the form of a slightly different unit cell, known as the Zincblende Structure. In this structure, G a and A s atoms are found at alternate locations inside the unit cell. O n close study of the unit cell lattice of Fig. 1.1, we note that each atom is attached to the four nearest neighboring atoms. This attachment of atoms, in fact, represents the force that holds the lattice atoms together and is known as covalent bonding. E a c h a t o m of C o l u m n I V elements has four electrons in its outermost shell. T h e covalent bonding results from the sharing of electrons b e t w e e n atoms. W h e n each atom, say A, is shown b o n d e d to four neighboring atoms, each of t h e four neighboring atoms contributes one electron to the b o n d with A. A t o m A, therefore, contributes one electron to each bond, so that two electrons are shared by a t o m A with each of the four atoms. This sort of bonding accounts for some physical proper­ ties of these solids. 1.2 W H A T A R E W E L O O K I N G FOR N O W ? Eventually, we are interested in determining the current in a semiconductor device in response to the application of a source of energy, such as an electric source or light source. To d o that, we n e e d to k n o w t h e types and densities of t h e current carri­ ers and their masses.

4

Chapter 1

Atomic Structure and Quantum Mechanics

In a semiconductor, there are b a n d s of energy in which electrons can exist and other b a n d s in which they cannot. We n e e d to establish the bases for the formation of these bands. The existence of b a n d s results from t h e allocation of specific energy levels to an electron in an a t o m and the consequent displacement of these energy levels by introducing t h e effect of t h e forces that atoms exert on each other. A s a result, each electron in t h e solid has a specific energy level; the combinations of these levels form bands. Before we study the formation of bands, we will present a perspective of the theories that have evolved and have a t t e m p t e d to explain experimentally observed properties of h e a t e d materials. Classical mechanics' concepts fail to explain these properties as well as all other p h e n o m e n a that take place at the atomic levels. Such explanations required the evolution of the science of Wave Mechanics, which treats electrons as particles that have wave-like properties.

1.3

F A I L U R E O F C L A S S I C A L M E C H A N I C S AT A T O M I C L E V E L S Two experimentally observed p h e n o m e n a evolved that could not be explained by the theories of Classical or N e w t o n i a n mechanics. These were: blackbody radiation and the sharp discrete spectral lines emitted by h e a t e d gases. First, we will consider blackbody radiations. A blackbody is a h e a t e d solid labeled as an ideal radiator of electromagnetic waves. W h e n a solid is heated, it emits radiation over a certain b a n d of frequencies. The h e a t e d atoms vibrate so that the amplitudes and frequencies of the vibrations seem to resemble the radiating a n t e n n a of a broadcast station. Classical mechanics predict that this h e a t e d solid emits radiation over a continuous b a n d of frequencies. M e a s u r e d responses, however, indicate that this is not true. A s a m a t t e r of fact, radi­ ation takes place only over a certain b a n d of frequencies. F u r t h e r m o r e , t h e fre­ quency spectrum of the radiation changes as the t e m p e r a t u r e of the solid is varied, as shown in Fig. 1.2. A similar p h e n o m e n o n occurs w h e n certain gases are heated. It was experi­ mentally observed that h e a t e d gases emit radiation in small discrete quantities at certain discrete spectral wavelengths. Again, Classical mechanics had n o explana­ tion for this. The conclusion that was reached confirmed that Classical mechanics could not predict p h e n o m e n a that occur at microscopic scales or at atomic levels.

1.4 P L A N C K ' S H Y P O T H E S I S The first b r e a k t h r o u g h came about with the work of Max Planck. Planck's hypothe­ sis, presented in 1901, was that light is emitted or absorbed in discrete units of energy called photons. The vibrating atoms of a h e a t e d body emit radiation at a fre­ quency ν and the energy is restricted to certain discrete, or quantized values, given by Ε = nhv for η = 0,1,2,3,.

(1.1)

Section 1.5

Bohr's Classical Model of the Atom

p Wavelength in mm

-

5

Figure 1.2 Relative intensity of radiation from a blackbody versus wavelength at various temperatures. From E. Uiga, Optoelectronics, Prentice Hall (1995)

where ν is the frequency of radiation and h is a factor that Planck obtained by a the­ oretical fit to the experimental results. This is labeled Planck's constant and its value is 6.62 X l f r J - s . Planck's hypothesis confirmed that, in addition to the quantizing of electro­ magnetic waves, light may also b e viewed as consisting of particles labeled wave packets, or photons. This laid the foundation for t h e theory of t h e dual n a t u r e of light. It also paved the way in explaining the energy levels of electrons in a solid, as it considered electrons to be particles of m a t t e r while at the same time possessing wave-like properties. A t this time, Rutherford explained electron behavior as resulting from a circu­ lar motion a r o u n d the nucleus. Following u p o n t h e conclusions of R u t h e r f o r d and Planck's hypothesis, Niels Bohr, in 1913, put forward a m o d e l confirming the planetary-like motion of elec­ trons a r o u n d an a t o m and included the quantum-like theory of Planck. We will now consider B o h r ' s theories beginning with his classical m o d e l of t h e atom. 34

1.5 B O H R ' S C L A S S I C A L M O D E L O F THE A T O M The work of R u t h e r f o r d c o m p a r e d the nucleus of an a t o m to t h e sun and t h e elec­ trons that revolve a r o u n d t h e nucleus to t h e planets that revolve a r o u n d the sun. U n d e r the influence of gravitational forces, the planets move a r o u n d the sun in a quasi-circular orbit. B o h r suggested that within t h e atom, gravitational forces (due to masses) b e c o m e negligible w h e n c o m p a r e d to the electrostatic forces (due to charges) that

6

Chapter 1

Atomic Structure and Quantum Mechanics

control the orbit of the electron. The electrical forces are d e t e r m i n e d by Coulomb's Law, which relates t h e force exerted by one charged object on another. In accordance with B o h r ' s model, we will use Coulomb's L a w for t h e oneelectron hydrogen a t o m and d e t e r m i n e an expression for the total energy that the electron possesses, assuming that it is rotating in a fixed circular orbit about the nucleus.

E n e r g y of H y d r o g e n Electron In a hydrogen atom, t h e electrostatic attractive force b e t w e e n the nucleus, consisting of t h e proton, having a charge +q and the electron having a charge of —q is given by

4ire r

(1-2)

2

0

where ε is the permittivity of free space in F a r a d s / m e t e r , r is t h e separation b e t w e e n the electron and t h e nucleus in meters, and F i s in newtons. We assume that the mass of the p r o t o n is much greater than that of the electron so that the p r o t o n is fixed at the center of the system. To maintain motion in a fixed circular orbit having radius r, the electrostatic force is balanced by an equal and opposite centripetal force. We assume that the positive reference direction for the force is o u t w a r d so that t h e attraction force in E q . (1.2) is negative. Because of the presence of the positively charged nucleus, an electric field is created a r o u n d it. Electric field intensity, %, defined as t h e force on a unit positive charge, is directed o u t w a r d and is thus positive. F r o m E q . (1.2), we obtain the expression for % at t h e location of the electron as 0

4ττε /

(1.3)

2

(

where, for the units used in E q . (1.2), % is in v o l t s / m e t e r . T h e electric field intensity approaches zero as r approaches infinity. The electrostatic potential, V, due to t h e electric field, is defined by V = - \

% dx

(1.4)

ο

Assuming that the reference for zero potential is at infinity, and since t h e sys­ tem is assumed to possess spherical symmetry, we determine the expression for the potential of a point at r as

V

~ 1

2

4ττε Γ " 4ττε Γ 0

0

( 1

'

5 )

The potential energy of t h e electron at a distance r, defined as t h e product of the potential at that point and t h e charge, becomes W(P.E.) = V{-q)

= ^

(1.6)

T h e negative sign implies that in moving an electron from infinity to r, nega­ tive work is d o n e in contrast to t h e positive work that needs to be e x p e n d e d in mov­ ing a positive charge towards the nucleus. This implies that an electron at point r has to b e restrained to prevent it from moving towards the nucleus. A sketch of the potential energy is shown in Fig. 1.3. We n o t e that the actual drawing has spherical symmetry so that it has t h e shape of a funnel. We are interested in t h e total energy that t h e hydrogen electron possesses, which, in addition to potential energy, includes the kinetic energy resulting from t h e velocity of the electron. A s it moves in t h e circular orbit, t h e electron is subjected to a centripetal force (mv /r), which exactly balances the attraction force of the nucleus, causing an angular acceleration v /r. By setting t h e m a g n i t u d e of t h e force in E q . (1.2) equal to t h e m a g n i t u d e of the centripetal force, we have 2

2

—

=

(17)

2

After solving for mv , the expression for the kinetic energy becomes K.E.

2

=

(1-8)

θΊτε Γ 0

The total energy, E, of the electron is 2

2

Ε = K.E. + VT(P.E.) = — -q2 - +, -2— = — 3 q 4ττε Γ 8ττε Γ 8ττε ?0

0

(1.9)

0

In accordance with t h e references established earlier, the electron has the highest energy (least negative) as its distance from the nucleus approaches infinity.

T h e H y d r o g e n A t o m as a R a d i a t i n g A n t e n n a The relations we just derived indicate that the electron moves in a circular orbit having a radius r, velocity v, and total energy E. If a hydrogen a t o m is heated, the electron absorbs energy and moves to a higher level, corresponding to a new total energy E. It is thus possible for the electron to occupy any orbit, depending u p o n

8

Chapter 1

Atomic Structure and Quantum Mechanics

the energy it gains and, t h e orbital radius is d e t e r m i n e d from E q . (1.8.) Hence, according to this classical theory, the electron can take on continuous values of energies E, depending u p o n t h e radii of the orbit. This, however, contradicts the results of diffraction experiments. Diffraction is defined as the scattering that light rays u n d e r g o when they pass t h r o u g h slits of a solid. T h e light rays emerge deflected, forming fringes of parallel light, dark, and colored bands. The diffraction experiments, using electrons, indicated that electrons can only occupy certain dis­ crete energy levels as they emit radiation only at certain discrete frequencies. The question is then: H o w does the electron emit radiation? E l e c t r o n motion at constant speed results in a constant electric current. Since the electron that is orbiting about a p r o t o n is subjected to centripetal acceleration, its velocity increases with time, resulting in a time-dependent current. F r o m the laws of electromagnetic theory, a time-varying current generates a time-varying magnetic field. By Maxwell's equation, a time-varying magnetic field induces a time-depen­ dent electric field. Electromagnetic fields and electromagnetic waves are generated, leading to radiation that is similar to that emanating from a television or radio transmitting a n t e n n a . Such radiation of energy may be possible over a wide b a n d of frequencies. If the hydrogen electron is to emit radiation, it must continuously lose energy, revolve in smaller helical orbits with decreasing velocity, and get closer and closer to t h e nucleus. H e n c e , if a charge is accelerated, it radiates energy and therefore expe­ riences a loss of energy. This dilemma, faced by Bohr's classical model, led him to qualify it by his quantized model.

1.6 B O H R ' S Q U A N T I Z E D M O D E L The classical theory proposed by B o h r obviously failed w h e n it was applied to the hydrogen atom. In trying to explain the dilemma h e faced, B o h r p r o p o s e d a model that confirmed the concept of the quantization of energy. The m o d e l was without proof, as e m b o d i e d in his two proposals k n o w n as postulates. His first postulate was based on his classical theory, with the exception that an electron could remain in a circular orbit without radiating any energy. His second postulate stated that a q u a n t u m of radiation is emitted or absorbed when an elec­ tron moves from one energy level to another. We summarize the substance of the model as follows: 1. Electrons revolve a r o u n d t h e nucleus only in certain definite circular orbits, every orbit corresponding to a certain level of energy. 2. These orbits are labeled stationary orbits. Electrons could stay in these station­ ary orbits without radiating any energy. 3. T h e transfer of an electron from an orbit of lower energy to an orbit of higher energy requires absorption of radiation by the atom, while a fall from a higher energy level to a lower energy level orbit results in the emission of radiation. This energy difference, whether absorbed or emitted, is

Section 1.6 E -E 2

Bohr's Quantized Model

= hv

L

9

(1.10)

4. The only stationary orbits of the electrons are those for which the angular m o m e n t u m is also quantized. Applying conclusion n u m b e r 4 above to the single hydrogen atom, we have mvr

n

= nh/2ir

η = 1,2,3,...

(1.11)

where m is t h e electron mass, ν is the linear electron velocity, r is the orbital radius for a given value of n, and η is k n o w n as a quantum number. Since we are still refer­ ring to circular orbits of t h e electron, E q . (1.7) applies. W h e n solved for t h e velocity it gives n

I ν

q

2

\i/2

\4Tte rm 0

Substituting t h e above expression for ν in E q . (1.11) and solving for r , we get 2

2

nhs

n

S

°

(1.12)

T h e expression for the velocity becomes

2s nh 0

^

^

The electronic orbits assumed in t h e B o h r theory are circular so that the expressions derived earlier for the potential energy and total energy, Eqs. (1.6) and (1.9), are valid. By substituting the relation for r from E q . (1.12), as developed from Bohr's theory, in the expression for total energy, E, of an electron, as given by E q . (1.9), we obtain an expression for the quantized energy of the hydrogen electron as E

- =

<

U

4

)

If we use rationalized M K S units, as found in A p p e n d i x A , for the constants in Eq. (1.14), the unit of energy will be t h e joule. We now define a unit of energy k n o w n as t h e electron-volt, eV, where one electron-volt is the energy acquired by an electron w h e n elevated through a potential difference of one volt. O n e e V = 1.6 X 1 0 ~ joules. Replacing all t h e constants by their values, E q . (1.14) becomes 19

2

£ „ ( e V ) = - 1 3 . 6 / n , η = 1,2,3,...

(1.15)

where Ε is t h e energy of the allowed energy level of the discrete orbit, which is obtained by replacing η with t h e relevant integer. It is obvious that t h e electron can exist only at certain discrete energy levels. It is thus possible for an electron to move b e t w e e n any two such levels with the consequent release of radiation, going from a higher (higher n) to a lower (lower n) energy level. T h e corresponding frequency of radiation is d e t e r m i n e d from E q . (1.10). The electron can also b e m o v e d to a higher

10

Chapter 1

Atomic Structure and Quantum Mechanics

E{eV), 0 -0.56 I -0.87 |

5 4

-1.53

-3.41

13.60

1

•- 1

Figure 1.4 Energy level diagram for the hydrogen atom. Six of the infinite number of possible transitions are indicated.

level by the absorption of radiation, again in accordance with E q . (1.10). In addition to the energy, both the orbital radius and velocity are quantized, as expressed by Eqs. (1.12) and (1.13) respectively. The allowed energy levels in the hydrogen atom, as d e t e r m i n e d from E q . (1.15), are usually represented by an energy level diagram, as shown in Fig. 1.4. Later, Sommerfeld, following u p o n Bohr's conclusions, verified that the elec­ tron orbits w e r e not circular but elliptical. This was later confirmed by others. Planck's hypothesis provided an explanation for the discreteness of radiation and the particle-like theory of light. B o h r confirmed the quantization of t h e energy, the orbital radius, and the orbital velocity of the hydrogen electron. Thus, the quan­ tum concept was extended to p h e n o m e n a taking place at atomic dimensions. The B o h r theory constituted a big step forward in the explanation of the hydrogen spectra but it did not provide complete answers. While the emission fre­ quencies predicted by Bohr, for the hydrogen atom, are quite close to those that were experimentally observed, a major weakness of Bohr's theory is that it cannot be e x t e n d e d to t h e modeling of atoms that are m o r e complex than the hydrogen atom. M o r e complete answers were provided by de Broglie and Schrodinger, who provided the basis of t h e new science of Wave Mechanics. REVIEW QUESTIONS Q l - 1 Give examples of physical systems to which the theories of classical mechanics apply. Ql-2 On what was Planck's hypothesis based? Ql-3 What was Bohr's contribution to the science of Wave Mechanics and what were the weaknesses of his postulates?

Section 1.7

W a v e Mechanics

11

Ql-4 What prompted Bohr to quantize the energy of an electron? Ql-5 Why was the hydrogen atom selected for consideration?

HIGHLIGHTS • Classical mechanics could not provide explanations of the behavior of solids, such as elec­ tromagnetic radiation from certain bodies. • Max Planck hypothesized that it is possible for bodies to radiate energy at certain fre­ quencies, depending upon the energy that is imparted to them. He demonstrated that light has not only wave-like properties but particle-like properties, as well. • Neils Bohr first hypothesized that electrons can take on any value of energy in their orbit around the nucleus. He later modified this theory. • In his attempt to provide explanations of experimental results, Bohr further postulated that electrons can exist in certain orbits only, each orbit corresponding to a certain energy. He further stated that the transfer of an electron from one orbit to another was accompa­ nied by a release or a gain of energy. Bohr's explanation served to model the hydrogen atom but could not be extended beyond the one electron model.

EXERCISES El-1 a) Determine the binding energy for an electron in a hydrogen atom at a quantum number 3 level. Give the answer in eV and in joules. b) Repeat the above determination for an electron in a silicon coulombic potential, using the results of the hydrogen atom. The relative dielectric constant for silicon is 11.8 and the effec­ tive mass of a silicon electron is assumed to be 1.18 times that of the hydrogen electron. Ans:

a) Ε = -1.5eV

19

b) Ε = 0.02 X 10- J

El-2 a) Use Planck's hypothesis to calculate the wavelength associated with a leV photon, b) Repeat (a) for an electron. Ans:

9

b) λ = 1.22 X 10" m

1.7 W A V E M E C H A N I C S A t t e m p t s to explain electron motion in terms of wave motion were p r o m p t e d by the experimental observations of a n u m b e r of scientists. They observed that diffraction patterns could be observed and r e c o r d e d photographically w h e n b e a m s of electrons were passed through crystals. These results indicated that electrons h a d characteris­ tics usually associated with waves, and thus should obey wave-motion equations. It was de Broglie, in 1924, w h o laid the foundation of Wave Mechanics. H e confirmed B o h r ' s conclusion that t h e ratio of the energy of a wave to t h e frequency was a con­ stant, h, and postulated that the product of t h e momentum of the electron and t h e wavelength was also equal to this constant. This constant is, again, Planck's constant. This relationship b e t w e e n dynamic properties of the electron, such as m o m e n t u m , and wave properties, such as wavelength, forms the basis of wave mechanics. Thus,

12

Chapter 1

Atomic Structure and Quantum Mechanics

particle and wave properties w e r e related so that de Broglie's two conclusions, stated in equation form, are - = h, ν

(a)

Kmv = h

(b)

(1.16)

where h is Planck's constant and λ is the de Broglie wavelength. Wavelength, K, in meters, is the ratio of the velocity of light in m e t e r s / s e c o n d to the frequency in hertz and 1/v is t h e spatial period of the wave. The experimental observations of other scientists provided conclusive proof of the relationships formulated by de Broglie. It might be useful to emphasize t h e generality of de Broglie's hypothesis, which h e predicted from relativity theory. According to his theory, a hypothetical wave is associated with any particle such that Eqs. (1.16a) and (1.16b) hold. The magnitude of the intensity of the d e Broglie wave, at any point, is a m e a s u r e of the probability of finding t h e particle at that point. In essence, the m o t i o n of the particle is describable in probability terms r a t h e r than in terms of the deterministic laws of classical mechanics. This probability interpretation and the a t t e n d a n t uncertainty tie in immediately with Heisenberg's Uncertainty Principle, which will b e discussed shortly. T h e "plausibility" of Eqs. (1-16) may b e established as follows: 1. The relation Ε = hv is an expression of Planck's q u a n t u m hypothesis that energy is absorbed or emitted in quanta, t h e size of a q u a n t u m being hv. 2. A light q u a n t u m is labeled a p h o t o n , so that the energy of a p h o t o n of light of frequency ν is hv. 3. According to Einstein's theory, m a t t e r and energy are related by Ε = mc : Hence, a p h o t o n would have a mass of hv/c and m o m e n t u m hv/c or ρ = h/λ as given by E q . (1.16a), w h e r e ρ is the m o m e n t u m of the p h o t o n and c is the veloc­ ity of light. 2

2

We want to emphasize, at this point, that we are not implying the physical exis­ tence of electron waves. Rather, we are thinking of a moving particle as having wave­ like properties associated with it. That a particle has wave-like properties follows from electron diffraction experiments, which show that electron b e a m s b e h a v e like light beams, and thus must obey t h e same wave relations as light. Before carrying the analogy any further, let us summarize the accepted dual theories of light: a) Light has b o t h particle-like and wave-like properties. b) Light is c o m p o s e d of photons. Thus, a light ray consists of particles (i.e., p h o ­ tons) each of which possesses an energy, hv, where ν is the frequency of t h e light. c) T h e intensity of light at a location is the density of t h e p h o t o n s at that location. The analogy b e t w e e n light and electron b e a m s is extended so that t h e travel and diffraction of a light wave in a m e d i u m of varying refractive index is analogous

Section 1.8

Heisenberg's Uncertainty Principle

13

to the case of electron b e a m s traveling t h r o u g h a varying force field. B o t h p h e n o m ­ ena are mathematically analogous. It is found that when, in the optical case, the refractive index varies over dimensions of t h e same o r d e r of m a g n i t u d e as the wave­ lengths, classical theories cannot explain the p h e n o m e n a that occur. Similarly, classi­ cal mechanics fails w h e n a force field in a material varies over a distance of the order of a de Broglie wavelength. T h e r e a s o n we have to look to the wave theory for an explanation of t h e p h e n o m e n a occurring at atomic distances is partly a result of the H e i s e n b e r g Uncertainty Principle.

1.8 H E I S E N B E R G ' S U N C E R T A I N T Y PRINCIPLE Heisenberg's Uncertainty Principle, which was derived from rigorous mathematical physics, states that t h e m i n i m u m value of t h e product of the uncertainties in the val­ ues of two quantities, whose p r o d u c t has t h e dimension of action, ML /T(kg m / s e c ) , is given by Planck's constant, h. Two such sets of quantities are m o m e n t u m and position, and energy and time. The Principle states that t h e p r o d u c t of the uncertainties in m o m e n t u m ρ and position χ and those of energy Ε and time t are 2

2

Ap Ax > h/2-π

(a)

AEAt>h/2ir

(b)

(1.17)

A numerical example will clarify this result.

E X A M P L E 1.3 a) Determine the uncertainty in the position of a bullet of mass 20g travelling with an uncertainty in its velocity of lOcm/s. Planck's constant is h = 6.62 X 10~ J-s. 34

-3

b) Repeat part (a) for an electron that has an uncertainty in its velocity of 10 cm/s. Solution a) At constant mass m, AvAx

λ/2ττ 34

3

2

32

Ax = 6.62 X lCr /(2tr X 20 X 10" X 10 X 10~ ) = 5.26 X 10~ m b) Ax = 6.62 Χ 10- /(2ΤΓ X 9.1 Χ 10" x 10~ X 10~ ) = 11.57m 34

31

3

2

We note, while Ax for the bullet is negligible compared to its dimensions, the Ax for the electron is many orders of magnitude larger than the diameter of the atom.

In accordance with the Uncertainty Principle, it is thus meaningless to speak of t h e motion of an electron in a circular or elliptic orbit because such a statement implies that we can trace t h e p a t h of the electron by making at least two successive observations of its position. In fact, for this example, if t h e position of t h e electron is d e t e r m i n e d very accurately, so that Δχ is small, there is a lack of precision in the uncertainty of the m o m e n t u m and Ap increases. If the limit of Ax is zero, t h e n the uncertainty in t h e m o m e n t u m is infinite. The best that one can h o p e for is to set u p a

14

Chapter 1

Atomic Structure and Quantum Mechanics

probability p a t t e r n a r o u n d the nucleus, showing t h e region in which there is an appreciable probability of finding t h e electron. Thus, t h e Uncertainty Principle highlighted a n o t h e r difficulty with t h e B o h r theory, as it is not possible to assign a certain orbit radius or m o m e n t u m to a parti­ cle. A s was shown in the example, for large-scale p h e n o m e n a where Ax for the bullet is negligible c o m p a r e d to the dimensions of t h e bullet, the Uncertainty Principle does not pose any serious difficulties, whereas for the electron, t h e posi­ tion is indeterminate. Because t h e position, velocity, and m o m e n t u m of an electron cannot b e d e t e r m i n e d using classical mechanics, one has to revert to wave mechan­ ics to study t h e behavior of the electron.

1.9 S C H R O D I N G E R ' S E Q U A T I O N It was Schrodinger who, in building on d e Broglie's work, first p r o p o s e d t h e wave equation k n o w n by his n a m e . H e incorporated the quantization theory proposed by Planck and t h e wave-like n a t u r e of m a t t e r as p r o p o s e d by de Broglie. This equation is as basic to wave mechanics as Newton's laws are to classical mechanics and as Maxwell's equations are to electromagnetic theory. Schrodinger's equation in the steady-state expresses the probability of locating a particle at a point in space where the wave function Ψ(χ,ν,ζ) is a m e a s u r e of that probability and is given by ΨΨ

+ —^-(E

~ \Υ)Ψ

= 0

(1.18)

w h e r e m is the mass of the particle, W is t h e potential energy of the particle, and Ε is its total energy. This is t h e equation of a wave and w h e n applied to electron waves, where m is the mass of t h e electron, it gives a m e a s u r e of t h e probability of finding a given electron at a certain location. I n d e e d , t h e quantity | ψ | ί ί ν is the probability that a particle with potential energy W and total energy Ε will b e located in the spa­ tial volume dV at t h e point resulting from expressing Ψ in terms of x, y and z. Stated in a n o t h e r m a n n e r , | ψ | d V is the probability that a particle having potential energy W will b e located within dV and, if it is there, that it will then have total energy E. The wave function is allowed to b e a complex quantity and will, in general, for the steady-state, be a function of x,y and z. Wave Mechanics is, in effect, t h e fundamental branch of mechanics, and ordi­ nary N e w t o n i a n mechanics is derived from it. Thus, w h e n a particle moves in a force field, such that the change of potential occurs over distances that are very large c o m p a r e d with t h e wavelength associated with the particle, the laws of N e w t o n i a n mechanics are applicable. H o w e v e r , if the change in potential occurs over distances comparable to t h e wavelength, as in t h e case of t h e periodic potential of a crystal, then N e w t o n i a n mechanics does not apply and the wave n a t u r e of the particle gives rise to completely n e w p h e n o m e n a u n a c c o u n t e d for. We summarize the preceding discussion as: 2

2

1. The electron is not to be r e g a r d e d as a wave but as a particle, with an associated hypothetical de Broglie wave.

Section 1.9

Schrodinger's Equation

15

2. Over a small region where the electron is k n o w n to exist, a probability wave packet could be set up. The wave packet is a q u a n t u m mechanical m e a s u r e that can b e assumed to b e localized at a given point in space. It can b e said to be analogous to a classical particle that can be located in a given point in space. T h e wave packet consists of the summation of constant energy wave functions, Ψ, assembled about a center frequency so that this packet can be localized in a given point in space. A sketch of a wave packet is shown in Fig. 1.5. Over a dis­ tance that is largely c o m p a r e d with the wavelength, the motion of the electron can be described by the m o t i o n of t h e wave packet, such motion being governed by the laws of classical mechanics, although there is some uncertainty in the simultaneous values of position, m o m e n t u m , velocity, etc. 3. W h e n considering the motion of the electron over distances that are small com­ p a r e d with the wavelengths, such as inside the wave packet or a r o u n d the I nucleus, n o a t t e m p t should b e m a d e to locate the position of the electron along its path at two successive instants of time. A n y a t t e m p t to determine the first position will disturb the electron and cause it to trace out a different path. U n d e r such circumstances, it is appropriate to speak only of t h e probability of locating the electron at a certain point. This probability can be d e t e r m i n e d by a solution of Schrodinger's equation. Calculations in q u a n t u m mechanics use the probability function to d e t e r m i n e the position, velocity, and m o m e n t u m of a system. A s we shall see later, finding Ψ is not the ultimate result we are after. This quantity is a hypothetical one; a m a t h e m a t ­ ical tool used to obtain, mainly, the allowed system energies. In finding Ψ, two pos­ tulates are of considerable importance. These are: Postulate 1.

T h e function Ψ and its first derivative are finite, continuous, and single-valued.

Postulate 2.

The probability per unit length (per unit volume in the three-dimensional case) of finding a particle at a particular position and at a certain instant of time is Ψ Φ * , where Ψ* is the complex conjugate of Ψ . Τ η ε reason for multi­ plying the function by its complex conjugate is that the probability must b e a positive real quantity, whereas Ψ is complex for t i m e - d e p e n d e n t cases. If we integrate ΨΨ* over the entire system (entire volume), the result must be unity. H e n c e ,

χ

Figure 1.5 The composition of a wave packet at a certain instant. Particle is most likely to be at x but a smaller probability is that it could be found somewhere in the range Δχ. Q

16

Chapter 1

Atomic Structure and Quantum Mechanics

ΨΨ

* dxdydz=

\Ψ\ dx dy dz = 1

J space

(1.19)

·* space

where t h e p r o d u c t dxdydz is t h e incremental volume. This is equivalent to saying that it is certain that t h e particle is somewhere in space. In t h e next section, we apply Schrodinger's equation to a simple structure in o r d e r to illustrate t h e concepts of Wave Mechanics. W e then apply t h e solution to the study of t h e energy levels in t h e hydrogen atom. Application t o the Potential W e l l We shall assume, as shown in Fig. 1.6, that an electron is located in a potential well such that t h e potential energy W is zero b e t w e e n χ = 0 a n d χ = a, a n d infinite else­ where. T h e infinitely d e e p one-dimensional well is, in a way, typical of t h e Wave Mechanics p r o b l e m that o n e encounters in t h e study of t h e electron energies in a hydrogen atom. The total energy of t h e electron is kinetic, since W = 0. Therefore, t h e electron will b o u n c e b e t w e e n t h e walls, and, if perfectly elastic, will assume a stable state. We will d e t e r m i n e a n expression for Ψ in t h e potential well. Applying Schrodinger's equation in t h e steady-state (t—>°°) inside t h e o n e dimensional b o x w h e r e t h e potential energy W is zero, we have +

= ο

(1.20)

dx where k, labeled t h e wave vector, or wave n u m b e r , is given by k =

Vch?mE/r?

The solution t o E q . (1.20) is Ψ(χ,1ή = Asm kx + Bcos kx U7=c

(1.21) W~-

Electron in well has W = 0 and energy = Ε

x=0 Figure 1.6 Electron in a potential well.

Section 1.9

Schrodinger's Equation

17

The electron cannot p e n e t r a t e outside t h e potential well, so that the wavefunction is zero at χ = 0 and at χ = a. Thus, Ψ = 0 at χ = 0

and

at χ = a

from which we obtain, Β = 0, ka = m r

and

Ψ = Asm kx

(1.22)

w h e r e η = ± 1 , ± 2 , ± 3 , . . . . Then, setting the expression for k, which follows E q . (1.20), equal to ηττ/α, we have 2

2

a 8; equation becomes 2

E

Figure 1.10 Incident and reflected Region 1

x= 0

Region 2

waves together with penetrating wave.

The fact that Ψ has a n o n z e r o value implies that it is possible for an electron having energy Ε to exist in a region w h e r e t h e barrier energy, W, is greater t h a n t h e energy of the electron. This example can b e modified to include a third region, labeled Region 3, identical to Region 1 but placed to the right of R e g i o n 2. It can b e shown that it is possible for a parti­ cle having energy Ε in Region 1 to cross into R e g i o n 3 while going through a potential barrier of energy W, where W > E. The probability of this occurring is higher if the thick­ ness of Region 2 is reduced. H e n c e , the particle is said to tunnel through t h e barrier of Region 2. This concept is applied in t h e study of the E s a k i diode, also k n o w n as t h e tunnel diode, and in the formation of ohmic contacts. Having established the significance and plausibility of Schrodinger's equation, we will use it in the next chapter to establish the q u a n t u m n u m b e r s and, hence, t h e energy levels that a hydrogen electron can have. We will t h e n extend t h e results to t h e manyelectron sample, such as silicon. 2

REVIEW QUESTIONS Ql-6 Ql-7 Ql-8 Ql-9

What is the significance of Heisenberg's principle? What was de Broglie's contribution to Wave Mechanics? Briefly state the significance of Ψ in Schrodinger's equation. Interpret the solution of Schrodinger's equation to the deep potential well and how it relates to the science of Wave Mechanics.

HIGHLIGHTS • The properties of semiconductors are determined by phenomena on an atomic scale and, hence, require the explanations provided by quantum mechanics' concepts. • Wave Mechanics established the intimate link between the dynamic and the wave proper­ ties of the atom. Herein lies the analogy between a beam of electrons and a light beam. • De Broglie suggested that quantum-mechanical concepts in their duality applied not only to electromagnetic waves but to electrons as well.

22

Chapter 1

Atomic Structure and Quantum Mechanics

• Another link between particle-like properties and wave-like properties of matter was pro­ vided by the de Broglie relationship, which expresses wavelength in terms of momentum. • The most far-reaching step was taken when Schrodinger developed the extremely impor­ tant and complex equations that describe the properties of electrons in a physical system. It describes the large number of energies that an electron can have in a region where the energy is confined.

EXERCISES El-3 A tennis player serves a 75g ball at a speed of 200 km/h. If the uncertainty in the veloc­ ity of the ball is lOcm/s, determine the uncertainty in its position. Ans:

32

Δχ = 1.4 X 10" m

El-4 Use Schrodinger's second postulate to determine the expression for A in the solution of Schrodinger's equation, in one dimension, for the infinitely deep potential well. Ans: A =

Vl/a

El-5 Determine the energy in eV of an electron for η = 1 in the infinitely deep well when a = 50A. 3

Ans: Ε = 14.92 Χ ΙΟ" eV

PROBLEMS 1.1 Determine the density of GaAs given the lattice constant L = 5.65A, and the molecu­ lar weight is 144.63/mole. Avogadro's number is 6.02 Χ 10 atoms/mole. 1.2 Use the Bohr model for the hydrogen atom to plot potential energy and total energy as the radius r from the proton increases. Clearly identify the magnitudes of the kinetic energy at two separate radii. 23

1.3 Determine the velocity of an electron in the ground state of the hydrogen atom. 1.4 The laws of classical physics apply to the motion of a particle provided the dimensions of the system are much larger than the deBroglie wavelength. For the following elec­ trons, determine whether the laws of classical physics apply: a) An electron is accelerated in the beam of a cathode-ray tube that has an acceler­ ating voltage of 30KV. b) An electron that is accelerated by a potential of 100V in a device whose dimen­ sions are of the order of 2cm/s. c) An electron in a hydrogen atom. 1.5 Determine the energy of a photon having wavelengths λ = ΙΟ,ΟΟΟΑ and λ = 10A. Express the energy in eV and J. 1.6 The antenna of an AM radio station transmitter radiates 100KW of power at lOOOKHz. a) Calculate the energy of each radiated photon. b) Calculate the number of photons radiated per second. 1.7 An oscillator is operating at a frequency of 10MHz. a) Calculate the energy of the quantum of radiation of the oscillator.

Section 1.9

Schrodinger's Equation

b) Calculate the number of quanta in 10 J. a) Derive an expression for the wavelength of spectral lines emitted by the transi­ tions from the excited states to the ground state using the Bohr relation for the electron energy in the hydrogen atom, b) Calculate the wavelength of the first four spectral lines. The velocity of a certain free particle is 5 X 10 m/sec. The mass of the particle is 10~ kg. Determine: a) the particle energy. b) the de Broglie wavelength. In accordance with physics statistics, the average energy of an electron in a medium of free electrons at thermal equilibrium is 3kT/2, where k is Boltzmann's constant and Τ is in degrees kelvin. Determine, for the electron, a) its velocity, b) its momentum, c) the de Broglie wavelength at Τ = 300K. An electron is moving with a velocity of 10 m/s. Determine, for the electron: a) its momentum, b) its de Broglie wavelength in m and A, c) its energy in J and eV. An electron has a de Broglie wavelength of 100A. Determine: a) electron momentum, b) electron velocity. For an infrared radiation of 1 μηι, determine: a) the frequency of the radiation, b) the energy of the photon in eV. The uncertainty in the position of a particle having mass 10~ Kg is 10A. Determine the uncertainty in: a) the momentum of the particle, b) kinetic energy of the particle. For the electron in Problem 1.11, determine the wave vector k. For an infinitely deep potential well having a = 100A, determine for an electron the energy levels for « = 1,2 and 3. Calculate the energy in eV and Joules. An electron is located in a one-dimensional potential energy well having width of 3 A. Determine a) the kinetic energy of the electron in the ground state. b) the frequency of the spectral radiation of an electron that drops from the next higher state to the ground state. For a particle that has a mass of 2 grams and energy 1.5kT, determine the de Broglie wavelength at Τ = 300K. 6

1.8

1.9

5

30

1.10

1.11

1.12

1.13

1.14

1.15 1.16 1.17

1.18

23

5

30

chapter 2 ENERGY BANDS AND CURRENT CARRIERS IN SEMICONDUCTORS

2.0

INTRODUCTION We concluded in C h a p t e r 1 that in accordance with the Uncertainty Principle it is not possible to specify, at the same time, the location or the m o m e n t u m of an elec­ tron in a solid. This Principle points to one of the weaknesses of the B o h r hypothe­ sis, which assumed that electrons could b e assigned to certain orbits, which in essence implied that their position was known. We then decided that Wave Mechanics' concepts are n e e d e d to explain t h e behavior of electrons in solids. In this chapter, we will apply Schrodinger's equation to obtain information on the hydrogen atom. We then project the results, with the necessary modifications, to the many-electron solid. We establish the existence of discrete energy levels from which, because of their large n u m b e r and t h e closeness of these levels, energy bands result. T h e highest valence b a n d and t h e lowest conduction b a n d are of p a r a m o u n t interest. These b a n d s are separated by a region in which n o electrons of t h e semi­ conductor can exist: This is the forbidden band. A t each of these two t o p bands, a different carrier is said to exist. These carri­ ers are the electron and a vacant space k n o w n as t h e hole. We will use Schrodinger's equation to formulate the conditions existing in the hydrogen atom.

2.1 A P P L I C A T I O N O F S C H R O D I N G E R ' S E Q U A T I O N T O THE H Y D R O G E N ATOM The B o h r theory of the hydrogen a t o m is n o longer satisfactory for two major rea­ sons. First, it assumes that the electron rotates a r o u n d t h e nucleus in a given orbit.

24

Section 2.1

Application of Schrodinger's Equation to the Hydrogen Atom

25

This is not acceptable as it implies that t h e position of the electron at a given time is k n o w n and this violates Heisenberg's Uncertainty Principle. Second, it is not possi­ ble to apply B o h r ' s theory to an a t o m that has m o r e t h a n one o u t e r electron. We will therefore consider a quantum-mechanical solution since it is not constrained by these limitations. We will apply the results of the solution of the Schrodinger equa­ tion to d e t e r m i n e the possible energy states of the hydrogen electron and to study the many-electron problem.

Quantum Numbers T h e hydrogen atom is readily approximated by a nucleus of charge +q, fixed in position, in whose field the electron orbits. The electron, w h e n located at distance r from the nucleus has potential energy, given by E q . (1.6), as -q /Airs r. The poten­ tial energy distribution of the hydrogen a t o m is shown in the sketch of Fig. 2.1(a). The potential distribution is, in a way, similar to t h e potential well we consid­ ered in C h a p t e r l . T h e potential energy (negative) is highest at infinite r and lowest at t h e nucleus. A major difference b e t w e e n t h e two is the m o r e complex b o u n d a r y 2

0

Ε

Radial distance r

Atom core (a) ζ

y

Figure 2.1 (a) Illustration of potential well of hydrogen atom, (b) Spherical polar coordinates for Schrodinger's equation.

Chapter 2

Energy Bands and Current Carriers in Semiconductors

profile of the hydrogen atom potential distribution. F u r t h e r m o r e , the potential well is a one-dimensional p r o b l e m , whereas we wish to solve for the hydrogen a t o m in three dimensions. We will p r o c e e d with an outline of the general steps in the solution of Schrodinger's e q u a t i o n for the hydrogen a t o m and then present the results of this solution. We begin by substituting the expression for the potential energy in Schrodinger's equation, E q . (1.18), in t h r e e dimensions, so that we have, (2.1) We shall not solve this equation but we will outline the steps involved in the solution. First, the system, or the hydrogen atom, consists of a p r o t o n of charge q and an electron of charge —q and mass m . T h e position of the p r o t o n is fixed at the origin, as shown in Fig. 2.1(b), and t h e electron is at a point whose spherical coordi­ nates are (r, Θ, and φ ) . Second, E q . (1.18), written in three dimensions, is trans­ formed into the new coordinate system. Third, the m e t h o d of separation of variables is used to solve the Schrodinger equation. By this method, we assume a solution of the form (2.2)

Ψ (r, θ, φ) = /?(/•)/Γ(θ)/(φ)

W h e n t h e assumed solution is substituted in the spherical-polar form of Schrodinger's equation, three separate differential equations result; one in r only, one in θ only, and o n e in φ only. In the solution of the three equations, and to obtain physically acceptable results, three q u a n t u m n u m b e r s emerge; n, i and m , k n o w n as the principal, azimuthal, and magnetic quantum n u m b e r s respectively. This is analo­ gous to the single q u a n t u m n u m b e r n, obtained for the electron in the potential well example discussed in the previous chapter. Let us digress briefly to explain how a solution is obtained so as to identify the sources of these q u a n t u m n u m b e r s and their relation to the physical properties of the electron and to t h e electronic orbits. We will now consider a simplified Schrodinger equation and examine the results. This equation is i n d e p e n d e n t of θ and φ but is a function of r, h e n c e labeled the radial equation. T h e solutions to this equation have spherical symmetry. Before proceeding any further, we note that just as in the example of the potential well dis­ cussed in C h a p t e r 1, our eventual interest is not in the expressions for t h e wavefunc­ tions. For this work, we are interested in the conditions for which the wavefunctions satisfy the Schrodinger equation and the b o u n d a r y conditions. The radial form of the differential equation in Ψ has a series of solutions. The simplest and most basic solution, Ψ, expresses Ψ as a function of r and a factor identified as r . By substituting this solution in Schrodinger's radial equation, we obtain expressions for Ε and r (see prob. 2.6). W h e n higher order solutions for Ψ, such as Ψ and Ψ , are included, we obtain energy as a function of t h e quantum number, n. This n u m b e r is a m e a s u r e of the total energy of the electron in a particu­ lar state as well as being a m e a s u r e of t h e diameter of t h e major axis of the elliptical orbit, which is twice the radius of the circular orbit d e t e r m i n e d by Bohr. e

χ

n

Q

2

3

Section 2.1

Application of Schrodinger's Equation to the Hydrogen Atom

27

T h e q u a n t u m n u m b e r , n, is an integer that takes on t h e value 1, 2, and 3. Because of the association of η with t h e energy of the electron, the energy levels in η are also designated by the electronic shells K, L, and M, corresponding to η = 1,2, and 3. A quantum-mechanical interpretation of the orbital radius of the hydrogen a t o m is that it is n o longer a fixed n u m b e r (and t h e electron having a fixed orbit) but that there is a high probability of locating the electron at a distance identified by a quantum number. Two o t h e r q u a n t u m n u m b e r s result from the complete solution of t h e equa­ tions in each of θ and φ of the Schrodinger equations. The azimuthal number, i, results from the solution of t h e equation in θ such that its value determines b o t h the quantized angular m o m e n t u m as €/Ι/2ΤΓ and the d i a m e t e r of the minor axis of the quantized elliptic orbit of the electron as minor axis diameter n

major axis diameter

The values of t h e set of azimuthal q u a n t u m n u m b e r s cover t h e range € = 0to€ = n -

l

In the solution of the equation for φ, the magnetic orbital quantum number, m appears, determines t h e direction of t h e orbital angular m o m e n t u m , and has inte­ gral values that vary from — i to + € , including m = 0. The word magnetic results from the fact that an electron in an orbit represents an accelerating charge and, h e n c e an electric current that has a magnetic field. T h e existence of these three q u a n t u m n u m b e r s was predicted earlier by Sommerfeld. Later, G o u d s m i t and Uhlenbeck, guided by their knowledge of spectra m o r e complicated than that of hydrogen, predicted that the electron itself possesses a magnetic m o m e n t and an angular m o m e n t u m quite i n d e p e n d e n t of its rotation in an orbit a r o u n d the nucleus. F u r t h e r m o r e , it was observed that in a many-electron atom, the actual n u m b e r of q u a n t u m states is twice that predicted by the combination of n, i and m . It was then postulated that t h e electron, in its orbit a r o u n d the nucleus, spins on its own axis in two unique a n d opposite directions with respect to its orbital m o m e n t u m . Thus, each electron can exist in two spin states. A s a result, a fourth q u a n t u m n u m ­ ber, labeled the spin quantum number, m was included, which has a value of ± 1 / 2 . This intrinsic spin with respect to the orbital angular m o m e n t u m results in t h e dou­ bling of the n u m b e r of q u a n t u m states. Therefore, every q u a n t u m state is identified by four q u a n t u m numbers, designated by n, €, m , and m . (

(

(

s

(

s

T h e Pauli Exclusion Principle In 1925, Pauli formulated the principle that t h e r e can be only two electrons of o p p o ­ site spin in a q u a n t u m state. H e concluded this from a study of atomic spectra at about the same time that the extra q u a n t u m n u m b e r , m , was identified. H e , in effect, concluded that n o m o r e t h a n two electrons can have the same distribution in s

Chapter 2

Energy Bands and Current Carriers in Semiconductors

space. Thus, any q u a n t u m state, identified by the four q u a n t u m numbers, can accom­ m o d a t e n o m o r e t h a n o n e electron. Actually, the q u a n t u m n u m b e r s represent a n o t h e r way of defining the wavefunction of a given electron. To comply with t h e Pauli Exclusion Principle, it follows that each electron in the solid, regardless of the n u m b e r of electrons, is identified by four q u a n t u m num­ bers. In Table 2.1 are listed t h e various q u a n t u m n u m b e r s together with the maxi­ m u m n u m b e r of electrons that can be a c c o m m o d a t e d in an isolated atom. The r e a d e r will question our reference to m a n y electrons, whereas the origin of t h r e e of the four q u a n t u m n u m b e r s resulted from a solution of Schrodinger's equation for the one-electron m o d e l . In fact, the levels shown in Table 2.1 can be interpreted as the possible states that the hydrogen electron can have. F u r t h e r m o r e , if t h e results for t h e many-electron p r o b l e m are to b e derived from the single elec­ tron model, then the various levels defined by the q u a n t u m n u m b e r s η and € can a c c o m m o d a t e the n u m b e r of electrons shown in the table. TABLE 2.1 Shell

Quantum Numbers and Number of Electrons.

n>0

0< €s η- 1

-€ W . It is necessary to r e p e a t h e r e that our information is not obtained from the expressions for the wavefunctions but in the expressions that result from applying the b o u n d a r y conditions 0

0

Q

0

w

w , r

0

a

b X

0

L Figure 2.9 One dimensional Kronig-Penney model of the potential energy distribution.

Chapter 2

Energy Bands and Current Carriers in Semiconductors

and from the equations resulting from substitution of the expressions for the wave functions in Schrodinger's two equations. The two i m p o r t a n t solutions consist of an expression on t h e left-hand side of each of the equations, which include Ε and W , and one term on t h e right-hand side of each equation, given by cos k {a + b), whose value must lie b e t w e e n —1 and + 1 . Certain values of energy, E, m a k e the left-hand side of these equations greater t h a n 1 or less than - 1 . Those ranges of energies result in t h e forbidden bands. The o t h e r energy ranges are those that are either occupied by electrons or that could be occupied by electrons. A plot of these two equations is exhibited in Fig. 2.10(a). For the sake of comparison, we show in Fig. 2.10(b) the distribution of energy, E, for t h e potential well discussed in C h a p t e r 1. T h e following observations can b e m a d e from the plots: 0

•

T h e K r o n i g - P e n n e y m o d e l plot displays discontinuities and perturbations w h e n c o m p a r e d to the parabolic shape of t h e free particle solution. These modifications are greatest at t h e lower values of energy. A t the higher values of energy, the solution for this m o d e l is similar to that of t h e free particle. The implication is that the greater t h e electron energy, the less is the importance of the periodic potential in the crystal.

•

The K - P m o d e l plot has discontinuities at kL = ±BTT. O n e can conclude that at the values of energies corresponding to these discontinuities, electron waves cannot p r o p a g a t e in the solid. These values of k m a r k the boundaries of the energy zones that m a k e u p the forbidden bands.

•

A t higher values of E, t h e width of the permitted bands increases while the width of t h e forbidden bands is reduced.

•

T h e left-hand sides of the equations that are used to sketch Fig. 2.10 are not changed if kL changes by ±2ττ. For reasons related to the masses of t h e parti­ cles and that will b e c o m e m o r e clear later, it is convenient to shift the curves of t h e second and third allowed b a n d s by ± 2 T T / L along the jc-axis.The values

k

Figure 2.10 (a) Energy versus wave number k using the Kronig-Penney model. The dashed curve that is superimposed is the free particle solution, (b) Energy versus wave number for the potential well.

Section 2.4

Mathematical Model of Band Formation

39

of the second and third allowed b a n d for positive k are shifted by — 2TT/L and those for negative k are shifted by 2TT/L. In this manner, the sketch of Fig. 2.10 is limited to the region —ττ/L to ττ/L, as shown by the dotted curves m a r k e d c',d',e', a n d / ' , replacing the curves c, d, e, a n d / . T h e new curves are shown sketched in Fig. 2.11 and they represent the energy profile for the valence and conduction bands corresponding to η = 1,2, and 3. We have shown in Fig. 2.11 energy curves for three values of n. Obviously, by use of t h e two equations referred to earlier, it is quite possible to extend the sketches to higher values of n. H o w e v e r , the ones shown in the figures are a d e q u a t e to illustrate t h e relevance and benefits of the K r o n i g - P e n n e y model. It is important to realize that the one-dimensional K r o n i g - P e n n e y model bears a very general resemblance to the actual conditions existing in a crystal. In a real three-dimensional crystal, the E-k relationships are m u c h m o r e complicated than those obtained in Fig. 2.10. H o w e v e r , the K r o n i g - P e n n e y m o d e l results have exhibited two properties that are extremely important. First, bands exist in which electrons cannot exist and, second, the shape of the E-k curves at the locations of the forbidden b a n d s indicate that an opposite concavity exists b e t w e e n the shapes of t h e allowed b a n d s above and below a certain forbidden band. We will refer to this p r o p e r t y in relation to the effective mass later in this chapter. Direct a n d Indirect S e m i c o n d u c t o r The actual b a n d structures of semiconductors are m u c h m o r e complex than those shown in Fig. 2.11. O n e distinguishing feature of semiconductors is the location of the conduction b a n d energy minimum with respect to the valence b a n d m a x i m u m on t h e E-k diagrams. In silicon and germanium, and as shown in Fig. 2.12(b), the valence b a n d m a x i m u m does not occur at the conduction b a n d minimum. The valence b a n d m a x i m u m in all semiconductors occurs at k = 0, whereas the conduc­ tion b a n d m i n i m u m for Si and G e occurs at a different k, indicating a difference in m o m e n t u m b e t w e e n these two points. In gallium arsenide, the conduction b a n d minimum and the valence b a n d m a x i m u m occur at k = 0, as shown in Fig. 2.12(a).

η - 3 allowed band

Forbidden band η = 2 allowed band Forbidden band η = 1 allowed band

Figure 2.11 Energy distribution from the Kronig-Penney model for the lower three allowed bands.

40

Chapter 2

Energy Bands and Current Carriers in Semiconductors

Ε

Ε

Ε

Ε

s

k

Gallium arsennide (a)

K

Silicon

Figure 2.12 E-k sketches for (a) direct band gap and (b) indirect band gap semiconductors.

H e n c e , G a A s is known as a direct band gap semiconductor, and Si and G e are known as indirect band gap semiconductors. In a direct b a n d gap semiconductor, a p h o t o n of light energy, hv, can excite an electron from the t o p of the valence b a n d to the b o t t o m of the conduction band. Similarly, an electron in the conduction b a n d of a direct semiconductor can fall directly into an empty state in t h e valence b a n d and emit p h o t o n s of light having energy Ε . In indirect semiconductors, an electron in the conduction b a n d cannot fall directly into the valence b a n d because it must u n d e r g o a change in energy and a change in m o m e n t u m . A p h o t o n by itself cannot excite an electron from the t o p of the valence b a n d of an indirect semiconductor to the b o t t o m of t h e conduction b a n d because t h e p h o t o n has sufficient energy to cause the transition but does not possess the necessary m o m e n t u m for this transition. A n electron moving b e t w e e n the valence b a n d and the conduction b a n d of an indirect semiconductor can occur through a defect in the semiconductor or by the action of phonons, which can provide sufficient m o m e n t u m to assist indirect transi­ tions. The application to which the direct semiconductors b e c o m e i m p o r t a n t is the optical device. In this case, G a A s is a principal semiconductor used in semiconduc­ tor lasers and light-emitting diodes.

2.5 C O V A L E N T B O N D M O D E L A representation that complements the energy b a n d diagram known as the covalent bond model is shown in Fig. 2.13. This diagram is a two-dimensional form of the dia­ m o n d lattice structure shown in Fig. 1.1 in which each a t o m is b o n d e d to its four nearest neighbors. The bonding is a result of the fact that each a t o m shares four outermost-orbit electrons with four adjacent atoms. These four electrons occupy the 3s and 3p levels in the energy b a n d representation, shown in Fig. 2.6. A t Τ = OK, t h e r e will be AN

Covalent Bond Model

Section 2.5

\/ \/ \/ '

"

'

W

V

V

\

/

V

V

V

V

\

/

\/ %

%

"

V

\

/

V

\

V

41

/

XXX-X //

%

ν

ν

/

~

\

^

\

Free electrons

(ft) Figure 2.13 Covalent bond model of a semiconductor (a) at low temperature and (b) at room temperature.

lines in Fig. 2.13(a), Ν being the n u m b e r of atoms in the crystal, which normally fill the lower b a n d (the valence b a n d ) shown in Fig. 2.6 and Fig. 2.7(b). T h e lines shown in the figure represent one electron each and the circles rep­ resent the a t o m core, which includes t h e nucleus and all other electrons of the atom except the ones in the outermost orbits, such as t h e 3s and the 3p levels in silicon. It is of interest to realize that it is the covalent bonding that imparts hardness to G r o u p I V semiconductors, such as G e and Si. Of course, carbon is the hardest m a t e ­ rial. In III-V semiconductors, such as G a A s , t h e binding forces represent the cova­ lent bonding in addition to t h e ionic forces so that the strength of the bonding is high. This translates into a higher b a n d gap energy and a higher melting point w h e n c o m p a r e d to G r o u p I V semiconductors having the same atomic n u m b e r and lattice constant. A t the n o r m a l atomic spacing, R , and for a semiconductor at OK, all electrons are attached to their cores and are said to occupy their lowest possible levels of energy in the b a n d diagram. A n increase of t e m p e r a t u r e imparts energy to the elec­ trons, and at 300K ( r o o m t e m p e r a t u r e ) , m a n y electrons are m o v e d from the top of the valence b a n d to the b o t t o m of the conduction band. If the imparted energy is exactly equal to t h e b a n d gap energy, the electrons at t h e top of t h e valence b a n d are transferred to the b o t t o m of the conduction b a n d and possess potential energy E For energy greater than t h e b a n d gap Ε , t h e electrons at the t o p of the valence b a n d m o v e u p to t h e conduction b a n d and acquire kinetic energy; their total energy places t h e m above E A t Τ = 300K, approximately one in 1 0 silicon electrons escapes a covalent b o n d and becomes what is k n o w n as a free electron able to travel t h r o u g h o u t the structure. In the energy b a n d model, an electron becomes free w h e n it has b e e n m o v e d from the t o p of t h e valence b a n d to the b o t t o m of the conduction band. With this in mind, we will n o t e one distinction b e t w e e n metals and semicon­ ductors. T h e conductivity of metals decreases with increasing t e m p e r a t u r e , whereas, 0

c

c

12

42

Chapter 2

Energy Bands and Current Carriers in Semiconductors

as we shall see later, the conductivity of semiconductors, under certain conditions, may increase with increasing t e m p e r a t u r e . In metals, the conduction (free) electrons are distant from their nucleus, move freely a m o n g atoms, and are b o n d e d to different atoms at different times. A metal is thus said to consist of an array of positive ions s u r r o u n d e d by closed-shell electronic orbits immersed in a gas of free electrons. T h e r e is n o forbidden gap that the elec­ trons have to overcome. While there are approximately 1 0 free electrons per c m in silicon at r o o m t e m p e r a t u r e , a metal will have approximately 1 0 electrons per cm . 10

3

20

3

2.6 C U R R E N T C A R R I E R S — E L E C T R O N S A N D H O L E S A unique feature of semiconductors that results from their particular energy b a n d diagram is that two types of carriers exist: electrons in the conduction b a n d and holes in the valence band. In metals, conduction consists of the controlled motion of electrons only. So, where d o t h e holes c o m e from a n d h o w d o they b e h a v e as current carriers? T h e Hole W h e n a p u r e semiconductor, such as silicon, which is initially at Τ = OK, acquires t h e r m a l energy equal to or greater than the b a n d gap energy, Ε , electrons are excited from the top of the valence b a n d and into the conduction band. They b e c o m e free electrons. For every electron that leaves the valence band, a vacancy in the covalent bonding is left behind into which another electron in that valence b a n d may move. W h e n an electric field is applied to the silicon, the electrons in the con­ duction b a n d acquire velocity in a direction opposite to that of the field. Similarly, the electrons in the valence band, which m o v e d to fill the vacancies, gain velocity. By having the vacancy occupied by a n o t h e r electron, the vacancy moves in t h e direc­ tion of the field. This vacancy is the hole. Thus, b o t h the electron and the hole cause electric current in the same direction with the hole moving in the direction of the field and the negatively charged electron moving opposite to the direction of the electric field. O n e crude analogy is that of a two-level parking garage where, initially, all the cars are arranged in a single row in the ground floor and the u p p e r floor is com­ pletely empty. Until a car is m o v e d to the first floor from the ground floor, there can be n o motion of cars in either floor since the ground floor is full and the first floor is empty. W h e n the front car is m o v e d upstairs, then all the cars behind the space that the front car occupied can now move one car-length forward, thus causing the empty space to move towards the back. The car that was moved upstairs is available for motion and it is analogous to our electron in t h e conduction band. The vacancy that was created on the ground floor is analogous to the hole and it moves in a direction opposite to the motion of the cars. A n o t h e r analogy is represented by the cylinders shown in Fig. 2.14. The b o t t o m cylinder in Fig. 2.14(a) is completely filled with a liquid and the top cylinder is empty. These cylinders represent the valence and conduction bands

Current Carriers—Electrons and Holes

Section 2.6

43

Liquid motion

(a)

(b)

Figure 2.14

Analogy to bands in silicon.

respectively of silicon at absolute zero. In this state, there can b e n o motion of liquid in t h e b o t t o m cylinder and, obviously, n o n e in the top cylinder. W h e n a small vol­ u m e of liquid is transferred from t h e b o t t o m cylinder to the t o p cylinder, a possibil­ ity exists for motion of liquid in b o t h cylinders: the small volume of liquid in the t o p and the b u b b l e left behind in t h e b o t t o m cylinder. By tipping b o t h cylinders to the right side in Fig. 2.14(b), gravity forces the liq­ uid in the t o p cylinder to move right and the bubble in the b o t t o m cylinder to m o v e left. T h e small volume of liquid and t h e bubble move in opposite directions. The force of gravity is analogous to an applied electric field, t h e small volume is analo­ gous to the free electron and t h e b u b b l e is analogous to t h e hole. A n a l y t i c a l Description of t h e Hole A m o r e analytical description of the hole is d e t e r m i n e d as follows: Electrons are thermally excited from the valence b a n d to the conduction band, leaving empty states in the valence band. W h e n an electric field is applied, electrons in t h e conduction b a n d are accelerated, and so are the electrons in the valence b a n d as they m o v e into the empty states. T h e current density ( a m p s / m ) of electrons in the valence band, J , can b e d e t e r m i n e d by a s u m m a t i o n of t h e motion of all the electrons in the valence b a n d as: 2

VB

Kb =

Σ

v

" 1d

(2-3)

w h e r e ( — q) is the charge of an electron, v is its drift velocity, and t h e negative sign indicates that t h e direction of current is opposite to t h e direction of motion (v ) of the electrons. Mathematically, one can also state that the current density in the valence b a n d is m a d e u p of two c o m p o n e n t s : t h e s u m m a t i o n of t h e motion of all the electrons in a completely filled b a n d (no vacancies) minus that current associated with t h e miss­ ing electrons as, d

d

^~qv filled band d

0 0

_^~qv empty states d

Since t h e current in a filled b a n d is zero, the current resulting from t h e avail­ ability of empty states is the current in t h e valence b a n d given by

Chapter 2

Energy Bands and Current Carriers in Semiconductors

J = VB

Σ

Φ*

(2-5)

Because t h e direction of motion of t h e carriers is in response to an electric field, we conclude that t h e hole has a positive charge and the free electron has a negative charge since the electric field causes the holes and the electrons to m o v e in opposite directions. Electron a n d Hole Energies The generation of an electron-hole pair in the bands is shown in Fig. 2.15(a) and in t h e covalent picture is shown in Fig. 2.15(b). In the b a n d picture, the electron has m o v e d u p in energy, and in the covalent b o n d illustration, t h e electron is free to r o a m with its place becoming available for occupancy. In the energy b a n d diagram, the word " e n e r g y " refers to the energy of the electrons. W h e n t h e energy of t h e electron is increased, that electron occupies a higher energy level in t h e band. This applies equally well to conduction b a n d and valence b a n d electrons. A n increase in the energy of a hole is caused by the raising of the energy level of the electron. A s electrons move u p within the valence band, holes b e c o m e avail­ able and m o v e d o w n t h e b a n d occupying lower levels of t h e electron energies. Thus, an increase in the energy of a hole is associated with downward motion in the valence band. A n electron located at the level of the b o t t o m of the conduction b a n d at rest has potential energy E , and zero kinetic energy. Similarly, a hole located at the top of the valence b a n d E at rest, has potential energy E , and zero kinetic energy. c

v

v

f

\/ \/ V

Ο (a) P.E. of electron

1 K.E. of electron on on .a υ a

K.E. of hole

m

U

Ά

a

Ο

"-ι

xt

Increasing electron energy

%

o

/y

vYyv vVV v Hole

Electron

(b)

Ο

P.E. of hole (c)

Figure 2.15 Generation of electron/hole pair; (a) energy band model, (b) covalent bond model, and (c) directions of energy increases of electrons and holes.

Section 2.7

Effective Mass

45

Electrons located above E and holes below E , as shown in Fig. 2.15(c), are said to have kinetic energy given by the difference b e t w e e n their energy location and the respective b a n d edge energy. c

y

2.7 EFFECTIVE M A S S We have just concluded that a hole possesses a positive charge, q, and t h e electron's charge is — q where q = 1.6 X ICC coulombs. H o w e v e r , the mass of t h e electron in t h e conduction b a n d and t h e mass of the hole in the valence b a n d are quite different and b o t h differ from the mass of an electron in a vacuum. The mass of a particle will b e defined as the ratio of t h e net force on t h e parti­ cle to t h e acceleration that t h e particle experiences. In using t h e mass of an electron in v a c u u m or free space, one completely ignores the effect of the crystal in which the electron is immersed. In fact, the forces within the crystal are much greater than t h e force exerted by an electric field of the strength normally applied to an electron. The forces in t h e crystal are k n o w n as lattice forces. In o r d e r to account for the lattice forces in the equation of motion, we intro­ duce t h e following two equations: 19

dv F + lattice forces = m — "dt dv n

F=

m

-

t

(2.6)

where F is t h e externally applied force, as by the electric field, m is t h e mass of an electron in a space free of lattice forces, and m* is t h e mass of the electron that includes t h e effect of t h e forces in the crystal. Such a mass, m*, is labeled the effec­ tive mass of an electron and m* is t h e effective mass of a hole. We will now deter­ mine expressions and definitions for the effective masses. In C h a p t e r 1, we d e t e r m i n e d that for b o t h a free particle and for t h e particle in the well, the energy is given by Q

2

P

E = f~

2m

(2.7) 0

where ρ is the particle m o m e n t u m and m is the free-electron mass. The motion of an electron in t h e conduction b a n d of a semiconductor is analogous to that of a free particle, except that t h e conduction b a n d electron is subjected to lattice forces. In this case, we label this mass as t h e effective mass of an electron and we define the energy as 0

where m* is the effective mass of an electron and ρ is labeled t h e crystal m o m e n t u m in reference to t h e forces of the crystal that act on t h e electron. In an o p e r a t i o n sim­ ilar to Eqs. (1.27) and (1.29), m o m e n t u m is also written as

Chapter 2

Energy Bands and Current Carriers in Semiconductors ρ =

hk/2ir

(2.9)

w h e r e h is Planck's constant and k is t h e wave n u m b e r . The expression for t h e energy of an electron in the lattice is obtained by using E q . (2.9) in E q . (2.8), Ε =

2

2

2

h k /8TT m*

(2.10) 2

The important conclusion is that the energy is directly proportional to k and inversely p r o p o r t i o n a l t o t h e effective mass of an electron. It is obvious from Fig. 2.11 that t h e effective mass is not necessarily constant, it is a constant only if the relation of Ε to k is purely parabolic. Based on E q . (2.10), the effective mass of an electron is defined as (2.11) O n close study of the E-k plot of Fig. 2.11 for an electron in a varying field, it is a p p a r e n t that the second derivative of Ε with respect to k is negative w h e n the E-k curve is concave d o w n w a r d s and it is positive w h e n t h e curve is concave upwards. T h e curves n e a r the m i n i m u m of the conduction b a n d and n e a r t h e m a x i m u m of the valence b a n d are nearly parabolic so one can safely assume that the mass of the electron is constant in those two regions. H o w e v e r , n e a r t h e m i n i m u m of the con­ duction band, t h e E-k curve is concave upwards so that t h e mass of an electron is positive, whereas near the m a x i m u m of the valence band, the curve is concave downwards so that the mass of an electron is negative. T h e direction that a particle will b e accelerated w h e n an electric field is applied is d e t e r m i n e d by t h e sign of t h e mass and the charge as a =

q%/m

(2.12)

A positive mass electron (one n e a r the b o t t o m of t h e conduction b a n d ) having a negative charge will b e accelerated in a direction opposite to that of the field, in accordance with E q . (2.12). The motion of an electron opposite to the direction of the electric field constitutes a current in the direction of the field since current is defined as being opposite to t h e direction of t h e motion of an electron. A n electron (negative charge) n e a r the t o p of the valence band, which has a negative mass, m a y b e regarded, as far as its motion is concerned and in accordance with E q . (2.12), as a positive mass-positive charge particle as t h e two negative signs in E q . (2.12) are replaced by a positive sign. Therefore, the electron n e a r t h e t o p of the valence b a n d is accelerated and causes current in the direction of the electric field. In conclusion, an electric field applied to a semiconductor accelerates elec­ trons in t h e conduction b a n d in an opposite direction to t h e electric field and accel­ erates positive particles in the valence b a n d in the same direction as the electric field. The result is a current in t h e direction of the electric field consisting of elec­ trons (positive mass, negative charge) moving in t h e conduction b a n d and the parti­ cles (positive mass, positive charge) moving in the valence band. These particles are the holes.

Section 2.8

Conductors, Semiconductors, and Insulators

47

A p p r o x i m a t e values of the effective masses of electrons m* and holes m* rel­ ative to t h e masses in a vacuum, are listed in Table 2.3. We n o t e from the table that the effective masses of electrons and holes in gallium arsenide are smaller than those in silicon. We refer to the E-k diagrams of Fig. 2.12 and n o t e that t h e curvature of the Ε vs. k diagram, in particular near the b o t t o m of the conduction b a n d in G a A s , is considerably greater t h a n that in Si. Consequently, and in accordance with Eq. (2.11), electrons in G a A s have a m u c h smaller effective mass than in Si. TABLE 2.3 Density of states effective masses of electrons and holes in Si, Ge, and GaAs and 300K

m*/m m*/m

0

n

Si

Ge

GaAs

1.18 0.81

0.55 0.37

0.065 0.52

These masses a r e labeled as density of states effective masses. A different def­ inition for masses with different values is used for calculations of t h e mobilities of electrons a n d holes. They are labeled conductivity effective masses. We will define mobility of a particle as t h e ratio of t h e velocity of the particle to t h e elec­ tric field. In this b o o k , all references for effective mass will b e to that of t h e density of states.

2.8 C O N D U C T O R S , S E M I C O N D U C T O R S , A N D I N S U L A T O R S The feature that distinguishes these three types of materials is the extent of their ability to conduct electrical current. For conduction of current to take place, t h e fol­ lowing r e q u i r e m e n t s must b e met: 1. T h e r e must b e energy b a n d s that are partially filled with electrons. Since elec­ trons occupy the lowest b a n d s first, these partially filled b a n d s are located at or near the top of an energy band. 2. A n electric field must b e applied to accelerate the electrons in the partially filled bands. In being accelerated, electrons gain energy, but c o m p a r e d to the energies separating the bands, if a forbidden b a n d exists, this energy is very small. For energy bands to b e c o m e partially filled with electrons, either electrons are lifted, by acquiring energy, from a completely filled b a n d to a completely empty band, or there are b a n d s of energy that are empty and overlap filled b a n d s without the existence of forbidden bands. The structural difference b e t w e e n insulators and semiconductors is that insu­ lators have a very wide b a n d separating the valence b a n d from the conduction band, whereas semiconductors have a m u c h smaller b a n d gap energy. In conductors, the t o p occupied b a n d is only partially filled with electrons. We show in Fig. 2.16 the relationship b e t w e e n conduction and valence b a n d s in t h e three materials.

Chapter 2

Energy Bands and Current Carriers in Semiconductors Metals

Overlap

Overlap

Filled

Filled

Partially filled

I'anialK filled

Filled

Filled

Empty

Partially filled

Metals

Semiconductors

E 3eV g

Filled ( a ) T = 0K

Filled (6)Γ=300Κ

Figure 2.16 Conditions of energy bands in metals, semiconductors, and insulators at (a) very low temperatures and (b) at Τ = 300K. A t low temperatures, the conduction b a n d of a semiconductor is practically empty. Applying an electric field to the solid accelerates the electrons in the valence b a n d and increases their kinetic energy. This increase in energy is n o w h e r e sufficient to m o v e the electron across the gap into t h e conduction band. Therefore, n o current flows. By raising the t e m p e r a t u r e of t h e semiconductor, some electrons at the t o p of the valence b a n d gain enough t h e r m a l energy to m o v e into t h e conduction band. B o t h t h e electrons in t h e conduction b a n d and t h e holes in the valence can now b e accelerated, gain kinetic energy, and therefore carry a current w h e n an electric field

Section 2.8

Conductors, Semiconductors, and Insulators

49

is applied. The conductivity at 300K is small, c o m p a r e d to metals, hence the n a m e semiconductor. The b a n d gap energy of an insulator, such as d i a m o n d or silicon dioxide, is sev­ eral times greater than that of semiconductors, such as silicon or gallium arsenide. E v e n at very high t e m p e r a t u r e s (high enough to a p p r o a c h the melting point), very few electrons acquire enough t h e r m a l energy to be raised to the conduction band. Therefore, the resistivity of insulators is very high and their conductivity is very low. In metals, the top b a n d is only partially filled with electrons. Bands b e c o m e partially filled if either the n u m b e r of electrons is not sufficient to fill the b a n d or, m o r e commonly, a completely filled valence b a n d overlaps an empty conduction band. A n electric field applied to the m e t a l at a t e m p e r a t u r e of 300K easily moves electrons to higher empty levels and causes a current to flow. Thus, the conductivity of metals is very high.

REVIEW QUESTIONS Q2-7 Q2-8 Q2-9 Q2-10 Q2-11 Q2-12 Q2-13

Distinguish between a direct and an indirect semiconductor. Give an example of each. Suggest an application for indirect semiconductors in relation to their absorption and emission of photons. Trace on Fig. 2.12(b) the motion of a hole in a semiconductor to which an electric field is applied. Why is the mass of an electron negative for an electron located at the top of the valence band? Why are the effective masses of holes and electrons smaller in GaAs than in Si? Why is the ability to conduct electricity, or conductivity, higher in metals than in semi­ conductors at room temperature? Why is the conductivity of insulators, when compared to that of a semiconductor, neg­ ligible?

HIGHLIGHTS • •

•

No forbidden band exists in a metal; a wide gap (several eV) exists in insulators. The gap is 1.12eV in silicon and 1.41eV in gallium arsenide. The effective mass of an electron is inversely proportional to the second derivative of the expression for energy as a function of the wave vector. Thus, at the top of the valence band, the mass is negative; it is positive at the bottom of the conduction band. A negative-mass, negative-charge electron that is subjected to an electric field is accel­ erated in the direction of the electric field, opposite to that of a negative-charge, posi­ tive-mass electron. The negative-mass, negative-charge electron is interpreted to be a hole, which has positive charge and positive mass.

EXERCISES E2-4

The energy gap of GaAs is 1.42V. Determine: a) the minimum frequency of light that will cause the transition of an electron from the valence band to the conduction band.

50

Chapter 2

Energy Bands and Current Carriers in Semiconductors b)

The wavelength of this light. 14

Ans: a) / = 3.43 X 10 Hz

7

b) λ = 8.74 X 10" m 10

3

E2-S At room temperature, it is calculated that there are 10 electrons per cm that have moved from the valence band to the conduction band of silicon. a) What is the density of holes? b) What fraction of the electrons have moved into the conduction band? Ans: a) ρ

10

10 cm"

=

3

b) 1 in 10

13

PROBLEMS 2.1 A certain atom has the following subshells: 2/,3d,3/,3g,5g Determine: a) The values of the quantum numbers η and I that correspond to each of the subshells. b) Which of the subshells are allowable? 2.2 For the subshells of Problem 2.1, determine the number of electrons in each. 2.3 The expression for the potential energy of an electron in a one-dimensional crystal lattice is given by Eq. (1.6). Plot the potential energy of the electron inside the lattice considering three atomic cores (Zq) located at χ = 0, χ = a and χ = 2a. Choose a suit­ able scale for potential energy and cover the region from χ = 0 to χ = 3a. 2.4 The E-k diagram for a particular energy band of a certain material is shown in the fig­ ure below. An electric field is applied to the material in such a direction as to cause a force in the negative k-direction. Determine: a)

The signs of the effective masses of the wavepackets made Up of groups of states near A, B, C and D.

b) c)

The direction of the velocity of each of the wavepackets. The direction of acceleration of each wavepacket.

Ε

Chapter 2

Problems

51

2.5 The E-k diagram for a certain material is shown in the figure below. The diagram for a free electron is shown dotted. Sketch as accurately as possible: a) dE/dk versus k. b) d E/dk versus k. 2

2

Ε

k

2.6 The wavefunction represented by the time-independent equation for the hydrogen atom, in spherical coordinates, is given by: 1 d I^ 3Ψ\ fdr { drj

a) b)

1 J)_ I Λ ί η θ θθ V

+

3Ψ\

3Θ J

1 βΨ r sin 6 δφ 2

2

2

2

Reduce the equation to one in r only, independent of θ and φ. For the equation determined in (a), first, assume a solution of the form *Pj = C exp (—r/r ) and, second, use Postulate 2 in Chapter 1 for the volume of a spherical shell of radius r and thickness dr in order to determine an expression forC. Determine the expression for r by substituting the solution for Ψ in the dif­ ferential equation found in (a). Determine the expression for the total energy of the electron in the lowest energy state. Given: 0

c) d)

Q

η —ax

x e Jo

dx = n\/a

1

n+1

chapter 3 INTRINSIC AND EXTRINSIC SEMICONDUCTORS

3.0 I N T R O D U C T I O N We learned in the last chapter that each electron in a solid is assigned a specific level of energy and that these levels are so close to each other that they m e r g e into bands of energy. We also found that these b a n d s of energy that electrons may occupy are separated by forbidden energy gaps w h e r e n o electron of t h e solid can exist. In semiconductors, the t h r e e t o p b a n d s are the valence band, t h e forbidden gap, and the conduction band. In this chapter, we will d e t e r m i n e expressions for t h e densities of electrons and holes in the conduction b a n d and valence b a n d respectively, for a semiconduc­ tor into which n o impurities have b e e n deliberately introduced. We will t h e n deter­ mine t h e densities w h e r e small traces of impurities are added to the semiconductor. So that we can d e t e r m i n e current-voltage relationships, we n e e d to explain the processes that cause electrons and holes to move. These we will study in C h a p t e r 4.

3.1 D E N S I T Y O F STATES The densities of electrons and holes in the conduction and valence b a n d s respec­ tively are d e p e n d e n t on two factors. First, we n e e d to k n o w the density of states available for occupancy. Then, we d e t e r m i n e the probability of occupancy of the various states at their respective energy levels. The density of states refers to the n u m b e r of available electron states for unit volume per unit energy at a certain energy level. The expression for t h e density of states in a metal, derived in A p p e n d i x C, is given by

52

Section 3.1

M*) = i(ffV

Density of States

53

(3.1)

where N(E) is t h e density of states function for free electrons in a metal m e a s u r e d per unit volume, per unit energy, located about the energy level E. Its significance is clearer if we state the following: To obtain the total n u m b e r of energy states per unit volume in a given energy range, dE, about E, we multiply N(E) by dE. In fact, it b e c o m e s an integration process. We will assume that the function given by E q . (3.1) is valid for semiconduc­ tors, provided we use the effective mass for t h e electron, since t h e electron moves in the periodic potential of the crystal lattice and it may b e assumed to b e a free elec­ tron. The densities of electrons and holes will be d e t e r m i n e d by using this function t o g e t h e r with a probability of occupancy of a state function. The energy Ε corresponds to a set of q u a n t u m n u m b e r s and can therefore take on only certain discrete values. A sketch of the distribution of states is shown in Fig. 3.1. This curve is not continuous but is m a d e u p of a set of discrete points with the adjacent states so close to each other that, for all practical purposes, o n e can safely consider it continuous. A n analogy should help clarify the significance of the variation of N(E) and E. Let us consider an oval-shaped stadium, with seats at all elevations a r o u n d t h e field, and label its volume as a unit of volume. A t the higher elevations, o n e can count m o r e seats a r o u n d the stadium for a given level t h a n at t h e lower elevations. Thus, we can say that the n u m b e r of seats per unit volume (the whole stadium) per m e t e r of elevation corresponds to N(E). The n u m b e r of seats per m e t e r of elevation at the higher elevations is larger than t h e n u m b e r of seats per m e t e r of elevation at the lower elevations. Similarly, the density of states N(E) is larger at higher energy levels E. Classical mechanics states that at absolute zero all electrons have zero energy, and that w h e n a material is h e a t e d from absolute zero, each particle absorbs an a m o u n t of energy equal to kT, w h e r e k is Boltzmann's constant and Τ is the absolute t e m p e r a t u r e . H o w e v e r , Wave Mechanics theories obviously do not agree with those of classical mechanics. First, only very few electrons have zero energy at zero t e m p e r a t u r e . Second, w h e n the material is heated, only those electrons close to a certain energy level, k n o w n as the Fermi level, can be excited to higher unoccupied levels. T h e significance and relevance of the Fermi energy level will be clarified in

Ε

N(E) 0

Figure 3.1 Distribution of energy states; Ε versus N(E).

54

Chapter 3

Intrinsic and Extrinsic Semiconductors

the following section. Therefore, w h e n a material is h e a t e d from absolute zero, not every electron gains energy kT, as classically happens. According to Wave Mechanics, only those electrons within an energy range of the o r d e r of kT of the Fermi level can b e thermally excited. F u r t h e r m o r e , in t h e study of energy b a n d for­ mation, we found that t h e solid has states of varying energy, even at absolute zero, and that is because of t h e Exclusion Principle.

3.2 F E R M I - D I R A C D I S T R I B U T I O N F U N C T I O N The relations derived in A p p e n d i x C describe conditions occurring at absolute zero. A s we raise t h e t e m p e r a t u r e , we expect some electrons to move to higher energy levels and thus to energy levels previously unoccupied. The question then is: W h a t is the distribution of t h e electrons with respect to energy as t h e t e m p e r a t u r e is raised above zero, subject to t h e assumption that the density of states is i n d e p e n d e n t of temperature? We also n o t e d that t h e Pauli Exclusion Principle states that a q u a n t u m state can b e occupied by only two electrons that have opposite spin. A t absolute zero, the electrons therefore occupy t h e lowest energy levels possible. Because of the Pauli Exclusion Principle, the distribution of electrons is governed by Fermi statistics. The Fermi statistics describe what occurs as t h e t e m p e r a t u r e is raised above zero. It describes t h e probability that a state of energy Ε is filled by two electrons of o p p o ­ site spin. T h e Fermi-Dirac distribution predicts that as t h e t e m p e r a t u r e increases, an energy state corresponding to energy Ε will have a higher probability of being occu­ pied t h a n at a lower t e m p e r a t u r e . The Fermi function is given by 1 A

E

)

=

e x p [ ( £ - E )/kT] F

( 3

+ 1

-

2 )

where f(E) is t h e probability that a state with energy Ε is occupied, E is the Fermi energy, k is B o l t z m a n n ' s constant, and Γ is the absolute t e m p e r a t u r e . We will discuss shortly t h e physical significance of E . The above expression can also b e interpreted to m e a n that at a certain tem­ p e r a t u r e T, t h e probability of occupancy of a state is lower, the higher the energy level Ε of that state. Thus, the states at higher energy levels are less likely to b e occu­ pied than states at lower energy levels. Since we observe from E q . (3.1) that t h e den­ sity of states is higher at the higher energy level, we can conclude that at the higher energy levels, w h e r e the states are m o r e numerous, the probability of occupancy of a state is m u c h lower than at t h e lower-level, less-numerous states. T h e r e a s o n for this seemingly a b n o r m a l behavior results from the fact that electrons initially occupy the lowest-level states w h e r e the Pauli Exclusion Principle allocates only two elec­ trons of opposite spin to each state. A l t h o u g h E is defined in A p p e n d i x C, we shall relate its significance in this context as well. A t t e m p e r a t u r e s approaching absolute zero and for Ε < E , E q . (3.2) indicates t h a t / ( £ ) will b e equal to 1 and, for Ε > E ,f(E) will b e zero. Thus, at absolute zero, E is t h e dividing energy below which all states are occupied with p

p

p

p

p

p

Section 3.2

Fermi-Dirac Distribution Function

55

f(E)

Figure 3.2 Fermi-Dirac distribution function for three temperatures. Temperature is measured in K.

electrons and above which all states are vacant. This is the same definition for E we arrived at in A p p e n d i x C. A s t h e t e m p e r a t u r e increases, some electrons acquire enough energy to move into states above E . Figure 3.2 shows t h e Fermi-Dirac distribution function for three temperatures. N o t e t h e rounding-off of the curve at the higher energy levels. Also n o t e that at Ε = E , f(E) is always 1/2. We can therefore define E as being that energy level for which the probability of occupancy is 1/2; or we can say that over a long period of time, the probability is that half t h e states are filled at the level E. F

p

p

p

P

We have seen from E q . (3.1) that N(E) gives the density of the states at energy E. A t t h e r m a l equilibrium, the density of electrons, dn, having energy b e t w e e n Ε and Ε + dE is t h e n dn = f(E)

X

N(E) dE

(3.3)

By substituting f o r / ( £ ) from E q . (3.2) and for N(E) from E q . (3.1) and using the effective mass for a semiconductor, we can then write E q . (3.3) as 2

2

_^/2)(8m*/h r E^dE a

n

{

1 + e x p [ ( £ - E )/kT]

±

V

F

F r o m E q . (3.4), we can then calculate t h e incremental density of free electrons corresponding to a specific interval of energy dE, for a fixed Fermi level, and as a function of t h e t e m p e r a t u r e T. A word of caution is in o r d e r here concerning the location of E in semicon­ ductors. In a later chapter, we shall n o t e that t h e level E lies in t h e forbidden band. You may t h e n question our statements h e r e concerning occupancy of E , even p

p

p

56

Chapter 3

Intrinsic and Extrinsic Semiconductors

though it is in t h e forbidden gap. The answer is that energy states obviously exist in the forbidden b a n d but these states do not accommodate electrons of t h e semicon­ ductor. Since t h e probability of occupancy of a state i s/ ( / ? ) , the probability of vacancy of a state b e c o m e s [1-/(£)]. In the following example, we illustrate the application of the Fermi distribu­ tion function to a metal.

E X A M P L E 3.1 (a) Determine the probability of occupancy of a state that is located at 0.259eV above E at: F

i) Τ = OK

ii) Τ = 300K

iii) Τ = 600K

(b) Determine the probability of vacancy of a state that is located at 0.4eV below E atT= p

300K.

(c) Repeat part (b) if the state is at O.OleV above /·.', at I · 300K. Solution (a) i) At T= OK M

Sx259 = °

=

1 + exp- 0 ii) A t T = 300K Λ

Ε

=

)

=

1

+

eXP

4.54

Χ

1 0

-5

075z5v

iii) At Τ = 600K jtE) J X

'

=

—- = 6.69 X 10~ 0.259 1 + exp . . 0.0518

3

n r

F

(b) The probability of vacancy is [1 — f(E)] so that at 0.4V it becomes ί ^ ^ 1 + exp0.0259

1-/( ) = 1 £

ο

(c) The probability of vacancy at 0.01 V becomes 1

' o.o, - ° 1

+

6 Χ Ρ

5 9 5

0Λ25^

3.3 C A R R I E R DENSITIES Semiconductors b e c o m e useful only after special impurities are a d d e d to t h e m . In its almost p u r e form, and w h e n n o impurities are added, the semiconductor is labeled as intrinsic. It is extrinsic w h e n selected impurities are a d d e d and t h e semi­ conductor is said to be doped with impurities.

Section 3.3

Carrier Densities

57

Densities of S t a t e s in t h e Conduction a n d V a l e n c e B a n d s Initially, we will d e t e r m i n e expressions for t h e electron and hole densities in an intrinsic semiconductor. We will assume that the expression for the density of states given by E q . (3.1) applies for all conduction b a n d and valence b a n d energies with the a p p r o p r i a t e effective masses substituted. T h e r e are about 5 Χ 1 0 atoms/cm in silicon. A t Τ = OK, the AN q u a n t u m states in t h e valence b a n d are filled with electrons and the AN states in t h e conduc­ tion b a n d are empty. These states are distributed t h r o u g h o u t t h e bands. The states of i m p o r t a n c e in semiconductor devices, as we shall see later, are the ones near the top of t h e valence b a n d and t h e states near t h e b o t t o m of the conduction band. To obtain the densities of states n e a r the b a n d edges, we let E represent t h e m i n i m u m electron energy in the conduction b a n d and E represent the m a x i m u m hole energy in the valence band. We recall that an increase of t h e energy of an elec­ tron in the conduction b a n d corresponds to t h e electron moving u p on t h e energy scale in the conduction band. A n increase of the energy of a hole in the valence band, then, corresponds to the hole moving down into the valence band. Thus, we change t h e variable for integration so that E q . (3.1) b e c o m e s (E — E ) for t h e con­ duction b a n d and it b e c o m e s (E — E) for t h e valence b a n d states. Therefore, the functions for t h e densities of states in t h e conduction and valence bands, as shown in Fig. 3.3, b e c o m e 22

3

c

v

c

v

(3.5) where Ν (E) is t h e n u m b e r of states per unit volume per unit energy at Ε in t h e con­ duction band. The corresponding density of hole states in t h e valence b a n d at Ε is (3.6) where Ν (E) is t h e density of states in t h e valence b a n d and is assumed to be located at E. Using the above expressions, we d e t e r m i n e t h e electron density in the conduction b a n d from (3.7) Ε

N(E)

Figure 3.3 Distribution of the density of states in the conduction and valence bands.

Chapter 3

Intrinsic and Extrinsic Semiconductors

where f (E) is given by E q . (3.2). The hole density, which represents the density of vacant states, is given by c

i where f (E) v

Ey

N (E)f (E)dE p

(3.8)

v

-oo(BOTTOM)

is t h e probability of vacancy of a state in the valence b a n d and is

The Boltzmann Approximation To simplify t h e integration in Eqs. (3.7) and (3.8), we m a k e an important assump­ tion that is related to t h e location of t h e Fermi level. This assumption, k n o w n as the Boltzmann approximation, consists in dropping t h e unity t e r m in t h e d e n o m i n a t o r of the expression for the Fermi function given by E q . (3.2). Based on this approxi­ mation, E q . (3.2), for the probability of t h e occupancy of a state by an electron, becomes f (E) c

= exp [ - ( £ - E )/kT]

for Ε > E

F

(3.9)

p

T h e reason for restricting the energy level to values greater than E is that an energy level Ε < E in E q . (3.9) makes the probability greater than unity, which is meaningless. Let us briefly digress to investigate the condition for which t h e approximation of E q . (3.9) is fairly valid. U p o n comparing f(E) from Eqs. (3.2) and (3.9) at various values of (E - E ), we n o t e that for (E - E ) = 3kT,f(E) calculated from E q . (3.9) differs from that of E q . (3.2) by about 5 percent and at (E - E ) = 4kT t h e differ­ ence is a b o u t 1.8 percent. We will arbitrarily state that E q . (3.9) is valid provided the Fermi level is at least 3kT below the b o t t o m of the conduction b a n d where all free electrons reside. For energy levels below E , the probability of a vacancy (or hole occupancy) can b e written from E q . (3.2) as F

F

F

F

F

F

Ι _ f) (F

U

=

Λ

)

1 exp [(E - E )/kT] F

=

+ 1

ex [(E E )/kT] 1 + exp [(E - E )/kT] V

F

F

(

'

A t values of (E — E) = 3kT, t h e exponential term in t h e d e n o m i n a t o r of E q . (3.10) becomes very small so that the validity of this equation is restricted to values of (E - E) > 3kT. E q u a t i o n (3.10) b e c o m e s F

F

1 - f (E) c

= exp - [(E

F

- E)/kT]

(3.11)

The restriction to E q . (3.11) implies that t h e Fermi level is at least 3kT above the t o p of the valence band. Since the electrons occupy states in the conduction b a n d and holes represent unoccupied states in t h e valence band, we conclude that t h e validity of the B o l t z m a n n approximation is restricted to t h e range of Fermi energies extending from 3kT above the top of t h e valence b a n d to 3kT below the b o t t o m of the con­ duction band.

Section 3.3

Carrier Densities

59

It is possible that in a highly d o p e d semiconductor, as we shall see later, t h e Fermi level moves to within less than 3kT from the b a n d edges and even into the bands. Such semiconductors are said to b e degenerate. We conclude that for a n o n - d e g e n e r a t e semiconductor, t h e Fermi-Dirac equa­ tion r e p r e s e n t e d by f for electrons and f, for holes respectively becomes f (E)

= exp - ( E -

f (E)

= exp-(E -E)/kT)

c

v

Ep/kT)

(a) (b)

F

(3.12)

Expressions f o r Electron a n d Hole Densities Using t h e expressions in Eqs. (3.5) and (3.6) together with Eqs. (3.11) and (3.12) in Eqs. (3.7) and (3.8), we have for electrons

For holes, t h e expression is

Sketches graphically describing the operations in Eqs. (3.7) and (3.8) are shown in Figs. 3.4 and 3.5 for t h e electron and hole distributions in the conduction and valence b a n d s respectively, for different locations of E . Before we p r o c e e d with evaluating the integral, we will clarify two questions that an interested r e a d e r may raise. T h e first concerns changing the limits of inte­ gration. We have replaced the t o p of t h e conduction b a n d by plus infinity and the b o t t o m of the valence band by minus infinity. The reason is to simplify the integration. However, t h e change is quite justifiable since t h e exponential functions f {E) and f(E) decrease so rapidly as we m o v e away from the b a n d edges that t h e densities of the carriers b e c o m e negligible as we move a few kTs away from the edges of the conduction and valence band edges. The second question concerns the effective masses We confirmed at t h e end of C h a p t e r 2 that the effective mass, m*, is a function F

c

)xfc(E)

)[i-/c(£

Figure 3.4 The product of the distribution of energy states and the Fermi function gives the distribution of electrons and holes for E midway between E and E . F

c

v

60

Chapter 3

Intrinsic and Extrinsic Semiconductors

N (E)xf (E)

N„(E)

n

N {E)

c

N (E)x[l-f (E

P

p

Ey

ΝΛΕ)

c

N (E)xf (E) n

c

Ei E F

N„(E)x[l-f (E c

N (E)

~l-fc(E)

P

Figure 3.5 This is Fig. 3.4 repeated except that (a) the Fermi level is assumed to be above the middle of the gap and (b) the Fermi level is assumed to be below the middle of the gap. Note that at Ε = E f(E) = 0.5. p

of the energy level. In Eqs. (3.13) and (3.14) we have, however, assumed that this mass is constant and we placed it outside the integrand. The justification is that since the exponential in t h e expressions decays very rapidly as we move away from the conduction and valence b a n d edges, E and E , our interest is in the carrier densi­ ties over t h e n a r r o w portions above E a n d below E . In these small regions, t h e effective mass is relatively constant as we indicated earlier. To integrate E q . (3.13), we l e t x = ( E — E )/kTso that t h e integral becomes c

y

c

y

c

v _ «

3

/

2

{xkf)

kT e x p ^

(xkT

+ E

c

—

kT

dx

(3.15) The integral has a value of

V

6

/ 2

and η becomes 2 1/2

3

n=im2\ \^4^^){kT)^ h 2

kT

2~

(3.16)

Section 3.3

Carrier Densities

61

A similar expression is obtained for ρ as l2m*kWn P = 2(—j -j

(E

F

-

Ey\

(3-17)

F

T h e expressions for η and ρ in Eqs. (3.16) and (3.17) can be written as η = Njxp-^j?*)

(3-18)

(3.19) where TV and TV are given by (3.20)

(3.21) 3

For m in kg., k in J / K and Λ in J-s, TV is in m~ . The terms TV. and TV^ are referred to as the effective density of states of t h e conduction and valence bands respectively. The values for TV., and N are shown in Table 3.1 where the difference b e t w e e n TV and TV result from t h e different values of v

C

V

the effective masses of electrons and holes. Relative masses of electrons and holes are shown in Table 2.3. We r e m i n d t h e r e a d e r that t h e above relations were derived subject to the approximation that E was n o closer than 3kT units of energy to either the conduc­ tion b a n d or valence b a n d edges. F

TABLE 3.1 Effective density of states and band gap energy at Τ = 3Q0K. Semiconductor Si Ge GaAs

3

3

iV (cm~ )

N (cm~ )

c

E (eY)

v

19

3.22 x 10 1.03 x 10 4.21 Χ 1 0 " 19

g

1.83 x 10" 5.35 x 10 9.52 Χ 10 18

18

1.12 0.66 1.42

By taking the product of η and ρ from Eqs. (3.18) and (3.19), we obtain the interesting result np = N N e x p - ( c

v

£ c

~

£ y r

)

(3-22)

Intrinsic Carrier Density In an intrinsic semiconductor, o n e to which n o impurities have b e e n a d d e d , at Τ > OK, the density of electrons in the conduction b a n d must equal the density of holes in the valence b a n d because for every electron that is excited to the conduction

62

Chapter 3

Intrinsic and Extrinsic Semiconductors

band, a hole is created in t h e valence b a n d . We define this density as n and nf, from E q . (3.22) b e c o m e s i

np = nf = N N e x p - ( ^ ^ ) c

(323)

v

The intrinsic carrier density, « , assuming that t h e effective masses d o not change with t e m p e r a t u r e and using Eqs. (3.20) and (3.21) in E q . (3.23), becomes ;

n = K 7/ t

x

3 / 2

exp^|

(3.24)

where Κ is a constant i n d e p e n d e n t of t e m p e r a t u r e and Ε , equal to (E — E ), is the b a n d gap energy of the semiconductor. High t e m p e r a t u r e s and smaller b a n d gaps favor large values of intrinsic carrier density. It is i m p o r t a n t to n o t e that the Τ in t h e e x p o n e n t has a m u c h greater effect t h a n the factor Τ . We will illustrate these effects in E x a m p l e 3.2. By substituting for the constants in E q . (3.24) and at Τ = 300K, we find that t h e intrinsic carrier density in silicon is approximately 1 X 1 0 c m , signifying that this is the density of electrons in t h e conduction b a n d and the density of holes in the valence band. Mainly because of t h e larger b a n d gap, the intrinsic carrier density in gallium arsenide at Τ = 300K is 2.49 X 1 0 c m . T h e r e are 5 Χ 1 0 atoms per cubic centimeter in silicon and each a t o m has four electrons in t h e covalent b o n d representing t h e 3s and 3p levels. These same levels form t h e conduction and valence b a n d s in silicon. This results in a density of 2 Χ 1 0 electrons p e r c m available for conduction, all residing in t h e valence b a n d at Τ = OK. A t Τ = 300K the density of electrons (in the conduction b a n d ) is 1 X 1 0 c m ~ . Approximately, therefore, o n e in 1 0 of t h e electrons available at Τ = OK at the t o p of the valence b a n d is elevated at r o o m t e m p e r a t u r e from the t o p of the valence b a n d to t h e conduction band. This represents a very small propor­ tion of t h e available electrons. By substituting for t h e constant K in Eq. (3.24) from t h e expressions in Eqs. (3.20) and (3.21), we have λ

c

1 0

6

v

- 3

- 3

22

23

10

3

3

13

t

». = f f f OWT expg§)

(3.25)

We observe that t h e intrinsic carrier density is i n d e p e n d e n t of the location of the Fermi level and d e p e n d s strongly o n the band gap energy, Ε , and o n the tem­ perature. T h e strong d e p e n d e n c e o n t e m p e r a t u r e is a result of trie rapidly varying t e r m in t h e exponent since the n u m b e r of carriers that acquire t h e r m a l energy is increased as m o r e of t h e electrons that are d e e p e r in t h e valence b a n d are able to move u p to t h e conduction band. This d e p e n d e n c e is illustrated in Fig. 3.6. The intrinsic carrier density decreases exponentially with an increase of Ε , accounting for the very low conductivity of insulators. Calculations of t h e intrinsic carrier density are illustrated by E x a m p l e 3.2. E X A M P L E 3.2 Calculate the intrinsic carrier density of silicon at: a ) Τ = 300K b ) Τ = 600K Include effect of variation of Ε with temperature.

Section 3.3

Carrier Densities

Solution a) n? - N N e~ g Using the values of N and /V. from Table I and using an Ε of l.lleV, we determine nf as E

c

/kT

v

c

2

nf = 0.9772 X 1 0 W ,J

n. = 9.885 X 10 cm-

6

3

b) At Τ = 600K, we determine E , from the relation in Table 2-2, to be 1.032eV. Using Eq. (3.24) and «(300K) = 10'°cm " , we have g

3

»r(600K) _ / 6 0 0 \ e x p ( - 1 . 0 3 2 / ( 8 . 6 2 x 1 0 " X 600) 5

2

« (300K)

_

UoO/

exp ( - 1 . 1 2 / ( 8 . 6 2 Χ 10"" X 300) . 5

n,.(600K) = 3.337 x 10 cm 15

3

= 1.137 Χ 10

1

63

64

Chapter 3

Intrinsic and Extrinsic Semiconductors

We highlight the influence of Τ in the exponential by (a) assuming that /. does not change with temperature and (b) comparing the factors by which each of the temperature terms, in the expression for n , affects n For an increase of Γ from 300K to 600K, the T term increases n by a factor of 2.8, whereas the same increase in Γ causes the exponential tei m to increase //. by about a factor of 50,000. It is important to indicate that there are discrepancies among various sources in the values of the intrinsic carrier density. The reason for this uncertainty in the calculated value is a result of the very complex relations from which the values of the effective masses are determined. In this book and for silicon at / • 300K. we will use a value of 1 X 10'° carriers per cubic cen­ timeter. This value agrees fairly well with the calculation using the values of Ν and \ listed earlier. There is no sacrifice of accuracy in this approximation. In fact, the advantage is that it is a number that can easily be remembered. 2

1

r

t

Location o f Fermi Level O n e major assumption that we m a d e in deriving t h e expressions for t h e electron and hole densities for t h e intrinsic semiconductor is that t h e Fermi level is at least 3 kT e V distant from t h e edges of t h e conduction and valence bands. We investigate t h e strength of t h e validity of this assumption for an intrinsic semiconductor by equating t h e expressions for t h e intrinsic electron and hole densities in Eqs. (3.18) and (3.19) and solving for E , so that F

We use t h e expressions for N a n d N from E q . (3.20) a n d (3.21) t o obtain c

v

TV, The expression for E E

F

\m*l

(3.27)

\m*

becomes

, _ ^2

+

WS_SL±S-W!S) 4

m*

2

4

< 3

.

2 8 )

\m*j

w h e r e (E + E )/2 represents t h e middle of t h e b a n d gap. T h e Fermi level for an intrinsic semiconductor is labeled E so that c

v

i

E = E =E i

F

v

+

3

f- -kT®< < ©

Figure 3.9 A boron atom accepts an electron to complete the covalent bond and hence creates a vacancy, a hole, into which another electron can move.

Chapter 3 E

Intrinsic and Extrinsic Semiconductors

c

s

0.045eK boron

1

Ε

Λ

E

v

Ο

Ο

Ο

Ο

ο

ο

ο

f Τ=0

tttt

Γ=100Κ

Γ=300Κ

Figure 3.10 The acceptor energy level in the energy band diagram of silicon and the effect of temperature on the ionization of the acceptor atoms. At Τ = 300K all acceptor atoms are ionized by the addition of electrons from the valence band.

r o o m t e m p e r a t u r e , all d o n o r atoms are positively ionized, each a t o m having lost an electron to the conduction band. Similarly, each acceptor a t o m is negatively ionized at r o o m t e m p e r a t u r e . Listed in Table 3.2 are the results of m o r e accurate calcula­ tions of t h e ionization energies of donors and acceptors in silicon. The ionization energy is defined as the energy required to ionize a donor or an acceptor atom. T A B L E 3.2 Silicon

Ionization Energies of Impurities in

Element

Function

Ionization energy (eV)

Boron Aluminum Gallium Phosphorus Arsenic Antimony

Acceptor Acceptor Acceptor Donor Donor Donor

0.045 0.057 0.065 0.044 0.049 0.039

Source: Reprinted from S. M. Sze and J. C. Irvin, "Resistivity, Mobility and Impurity Levels in GaAs, Ge and Si at 300K," Solid State Electronics 11,599 (1968), with kind permission from Elsevier Science Ltd.,The Boulevard, Langford Lane, Kidlington 0X5#1GB,UK.

In addition to t h e controlled addition of donor and acceptor impurities, other impurities may exist at energy levels m o r e distant from the b a n d edges than the controlled impurities. These impurities are randomly available and usually act as centers of recombination generation or traps. Because they exist a r o u n d t h e middle of the energy gap, they can act as energy levels at which electrons and holes m e e t to recombine. M o r e will be said later about recombinations.

Section 3.6 3.5

Densities of Carriers in Extrinsic Semiconductors

71

THERMAL EQUILIBRIUM In an earlier section, we derived an expression for the intrinsic carrier density, the electron density, and t h e hole density, as a function of the t e m p e r a t u r e and the band energy gap. We concluded that at r o o m t e m p e r a t u r e t h e intrinsic carrier density in silicon is approximately 1.0 X 1 0 c m . T h i s carrier density results from the transfer of electrons to the conduction b a n d as they receive thermal energy. T h e question that comes to mind is: A s long as t h e thermal energy is available, why is it that the density of the carrier does not continuously increase, but rather reaches a fixed value? The answer is that as carriers are continuously generated, and w h e n t h e r e is n o other source than t h e r m a l energy, electrons and holes continuously recombine and pairs of carriers disappear. In moving through the lattice, they encounter obstacles causing t h e m to lose energy and disappear. Thus, the rate of generation is accompa­ nied by the equal and opposite r a t e of recombination. This condition exists at ther­ mal equilibrium. Thermal equilibrium is therefore defined as the state in which a process is accompanied by an equal and opposite process, while the system is held at constant temperature, and no external source of energy acts on it. In t h e next chapter, we will present analytical relationships for the processes of carrier generation and recombinations. 10

3.6

_3

D E N S I T I E S O F C A R R I E R S IN E X T R I N S I C S E M I C O N D U C T O R S W h e n dealing with d o p e d semiconductors, we refer to electrons in N-type semicon­ ductors, t h e m o r e n u m e r o u s carriers, as the majority carriers. Holes in N-type semi­ conductors are minority carriers. In P-type semiconductors, holes are the majority carriers and electrons are t h e minority carriers. We stated earlier that the density of electrons, w h e n donors are added to sili­ con, is not equal to the sum of the intrinsic electron density and the density of the ionized d o n o r atoms. What, then, is t h e resulting density? In an earlier section, and prior to determining the intrinsic carrier density, we found that the product of η and p, n\, is a constant for a certain semiconductor at a certain t e m p e r a t u r e and is given by np = nj = N N exp (-E /kT). T h e r e is n o con­ dition in the expression that restricts it to intrinsic semiconductors because E does not change with impurity concentration and N and N are constants. This product is therefore a constant equally valid for intrinsic as well as for d o p e d semiconductors, provided that we refer to the carrier densities at t h e r m a l equilibrium (a condition w h e n there are n o sources of energy other than thermal energy). We label the extrinsic values of the electron and hole density at t h e r m a l equilibrium as n and p respectively so that we can write, c

v

g

g

c

v

Q

n

oPo

=

n

i

Q

(3.30)

In order to determine the values of n and p for a d o p e d semiconductor, we n e e d a n o t h e r relationship that relates them. 0

Q

Chapter 3

Intrinsic and Extrinsic Semiconductors

C h a r g e Neutrality We consider the general case w h e n both donors and acceptors are a d d e d to a semi­ conductor. Assuming t h e r m a l equilibrium, t h e material includes particles that have positive charges and others that have negative charges. The negative charges are m a d e u p of t h e electrons in the conduction b a n d and the acceptor atoms which are ionized. T h e holes in the valence b a n d and the ionized donor atoms constitute the positive charges. Since the intrinsic silicon is charge neutral and the a d d e d impuri­ ties are neutral, t h e resulting mix is also neutral so that, charge neutrality requires, q(n

+ N~) = (p

0

0

+

+ N )q

(3.31)

D

where N denotes the density of acceptor atoms and Np refers to the density of d o n o r atoms, assuming they are all ionized and n and p are t h e t h e r m a l equilib­ rium values of electron and hole densities respectively. By replacing p in E q . (3.30) by its equivalence from E q . (3.31), a quadratic equation in n is obtained. The solu­ tion to the quadratic equation is given by E q . (3.32) where the positive sign has been selected for the t e r m u n d e r t h e square root because that t e r m has a larger magni­ t u d e than the first term and n cannot be negative. A

Q

Q

0

Q

Q

1/2

ND-NA

(3.32)

We have d r o p p e d the positive and negative superscripts from the symbols for impu­ rity densities since we have already assumed that all impurity atoms are ionized at room temperature. For an N-type semiconductor, either N » N or N is zero. F r o m E q . (3.32), we find n for an TV-type semiconductor to be, D

A

A

Q

1/2

2

n

0

= — 2

N\ D

+

(3.33)

The density of the holes is t h e n calculated from E q . (3.30) to be, Po = Z-

(3-34)

ο In solving Eqs. (3.30) and (3.31) simultaneously and w h e n numerical values are given for N and N , it is always advisable to obtain a quadratic equation for the larger expected carrier density, the expected majority carrier density. The r e a d e r should try solving for the minority carrier density to appreciate this caution. If the d o n o r density N is m u c h greater than « , which is quite c o m m o n in most extrinsic semiconductors, E q . (3.33) can b e simplified so that D

A

D

2

n = N 0

and p

D

0

n = -j-

(3.35)

D 16

For a d o n o r density of 1 0 c m ~ 1 0 c m ~ , we find, 10

3

in silicon at Τ = 300K, where n. = 1 X

3

1 6

n = 10 cm Q

- 3

4

3

and p = 1 X 1 0 c m " . Q

Section 3.6

Densities of Carriers in Extrinsic Semiconductors

73

in η

Extrinsic region full ionization

Intrinsic Extrinsic region all donors ionized

Extrinsic regie partial ionizatii

in Ν r Intrinsic region

100

_L 200

300

400

_L 600

500

T(K)

(l/:

Figure 3.11 Variation of majority carrier density with temperature. In the logarithmic sketch, the slope of the line in the intrinsic region is a measure of the band gap energy Ε .

We note, therefore, that the density of majority carriers is m u c h greater than the density of minority carriers. A n interesting supplement to t h e results of this section is obtained by plotting the ratio of the majority carrier density to the d o n o r density as a function of temper­ ature, as shown in Fig. 3.11. We assume that a d o n o r impurity having density N of 1 0 c m ~ is a d d e d to a silicon sample. A t t e m p e r a t u r e s n e a r zero K, the electron density in the conduction b a n d is zero since t h e thermal energy at that t e m p e r a t u r e is not sufficient to ionize any d o n o r atoms and certainly not enough to excite electrons from t h e valence b a n d into the conduction band. A s the t e m p e r a t u r e is increased, some d o n o r atoms b e c o m e ionized, losing their electrons to the conduction band, while t h e r e is still not sufficient energy to m o v e electrons out of t h e valence band. We notice that u p to about 150K, the electron density is smaller t h a n the d o n o r density because not all d o n o r atoms are ionized and very few electrons have b e e n excited from t h e valence b a n d to the conduction band. The region 0-150K is k n o w n as the freeze-out region. A t t e m p e r a t u r e s b e y o n d 150K, all d o n o r atoms are assumed to b e c o m e ionized so that η = TV^.This concentration of electrons remains constant at the value of N u p to r o o m t e m p e r a t u r e , at which N » n The range from η = Np = N u p to where n begins to increase b e y o n d its r o o m t e m p e r a t u r e value is k n o w n as the extrinsic region. This range extends from about 150 to 400K. A b o v e 400K the intrinsic carrier density begins to increase rapidly and eventually reaches a value greater than N . The material begins to b e c o m e intrinsic again and the electron density b e c o m e s m u c h larger than N because of t h e large n u m b e r of electrons that are thermally excited from the valence b a n d to the conduction band. For t e m p e r a t u r e s exceeding 400K, t h e region is labeled the intrinsic region. In the intrinsic region of Fig. 3.11, the densities of b o t h holes and electrons increase, how­ ever, their thermal equilibrium values are still d e t e r m i n e d by Eqs. (3.30) and (3.31). 16

3

D

D

D

D

;

D

D

r

74

Chapter 3

Intrinsic and Extrinsic Semiconductors

Calculations of electron and hole densities in extrinsic silicon are shown in Examples 3.4 and 3.5.

E X A M P L E 3.4 17

16

To a sample of intrinsic silicon we add 10 atoms per cc of phosphorus and 9 Χ 10 atoms per cc of boron. Calculate the density of the majority and of the minority carriers at: a) 1 = 300K,

b) T= 600K

Solution a) at Τ = 300K, we assume that all impurities are ionized so that, np = n] = 1 Χ 10

20

(1)

Neutrality required that η +N η +9

A

=ρ + N

D

x 10 = ρ + 10" 16

(2)

Solving equations 1 and 2 simultaneously, we have 16

η = 10 cm~

3

— 10 c ,-~3 « We observe that the electrons supplied by the donor atoms dominate the resulting electron density. 4

ρ —

b) at Τ = όΟΟΚ.η , = 11.137 X 10 cm 2

30

6

so that np = 11.137 X 103°

η +9

x 10 = ρ -+ 10 16

(3) 17

(4)

Solving equations 3 and 4 simultaneously, we have 16

« = 1.1 x 1 0 c n r M

3

2

„ = _;-= 1.014 Χ 1 0 ' W

3

Η

Comparing the results to those of part (a), we note the effect of the thermally generated elec­ trons and holes.

E X A M P L E 3.5 17

-3

A sample of intrinsic silicon is doped with 10 atoms/cm of phosphorus. Calculate the den­ sity of the electrons and the holes at:

Section 3.6

Densities of Carriers in Extrinsic Semiconductors

a) Τ = 300K

75

b) Τ = 600K

Solution a) At Τ = 300K, we find that 10

n- = 1 X 10 crrr

3

t

np

-

ιή

1 x HPcm *

5

(1)

Neutrality requires that η~ρ

+ Ν «ρ

+10"

ν

(2)

Solving equations 1 and 2 simultaneously, we have 17

:

3

n * 10 cm \p = - = l x

10 cm"

3

b) At Γ = 600K, 2

30

η = 11.137 X 10 cnT

6

m:™4

Φ)

« = /> + 1 0

1 7

(4)

Solving equations 3 and 4 simultaneously, ,7

n = 1.0011 X 10 cm-

3

The hole density is calculated to be P=jp=

1.11 X 10"cm-

J

Note the large increase in the hole density while the electron density remained practically constant.

A d d i t i o n a l expressions f o r n a n d p 0

0

We have d e t e r m i n e d in Section 3.3 that, for an intrinsic semiconductor, the Fermi level E lies approximately halfway b e t w e e n the b o t t o m of the conduction b a n d and the top of t h e valence b a n d at (E + E )/2. We have d e n o t e d this intrinsic Fermi level by t h e symbol E . Since n is useful in calculating carrier densities in extrinsic semiconductors, we will find E useful as a reference level when dealing with extrinsic semiconductors. E q u a t i o n (3.18) can therefore b e modified by replacing E by E. and n by , so that F

v

{

i

{

f

n, =

Ν^χρί^η^ή

(3.36)

Similarly, and since we are referring to an intrinsic semiconductor. Eq. (3.19) becomes

76

Chapter 3

Intrinsic and Extrinsic Semiconductors

N zx (^-B

(3.37)

/V exp^ = n exp^

(3.38)

7V exp^ = n exp^;

(3.39)

n =

v

i

V

j

E q u a t i o n s (3.36) and (3.37) give c

;

and t )

I

Replacing t h e terms on t h e left-hand side of Eqs. (3.38) and (3.39) by their equivalence from Eqs. (3.18) and (3.19), and solving for η and p, we have 3

n = n e x p p T ^ ) = «ο Ε

η εχρ(^ ή=ρ

Ρ =

ί

40

(· )

t

(3.41)

0

For an extrinsic s e m i c o n d u c t o r in t h e r m a l equilibrium, t h e η and ρ in Eqs. (3.40) and (3.41) h a v e b e c o m e n and p . I n each of these two equations, t h e vari­ ables a r e t h e t h e r m a l equilibrium carrier densities a n d t h e energy distance b e t w e e n E a n d E . Thus, for an intrinsic semiconductor, E is at E., w h e r e a s for an extrinsic s e m i c o n d u c t o r a n d as n and p change, t h e distance b e t w e e n E a n d E changes. T h e location of E. with respect to t h e c o n d u c t i o n and valence b a n d edges has n o t changed. Therefore, E provides a useful reference for extrinsic semiconductors. 0

i

Q

F

p

Q

0

i

F

i

3.7 F E R M I L E V E L IN E X T R I N S I C S E M I C O N D U C T O R S F r o m E q . (3.40), we find t h e location of t h e Fermi level in an N-type semiconductor, where η = n = N , as Q

D

E

F

- Ei = kT£n^ n

i

= kT in — i

2

(3.42)

n

For a P-type s e m i c o n d u c t o r , ρ = p — N , and from E q . (3.41) we have Q

Ei - E = F

A

kTln^

= kT In ^

(3.43)

We observe from Eqs. (3.42) and (3.43) that for an TV-type semiconductor the Fermi level is higher t h a n E and moves u p closer to the conduction band. For a P-type semiconductor, t h e Fermi level moves d o w n towards the valence b a n d since E is less t h a n E.. We recall that E the Fermi level for an intrinsic semiconductor, is approximately midway b e t w e e n the conduction and valence bands. Calculations for the locations of the Fermi level in extrinsic silicon are carried out in t h e following example. i

F

p

Section 3.7

Fermi Level In Extrinsic Semiconductors

77

E X A M P L E 3.6 16

(a) Determine the location of the Fermi level, with respect to E and £,„ when 10 phosphorus atoms per cc are added to a sample of intrinsic silicon at Τ - 300K. (b) Repeat part (a) if 10 boron atoms per cc replace the phosphorus atoms. c

15

(c) Repeat part (a) if Τ = 600K. 18

1

(d) Repeat part (a) if 10 atoms replace the 10 * atoms of phosphorus. Solution From Example 3.3, we have E at 300K to be 0.0073eV below the middle of the band gap. (a) By using Eq. (3.42) for phosphorus donor atoms, where n = N . we have i

0

D

AT

t,

Ε

I, I ni - - - 0.025N5 i » n I X 10 .358et/ =

0.35S.-1

10

t

E

c + E

=

v

E

0

i

0

0

7

3

e

l

/

a

n

d

E

=

E

+

c

+

E

y

+ 0.3507eV

0

From Table 2.2, Ε (at 300K) = l.lleV - E . - F. . by using this in the expression for Ε above, we obtain E = E ~~ 0.209eF = E + 0.9107. (

p

c

v

y

(b) We use Eq. (3.43) for boron acceptor atoms, w h e r e ρ = N , as ϋ

A

2

Ε - E = 0.02583 in 7 - ^ 0 = °- 98 V F

e

and from Example 3.3 we have E + E £,· = — r

v1

Λ

Λ Λ Μ Λ

c

y

+ 0.255eV and /Γ,.. ~ /·,,. - 0.865cK

E **E F

, - 0.0073, also E = E + 1.12, we find

Y

15

(c) At Τ = 600K, Η. is determined in Hxample 3.2 as 3.337 x 10 cm \ E, = 1.032cV and 1 2 X 10'* n = 1.2 Χ 10 cm"" , so that E = 0.05166 €« 3 i5- F / 0.065eK 16

3

£

0

F

3 3 7

Assuming that E. is approximately at (E + Ε )/2, of the band gap. c

18

χ

=

£

+

1 Q

E is located approximately in the middle

γ

F

10

(d) E - E, = 0.02583 Cn (10 /10 ) = 0.475eF, E = £ . + 0.475. E + Ey , .. . Using = — ^ — \ we have E = E + 0.56 + 0.475 = E + 1.035 = £ F

F

(

r

F

v

v

c

- 0.085e V

This example shows that the Fermi level moves toward the conduction band level, E , when donor atoms are added and moves towards the valence band when acceptors are added. As the temperature is increased, for both donors and acceptors, the Fermi levels move back towards the middle of the gap as the semiconductor approaches the intrinsic state. Keeping in mind that the Fermi level for an intrinsic semiconductor is located approximately midway between the bottom of the conduction band and the top of the valence band, the following conclusions can be drawn from the example: c

• The Fermi level moves toward E when donor impurities are added. It moves toward E when acceptors are added. (

• Increasing the temperature beyond a certain value causes the semiconductor to become intrinsic and the Fermi level to move toward the middle of the band gap.

v

78

Chapter 3

Intrinsic and Extrinsic Semiconductors

• Increasing the doping density in part (d) caused E. to move yet closer to E Further increase of the doping to 1 0 cm~ causes the conductor to become degenerate, since E is less than 3ATfrom E„. In conclusion, we can stale that, at a fixed temperature, the location of /.', is a measure of the doping of a semi-conductor. r

19

3

r

REVIEW QUESTIONS Q3-8 Q3-9

In an intrinsic semiconductor, explain why E is not located exactly midway between P

the valence and conduction bands. As the temperature of a sample of semiconductor that is doped with donor impurities is increased, in what direction does E move? Briefly define the phrase "thermal equilibrium"? What is a degenerate semiconductor? The configuration of electrons in a certain element is 4s 4p . Is it a donor or an accep­ tor element? By referring to the periodic table, determine which element is represented by the configuration 4s 4p . The density of atoms in silicon is 5 X 10 cm~ . Since each atom has four valence elec­ trons, the density of states becomes 2 Χ 10 . Why are N and N so much smaller than this number? P

Q3-10 Q3-11 Q3-12 Q3-13

2

2

Q3-14

3

i

22

3

23

v

c

HIGHLIGHTS •

•

•

• •

By adding controlled amounts of impurities to semiconductors, the densities of holes and electrons can be markedly changed. These impurities may be elements either from Column V of the periodic table, known as donors, or they may be from Column III, known as acceptors. Phosphorus and arsenic are donors; boron and gallium are accep­ tors. When donor impurities are added to an intrinsic semiconductor, the more numerous carriers, electrons, are called majority carriers and the holes are minority carriers. When both donors and acceptor impurities are added, the one with the larger density determines which carrier becomes the majority. A unique property of semiconductors is the large changes that are made in the densi­ ties of the carriers by the addition of impurities, also known as dopants. The changes in the carrier densities are translated into changes in conductivity. A semiconductor that has been doped is known as an extrinsic semiconductor. Donor and acceptor atoms occupy energy states in the forbidden band, donor atoms are close to E , and acceptor atoms are close to E . In these locations it takes very lit­ tle energy to excite a donor electron to go into the conduction band and also to excite an electron from the valence band to an acceptor level, thus generating a hole. A donor atom that loses an electron and an acceptor atom that gains an electron are said to be ionized. In intrinsic semiconductors, the Fermi level E is located close to the middle of the band gap. The Fermi level moves up towards the conduction band when donors are added and moves closer to the valence band when acceptors are added. c

•

y

F

Which Semiconductor?

Section 3.8

79

EXERCISES E3-4 A sample of semiconductor is doped with N is 10 cm~ . Determine η and p.

D

13

13

3

13

3

= 10 cm~ . The intrinsic carrier density

3

Ans: Μ = 1.61 X 10 cm~ 0 5

E3-5 If (N N ) - is theoretically the largest possible value of intrinsic carrier density that can be generated, determine as a fraction of this value, the number of electrons that can be thermally excited into the conduction band at 300K in germanium. Neglect changes of Ε with temperatures given Ε = 0.66eV. c

r

Ans:

6

3 χ 1ΓΤ

E3-6 Repeat Ex. 3-6 for silicon at 300K Ans: 4 Χ ΙΟ" 3

10

3

E3-7 Determine the density in cm and m of free electrons in silicon if the Fermi level is 0.2eV below E at 300K. c

16

Ans: η = 1.08 X 10 cm16

3

3

E3-8 A sample of silicon is doped with N = 5 X 1 0 c n r at room temperature. Determine the location of the Fermi level relative to the conduction band edge. D

Ans: E c

E = 0.16eV F

3.8 W H I C H S E M I C O N D U C T O R ? T h r e e s e m i c o n d u c t o r s h a v e b e e n used in t h e fabrication of devices since t h e early days of t h e s e m i c o n d u c t o r industry. T h e y are g e r m a n i u m , silicon, a n d gallium arsenide. G e r m a n i u m was used early on but since t h e 1960s silicon has b e e n t h e d o m i n a n t semiconductor. G a l l i u m arsenide has just recently acquired great i m p o r t a n c e for certain devices. T h e c o m p a r i s o n a m o n g t h e t h r e e throws a light on t h e particular p r o p e r t i e s that m a k e a s e m i c o n d u c t o r attractive for device fabrication. Semiconductors can b e classified into two major categories: elemental and c o m p o u n d semiconductors. E l e m e n t a l semiconductors, such as g e r m a n i u m and sili­ con, are classified in G r o u p I V of the Periodic Table of elements and have four valence electrons. C o m p o u n d semiconductors are composed of a combination of G r o u p III and G r o u p V elements or G r o u p II and G r o u p V I elements. E x a m p l e s of important c o m p o u n d semiconductors are gallium arsenide, gallium phosphide, indium phosphide, and indium arsenide. G e r m a n i u m has two distinct attractive qualities: In the first place, it can b e refined and processed m o r e easily than the others. In t h e second place, b o t h elec­ trons and holes have higher mobilities than the corresponding carriers in silicon, as can be seen in Table 3.3. The higher mobility translates into faster switching and higher operating frequency limits. A disadvantage of g e r m a n i u m is its high sensitivity to t e m p e r a t u r e because of its relatively n a r r o w b a n d gap, which may cause instability in a device. Instability

Chapter 3

Intrinsic and Extrinsic Semiconductors

results because a higher t e m p e r a t u r e increases t h e density of electrons that are excited to the conduction band, thus increasing the current that results in higher heat dissipation, which increases t h e t e m p e r a t u r e . The m o r e serious p r o b l e m is the difficulty of introducing controlled a m o u n t s of impurities into small selected areas. Because of this, one has to work with large areas. This results in carriers having to take longer times in covering distances within t h e device. This longer travel time m e a n s slower switching speed. Silicon has important advantages that are u n m a t c h e d by germanium. It is a b u n d a n t in n a t u r e in t h e form of sand and quartz. Thus, t h e cost of t h e starting material is negligible. Because silicon has a wider energy gap (forbidden b a n d ) than germanium, it can b e used at higher temperatures, thus greatly reducing a cause of instability. Silicon devices can b e o p e r a t e d safely at t e m p e r a t u r e s of about 200°C, whereas g e r m a n i u m devices are limited to a safe operating t e m p e r a t u r e of 80°C. Silicon has a major processing advantage in that it forms a stable oxide, silicon dioxide. The silicon dioxide offers a t o p quality insulator and w h e n used in device processing, it provides a good barrier for the diffusion of impurities in certain selected areas. F u r t h e r m o r e , it b e c o m e s possible to operate with very small dimen­ sions of the o r d e r of one micron or less. Faster switching and higher frequency limits are the benefits. Thus, silicon is a t o p quality semiconductor while providing an insulator with excellent properties. Gallium arsenide has t h e major advantage of a mobility that is approximately five times that of silicon, since a higher mobility is translated into a higher velocity of t h e carriers for a fixed electric field. Of course, this property offers possibilities of faster switching. It is, however, m o r e difficult to process and is m o r e expensive than silicon. O n e of its major applications, as we shall see later, is in optical devices. We have listed in the accompanying table some properties of t h e t h r e e major semiconductors. TABLE 3.3 Properties of Silicon, Germanium and Gallium Arsenide at Τ = 300K Property

Unit

Density of atoms Energy gap Effective mass m*/m electron hole Effective density of states conduction band valence band Intrinsic carrier density Mobility at low doping electron hole Breakdown field Relative permittivity Melting point

cirr eV

Si

3

5 x 10 1.12

Ge 22

4.4 x 10 0.66

GaAs 22

2.2 Χ 10 1.42

22

Q

1.182 0.81 cm

0.0655 0.524

- 3

3.22 Χ 10 1.83 Χ 10 1 Χ 10

3

cnr cm (V - s ) 2

0.553 0.357

10

19

19

1.03 Χ 10 5.35 Χ 10 2.17 x 10

19

18

13

4.21 Χ 10 9.52 x 10 2.49 x 10

- 1

V/cm dimensionless °C

1350 480 3 Χ ΙΟ 11.8 1410

5

3900 1900 10 15.8 940 5

8800 400 4 χ 10 13.1 1240

5

17

18

6

Chapter 3

Problems

81

PROBLEMS 3.1 Draw the energy band diagrams, showing E , E , E , and E , for the following; assum­ ing all impurities are ionized, a) Intrinsic silicon at 300K. b) Silicon doped with 10 boron atoms c m at 300K. c

v

17

F

t

- 3

16

- 3

3.2 A silicon sample is doped with 10 c m of phosphorus atoms. Assume that all phos­ phorus atoms are ionized at 300K and determine: i) The electron density n . ii) The hole density p . iii) The location of the Fermi level with respect to E Q

0

r

18

3

3.3 Repeat Problem 3.2 if the phosphorus doping is at a level of 10 cm~ . 16

3

3.4 Repeat Problem 3.2 for a boron doping of 10 cm~ . 3.5 The Fermi level in a silicon sample at equilibrium is located QAOeV below the middle of the band gap. At Τ = 300K, a) Determine the probability of occupancy of a state located at the middle of the band gap. b) Determine the probability of occupancy of the acceptor states if the acceptor states are located at 0.04eV~ above the top of the valence band. c)

Check if the assumption of complete ionization in part (b) is valid. 19

3

3.6 The effective conduction band density of states is N = 3.22 X 10 cm~ and the effec­ tive valence band density of states is N = 1.83 X 10 cm~ . Assume that N and N are located at the conduction band and valence band edges respectively. Let Τ = 300K. c

19

3

v

c

v

a) b) c)

For Problem 3.5, determine the thermal equilibrium electron density. Determine the thermal equilibrium hole density. Calculate the p n product. 0

a

3.7 In a silicon sample at equilibrium the Fermi level is located above the middle of the band gap by 0.38eV. The phosphorus donor states are located 0.04eV below the conduction band. Determine the percentage ionization of the phosphorus atoms at Τ = 300K. 15

3

3.8 A silicon sample at 300K has an acceptor density of 10 cm~ . Determine: a) The hole density. b) The electron density. 10

3

3.9 A certain silicon sample at 300K has p = 4 X 1 0 c n r . Determine: a) The electron density. b) The acceptor density if the donor density is 1.25 Χ 10 c m . 0

10

16

3

3.10 A certain silicon sample at 300K has 7V = 1 X 10 cm- and N Determine: a) The majority carrier density. b) The minority carrier density. D

-3

16

A

3

= 0.8 X 10 cm- .

Chapter 3

Intrinsic and Extrinsic Semiconductors

3.11 Determine the electron and hole thermal equilibrium densities and the location of the Fermi level, with respect to E , for a silicon sample at 300K that is doped with a) 1 X 10 cm" of boron. b) 3 X 10 cm~ of boron and 2.9 X 10 cm" of phosphorus. 3.12 Determine the approximate donor binding energy for silicon given e = 11.8 and m*/m = 0.26. 3.13 An TV-type silicon sample has an arsenic dopant density of 10 cm~ . Determine: a) The temperature at which half the impurity atoms are ionized. b) The temperature at which the intrinsic density exceeds the dopant density by a factor of 10. Assume E does not change with T. c) Assuming complete ionization, calculate the minority carrier density at 300K and the location of the Fermi level referred to E 3.14 A silicon sample has the energy band diagram shown in the figure below. Given Ε = 1.12eV,and«. = 10 cm- . a) Sketch the potential V as a function of x. b) Sketch the electric field % as a function of x. c) Determine the values of « and p at χ = Xj and χ = χ 3.15 A silicon sample is doped with 10 donor atoms/cm . Draw an energy level diagram showing the location of the Fermi level with respect to the middle of the band gap for: a) 77K (liquid nitrogen). c

15

3

16

3

16

3

r

0

17

3

g

c

10

3

0

Q

χ

16

3

b) 300K. c) 600K. Neglect the change of £ with temperature. 3.16 a) At Τ = 300K, what percentage of the electrons in a cm of silicon are located in the conduction band? b) Repeat for Τ = 500K. 3.17 On the energy band diagram, electron energy is measured upwards. Explain why hole energy is measured downwards. g

3

Ε

χ =

0

X\

*

2

•

χ

Chapter 3

Problems

83

3.18 In the band diagram shown below, which carrier has the larger kinetic energy? Explain the reason for your answer. 1I

leV

2eV

4eV

ο

chapter 4 CARRIER PROCESSES: DRIFT, DIFFUSION AND GENERATIONRECOMBINATION 4.0 I N T R O D U C T I O N In the last chapter, we identified electrons and holes as the current carriers in a semiconductor. T h e electrons exist in t h e conduction b a n d and t h e holes are in the valence band. In an intrinsic semiconductor, at t h e r m a l equilibrium, the density of electrons is equal to the density of holes because the r a t e of generation of electronhole pairs is balanced by the r a t e of recombination of electrons and holes. The addition of small traces of controlled impurities increases the density of one carrier while decreasing the density of the other. The density of electrons is e n h a n c e d by the addition of d o n o r impurities while the addition of acceptor impuri­ ties increases t h e density of holes. In o r d e r to d e t e r m i n e expressions for t h e carrier currents in terms of an applied voltage and other factors, it is necessary to discuss and define the processes by which carriers are caused to move. Carrier m o t i o n is caused by two conditions; the application of an electric field whose force accelerates the carriers and, second, a difference of carrier concentra­ tion b e t w e e n two points causes carriers to move by diffusion from regions of high concentration to regions of low concentration. F u r t h e r m o r e , t h e processes of carrier generation and recombination affect t h e resulting densities of t h e carriers. It is by considering t h e combined effects of carrier motion, generation, and recombination, o n t h e distribution and motion of the carriers, that we can proceed to set u p t h e expressions from which we d e t e r m i n e the relations describing the o p e r a t i o n of semiconductor devices.

84

Section 4.2

Thermal Velocity

85

4.1 V E L O C I T Y LIMITATIONS In the last chapter, we derived relationships for t h e densities of carriers in an extrin­ sic semiconductor. W h e n we consider a bar of semiconductor, to which an electric field is applied, t h e current in the bar will d e p e n d on t h e density of the carriers, the cross-sectional area of the bar and t h e velocity with which the carriers move. For an /V-type semiconductor, w h e r e n » p , the current is given by Q

/

T

mpS =

couls)

= H

0

/electro)

q

Ί^) ° {-^rj

/ couls \

lele^irj

V

/cm) ) *

, A

(

c

m

9 X }

The question we n e e d to answer is: H o w d o we d e t e r m i n e the m e a n velocity of the carriers? Obviously, t h e velocity d e p e n d s on the acceleration, which in turn d e p e n d s on t h e electric field intensity. For a constant electric field, w h a t is t h e limi­ tation to the velocity that t h e carriers can acquire? In our discussions in earlier chapters, we assumed that the solid possesses a perfectly periodic lattice and disregarded any irregularities in it. In a metal at Τ > OK, electrons are continuously in motion due to the t h e r m a l energy they acquire. If a constant force is applied to the electrons, as by an electric field, t h e electrons in the perfectly periodic crystal would b e accelerated, gain energy, move to empty states at higher energy levels within the same band, and continue to gain in velocity and energy. Electric current is proportional to the n u m b e r and velocity of the electrons and, therefore, the current in a material to which an electric field is applied, in the absence of o t h e r p h e n o m e n a that might limit its value, will tend to increase without limit as t h e velocity increases. However, we k n o w from O h m ' s law that the current reaches a fixed value indicating a constant velocity. The velocity we are referring to is a mean velocity, averaged over t h e velocities of all t h e free electrons. T h e m e a n velocity is constant because of a resistive force that prevents further acceleration. This resistive force is a direct result of t h e collision of t h e electrons with atoms or with the regions in t h e lattice of t h e material that disturb t h e motion of the electrons. T h e collison could also b e with foreign atoms. Before further discussion of t h e n a t u r e or effect of t h e collisions, let us present the condition of the electrons in a semiconductor w h e n they are not u n d e r t h e influ­ ence of an electric field.

4.2 T H E R M A L V E L O C I T Y Electrons and holes in semiconductors are in constant motion because of t h e ther­ mal energy they receive. Since they are in motion, they are not associated with any particular lattice position. A t any o n e time, the electrons and holes m o v e in r a n d o m directions with a m e a n r a n d o m velocity. This velocity, k n o w n as the thermal velocity, is of t h e o r d e r of 1 0 c m / s for electrons in silicon. Because of the motion in ran­ d o m directions, the current resulting from t h e m o t i o n of all t h e carriers in any one direction is zero. 7

86

Chapter 4

Carrier Processes: Drift, Diffusion and Generation Recombination

A s a m e a s u r e of t h e energy of the electron, we establish t h e level of the energy of an electron at rest to b e at the b o t t o m of t h e conduction b a n d E . The kinetic energy of an electron is m e a s u r e d by the energy separation above E , as Ε — E It has b e e n established that, at thermal equilibrium, the m e a n - s q u a r e thermal velocity of the electron is related to t e m p e r a t u r e by t h e relation* c

c

c

(4.1) w h e r e m* is the conductivity effective mass of t h e free electron, v its m e a n t h e r m a l velocity, k is Boltzman's constant, and Τ is the t e m p e r a t u r e in degrees Kelvin. For silicon at Τ = 300K, this velocity has b e e n calculated to be approximately 2.3 X 1 0 c m / s . T h e m e a n kinetic energy for all the electrons at t h e r m a l equilibrium is £ — E = (3/2)kT, and this translates into 0.04eV at 300K, which is slightly above the conduction b a n d edge E , and for silicon represents approximately 1/25 of the b a n d gap energy E . Electrons traveling in a solid u n d e r the influence of a small applied electric field collide with the lattice, exchange energy with the lattice, and start all over. D e p e n d i n g u p o n the magnitude of the field, the electron gives u p a certain a m o u n t of heat to the lattice. th

7

c

c

g

4.3 C O L L I S I O N S A N D SCATTERING W h e n a relatively low-intensity electric field is applied to a metal or a semiconduc­ tor, electrons acquire a m e a n velocity in accordance with O h m ' s law. This new velocity is superimposed on t h e t h e r m a l velocity and its magnitude is m u c h smaller than t h e m e a n t h e r m a l velocity. This new velocity c o m p o n e n t refers to the average rate of m o t i o n of the electron population in the direction of the force of the electric field. This velocity is k n o w n as the drift velocity, v . Sketches comparing the r a n d o m field-free motion of electrons with the fielddirected motion are shown in Fig. 4.1. It is i m p o r t a n t to point out that since the drift velocity is, in most devices, much smaller t h a n t h e t h e r m a l velocity, t h e drift velocity can be considered a p e r t u r b a t i o n of the t h e r m a l velocity. The electron that is moving in a solid, subjected to an electric field, collides with the atoms and with t h e nonideal lattice structure, loses most of its velocity, and then r e p e a t s the process. This process of collision is m o r e accurately k n o w n as scat­ tering. It is important to point out that quantum-mechanical calculations indicate that scattering does not take place, and h e n c e n o energy exchange occurs, w h e n an electron is traveling in a perfectly periodic and stationary lattice. A non-stationary lattice results w h e n the atoms vibrate by acquiring heat. Scattering does not refer only to a direct collision b e t w e e n the moving elecd

*From Halliday, Resnick, and Walker, Fundamentals of Physics, p. 582; copyright © Wiley (1993). Reprinted by permission of John Wiley & Sons, Inc.

Section 4.3

Collisions and Scattering

87

trons and t h e atomic core. It is caused m o r e by t h e variation in the potential distri­ butions in the solid w h e n the crystal structure is not perfectly periodic. Such a solid is labeled aperiodic. Aperiodicity is caused by t h r e e factors, one of which is a tem­ p e r a t u r e of the solid above zero. Aperiodicity may also result from imperfections in t h e solid and from the presence of impurities. T h e r e are therefore t h r e e p h e n o m e n a that individually or jointly cause departures from t h e periodic potential. A t n o r m a l operating temperatures, t h e lattice atoms vibrate about their equi­ librium positions. These vibrations disturb t h e periodicity and change the periodic potential, which generates an electric field that scatters t h e carriers. A n electron travelling through t h e lattice undergoes scattering, which causes changes in the magnitude and direction of its velocity. T h e scattering results b o t h from collisions with the vibrating atoms themselves and with t h e varying potential field that results from the displacement of the vibrating atoms. The higher t h e t e m p e r a t u r e , the larger is t h e amplitude of the atomic vibrations, the larger is the scattering cross sec­ tion, the larger is the disturbance of t h e potential field, and therefore t h e higher the probability of the scattering taking place. A second mechanism that causes scattering results from the presence of ion­ ized impurity atoms. These atoms p r o d u c e an electric field that changes t h e poten­ tial distribution in the crystal. This type of scattering is m o r e important at lower t e m p e r a t u r e s w h e n the t h e r m a l scattering is weaker. The higher the impurity con­ centration, t h e higher is the probability of this type of scattering. A third cause of scattering, although less important t h a n thermal or impurity scattering, results from imperfections such as vacancies in the crystal structure and crystallographic defects. Crystal imperfections cause a change in t h e periodic potential and thus create an electric field. A n electron passing near an imperfection interacts with the field resulting in a change of the direction of motion. We conclude that at lower temperatures, impurity scattering is m o r e domi­ nant, while at the higher temperatures, t h e r m a l (lattice) scatterings is m o r e effective.

88

Chapter 4

Carrier Processes: Drift, Diffusion and Generation Recombination

4.4 C O L L I S I O N S EFFECTS Drift Velocity T h e effect of t h e scattering, or the collisions, on the motion of a carrier is equivalent to a frictional resistive force that controls the acceleration of the particle and limits its velocity to t h e drift velocity. The m o t i o n of an electron that is subjected to an electric field, in the x-direction, obeys t h e equation, d χ m

n

d

t

„

^

2

-

X

dx -

W

(4-2)

where % is t h e electric field intensity, m* is the electron effective mass, q is the elec­ tronic charge, and the second t e r m on t h e right h a n d side of E q . (4.2) is the frictional resistive force where w is a constant d e p e n d e n t on the solid. The electric field does not accelerate the electron continuously because the resistive force increases as t h e velocity increases until such time that t h e resistive force balances the force of t h e electric field until t h e steady-state, w h e n t h e velocity becomes the drift velocity. T h e drift velocity can b e found by setting t h e impulse (force X time), applied to an electron during the time T b e t w e e n collisions, equal to the m o m e n t u m gained by t h e electron in that period. In t h e steady state, t h e drift velocity is d e t e r m i n e d from x

c

-xtc

=

m

> d

w h e r e T is the m e a n time b e t w e e n collisions, k n o w n as the m e a n scattering time, and the drift velocity is given by c

(43)

"-- τ , the particle velocity will have a steady-state value given by t h e drift velocity, the negative sign indicating that the velocity is in a direction opposite to that of the electric field. Suppose that, as t h e electron is being accelerated, the electric field is r e m o v e d and at this time assume its velocity is v„. For this condition, the velocity in accor­ dance with the solution to E q . (4.2) b e c o m e s

«(i) = « exp(-r/T ) 0

C

(4.6)

Section 4.4

Collisions Effects

89

The time constant T may also b e interpreted as the factor that controls the rate at which t h e velocity and t h e current decay to zero after the field is removed. This time is k n o w n as the relaxation time and it is of the o r d e r of l ( F s . The steady-state value of the drift velocity in t h e presence of the electric field is also found by setting t > > τ , in E q . (4.4).This time has also b e e n r e p r e s e n t e d to refer to a m e a s u r e of the time it takes for carriers in a semiconductor to move and neutralize charges. The r e a d e r may have concluded that w h e r e we speak of the velocity of an electron, we are implying that it is possible to isolate the particle. In fact, t h e velocity h e r e refers to the m e a n of t h e velocities of all the electrons that are free. We defined earlier scattering in t h e presence of an electric field, as the process of collision b e t w e e n the field-directed electron and t h e lattice structure. We will now illustrate this operation on an energy b a n d diagram. c

15

Collisions a n d E n e r g y Exchanges In Fig. 4.2(a), we show an N-type semiconductor with metal contacts attached to both ends. A t point A , a voltage + V is applied with respect to contact B, which is at g r o u n d potential. We will study the effect on the energy of an electron released from rest at point B. Before we do that, we will review t h e energy relations w h e n t h e bar is replaced by a vacuum enclosure. A t terminal Β of the vacuum enclosure, an elec­ tron is released from rest, and since there are n o obstacles and hence n o collisions along the way, t h e electron accelerates and strikes plate A . A t point B, t h e electron lost potential energy qV; at impact with terminal A , all the potential energy is con­ verted to kinetic energy and hence heat at point A. In the energy b a n d diagram of the semiconductor bar of Fig. 4.2(b), an elec­ tron held at point Β has potential energy qV and its potential energy w h e n it arrives

A

Β

(b)

Figure 4.2 Illustrating conduction in a bar and by use of the energy band diagram (a) semiconductor bar with a voltage applied; (b) motion of electron on energy band diagram.

90

Chapter 4

Carrier Processes: Drift, Diffusion and Generation Recombination

at A is zero. Since t h e b o t t o m of t h e conduction b a n d at E is t h e level of the poten­ tial energy of an electron at rest, t h e n the energy b a n d diagram for an electron in this illustration shows a decreasing potential energy and hence a decreasing level of E as we p r o c e e d from Β to A in Fig. 4.2(b). This explains t h e slope of t h e energy b a n d diagram. A n e l e c t r o n r e l e a s e d from rest at p o i n t Β in t h e bar is accelerated by t h e electric field in Fig. 4.2(a), acquires kinetic energy, a n d just before p o i n t C, its total e n e r g y is u n c h a n g e d as it replaces t h e p o t e n t i a l energy it lost by kinetic energy. W h e n it collides with t h e lattice structure at p o i n t C, it loses all t h e kinetic energy it acquired, its total energy is p o t e n t i a l , a n d t h e electron d r o p s to t h e level E c o r r e s p o n d i n g to p o i n t C. It is accelerated again, collides with t h e lattice, and d r o p s b a c k to t h e lower E level at p o i n t D. This process continues until t h e elec­ t r o n r e a c h e s p o i n t A , w h e r e it has lost all t h e qV p o t e n t i a l energy that it h a d at p o i n t B. This p o t e n t i a l energy has b e e n c o n v e r t e d , along t h e way, into h e a t in t h e lattice. c

c

c

c

4.5 M O B I L I T Y Referring to E q . (4.3), we observe that the drift velocity of electrons (or holes) is given by the p r o d u c t of t h e electric field intensity and a factor that d e p e n d s on the electronic charge, the effective mass, and t h e relaxation time of t h e semiconductor. We label that factor the mobility of a carrier, μ, so that for an electron we have

2

where v is the drift velocity of an electron in c m / s , μ is its mobility in c m / v o l t - s e c and % is t h e electric field intensity in V / c m . The negative sign denotes that, for an electron, the direction of the velocity is opposite to the direction of the electric field intensity. By using E q . (4.3), in E q . (4.7), the mobility of t h e electron becomes d

η

< 4

·

8 )

The analogous expressions for hole m o t i o n are = ^

(4-9)

and the mobility of the hole is

- %

410

< >

Because the time interval b e t w e e n collisions depends mainly on t e m p e r a t u r e and impurity concentration, the mobility depends on those two p a r a m e t e r s as well as on the material through the effective mass. With reference to Fig. 2.12, which showed a m u c h greater curvature near the b o t t o m of the conduction b a n d of G a A s t h a n that of Si, the effective mass of an electron in G a A s is m u c h smaller than in Si. A s a result, t h e electron mobility in G a A s is four to five times greater than in silicon.

Section 4.5

Mobility

91

GaAs electrons

I0

1(P

:

10

4

5

LO

6

io

Electric field (V/cm) Figure 4.3 Variation of drift velocity with electric field intensity for gallium arsenide and silicon. Source: Reprinted with permission from Ruch and Kino, "Measurement of the Velocity-Field Characteristics of Gallium Arsenide," Applied Physics Letters 10,40 (1967) copyright American Institute of Physics; and Caughey and Thomas "Carrier Mobilities in Silicon Empirically Repated to Doping and Field," Proc. IEEE 55,2192 (1967) © IEEE.

The above t r e a t m e n t assumes that the time interval between collisions is inde­ p e n d e n t of the magnitude of the applied electric field intensity. This is valid p r o ­ vided the drift velocity is m u c h smaller t h a n t h e t h e r m a l saturation velocity, v , which is approximately 10 c m s / s for electrons in silicon. E x p e r i m e n t a l results of t h e m e a s u r e m e n t s of t h e drift velocity of electrons and holes are shown in Fig. 4.3. It is quite evident that at very large values of electric field intensity, the drift velocity approaches the saturation velocity a n d h e n c e the mobility decreases with increasing electric field intensity. We also note, in Fig. 4.3, the d e p e n d e n c e of the drift velocity, and hence the mobility in the linear region, on the material through its d e p e n d e n c e on the effec­ tive masses of the electrons and the holes. &

7

Effects of I m p u r i t y C o n c e n t r a t i o n a n d T e m p e r a t u r e o n Mobility

The d e p e n d e n c e of the mobility on the impurity concentration, through the m e a n scattering time T , results from the scattering of the carriers by the coulombic effect of the ionized impurities as t h e carriers travel in the solid. High doping density c

92

Carrier Processes: Drift, Diffusion and Generation Recombination

Chapter 4

50 I 100

I

,

200

,

I 500

,

,

. . I 1000

T(K) Figure 4.4 Variation of electron mobility with temperature and doping in silicon. Source: S.M. Sze, Semiconductors: Devices, Physics and Technology, p. 33; copyright © Wiley (1985). Reproduced by permission of John Wiley & Sons, Inc.

m e a n s m o r e scattering, lower drift velocities, and mobilities. The variation of mobil­ ity with scattering through its d e p e n d e n c e on impurity concentration and t e m p e r a ­ ture is shown in Fig. 4.4 and highlighted in the inset. The question t h e n is: Why is it that, in the inset of Fig. 4.4, the mobility is seen to increase with an increase of t e m p e r a t u r e in a certain t e m p e r a t u r e range, yet decreases with an increase of t e m p e r a t u r e in a n o t h e r t e m p e r a t u r e range? A t low t e m p e r a t u r e s , lattice vibrations are so small so that as t h e t e m p e r a t u r e increases, t h e carriers receive t h e r m a l energy that increases their velocity, thus reducing the time that the electron spends in the vicinity of the lattice and impurity atoms. This reduces t h e scattering cross section, which causes the mobility to increase with t e m p e r a t u r e . T h e increase of mobility with t e m p e r a t u r e continues until, at a certain temper­ ature, t h e lattice vibrations d o m i n a t e and the scattering cross section increases, thus increasing the possibility of collision with t h e lattice and decreasing the mobility with an increase of t e m p e r a t u r e .

Section 4.5

Mobility

93

Expressions f o r t h e M o b i l i t y W h e n two scattering processes occur at t h e same time, the combined effect on the mobility is d e t e r m i n e d from the probabilities of the two processes. If in an incre­ mental time, dt, t h e probability of scattering by process 1 is dt/j where τ, is the time intervals b e t w e e n two collisions and the probability of scattering by process 2 is άί/τ , t h e n t h e total probability dt/j is the sum of the two probabilities so that we can write v

2

c

dt/τ

= dt/j

+

1

άί/τ

2

(4.11)

or

W h e r e τ. and τ refer to the collision times due to impurities and t e m p e r a t u r e respectively. The mobility μ can be expressed in terms of the mobilities due to impu­ rities and d u e to t e m p e r a t u r e μ and μ, respectively as (

;

1

=

1

+

1

(4.12)

It has b e e n shown that the mobility due to lattice scattering ( t e m p e r a t u r e ) , μ , is proportional to 1/T . The mobility caused by impurity scattering, μ , has b e e n d e t e r m i n e d to be proportional to T /N, where Ν is the total impurity density (

312

;

3I2

+

(N

D

N ). A

Empirical expressions for the mobilities of electrons and holes in silicon in c m / V - s , as a function of doping density and t e m p e r a t u r e have b e e n derived,* and shown below, w h e r e Τ is in Κ, Τ = Γ/300, and Ν is in c m . 2

- 3

8

2

7.4 Χ 1 0 Γ " ·

m = 887y 1 +

3 3

Ν 17

2

4

(1.26 Χ 10 Γ„ · ) 8

(4.13) 0.88Γ

- 2

1.36 Χ 1 0 Γ -

54.37Λ

Ν

1

.(2.35 χ

2 3

(4.14) 0.887Λ

i o

1

7

r

2

4

-0.146

-0.146

)

A t Τ = 300Κ these expressions reduce to: 1252 μ

"

μ-»

(4.15) 17

1 + 0.698 X 10- 7V 54.3 +

407 1 + 0.374 Χ

17

10~ Ν

These relations agree fairly well with experimental results. *From Arora et at, "Electron and Hole Mobilities in Silicon as a Function of Concentration and Tem­ perature," IEEE Transactions on Electron Devices, Vol. 29, No. 2, p. 2192, February 1982. © 1982, IEEE.

(4.16)

Chapter 4

Carrier Processes: Drift, Diffusion and Generation Recombination

3

Impurity concentration (cm ) Figure 4.5 Mobilities of electrons and holes for Si, Ge and GaAs at Τ = 300K, as a function of total impurity density (N + N ). Source: Reprinted from S. M. Sze and J. C. Irvin, "Resistivity, Mobility and Impurity Levels in GaAs, Ge and Si at 300K," Solid State Electronics, 11, 599 (1968) with kind permission from Elsevier Science Ltd., The Boulevard, Langford Lane, Kidlington 0X5 1GB, UK. D

A

Plots of m e a s u r e m e n t s of the mobility versus total impurity concentration (N + N ), at Τ = 300K, are shown in Fig. 4.5. We observe that at low concentration levels t h e mobility is i n d e p e n d e n t of the concentration. We conclude this section by summarizing the factors that influence the mobility: D

A

1. T h e mobility d e p e n d s very strongly on the t e m p e r a t u r e and on irregularities in the crystal, which may b e d u e to the growing of the crystal or t h e impurities. 2. So long as the drift velocity is less than t h e thermal velocity, the mobility varies directly with, and is proportional to, t h e time b e t w e e n collisions. 3. If t h e drift velocity approaches t h e saturation velocity through an increase of the electric field intensity, the mobility decreases with an increase of t h e electric field.

Drift

Section 4.6

95

Current and Conductivity

4. The mobility, therefore, is affected to varying degrees by: irregularities in the lattice structure, t e m p e r a t u r e , concentration of added impurities, and the elec­ tric field intensity.

4.6 DRIFT C U R R E N T A N D C O N D U C T I V I T Y W h e n an electric field is applied to a semiconductor, the electrons and holes acquire drift velocities and the c o n s e q u e n t m o t i o n of carriers results in an electric current. This current is k n o w n as drift current. The electron drift current is given by: I =-Aqnv n

(4.17)

d

2

w h e r e A is the cross sectional area in c m , n o r m a l to the direction of current flow, q is t h e charge in coulombs, η is t h e electron density in c m " , v is the electron drift velocity in c m / s e c , and I is in amperes. T h e direction of t h e current is opposite to that of the motion of t h e electrons. By replacing for the velocity from E q . (4.7) into E q . (4.17), we have 3

d

l =Aqn^ % n

(4.18)

n

The current due to electron motion is thus in t h e same direction as the electric field intensity. The electron current density in a m p s / c m is 2

The factor multiplying % is k n o w n as the conductivity

σ so that η

J = σ%

(4.19)

n

1

w h e r e the unit of the conductivity is ( o h m - c m ) " . Analogously the current density for holes becomes J = p

p%

= σ%

qVlp

(4.20)

ρ

The total drift current density, / , is (

J = J +J t

η

=σ%+σ% ρ

η

(4.21) ρ

κ

'

The total conductivity is σ

ί =

σ

η+

σ Ρ

=