Vật lý điện tử GS Le Tuan

Using the de Broglie relation p = Ñ k the ‘volume’ of phase space can be written as 1.2 Dx Dy Dz Dk xDkyDkz = 2 p 3 The density of states per unit energy and per unit volume, which is

FUNDAMENTAL QUANTUM CONCEPTS

1.1 The Bohr model

The results of emission spectra experiments led Niels Bohr (1913) to construct a model for thehydrogen atom, based on the mathematics of planetary systems If the electron in the hydrogenatom has a series of planetary-type orbits available to it, it can be excited to an outer orbit and thencan fall to any one of the inner orbits, giving off energy corresponding to one of the lines of Fig.2.1.To develop the model, Bohr made several postulates

è Electrons exist in certain stable, circular orbits about the nucleus This assumption implies that the

orbiting electron does not give off radiation as classical electromagnetic theory would normallyrequire of a charge experiencing angular acceleration; otherwise, the electron would not be stable

in the orbit but would spiral into the nucleus as it lost energy by radiation

è The electron may shift to an orbit of higher or lower energy, thereby gaining or losing energy

equal to the difference in the energy levels (by absorption or emission of a photon of energy hw)

è The angular momentum pq of the electron in an orbit is always an integral multiple of Planck’sconstant divided by 2p (h/2p is often abbreviated Ñ for convenience), pq = n Ñ, (n = 1, 2, 3, 4, ).Although Bohr proposed this ad hoc relationship simply to explain the data, one can see that this

is equivalent to having an integer number of de Broglie wavelengths fit within the circumference

of the electron orbit (Prob 2.2) These were called pilot waves, guiding the motion of theelectrons around the nucleus The de Broglie wave concept provided the inspiration for theSchrödinger wave equation in quantum mechanics discussed later

Fig 2.1 Electron orbits and transitions in the Bohr model of the hydrogen atom Orbit spacing is not drawn to scale.

Bohr supposed that the electron in a stable orbit of radius r about the proton of the hydrogenatom, we can equate the electrostatic force between the charges to the centripetal force:

pq= m v r = n Ñ Since n takes on integral values, r should be denoted by r n to indicate the nth orbit Then Eq.

with the ionization energy for hydrogen atom E1 º - 13.6 eV

The transitions in hydrogen emission spectra have been explained (see Fig, 2.1), for example,

between the orbits n1 and n2

with R is the Rydberg constant, which is frequently serving as an unity of energy in atomic physics.

Although the Bohr model was immensely successful in explaining the hydrogen spectra, ous attempts to extend the "semi-classical" Bohr analysis to more complex atoms such as hcliumproved to be futile Success along these lines had to await further development of the quantum

numer-mechanical formalism Nevertheless, the Bohr analysis reinforced the concept of energy

quantiza-tion and the attendant failure of classical mechanics in dealing with systems on an atomic scale.

Moreover, the quantization of angular momentum in the Bohr model clearly extended the quantumconcept, seemingly suggesting a general quantization of atomic-scale observables

1.2 Black body radiaton

In this part we will very briefly recall some important moments in hystory of physics at the end

of the XIX century We know that heat transfer takes place through the processes of conduction,convection and radiation The first two processes can take place only in a medium, while the lastdoes not need a medium Robert Kirchhoff studied (1860) the problem of radiation and the proper-ties of perfect absorb materials He pointed out that when such a material is heated, it will emit

radiations of all wavelengths i.e it will be a perfect emitter Such a body is referred to as a black

body.For instance, a hollow body whose walls are at the same temperature behaves like a black

body They pointed out that it will be a perfect emitter emitting through a tiny hole in its surface,radiations of all wavelengths Further, any radiation entering such a cavity will undergo infinitelymany reflections inside and will lose energy at every reflection

Thus no incident radiation will emerge out of the tiny hole In other words, it is also a perfectabsorber Thus such a cavity, which can be easily constructed , is almost a perfect black body

In 1879, the Austrian physicist Josef Stefan reassessed the work of Dulong and Petit He madethe corrections and calculated the pure radiation component of heat transfer and the went to the lawfor the total emission power of black body

(1.8)

E = s T4with the Stefan constant s = 5.70 ä 10-8 W m-2 K-4 Then, Wien (1893) calculated the energy

u(l, T) emitted per unit volume per sec in unit wavelength interval.

(1.9)

u l, T = f l, T

l5 ï lp T = b where b = 2.898 ä 10-3 m.K is the Wien constant anf the nature of the function f(l, T) still is

unknown Based on Newtonian physics, by very adequate calculation of the emission within theblack body cavity to be made up of a series of standing waves, Ragleigh and Jeans came to thefamous formula:

(1.10)

u l, T = 8 p k T

l4

Fig 2.2 The black body emission spectra The outstanding violet catastrophe with Rayleigh-Jeans formula led Planck to

work out the concept of quantization of energy transfer.

However, it completely fails near the low wavelength end where it diverges to an infinite value

for u(l, T) This is often referred to as the Ultra-violet Catastrophe

The important step was taken by Max Planck around 1900 Planck suggested an interpolationformula by taking for Wien's function `f(l, T)' the form:

(1.11)

f l, T = 8p k b

Expl Tb  - 1

where k as the Boltzmann constant and b as an adjustable parameter

At that time, the physicists thought that the atoms of a material which absorbed or emittedradiation as harmonic oscillators Hence, oscillators were in equilibrium with radiation exchangingenergy with it Planck knew that radiations are electromagnetic waves Hence the cavity was filledwith these waves These waves were absorbed and emitted by the oscillators that behaved likeclassical `pendulums' Planck believed in the electromagnetic nature of radiation But the oscillatormechanism of absorption and emission of electromagnetic waves was more a theoretical model Hearriveed at his own formula for the black-body spectrum, suggesting that a harmonic oscillator cannot have any energy but only in integral multiples of a quantum of energy e0 = hn , where n is thenatural frequency of the oscillator and h a constant to be determined In other words, Planck quan-tized the permitted energies of an oscillator Thus, they were strictly non-classical in nature withenergies e = n hn , n being an integer These oscillators are in thermal equilibrium at any tempera-ture Therefore, Planck invoked the Boltzmann distribution to describe them Accordingly, theaverage oscillator energy e of the system of oscillators becomes:

(1.12)

Exphn

k T - 1

with the Planck constant h = 6.626 ä 10-34 J.s

Then, the total energy of radiation per unit volum e at l per unit wavelength interval is:

(1.13)

1.3 Basic Formalism

1.3.1 The five postulates of Quantum Mechanism

The formulation of quantum mechanics, also called wave mechanics focuses on the wavefunction, Y(x,y,z,t), which depends on the spatial coordinates x, y, z, and the time t In the following

sections we shall restrict ourselves to one spatial dimension x, so that the wave function depends solely on x An extension to three spatial dimensions can be done easily The wave function Y(x,t)

and its complex conjugate Y*(x,t) are the focal point of quantum mechanics, because they provide a

concrete meaning in the macroscopic physical world: The product Y*(x, t)dx is the probability to

find a particle, for example an electron, within the interval x and x + dx The particle is described

quantum mechanically by the wave function Y(x,t) The product Y*(x,t)Y(x,t) is therefore called the

window of quantum mechanics to the real world.

Quantum mechanics further differs from classical mechanics by the employment of operators rather than the use of dynamical variables Dynamical variables are used in classical mechanics,

and they are variables such as position, momentum, or energy Dynamical variables are contrastedwith static variables such as the mass of a particle Static variables do not change during typicalphysical processes considered here In quantum mechanics, dynamical variables are replaced byoperators which act on the wave function Mathematical operators are mathematical expressions

that act on an operand For example, (d / dx) is the differential operator In the expression (d / dx) Y(x,t), the differential operator acts on the wave function, Y(x,t), which is the operand Such

operands will be used to deduce the quantum mechanical wave equation or Schrödinger equation.The postulates of quantum mechanics cannot be proven or deduced The postulates are hypothe-ses, and, if no violation with nature (experiments) is found, they are called axioms, i e non-provable, true statements

Postulate 1

The wave function (x,y,z,t) describes the temporal and spatial evolution of a cal particle The wave function (x,t) describes a particle with one degree of freedom of motion

quantummechani-Postulate 2

The product Y*(x,t)Y(x,t) is the probability density function of a quantum-mechanical particle

Y*(x,t)Y(x,t)dx is the probability to find the particle in the interval between x and x + dx Therefore,

(1.14)

-¶

+¶

Y *x, t Y x, t „ x = 1

If a wave function (x, t) fulfills Eq (2.14), then (x, t) is called a normalized wave function

Equation (2.14) is the normalization condition and implies the fact that the particle must be

located somewhere on the x axis.

Postulate 3

The wave function (x, t) and its derivative (/ x) (x, t) are continuous in an isotropicmedium

(1.15)Limit x, t, x Ø x0 =  x 0 , t

(1.16)

Limit ∑

∑ x x, t, x Ø x0 = ∑

∑ x x, t

x =x0

In other words, (x, t) is a continuous and continuously differentiable function throughoutisotropic media Furthermore, the wave function has to be finite and single valued throughout

position space (for the one-dimensional case, this applies to all values of x).

Postulate 4

Operators are substituted for dynamical variables The operators act an the wave function (x,

t) In classical mechanics, variables such as the position, momentum, or energy are called

dynami-cal variables In quantum mechanics operators rather than dynamidynami-cal variables are employed.

Table 2.1 shows common dynamical variables and their corresponding quantummechanicaloperators

Table 2.1. Common physical dynamic variables and quantummechanic operatotors

x i— pxFunction of position, fx

1.3.2 Some properties of quantummechanical operators

Eigenfunctions and eigenvalues

Any mathematical rule which changes one function into some other function is called an tion Such an operation requires an operator, which provides the mathematical rule for the opera-tion, and an operand which is the initial function that will be changed under the operation Quan-tum mechanical operators act on the wave function Y(x, t) Thus, the wave function Y(x, t) is theoperand Examples for operators are the differential operator (d / dx) or the integral operator 

opera-…dx In the following sections we shall use the symbol x` for an operator and the symbol f(x) for

an operand The definition of the eigenfunction and the eigenvalue of an operator is as follows: Ifthe effect of an operator x`

operating on a function f(x) is that the function f(x) is modified only bythe multiplication with a scalar, then the function f(x) is called the eigenfunction of the operator x`,that is

(1.18)

x`Yx, t = l sYx, t

where ls is a scalar (constant) ls is called the eigenvalue of the eigenfunction For example, the

eigenfunctions of the differential operator are exponential functions, because

(1.19)

„

„ x e

ls x= ls els x

where ls is the eigenvalue of the exponential function and the differential operator

Linear operators and commutation law

Virtually all operators in quantum mechanics are linear operators An operator is a linear

operator if

(1.20)

x p x = p x x

However, in quantum mechanics the two linear operators, which correspond to x and p, do not

commute, as can be easily shown One obtains

(1.22)

Hermitian operators

In addition to linearity, most of the operators in quantum mechanics possess a property which isknown as hermiticity Such operators are hermitian operators, which will be defined in this section.The expectation value of a dynamical variable is given by the 5th Postulate according to (2.17).The expectation value is now assumed to be a physically observable quantity such as posi-tion or momentum Thus, the dynamical variable  is real, and  is identical to its complexconjugate

(1.24)

The definition of an hermitian operator is in fact more general than given above In general,hermitian operators satisfy the condition

(1.25)

-¶

There are a number of consequences and implications resulting from the hermiticity of an

operator Two more properties of hermitian operators will explicitly mentioned First, eigenvalues

of hermitian operators are real To prove this, suppose x` is an hermitian operator with tion y(x) and eigenvalue l Then

eigenfunc-and also due to hermiticity of the operator

Since Eqs (4.17) and (4.19) are identical, therefore l = l*, which is only true if l is real Thus,eigenvalues of Hermitian operators are real

Second, eigenfunctions corresponding to two unequal eigenvalues of an hermitian operator are

orthogonal to each other This is, if x` is an hermitian operator and y1(x) and y2(x) are tions of this operator and l1 and l2 are eigenvalues of this operator then

eigenfunc-Time-Independent Formulation

If the particle in the system under analysis has a fixed total energy E, the quantum mechanicalformulation of the problem is significantly simplified Consider the general expression for theenergy expectation value in the total volume space ý as deduced from Table 2.1 and Postulate 5:

(1.26)

Yr, t = Yx, y, z, t = y x, y, z e -i E tÑfor so-called , reminiscing about the main Newtonian equation in classical mechanics:

(1.30)

“ = ∑2

∑ x2 + ∑

2

∑ y2 + ∑2

∑ z2After substitution the general solution into the time-dependent , we can be dealed now with the

time-independent Schrödiger equation for a steady processes:

(1.32)

- Ñ2

2 m “ + U r yr = E yr

where eigen value E is total energy and eigen function y(r) for the patrticle in the steady state.

The function y(r) must be finite, continuous, and single-valued for all values of x, y, z and has thesimilar statistical meaning like Yr, t does.

1.4 Simple Quantum Mechanics Problem Solutions

1.4.1 The infinite square-shaped quantum well

The infinite square-shaped well potential is the simplest of all possible potential wells The

one-dimension infinite square well potential is illustrated in Fig 2.3(a) and is defined as

Fig 2.3 a) Schematic illustration of the 1-D infinite square quantum well (QW) The solutions of this QW are shown in the

terms of b) eigen-functions Yn x and eigen-state energies E n, and c) probability densities Yn*x Y n x

To find the stationary solutions for yn (x) and E n, we must find functions for yn (x), which

satisfy the Schrödinger equation The time-independent Schrödinger equation contains only the

differential operator d / dx, whose eigenfunctions are exponential or sinusoidal functions Since the

Schrödinger equation has the form of an eigenvalue equation, it is reasonable to try only tions of the differential operator Furthermore, we assume that yn (x) = 0 for | x | > L / 2, because the

eigenfunc-potential energy is infinitely high in the barrier regions Since the 3rd Postulate requires that thewave function be continuous, the wave function must have zero amplitude at the two potentialdiscontinuities, that is yn (x = ±L / 2) = 0 We therefore employ sinusoidal functions and differenti-

ate between states of even and odd symmetry in well region, like that

(1.33)

(1.34)

yn x = 0 , n = 0, 1, 2, and x > L

2The condition of continuity of wave functions yields the value of the constant A

(1.35)

A = 2 L

By inspection of the Schrödinger equation with the determined above wave functions, one candeduce the eigen values - energy levels for the QW:

(1.36)

E0 = Ñ

2p2

2 m L2when other levels are n-th excited state energies with n = 1, 2, 3, , respectively The spacing between two adjacent energy levels, that is E n – E n-1, is proportional to n Thus, the energetic spacing between states increases with energy The eigenstate energies are, as already mentioned,

expectation values of the total energy of the respective state Its easily to show, based on Postulate

5, that the energy of a particle in an infinite square well is purely kinetic The particle has nopotential energy

1.4.2 The 1-D asymmetric and symmetric finite square-shaped quantum well

In contrast to the infinite square well, the finite square well has barriers of finite height The potential of a finite square well is shown in Fig 2.4 The two barriers of the well have a different height and therefore, the structure is denoted asymmetric square well The potential energy is constant within the three regions I, II, and III, as shown in Fig 2.4 In order to obtain the solutions to the Schrödinger equation for the square well potential, the solutions in a constant

potential will be considered first.

Fig 2 4 1-D asymmetric finite quantum well with the width of L and barrier heights U I and U III.

Assume that a particle with energy E is in a constant potential U Then two cases can be

distin-guished, namely E > U and E < U In the first case (E > U) the general solution to the

timeindepen-dent one-dimensional Schrödinger equation is given by

(1.38)

yx = A cosk x + B sink x

where A and B are constants and

(1.39)

Ñ2Insertion of the solution into the Schrödinger equation proves that it is indeed a correct solu-

tion Thus the wave function is an oscillatory sinusoidal function in a constant potential with E >

U

In the second case (E < U), the solution of the time-independent one-dimensional Schrödinger

equation is given by

(1.40)

yx = C e k x + D e - k x where C and D are constants and

(1.41)

wave functions in a constant potential are either sinusoidal or exponential, the wave functions in

the three regions I (x 0), II (0 < x < L), and III (x L), can be written as

(1.42)

yI x = A ekI x

(1.43)

yII x = A cosk x + B sink x

(1.44)

yIII x = A cosk L + B sink L e- kIII x-L

When applying the boundary conditions for continuity of wave functions and their first

differen-tials, we have the first pair of them, i.e y I0 = yII0 and yII L = y III L, are already satisfied The second pair of equations,i.e y I'0 = yII'0 and y 'II L = y III' L , gives

AkI – B k= 0

(1.46)

AkIIIcosk L – k sin k L + B k IIIsink L + k cos k L = 0

This homogeneous system of equations has solutions, only if the determinant of the systemvanishes, so one obtains

(1.47)

tank L = k LkI L + kIII L

k2L2- kI LkIII L

which is the eigenvalue equation of the finite asymmetric square well.

For the finite symmetric square well, which is of great practical relevance, the eigenvalue equation is given by

(1.48)

tank L = 2 k L k L

k2L2- k2L2

where k = I= III If is expressed as a function of k, then solving the eigenvalue equation

yields the eigenvalues of k and, the allowed energies E and decay constants , respectively The

allowed energies are also called the eigenstate energies of the potential We must give up a trivial solution kL = 0 (and thus E = 0) which possesses no practical relevance Non-trivial solutions of

the eigenvalue equation can be obtained by a graphical method Fig 2.5 shows the graph of the

left-hand and right-left-hand side of the eigenvalue equation The dashed curve represents the right-left-handside of the eigenvalue equation The intersections of the dashed curve with the periodic tangentfunction are the solutions of the eigenvalue equation The quantum state with the lowest non-trivial

solution is called the ground state of the well States of higher energy are referred to as excited states.

Fig 2 5 Graphical solution of the eigen-value equation for a symmetric QW The function 2 k L k L/(k2L2 -k 2L2 ) is expressed by the dashed curve The crossing points of the tangent function and the dashed curve are solutions for the

eigen-value equation The solution k L = 0 is a trivial solution having no practical relevance.

The dashed curve shown in Fig 2.5 has two significant points, namely a pole and an end point.

The dashed curve has a pole when the denominator of the right-hand side of the eigenvalue

equa-tion vanishes, i e., when kL = L

k LPole = m U

Ñ2 L

and end point of the dashed curve, ensuring that k is still staying as a real value,

(1.50)

k LEnd point = 2 m U

Ñ2 L

There are no further bound state solutions to the eigenvalue equation beyond the end point

Now once the eigenvalues of k and are known, one is allowed for the determination of the constants A and B and the wave functions It is possible to show that all states with even quantum numbers (n = 0, 1, 2 … are of even symmetry with respect to the center of the well, i e (x) = (–

x) All states with odd quantum numbers (n = 1, 3, 5 … are of odd symmetry with respect to the

center of the well, i e (x) = (–x) The even and odd state wave functions in the well are thus

of the form

(1.51)

V0 = Emax = E0nmax- 12Then,

(1.53)

nmax = Int 1 + V0

E0

Also from the special case where nmax = 2, one finds that there is only one bound state if V0 <

E0

1.4.3 1-D triangular quantum well

Consider a triangular well with constant electric field and an infinite barrier at x = 0, as shown

in Figure 2 6

Fig 2 6 1-D triangular quantum well with electric field ¶.

Schrödinger equation for this potential becomes:

(1.54)

- Ñ2

2 m

d2yx

dx2

+ q¶xyx = E nyx , x > 0 Since the Airy function is a solution to y'' - x y = 0 and approaches zero as x approaches infin-

ity, one can write the solution to the Schrödinger equation as:

(1.55)

where a n is the n-th zero of the Airy function The Airy function is shown in Figure 2 7

Fig 2 7 Airy function normalized to Ai(0) = 1.

Its zeros are approximately given by:

(1.57)

The description of the quantum states for particle in the one-dimension triangular quantum well

by the Airy function is mathematically adequate Unfortunately, the practical calculation of tum states dealing with Airy functions is extremely tedious That's why various approximationapproaches are being involved The most frequently used one with success among them is standardFang - Howard wave function:

quan-(1.60)

zx = = k3

2

1 2

x e-k

2x

2.1 Point A is at an eletrostatic potential of + 1 V relative to point B in vacuum An electron

initially at rest at B moves to A What energy (expressed in J and eV) does the electron have at A?What's its velocity in m/s?

2 2 Sketch an experimental setup in which a silver electrode (work function 4.73 eV) is sealed

in vacuum envelope with a second electrode and a voltage is applied between them to measure thephotoelectric effect If light of wavelength 2164 Å is used, what voltage is required to reducephotoelectric current between plates to zero? With the voltage turn off, what is the maximumvelocity of electrons (m/s) emitted into vacuum from the Ag surface?

2 3 Calculate the Bohr radius (Å) and energy (eV) for n = 1, 2, and 3

2 4 Show that the constant in the equation n21=  m q4

2 Ñ 2h4 p ¶0 2  1

n1 - 1

n2 is the Rydberg stant times the speed of light

con-2 5 Calculate l for the Lyman series to n = 5, the Balmer series to n = 7, and the Pasen series

to n = 10 Plot the result and find the wavelength limit for each of the three series?

2 6 The de Broglie wavelength of a particle l = h/m V describes the wave-particle duality for

small particles such as electrons What is the de Broglie wavelength (in Å) of an electron at 150eV?The same question for electrons at 10 keV, which is typical of electron microscope? What arethe advantage of electron microscopes compairing with the one working witj visible light

2 7 Rewrite the position–momentum form of the Heisenberg uncertainty relation into one of

energy-time form

2 8 Separate the variables in dependent Schrödinger equation in order to obtain the

independent one and write out proerties of their solutions Assuming Y(x) has a particular independent value yn x, show that the corresponding energy equals the separation constant E n

time-2 9 Calculate the first three energy levels for an electron in an infinite quantum well with the

width of 10 Å

is wavefunctions

2 10 Show that the choise of the magnetic quantum number m = , -3, -2, -1, 0, 1, 2, 3, is

proper, basing on the angular Schrödinger equation for electron in hydrogen atom

2 11 What do Li, Na, and K have in common? What do F, Cl, and Br have in common? What

are the electron configuratiosn for the ionized Na, Cl?

2 12 The exciton is a hydrogen atom-like entity encountered in advanced semiconductor work.

It consists of an electron bound to a +q charged particle (a hole) of approximately equal mass Bohratom results can be used in computing the allowed energy states of the exciton provided the

reduced mass, m r = m+ m-/(m+ + m_) ª m0 /2, replaces the electron mass in the Bohr atom

formula-tion In addition, the distance between the components of the exciton is always such that there areintervening semiconductor atoms Thus ¶ 0 in the Bohr formulation must also be replaced by K s,

where K s is the semiconductor dielectric constant Using K s = 11.8, determine the ground state (n =1) energy of an exciton in Si

2.13 Reflection High Energy Electron Diffraction (RHEED) has become a commonplace

technique for probing the atomic surface structures of materials Under vacuum conditions anelectron beam is made to strike the surface of the sample under test at a glancing angle (J < 100).The beam reflects off the surface of the material and subsequently strikes a phosphorescent screen.Because of the wave-like nature of the electrons, a diffraction pattern characteristic of the first fewatomic layers is observed on the screen if the surface is flat and the material is crystalline With adistance between atomic planes of d = 5 Å, a glancing angle of 1o, and an operating de Brogliewavelength for the electrons of 2dsinU, compute the electron energy employed in the technique

2.14 (a) Confirm, as pointed out in the text, that p x = 0 for all energy states of a particle in a

I-D box

(b) Verify that the normalization factor for wavefunctions describing a particle in a I-D box is

An = 2 a

(c) Determine (x) for all energy states of a particle in a I-D box

2.15 In examining the finite potential well solution, suppose we restrict our interest to energies

where x = E/Va § 0.01 and permit a to become very large such that aÑ a xmax>> p Present anargument that concludes the energy states of interest will be very closely approximated by those ofthe infinitely deep potential well

2.16 The symmetry of a problem sometimes allows one to simplify the mathematics leading to

a solution If, for example, the x = 0 point in the finite potential well problem is moved to themiddle of the well, it becomes obvious that the wavefunction solution must be symmetric about x =0; i.e., y(- x) = ± y(x)

DENSITY OF STATES AND EQUILIBRIUM CARRIER

CONCENTRATION

1.1 Density of states in bulk semiconductors (3D)

Carriers occupy either localized impurity states or delocalized continuum states in the tion band or valence band In the simplest case, each impurity has a single, nondegenerate state.Thus, the density of impurity states equals the concentration of impurities The energy of theimpurity states is the same for all impurities (of the same species) as long as the impurities aresufficiently far apart and do not couple The density of continuum states is more complicated and

conduc-will be calculated in the following sections Several cases conduc-will be considered including (i) a cal, single-valley band, (ii) an anisotropic band, (iii) a band with multiple valleys, and (iv) the

spheri-density of states in a semiconductor with reduced degrees of freedom such as quantum wells,

quantum wires, and quantum boxes Finally the effective density of states will be calculated.

1.1.1 Single-valley, spherical, and parabolic band

The simplest band structure of a semiconductor consists of a single valley with an isotropic (i.

e spherical), parabolic dispersion relation This situation is closely approximated by, for example,

the conduction band of GaAs The electronic density of states is defined as the number of electronstates per unit volume and per unit energy The finiteness of the density of states is a result of the

Pauli principle, which states that only two electrons of opposite spin can occupy one volume element in phase space The phase space is defined as a six-dimensional space composed of real

space and momentum space We now define a ‘volume’ element in phase space to consist of arange of positions and momenta of a particle, such that the position and momentum of the particleare distinguishable from the positions and momenta of other particles In order to be distinguish-

able, the range of positions and momenta must be equal or exceed the range given by the tainty relation The volume element in phase space is then given by

uncer-(1.1)

Dx Dy Dz Dp xDpyDpz = 2 p Ñ3where the "volume" element in phase space is 2 p Ñ3 When we go to one-dimensional case, wewill have agian the well-known Heisenberd's principle Dx Dpx = 2 p Ñ Using the de Broglie relation

(p = Ñ k) the ‘volume’ of phase space can be written as

(1.2)

Dx Dy Dz Dk xDkyDkz = 2 p 3

The density of states per unit energy and per unit volume, which is denoted by rDOS(E), allows

us to determine the total number of states per unit volume in an energy band with energies E1(bottom of band) and E2 (top of band) according to

In order to obtain the volume of k-space included between two energies, the dispersion relation

E = E (k ) will be employed For a given dE one can easily determine the corresponding length in

k-space The k-space length associated with an energy interval dE in one-dimensional case is simply

given by the slope of the dispersion relation While the one-dimensional dispersion relations can beillustrated easily, the three-dimensional dispersion relation cannot be illustrated in three-dimen-

sional space To circumvent this difficulty, surfaces of constant energy in k-space are frequently

used to illustrate a three-dimensional dispersion relation As an example, the constant energy

surface in k-space is illustrated in Fig 4.1 for a spherical, single-valley band A large separation of the constant energy surfaces, i e a large Dk for a given DE, indicates a weakly curved dispersion

and a large effective mass

Fig 4 1 Constant enerfy surface for a single-valley, isotropic band.

In order to obtain the volume of k-space enclosed between two constant energy surfaces, which correspond to energies E and E + dE, we (first) determine dk associated with E and (second) integrate over the entire constant energy surface The ‘volume’ of k-space enclosed between the

two constant energy surfaces shown in Fig 4 2 is thus given by

Fig 4 2 Constant enerfy surfaces with energy E and E + dE used to calculate volume in k-space enclosed between the

So finally, we obtain the density of states per unit energy and unit volume according to

rDOSE = 1

4p3 Surface

1

“k E k „s

In this equation, the surface element ds is always perpendicular to the vector grad k E (k) Note

that the surface element ds is in k-space and that ds has the dimension m-2

Next we apply the expression for the density of states to isotropic parabolic dispersion

rela-tions of a three-dimensional semiconductor In this case the surface of constant energy is a sphere

of area 4p k2 and the parabolic dispersion is E = Ñ2 k2 / (2m*) + Epot where k is the wave vector.

Insertion of the dispersion in the last equation yields the density of states in a semiconductor with asingle-valley, isotropic, and parabolic band

1.1.2 Single-valley, anisotropic, parabolic band

In an anisotropic single-valley band, the dispersion relation depends on the spatial direction.Such an anisotropic dispersion is found in III–V semiconductors in which the L- or X- point of theBrillouin zone is the lowest minimum, for example in GaP or AlAs The surface of constant energy

is then no longer a sphere, but an ellipsoid, as shown in Fig 4.3 The three main axes of the soid may have different lengths, and thus the three dispersion relations are curved differently If themain axes of the ellipsoid align with a cartesian coordinate system, the dispersion relation is

Fig 4 3 Ellipsoidal constant energy surface with a weakly curved dispersion relation along the k x axis and strongly curved

dispersion relation along the k y and k z axis.

The vector gradk E is given by grad k E = (Ñ2k x /m*x, Ñ2k y /m*y, Ñ2k z /m z*) Since the vector gradk E is

perpendicular on the surface element, the absolute values of ds and grad k E can be taken for the

integration Integration of Eq (4.7) with the dispersion relation of Eq (4.9) yields the density of

states in an anisotropic semiconductor with parabolic dispersion relations, i e.

(1.10)

p2Ñ3 m*x m*y m*z E - Epot

If the main axes of the constant-energy ellipsoid do not align with the k x , k y , and k z axes of the

coordinate system then m*x , m*y , and m*z can be formally replaced by m1*, m2*, and m3*

Frequently, the constant energy surfaces are rotational ellipsoids, that is, two of the main axes

of the ellipsoid are identical The axes are then denoted as the transversal and the longitudinal axesfor the short and long axes, respectively Such a rotational ellipsoid is schematically shown in Fig.4.3 A relatively light mass is associated with the (short) transversal axis, while a relatively heavy

mass is associated with the (long) longitudinal axis If the masses are denoted as m t* and m l* for thetransversal and the longitudinal mass, respectively, Eq (4.10) can be modified according to

(1.11)

p2Ñ3 m l*m t*2 E - EpotThe anisotropic masses m x*, m*y , m*z , m t*, and m l* are frequently used to define a density-ofstates effective mass This mass is given by

(1.12)

mDOS* = m x*m*y m*z1 3

(1.13)

mDOS* = m l*m t*21 3The density of states is then given by

(1.14)

accommodate carriers, since the minima occur at different k x , k y , and k z values, i e the Pauli

principle is not violated The density of states is thus obtained by multiplication with the number ofequivalent minima, that is

(1.15)

rDOSE = M c 2

p2Ñ3 m1*m2*m3* E - Epotwhere M c is the number of equivalent minima and m1*, m2*, and m3* are the effective masses formotion along the three main axes of the ellipsoid

Fig 4 4 Constant energy surfaces for the L-point of the Brillouin zone The band structure consists of eight equivalent

rotational ellipsoids.

Density of states in semiconductors with reduced

dimensionality (2D, 1D, 0D)

Semiconductor heterostructure allows one to change the band energies in a controlledway and confine charge carriers to two (2D), one (1D), or zero (0D) spatial dimensions

Due to the confinement of carriers, the dispersion relation along the confinement

direc-tion is changed The change in dispersion reladirec-tion results in a change in the density of

states

Confinement of a carrier in one spatial dimension, e g the z-direction results in the

formation of quantum states for motion along this direction Consider the ground state in

a quantum well of width L z with infinitely high walls The ground-state energy is

obtained from the solution of Schrödinger’s equation and is given by

(1.16)

The particle in the quantum well can assume a range of momenta in the z-direction;

the range is given by the uncertainty principle, i e

(1.17)

E = E0 for entire range of k z

The dispersion is flat, i e constant for all values of k z The z-component of the vector

gradk E (see Eq 4.9) is therefore zero and need not be considered.

We next consider the x- and y- direction and recall that the Schrödinger equation is

separable for the three spatial dimensions Thus, the kinetic energy in the x y - plane is

given by

E = Ñ

2

2 m*x k x2 + k2yfor a parabolic dispersion

Fig 4 5 Constant energy surfaces of a a) 3-dimensional, b) 2-dimensional, and c) 1-dimensional systems The

surfaces are a sphere, a circle, a point for 3D, 2D, and 1D systems, respectively.

The surface of constant energy for the dispersion relation given by Eq (4.18) is

shown in Fig 4.5, and is a circle around k x = k y = 0 The density of states of such a 2Delectron system is obtained by similar considerations as for the 3D case The reduced

phase space now consists only of the x y -plane and the k x and k y coordinates spondingly, the two-dimensional density of states is the number of states per unit-areaand unit-energy The volume of k-space between the circles of constant energy is given

Corre-by Eq (4.5) The equation is evaluated most conveniently in polar coordinates in which

k r = k x2 + k y21 2

is the radial component of the k vector The surface integral reduces to

a line integral and the total length of the circular line is 2p kr The volume of k-space

then obtained is

(1.20)

rDOS2 D E = m*

p Ñ2 , E ¥ E0

where E0is the ground state of the quantum well system For energies E ¥ E0, the 2Ddensity of states is a constant and does not depend on energy If the 2D semiconductorhas more than one quantum state, each quantum state has a state density of Eq (4.20).The total density of states can be written as

(1.22)

rDOS2 D E = m*

p Ñ2 

n

sE - E n

where E n are the energies of quantized states and s(E –En) is the step function

We next consider a one-dimensional (1D) system, the quantum wire, in which only

one direction of motion is allowed, e g along the x-direction The dispersion relation is

then given by E = Ñ2k x2 / (2m*) The ‘volume’ (i e length-unit) in k-space is obtained in

analogy to the

(1.23)

The volume in phase space of two electrons with opposite spin is given by 2p and

thus the 1D density of states is given by

(1.24)

rDOS1 D E = 1

p Ñ

m*

2 E - E0 , E ¥ E0

Note that the density of states in a 3-, 2- and 1-dimensional system has a functional

dependence on energy according to E12 , E0 , and E- 12 , respectively For more than

one quantized state, the 1D density of states is given by

(1.25)

Fig 4 6 Electronic density of states of semiconductor with 3, 2, 1 and 0 degrees of fredoom for electron

propaga-tion Systems with 2, 1 and 0 degrees of fredoom are reffered to as quantum wells, quantum wires, and quantum dot, respectively

Finally, we consider the density of states in a zero-dimensional (0D) system, thequantum box No free motion is possible in such a quantum box, since the electron is

confined in all three spatial dimensions Consequently, there is no k-space available

which could be filled up with electrons Each quantum state of a 0D system can therefore

be occupied by only two electrons The density of states is therefore described by a function

d-(1.26)

rDOS0 D E = 2 sE - E0

For more than one quantum state, the density of states is given by

(1.27)

1.3 Effective density of states in 3D, 2D, 1D, and 0D

semiconductors

The effective density of states is introduced in order to simplify the calculation of the tion of the conduction and valence band The basic simplification made is that all band states areassumed to be located directly at the band edge This situation is illustrated in Fig 4.7 for theconduction band The 3D density of states has square-root dependence on energy The effectivedensity of states is d-function-like and occurs at the bottom of the conduction band

popula-Fig 4 7 Energy dependent density od states, rDOS3 D

, and effective density of states, N c 3 D, at the bottom of the conduction band

An electronic state can be either occupied by an electron or unoccupied Quantum mechanicsallows us to attribute to the state a probability of occupation The total electron concentration in aband is then obtained by integration over the product of state density and the probability that thestate is occupied, that is

(1.28)

n = 

Ebottom

Etop

rDOSE f E „E

where f(E) is the (dimensionless) probability that a state of energy E is populated The limits of the

integration are the bottom and the top energy of the band, since the electron concentration in theentire band is of interest

As will be shown in a subsequent section, the probability of occupation, f(E), is given by the

Maxwell–Boltzmann distribution, also frequently referred to as the Boltzmann distribution, isgiven by

(1.29)

n = 

Ebottom

Etop

rDOSE fB E „E = N c f B E = E c

where N c is the effective density of states at the bottom of the conduction band and E c is theenergy of the bottom of this band Strictly speaking, the effective density of states has no physicalmeaning but is simply a mathematical tool to facilitate calculations For completeness, Eqs (4.26)and (4.28) are now given explicitly using the Boltzmann distribution and the density of states of anisotropic three-dimensional semiconductor:

n = N c e-E c -E F k B T

The upper limit of the integration can be taken to be infinity without loss of accuracy due to thestrongly converging Boltzmann factor Evaluation of the integral in Eq (4.29) and comparisonwith Eq (4.30) yields the effective density of states

(1.33)

N c = 1

2

m*k B T

p Ñ2 32

Note that the effective density of states given by Eq (4.31) applies to one minimum in the

conduction band If there are a number of M c equivalent minima in the conduction band, the

corre-sponding density of states must be multiplied by M c Furthermore, if the band structure is

anisotropic, the effective mass m* must be replaced by the density-of-states effective mass mDOS* For a degenerate valence band with heavy and light holes, the effective density of states is the sum

of both effective state densities, that is

(1.34)

The effective density of states in a two-dimensional system (i e a system with two degrees of

freedom) is obtained by the identical procedure as the three-dimensional effective density of states.The equations analogue to Eqs (4.29) and (4.30) then read

(1.35)

n 2 D = N c 2 D e-E c -E F k B T

where N c 2 D is the two-dimensional effective density of states The carrier concentration n 2 D sents the number of electrons per unit-area and is also referred to as the 2D density Evaluation ofthe integral yields

repre-(1.37)

n 1 D = N c 1 D e-Ec -E F k B T

The one-dimensional effective density of states is obtained as

(1.40)

The evaluation of a zero-dimensional density of states does not yield a simplification of thecarrier-density calculation, since the zero-dimensional density of states is d-function like Table 4.1summarizes the dispersion relation, the density of states, and the effective density of states ofsemiconductors with various degrees of freedom.

Table 4.1. The dispersion, densities of states, effective densities of states for 3-, 2-, 1-, and 0D syste

Degrees of freedom Dispersion

Systems are assumed to be isotropic.

1.4 Carrier concentration at thermal equilibrium

One of the most-important properties of a semiconductor is that it can be doped with differenttypes and concentrations of impurities to vary its resistivity Also, when these impurities are ion-ized and the carriers are depleted, they leave behind a charge density that results in an electric fieldand sometimes a potential barrier inside the semiconductor Such properties are absent in a metal

or an insulator

Figure 4 8 shows three basic bond representations of a semiconductor Figure 4 8a showsintrinsic silicon, which is very pure and contains a negligibly small amount of impurities Eachsilicon atom shares its four valence electrons with the four neighboring atoms, forming four cova-lent bonds Figure 4 8b shows an n-type silicon, where a substitutional phosphorous atom withfive valence electrons has replaced a silicon atom, and a negative-charged electron is donated tothe lattice in the conduction band The phosphorous atom is called a donor Figure 4 8c similarlyshows that when a boron atom with three valence electrons substitutes for a silicon atom, a positive-charged hole is created in the valence band, and an additional electron will be accepted to formfour covalent bonds around the boron This is p-type, and the boron is an acceptor

Fig 4 8 Three basic bond pictures of a semiconductor: a) intrisic Si with no impurity; b) n-type Si with donor (P); c) p-type

Si with acceptor (B).

These names of n- and p-type had been coined when it was observed that if a metal whiskerwas pressed against ap-type material, forming a Schottky barrier diode, a positive bias wasrequired on the semiconductor to produce a noticeable current Also, when exposed to light, apositive potential was generated with respect to the metal whisker Conversely, a negative bias wasrequired on an n-type material to produce a large current.

1.4.1 Carrier Concentration and Fermi Level

Firstly, let's to consider the intrinsic case without impurities added to the semiconductor The

number of electrons (occupied conduction-band levels) is given by the total number of states N(E) multiplied by the occupancy f(E), integrated over the conduction band,

(1.41)

N E = M c

2

p2Ñ3 mde32 E - E c

M c is the number of equivalent minima in the conduction band and mDOS is the density of state

effective mass for electrons determined by Eqs (4.12-4.13), where m1*, m2*, m3* are the effective

masses along the principal axes of the ellipsoidal energy surface For example, in silicon mDOS =

 m l* + m t* 213 The occupancy is a strong function of temperature and energy, and is represented

by the Fermi-Dirac distribution function

(1.43)

0.7 N c

Nondegenerate Semiconductors

By definition, in nondegenerate semiconductors, the doping concentrations are smaller than N c and the Fermi levels are more than several kT below E c (negative hF), the Fermi-Dirac integralapproaches

(1.47)

level as a function of carrier concentration is given by, for n-type semiconductor

(1.53)

E v - E F º k B Tln p

N v + 2-3 2 p

N v 

Intrinsic Concentration

For intrinsic semiconductors at finite temperatures, thermal agitation occurs which results incontinuous excitation of electrons from the valence band to the conduction band, and leaving anequal number of holes in the valence band This process is balanced by recombination of theelectrons in the conduction band with holes in the valence band At steady state, the net result is n

= p = n i , where n i is the intrinsic carrier density

The Fermi level for an intrinsic semiconductor (which by definition is nondegenerate) isobtained by equations Eqs (4.48) and (4.50):

(1.55)

(1.56)

p n = N c N vexp - E g

k B T = n i2

which is known as the mass-action law (for degenerate semiconductor, p n < n i2)

Using Eq (4.55) and E i as the reference energy, we will have another expression of carrier

concentration for n-type semiconductor:

(1.58)

p = n iexp E i - E F

k B T or E i - E F = k B T ln p

n i

1.4.2 Donors and acceptors

The impurrity ionization energy

When a semiconductor is doped with donor or acceptor impurities, impurity energy levels areintroduced that usually lie within the energy gap A donor impurity has a donor level which isdefined as being neutral if filled by an electron, and positive if empty Conversely, an acceptorlevel is neutral if empty and negative if filled by an electron

The ionization energy for a donor (E c - E D ) in a lattice can be obtained by replacing m o by theconductivity effective mass of electrons

Taking into account the permittivity of the semiconductor ¶s and of vacuum ¶0, the hydrogen

ionization energy E H = 13.6 eV, we can use the expression for the magnitude of the impurityionization energy

(1.61)

E, = (E A - E v) is 0.05 eV for Si and GaAs

Figure 4 9 shows the experimentally measured ionization energy for various impurities in a) Siand b) GaAs In practical application, the shallow impurities are frequently used while the deepimpurities have been being extracted from semiconductor materials as much as possible

Fig 4 9 The measured ionization energy for various impurities in a) Si and b) GaAs The solid bars under the number

represent donor levels, the hollow bars - acceptor levels Levels above the gap center are measured from E c, levels under

the gap center - from Em

Fermi level position in semiconductor energy band diagram

The Fermi level for the intrinsic semiconductor (Eq 4.56) lies very close to the middle of thebandgap Figure 4 l0a depicts this situation, showing schematically from left to right the simplified

band diagram, the density of states N(E), the Fermi-Dirac distribution function f(E), and the carrier

concentrations The shaded areas in the conduction band and the valence band represent electrons

and holes, and their numbers are the same; i.e., n = p = n i for the intrinsic case

When impurities are introduced to the semiconductor crystals, depending on the impurityenergy level and the lattice temperature, not all dopants are necessarily ionized The ionized concen-tration for donors is given by

(1.62)

1 + g AexpE A - E F   k B T

where the ground-state degeneracy factor g A is 4 for acceptor levels The value is 4 because inmost semiconductors each acceptor impurity level can accept one hole of either spin and the impu-

rity level is doubly degenerate as a result of the two degenerate valence bands at k = 0.

When impurity atoms are introduced, the total negative charges (electrons and ionized tors) must equal the total positive charges (holes and ionized donors), represented by the chargeneutrality

accep-(1.64)

n + N A- = p + N D+With impurities added, the mass-action law ( p n = n i2) still applies (until degeneracy), and the

pn product is always independent of the added impurities.

Fig 4 10 Schematic band diagram, density of states, Fermi-Dirac distribution, and carrier concentrations for (a) intrinsic,

(b) n-type, and (c) p-type semiconductors at thermal equilibrium Note that p n = n i2 for all three cases.

Consider the case shown in Fig 4.10b, where donor impurities with the concentration N D

(cm-3) added to the crystal The charge neutrality condition becomes

(1.65)

n = N D+ + p º N D+With substitution, we obtain

(1.66)

N c exp - E c - E F

1 + 2 expE F - E D   k B T

Thus for a set of given N c , E D , N D , and T, the Fermi level E F can be uniquely determined

implicitly Knowing E F the carrier concentrations n can be calculated Equation (4.66) can also be

solved graphically

For calculation of the Fermi level and hole concentration in p-type semiconductor we repeat

similar steps with Eqs (4.50, 4.63, 4.64 - 4.65) The similar set of the simplified band diagram, the

density of states N(E), the Fermi-Dirac distribution function f(E), and the carrier concentrations for

p-type material is presented in Fig 4.10c.

Without solving for Eq (4.66), it can be shown that for N Dp 1

2N c exp-E c - E D   k B T p

N A, the electron concentration can be estimated as

Fig 4 11 Electron density as a function of temperature for a Si sample with donor impurity concentration of 1015 cm-3.Figure 4.12 shows the Fermi level for Si and GaAs as a function of temperature and impurityconcentration, as well as the dependence of the bandgap on temperature At relatively high tempera-tures, most donors and acceptors are ionized, so the neutrality condition can be approximated bythe mass-action law (Eq 4.57)

Fig 4 12 Fermi level for (a) Si and (b) GaAs as a function of temperature and impurity concentration The dependence of

the bandgap on temperature is also shown

In an n-type semiconductor where N D > N A:

(1.69)

n n0 = 1

2 N D - N A  + N D - N A2 + 4 n i2  º N D ,

if N D - N A p n i , or N D p N A

(1.70)

p p0 = 1

2 N A - N D  + N A - N D2 + 4 n i2 º N A ,

if N A - N D p n i , or N A p N D

(1.73)

p p0 = N A = N vexp - E F - E v

k B T = n iexp E i - E F

k B T

The subscripts n and p refer to the type of semiconductors, and the sub-subscript o refers to the thermal equilibrium condition Thus, n n , p p and n p , p n are maiority and minority carrier concentra-

tions in n- and p-type semiconductors, respectively.

1.5 Problems

4 1 a) Explain why holes appear at the top of the valence band.

b) Explain why Si doped with 1014 cm-3 donors is n-type at 400K, but Ge is not.

4 2 The E-k relationship characterizing electrons confined to a two-dimensional surface layer

is of the form E  —2 k x

2 m 1

 —2 k y

2 m 2

, m 1 ∫ m 2 Assuming the side-lengths of the surface layer are much

larger than atomic dimensions, derive an expression for the density of states (g 2 D) for the electrons

in the two-dimensional surface layer Record all steps in obtaining your answer

4 3 Calculate the density-of-states effective mass associated with the X minimum for the given

Si band structure, using equation E = E0 - A cos( a k x  - B [cos(b k y) + cos(g kz)]

4 4 Calculate the number of electrons, holes in the unknown semiconductor (E g = 1.1 eV) with

E F 0.25eV below E i,

4 5 With E F located 0.4 eV above the valence band in a Si sample, what charge status wouldexpect for most Ga, Zn, Au atoms in the sample?

4 6 How much Zn must be added to exactly compensate a Si sample doped with 106 cm-3 Sb?

4 7 Show that the Eq 4 62 ND+ = N D

1+ g DexpEF - E D   k B T is valid [Hint: The probability of occupancy

is f(E) = 1 + h

g expE -E F

k B T -1, where h is the number of electrons that can physically occupy the level E, and g is the number of electrons that can be accepted by the level, also called the ground- state degeneracy of the donor impurity level (g = 2) ]

4 8 Calculate the average kinetic energy of electrons in the conduction band of an n-type

nondegenerate semiconductor

4 9 If a silicon sample is doped with 1016 phosphorous impurities/cm3, find the ionized donordensity at 77 K Assume that the ionization energy for phosphorous donor impurities and the

electron effective mass are independent of temperature (Hint: First select a N D+ value to calculate

the Fermi level, then find the corresponding N D+ If they don’t agree, select another N D+ value andrepeat the process until a consistent Nh is obtained.)

4 10 The Fermi-Dirac distribution function f(E) is given by Eq 4.43 The differentiation of

f(E) with respect to energy is f'(E) Find the width of f'(E), i.e., 2 [E(at fmax' ) - E(at 1

2 fmax' ) ],

where fmax' is the maximum value of f'(E).

4 11 Find the position of the Fermi level with respect to the bottom of the conduction band (E c

- E F) for a silicon sample at 300 K, which is doped with 2 ä 1010 cm-3 fully ionized donors

4 12 Gold in Si has two energy levels in the bandgap: E c - E A = 0.54 eV, E D - E v = 0.29 eV

Assume the third level E D - E v = 0.35 eV is inactive (a) What will be the state of charge of thegold levels in Si doped with high concentration of boron atoms? Why? (b) What is the effect ofgold on electron and hole concentrations?

4 13 For an n-type silicon sample doped with 2.86 x 10l6 cm-3 atoms P, find the ratio of the

neutral to ionized donors at 300 K (E c - E D) = 0.045 eV

4 14 The conduction band minima in GaP occur right at the first Brillouin zone boundary

along the <100> directions in k-space Taking the constant energy surfaces to be ellipsoidal with

m l /m 0 = 1.12 and m t /m 0 = 0.22, determine the density of states effective mass for electrons in GaP.

4.15 The valence band of InSb is a bit unusual in that the heavy-hole subband exhibits maxima

along (111) directions at a k-value slightly removed from k = 0 If the heavy-hole maxima are described by parabolic energy surfaces where m l and m t are the longitudinal and transverse effec-

tive masses, respectively, and if m lh is the effective mass for the light holes in a spherical subband

centered at k = 0, obtain an expression for the density of states effective mass characterizing the holes in InSb Your answer should be expressed in terms of m l , m t , and m lh

4.16 In Si, where m hh /m 0 = 0.537 and m lh /m 0 = 0.153, what fraction of the holes are heavy

holes? (For simplicity, assume that the quoted 4 K effective masses can be employed at anytemperature.)

4.17 The carrier distributions or number of carriers as a function of energy in the conduction

and valence bands were noted to peak at an energy very close to the band edges Taking the conductor to be nondegenerate, determine the precise energy relative to the band edges at whichthe carrier distributions peak

semi-4.18 (a) Establish a general expression (involving integrals) for the average kinetic energy,

(K.E.), of the conduction band electrons in a semiconductor

(b) Taking the semiconductor to be nondegenerate, simplify your general (K.E.) expression toobtain a closed-form result

4 19 Because m n* and thus N c are relatively small in GaAs, the donor doping at which the

material becomes degenerate is also relatively small Determine the N D doping at the

non-degenerat-e/degenerate transition point in GaAs at 300 K

4 20 Determine the degenerate doping limit for donors in Si as a function of temperature at 100

K increments between T = 200 K and T = 600 K Employ m n* / m 0 = 1.028 + (6.11 ä 10-4) T

-(3.09 ä 10-7) T2

4 21 Determine the equilibrium electron and hole concentrations inside a uniformly doped

sample of Si under the following conditions:

C H A P T E R 5

CARRIER TRANSPORT PHENOMENA

Carrier Drift and Mobility

The motion of a carrier drifting in a semiconductor due to an applied electric field Assuming that all the

carriers in the semiconductor move with the same velocity v d , the current I can be expressed as the total charge Q in the semiconductor divided by the time t r needed to travel from one electrode to the other with

the distance L, or:

t r =

Q

L  vÓ d

The current density jÓ

can then be rewritten as a function of either the charge density, Ρ, or the density of

carriers, n in the semiconductor through the cross-section A of the sample (q is the carrier's charge):

degenerate and non-relativistic electron gas have a thermal energy, which equals k B T 2 per particle perdegree of freedom A typical thermal velocity at room temperature is around 107cm/s, which exceeds thetypical drift velocity in semiconductors

In the absence of an applied electric field, the carrier exhibits random motion and the carriers move quicklythrough the semiconductor and frequently change direction When an electric field E is applied, the random

motion still occurs but in addition, there is on average a net motion along the direction of the field, withtaking into account the effective mass of carriers in the periodic potential of crystall semiconductor

*

m* âXvÓ d\

â t

The force consists of the difference between the electrostatic force and the scattering force due to the loss

of momentum at the time Τ of scattering This scattering force equals the momentum divided by the averagetime between scattering events, so that:

We now consider only the steady state situation in which the particle has already accelerated and hasreached a constant average velocity Under such conditions, the velocity is proportional to the applied

electric field and we define the mobility Μ as the drift velocity vÓ d to field E ratio:

where C l is the average longitudinal elastic constant of the semiconductor, E ds the displacement of the

band edge per unit dilation of the lattice, and m c* the conductivity effective mass

The mobility from ionized impurities Μi at the ionized impurity density N i, can be described by

Μ = 1

Μl +

1

Μi-1

In addition to the scattering mechanisms discussed above, other mechanisms also affect the actual mobility.For example, (1) the intravalley scattering in which an electron is scattered within an energy ellipsoid andonly long-wavelength phonons (acoustic phonons) are involved; and (2) the intervalley scattering in which

an electron is scattered from the vicinity of one minimum to another minimum and an energetic phonon(optical phonon) is involved For polar semiconductors such as GaAs, polar-optical-phonon scattering issignificant For the semiconductor structures, where carrier current can be affected by surface roughnessstatus, the surface roughness scattering must be involved Thus, the Matthiessen rule generally can bewritten as

Μ = q Τ m

q Λ m

3 m* k B T

The last term uses the relationship Λm = v thΤm , where thermal velocity is given by v th= 3 k B T  m*

Resistivity and Hall effect

Ÿ Electrical resistivity

Fig 5 1 Resistivity vs impurity concentration at 300 K for (a) silicon, and (b) GaAs

For semiconductors with both electrons and holes as carriers, the drift current jÓ

under an applied field E is