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Electronics Fundamentals

Chapter 1: Atomic Structure and Semiconductor Materials

Part 1- Atomic Structure

Evidence that scientists had in 1900 to indicate that the atom was not a fundamental unit:

1) There seemed to be too many kinds of atoms, each belonging to a distinct chemical element.

2) Atoms and electromagnetic phenomena were intimately related.

3) The problem of valence. Certain elements combine with some elements but not with others,

a characteristic that hinted at an internal atomic structure.

4) The discoveries of radioactivity, of x rays, and of the electron.

Thomson Model of the Atom:

J. J. Thomson - English physicist. 1897 Made a piece of equipment called a cathode ray tube.

It is a vacuum tube - all the air has been pumped out. Form when high voltage is applied across

electrodes in a partially evacuated tube. Rays originate at the cathode (negative electrode) and

move to the anode (positive electrode). Carry energy and can do work. Travel in straight lines in

the absence of an external field. Using a cathode ray tube, Thomson was able to deflect cathode

rays with an electrical field. The rays bent towards the positive pole, indicating that they are

negatively charged. Thomson used magnetic and electric fields to measure and calculate the ratio

of the cathode ray’s mass to its charge.

Figure 1: Schematic of J.J. Thomson's experiment.

As shown in the figure, the particle that comes off the filament is accelerated by the high voltage

V) and passed between two parallel plates across which an Electric field E exist. This field exerts

a force (FE) on the particle of charge (Q)and deflect it from its path according:

FE= Q*E in Newton’s.

Another force (FB) is exerted on the particle by a magnetic field (B) [FB = Q*B*v], where v is

the particle velocity. Thomson adjusted the directions of the electric and magnetic fields in such

a way that the two forces are opposite and equal. Under these conditions , emitted particle are no

longer deflected. Therefore: FE = FB thus: Q*E = Q*B*v, v= E/B.

The kinetic energy (Ek) received by the particle accelerated by V volt is: Ek = Q*V = 0.5*m*v2,

thus v = (2*Q*V/m)1/2

= E/B and Q/m = E2/ 2*V*B

2

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Knowing the electric field €, the magnetic field intensity (B) and the accelerating voltage (V),

the charge to mass ratio (Q/m) of the electron can be calculated.

He compared the value with the mass/ charge ratio for the lightest charged particle. By

comparison, Thomson estimated that the cathode ray particle weighed 1/1000 as much as

hydrogen, the lightest atom. He concluded that atoms do contain subatomic particles - atoms are

divisible into smaller particles. Since any electrode material produces an identical ray, cathode

ray particles are present in all types of matter - a universal negatively charged subatomic particle

later named the electron

J.J. Thomson discovered the electron and

knew that electrons could be emitted from

matter (1897). William Thomson proposed

that atoms consist of small, negative

electrons embedded in a massive, positive

sphere. The electrons were like currants in a

plum pudding. This is called the ‘plum

pudding’ model of the atom.

Ernest Rutherford (1871-1937) nuclear

physicist, Thomson’s student, New

Zealander teaching in Great Britain :Gold

Leaf Experiment Rutherford’s Experiments

(1910-11) Fired beam of positively-charged

alpha particles at very thin gold foil. Alpha

particles caused flashes of light when they

hit the zinc sulfide screen.

By Thomson’s model, mass and + charge of gold atom are too dispersed to deflect the

positively-charged alpha particles, so particles should shoot straight through the gold atoms.

Most alpha particles went straight through, and some were deflected, BUT a few (1 in 20,000)

reflected straight back to the source.

Expt. Interpretation: gold atom has small,

dense, positively-charged nucleus

surrounded by “mostly empty” space in

which the electrons must exist. Positively

charged particles called “protons” like tiny

solar system.

James Chadwick (1932) Discovered a

neutral (uncharged) particle in the nucleus.

Called it the “neutron”

The order of the elements is determined by their atomic number (= the number of protons). The

atomic mass of the elements is determined by the number of protons and neutrons. A given

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element can have different number of neutrons, and therefore different atomic masses. The

chemical properties of the elements are determined by the number of electrons in their outer

(valence) shells.

Property

Particle Mass in amu (g) Relative Charge(C)

Electron 0.00055 ( 9.110 x 10

-28

)

- 1(1.6*10-19

)

Proton 1.00728 (1.673 x 10

-24

+ 1(1.6*10-19

)

Neutron 1.00866 ( 1.675 x 10

-24

0

The Classical Atomic Model

Let’s consider atoms as a planetary model. The

force of attraction on the electron by the nucleus

and Newton’s 2nd law give

where v is the

tangential velocity of the electron.

atoms planetary mode

The potential energy is: The kinetic energy is:

𝑬𝒑 = ∫ 𝑬𝒆 ∞

𝒓dr=∫

𝒆𝟐

𝟒𝝅ɛ𝒐𝒓𝟐= −

∞

𝟎

𝒆𝟐

𝟒𝝅ɛ𝒐 𝒓

The total energy is

The total energy of electron is negative, so the

system is bound, which is good. Otherwise the

electron will leave the material if it possess

positive energy. From classical E&M theory, an

accelerated electric charge radiates energy

(electromagnetic radiation) which means total

energy must decrease. Radius r must decrease.

Electrons moving through the electrical field

generated by the protons in the nucleus would

radiate away energy and spiral down into the

nucleus. Calculations soon showed that a

“Rutherford atom” would last less than one

minute.

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The Photoelectric Effect:

A photoelectric effect is any effect in which

light energy is converted to electricity. First

explained by Albert Einstein in 1905. It

had long been know that when chemical

elements are heated, they gave off light of a

particular wavelength (or color) Sodium Potassium Lithium

Max Planck found experimental results that suggested light behaved more like a particle when he

was studying black body radiation. A black body is a theoretical object that absorbs all light that

falls on it. It reflects no radiation and appears perfectly black. Black body radiation is the

energy that would be emitted from an ideal black body.

In the year 1900, Planck published a paper on the electromagnetic radiation emitted from a

black object that had been heated. In trying to explain the black body radiation, Planck, in the

same year determined that the experimental results could not be explained with the wave form of

light. Instead, Planck described the radiation emission as discrete bundles of energy, which he

called quanta. A quantum (singular form of quanta) is a small unit into which certain forms of

energy are divided. These “discrete bundles of energy” once again raised the question of whether

light was a wav or a particle. Planck’s work also pointed out that the energy of a quantum of

light was related only to its frequency. Planck’s equation for calculating the energy of a bundle

of light is E= hν, where E is the energy of the photon in joules (J), ν is the frequency in hertz

(s), and h is Planck’s constant, 6:63*10-34

J. s or 4.14*10-15

eV.s.. The word quantum is used for

energy in any form; when the type of energy under discussion is light, the words quantum and

photon become interchangeable.)

In 1905, Einstein proposed that

electromagnetic radiation or light is made up

of photons. Thus the photon is the

elementary element of light or light is made

up of photons. Einstein show that- light

energy is not emitted continuously but it is

emitted by individual amount of energy

called as quantum of energy. Each photon of

a light wave of frequency ν has the energy

E is given by: E= hν

Properties of photon

1. A photon does not have any mass.

2. A photon does not have any charge and are not deflected in electric field or magnetic field.

3. All the quantum numbers are zero for a photon.

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4. In empty space, the photon moves at speed of light.

5. In the interaction of radiation with matter, radiation behaves as if it is made up of particles

called photons.

6. The energy and momentum of a photon are related as follows: E= p*c

where p- magnitude of momentum and c is the speed of light.

7. Photon is called as a virtual particles.

8. The energy of a photon is directly proportional to frequency and inversely proportional to its

wavelength.

Characteristic of photoelectric effect

Photoelectric effect- When light of suitable frequency is incident on metal surface then

electrons are emitted from surface called as Photoelectric effect.

1. Threshold frequency is different for different material.

2. Photoelectric current is directly proportional to intensity of light.

3. The K.E. of photoelectrons is directly proportional to frequency of light.

4. Stopping potential is directly proportional to frequency.

5. The process is instantaneous.

Einstein’s photoelectric function- According to quantum theory, radiation is considered as

shower of particles called photons. Energy of photon absorbed by the atom (hυ) is:

1. Used to detach the electron (W0) and

2. Kinetic energy (Ek) is given to electron.

hυ= W0 + Ek = W0 + 0.5mv2

Where, υ= Frequency of radiation,

W0= Photoelectric work function= hυ0,

m= Mass of electron v= Velocity of electron,

h= Planck’s constant and υ0= Threshold

frequency

0.5mv2 = hυ- W0 = hυ - hυ0 = h (υ- υ0)

Significance

1. If υ < υ0 - Kinetic energy is negative. i.e. No

emission.

2. If υ = υ0 - Kinetic energy is zero. i.e.

Emission just begins.

3. If υ > υ0 - Kinetic energy is positive. i.e.

Emission takes place.

Niels Bohr (1885-1962) Danish physicist: Bohr wondered why hydrogen emitted spectral lines,

and not just a continuous band of light.

The Bohr Model of the Hydrogen Atom (1913): Bohr’s general assumptions:

1. Stationary states, in which orbiting electrons do not radiate energy, exist in atoms and have

well-defined energies, En.

2. Transitions of electrons can occur between the stationary states, yielding light of energy:

E = En1

– En2

= h υ

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The electron will gain energy when it is moved from lower to higher state and loose energy vice

versa. Classical laws of physics do not apply to transitions between stationary states, but they do

apply elsewhere.

3. The angular momentum of the nth

state is: L = nh/2π. (Angular momentum is quantized!).

where n is called the Principal Quantum Number and h is plank constant. The angular

momentum is:

a0 is called the Bohr radius. It’s the radius of the Hydrogen atom (in its lowest-energy, or

“ground,” state). Where the Bohr radius is given by

The smallest diameter of the hydrogen atom is;

n = 1 gives its lowest energy state (called the

“ground” state)

The Hydrogen Atom

The energies of the stationary states where E0

= -13.6 eV.

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The amounts of total energy of the electron is very small. Therefore, instead of Joule we will use

the unit of electron volt (eV), which is the energy a body with a charge of one elementary

charge (e = 1.602 x 10-19

C) gains or losses when it is moved across electric potential difference

of 1 Volt (V).

E (in electron volt) = Q x V,

where Q is the electronic charge (1.602 x 10-19 C) and V is the voltage:

1 electron volt is equal to 1.602 x 10-19

Joule

Example1: Consider Q = -1.6x10-19

Coulomb, me = 9.18x10-31

Kg, εo=8.84x10-12

F/m, h=

6.63x10-34

J. sec or 4.12x10-15

eV.sec and 1 eV = 1.6x10-19

Joules. Calculate the radius and

velocity of the hydrogen electron moved to the second orbit (n=2).

Solution: r = ao*n2 = 0.53*10

-10* 2

2 = 2.12*10

-10 meters

v = [Q/(4π*εo*m*r)1/2

] = [ 1.6*10-19

/(4π*8.84x10-12

*9.18x10-31

*2.12*10-10

)]

v = 1.04*106 m/s

Example: Calculate the radius and velocity of the hydrogen electron having total energy:

(a) -13.6 eV, (b) -10 eV.

Solution:

(a) ra = (1.6*10-19

)2/ (8π*8.84x10-12

*13.6) = 5.3*10-11

m

va = [e/(4π*εo*m*r)1/2

]

= [ 1.6*10-19

/(4π*8.84x10-12

*9.18x10-31

*5.3*10-11

)] 6.9*106 m/s

(b) rb = (1.6*10-19

)2/ (8π*8.84x10-12

*10.4) = 7.19*10-11

m

vb = [e/(4π*εo*m*r)1/2

]

= [ 1.6*10-19

/(4π*8.84x10-12

*9.18x10-31

*7.19*10-11

)] = 5.93*106 m/s

Note: As the radii increase (the electron further away from the nucleus), the velocity of electron

decreases since the centrifugal force required to maintain the stable electron orbit is less.

Emission of light occurs when the atom is in an excited state and decays to a lower energy state

(nu → nℓ).

where υ is the frequency of a photon.

R∞ is the Rydberg constant.

The atom will remain in the excited state for a short time before emitting a photon and returning

to a lower stationary state. All hydrogen atoms exist in n = 1 (invisible). Bohr improved

uh E E

1 h

c hc

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Rutherford’s model by noticing that energy levels in atoms went up and down by specific, “pre-

set” amounts. He suggested that electrons move around the nucleus of an atom like planets

around the sun, and that they move from orbit to orbit as they gain and lose energy.

Louis de Broglie (1927) Particle/Wave Duality of electrons: for matter, just as much as for

radiation, in particular light, we must introduce at one and the same time the corpuscle concept and the

wave concept. In other words, in both cases we must assume the existence of corpuscles accompanied by

waves. All matters big or small behave like a wave. Electrons behave with wave and particle

properties at the same time.

λ= h/m*v, where h is Plank constant, m is the mass of the matter and v its velocity, but the

wavelength for photon is: λ =h/c, where c is the speed of light. Example: Calculate De Broglie wave length (λ) of: (a) A dust particle of mass 1x10

-5Kg

travelling at speed of 0.02m/sec. (b)An electron of mass 9.1x 10-31

Kg moving at 5x105 m/sec.

Compare the results in a and b. Suppose Plank constant h= 6.62x10-34

J.sec.

The Bohr model was a great step of the new quantum theory,

but it had its limitations.

1) Works only to single-electron atoms.

2) Could not account for the intensities or the fine structure of the spectral lines.

3) Could not explain the binding of atoms into molecules.

Quantum theory ; Planck’s work became the basis for quantum theory. Quantum theory is the

theory that energy can only exist in discrete amounts (quanta). The problem involved in

demonstrating this theory is that the scale of a quantum of energy is much smaller than the

objects we normally deal with. The results of the photoelectric effect indicated that if the

experimenter used low frequency light, such as red, no electrons were knocked off the metal. No

matter how many light waves were used and no matter how long the light was shined on the

metal, red light could not knock off any electrons. If a higher frequency light was used, such as

blue light, then many electrons were knocked off the metal. Albert Einstein used Planck’s

quantum theory to provide the explanation for the photoelectric effect. A certain amount of

energy was necessary for electrons to be knocked off a metal surface. If light were quantized,

then only particles of higher frequency light (and therefore higher energy) would have enough

energy to remove an electron. Light particles of lower frequency (and therefore lower energy)

could never remove any electrons, regardless of how many of them were used.

In all previous attempts to describe the electron’s behavior inside an atom, including in the Bohr

model, scientists tried to describe the path the electron would follow around the nucleus. The

theorists wanted to describe where the electron was located and how it would move from that

position to its next position.

In 1927, a German physicist named Werner Heisenberg, a German physicist stated what is now

known as the Heisenberg uncertainty principle. This principle states that it is impossible to

know both the precise location and the precise velocity of an electron at the same time. The

reason that we can’t determine both is because the act of determining the location changes the

velocity. In the process of making a measurement, we have actually changed the measurement.

The Heisenberg uncertainty principle treated the electron as a particle. In effect, the uncertainty

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principle stated that the exact motion of an electron in an atom could never be determined, which

also meant that the exact structure of the atom could not be determined. Consequently, Erwin

Schrodinger, an Austrian physicist, decided to treat the electron as a wave in accordance with de

Broglie’s matter waves. Schrodinger, in considering the electron as a wave, developed an

equation to describe the electron wave behavior in three dimensions. Unfortunately, the equation

is so complex that it is actually impossible to solve exactly for atoms and ions that contain more

than one electron.

Only at certain distances from the nucleus

would the electron complete an integer

number of wavelengths in its movement

around the nucleus.

Figure above is an example of a standing

wave. There are only certain energies

(frequencies) for which the wavelength of

the wave will fit exactly to form a standing

wave. These energies turn out to be the same

as the energy levels predicted by the Bohr

model, but now there is a reason why

electrons may only occupy these energy

levels. The equations of quantum mechanics

tell us about the existence of four quantum

numbers: principal energy levels, the

number of energy levels in any atom, and

more detailed information about the various

energy levels.

Solutions to Schrödinger’s equation involve four special numbers called quantum numbers.

These four numbers completely describe the energy of an electron. Each electron has exactly

four quantum numbers, and no two electrons have the same four numbers. The statement that no

two electrons can have the same four quantum numbers is known as the Pauli exclusion

principle.

Quantum numbers:

1. The principal quantum number (n) is a positive integer (1, 2, 3, . . . n) that indicates the

main energy level of an electron within an atom. According to quantum mechanics, every

principal energy level has one or more sub-levels within it. In any energy level, the maximum

number of electrons possible is Nmax = 2n2. Therefore, the maximum number of electrons that

can occupy the first energy level is 2 (2 * 12). For energy level 2, the maximum number of

electrons is 8 (2*22), and for the 3rd energy level, the maximum number of electrons is 18(2 *

32). The number of sub-levels in a given energy level is equal to the number assigned to that

energy level. That is, principal energy level 1 will have 1 sub-level, principal energy level 2 will

have two sub-levels, principal energy level 3 will have three sub-levels, and so on.

Table 1.1 lists the number of sub-levels and electrons for the first four principal quantum

numbers

Principal

Quantum

Number of

Sub-

Total

Number

Number Levels of Electrons

1 1 2

10

2 2 8

3 3 18

4 4 32

TABLE 1. : Number of Sub-levels and

Electrons by Principal Quantum Number.

Each energy level can have as many sub-

levels as the principal quantum number, as

discussed above, and each sub-level is

identified by a letter. Beginning with the

lowest energy sub-level, the sub-levels are

identified by the letters s, p, d, f, g, h, i, and

so on. The principal energy levels and sub-

levels are shown in the following diagram.

The principal energy levels and sub-levels

that we use to describe electrons are in red.

2. Orbitals: (L)

Quantum mechanics also tells us how many orbitals are in each sub-level, an orbital is defined as

an area in the electron cloud where the probability of finding the electron is high. The number

of orbitals in an energy level is equal to the square of the principal quantum number. Hence,

energy level 1 will have 1 orbital (12), energy level 2 will have 4 orbitals (2

2), energy level 3 will

have 9 orbitals (32), and energy level 4 will have 16 orbitals (4

2). The s sub-level has only one

orbital. Each of the p sub-levels has three orbitals. The d sub-levels have five orbitals, and the f sub-

levels have seven orbitals. If we wished to assign the number of orbitals to the unused sub-levels, g

would have nine orbitals and h would have eleven.

You might note that the number of orbitals in the sub-levels increases by odd numbers (1, 3, 5, 7, 9,

11, . . .). As a result, the single orbital in energy level 1 is the s orbital. The four orbitals in energy

level 2 are a single 2s orbital and three 2p orbitals. The nine orbitals in energy level 3 are a single 3s

orbital, three 3p orbitals, and five 3d orbitals. The sixteen orbitals in energy level 4 are a the single

4s orbital, three 4p orbitals, five 4d orbitals, and seven 4f orbitals. The maximum number of

electrons in each orbital is 2.

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Energy bands Theory

Energy bands consisting of a large number of closely spaced energy levels exist in crystalline

materials. The bands can be thought of as the collection of the individual energy levels of electrons

surrounding each atom. The wave functions of the individual electrons, however, overlap with those

of electrons confined to neighboring atoms. The Pauli exclusion principle does not allow the electron

energy levels to be the same so that one obtains a set of closely spaced energy levels, forming an

energy band. The energy band model is crucial to any detailed treatment of semiconductor devices. It

provides the framework needed to understand the concept of an energy bandgap and that of

conduction in an almost filled band as described by the empty states.

The energy levels are grouped in bands, separated by energy band gaps. The behavior of electrons

at the bottom of such a band is similar to that of a free electron. However, the electrons are

affected by the presence of the periodic potential. The combined effect of the periodic potential is

included by adjusting the value of the electron mass. This mass will be referred to as the effective

mass. The effect of a periodic arrangement on the electron energy levels is illustrated by

Figure shown are the energy levels of electrons in a carbon crystal with the atoms arranged in a

diamond lattice. These energy levels are plotted as a function of the lattice constant, a.

Isolated carbon atoms contain six electrons, which occupy the 1s, 2s and 2p orbital in pairs. The

energy of an electron occupying the 2s and 2p orbital is indicated on the figure. The energy of the 1s

orbital is not shown. As the lattice constant is reduced, there is an overlap of the electron wave

functions occupying adjacent atoms. This leads to a splitting of the energy levels consistent with the

Pauli exclusion principle. The splitting results in an energy band containing 2N states in the 2s band

and 6N states in the 2p band, where N is the number of atoms in the crystal. A further reduction of

the lattice constant causes the 2s and 2p energy bands to merge and split again into two bands

containing 4N states each. At zero Kelvin, the lower band is completely filled with electrons and

labeled as the valence band. The upper band is empty and labeled as the conduction band.

The energy band diagrams shown in the previous section are frequently simplified when analyzing

semiconductor devices. Since the electronic properties of a semiconductor are dominated by the

highest partially empty band and the lowest partially filled band, it is often sufficient to only

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consider those bands. This leads to a simplified energy band diagram for semiconductors as shown

in Figure below

A simplified energy band diagram used to describe semiconductors. Shown are the valence and

conduction band as indicated by the valence band edge, Ev, and the conduction band edge, Ec. The

vacuum level, Evacuum, and the electron affinity, c, are also indicated on the figure.

The diagram identifies the almost-empty conduction band by a set of horizontal lines. The bottom

line indicates the bottom edge of the conduction band and is labeled Ec. Similarly, the top of the

valence band is indicated by a horizontal line labeled Ev. The energy band gap, Eg, is located

between the two bands. The distance between the conduction band edge, Ec, and the energy of a

free electron outside the crystal (called the vacuum level labeled Evacuum) is quantified by the

electron affinity, χ multiplied with the electronic charge q. Evacuum = χ* q

Part 2- Semiconductors

Solid-state electronic materials

Electronic materials generally can be divided into three categories: insulators, conductors, and

semiconductors. The primary parameter used to distinguish among these materials is the resistivity

ρ, with units of Ω· cm. As indicated in Table 1.1, insulators have resistivity greater than 105

Ω·cm,

whereas conductors have resistivity below 10−3

Ω · cm. For example, diamond, one of the highest

quality insulators, has a very large resistivity, 1016

Ω·cm. On the other hand, pure copper, a good

conductor, has a resistivity of only 3 × 10−6

Ω·cm. Semiconductors occupy the full range of

resistivity between the insulator and conductor boundaries; moreover, the resistivity can be

controlled by adding various impurity atoms to the semiconductor crystal.

Elemental semiconductors are formed from

a single type of atom (column IV of the

periodic table of elements; see Table 1.2),

whereas compound semiconductors can be

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formed from combinations of elements from

columns III and V or columns II and VI.

These later materials are often referred to as

III–V (3–5) or II–VI (2–6) compound

semiconductors.

Covalent bond model

Atoms can bond together in amorphous, polycrystalline, or single-crystal forms. Amorphous

materials have a disordered structure, whereas polycrystalline material consists of a large number of

small crystallites.

Most of the useful properties of semiconductors,

however, occur in high-purity, single-crystal

material. Silicon—column IV in the periodic

table—has four electrons in the outershell.

Single-crystal material is formed by the

covalent bonding of each silicon atom with its

four nearest neighbors in a highly regular three-

dimensional array of atoms, as shown in Fig.

1.8.

Figure 1.8: Silicon crystal lattice structure.

Diamond lattice unit cell.

Much of the behavior we discuss can be visualized using the simplified two-dimensional covalent

bond model of Fig. 1.9. a. At temperatures approaching absolute zero, all the electrons reside in the

covalent bonds shared between the atoms in the array, with no electrons free for conduction. The

outer shells of the silicon atoms are full, and the material behaves as an insulator. As the

temperature increases, thermal energy is added to the crystal and some bonds break, freeing a small

number of electrons for conduction, as in Fig. 1.9 b. The density of these free electrons is equal to

the intrinsic carrier density ni (cm−3

), which is determined by material properties and temperature:

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(a) (b)

Figure 1.9 (a) Two-dimensional silicon lattice with shared covalent bonds. At temperatures

approaching absolute zero, 0 K, all bonds are filled, and the outer shells of the silicon atoms are

completely full. (b) An electron–hole pair is generated whenever a covalent bond is broken.

where EG = semiconductor bandgap energy in eV

(electron volts), k =Boltzmann’s constant (8.62* 10-5

eV/K), T =absolute temperature, oK, B =

material parameter, 1.08 × 1031

K−3

cm−6

for Si.

Bandgap energy EG is the minimum energy needed to break a covalent bond in the semiconductor

crystal, thus freeing electrons for conduction. The density of conduction (or free) electrons is

represented by the symbol n (electrons/cm), and for intrinsic material n = ni. The term intrinsic

refers to the generic properties of pure material. Although ni is an intrinsic property of each

semiconductor, it is extremely temperature-dependent for all materials.

Example: Calculate the value of ni in pure silicon at room temperature (300 K). EG = 1.12 eV,

assume T = 300 oK at room temperature.

A second charge carrier is actually formed when the covalent bond in Fig. 1.9 is broken. As an

electron, which has charge −q equal to −1.602 × 10−19 C, moves away from the covalent bond, it

leaves behind a vacancy in the bond structure in the vicinity of its parent silicon atom. The vacancy

is left with an effective charge of +q. An electron from an adjacent bond can fill this vacancy,

creating a new vacancy in another position. This process allows the vacancy to move through the

crystal. The moving vacancy behaves just as a particle with charge +q and is called a hole. Hole

density is represented by the symbol p (holes/cm3). As already described, two charged particles are

created for each bond that is broken: one electron and one hole. For intrinsic silicon, ni = n = p,

and the product of the electron and hole concentrations is:

pn = ni2 . The pn product is given by Eq. whenever a semiconductor is in thermal equilibrium.

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Drift currents

Electrical resistivity ρ and its reciprocal, conductivity σ, characterize current flow in a material

when an electric field is applied. Charged particles move or drift in response to the electric field,

and the resulting current is called drift current. The drift current density j is defined as:

j = Q*v: (C/cm3)(cm/s) = A/cm

2. where j = current density, Q = charge density, the charge in a

unit volume, the charge in coulombs moving through an area of unit cross section, v =velocity of

charge in an electric field.

Mobility

We know from electromagnetics that charged particles move in response to an applied electric field.

This movement is termed drift, and the resulting current flow is known as drift current. Positive

charges drift in the same direction as the electric field, whereas negative charges drift in a direction

opposed to the electric field. At low fields carrier drift velocity v (cm/s) is proportional to the

electric field E (V/cm); the constant of proportionality is called the mobility μ:

vn=−μn*E : where vn = velocity of electrons (cm/s), vp= velocity of holes (cm/s), μn =electron

mobility, 1350 cm2/V · s in intrinsic Si, μp =hole mobility, 500 cm

2/V · s in intrinsic Si.

Exercise: Calculate the velocity of a hole in an electric field of 10 V/cm. What is the electron

velocity in an electric field of 1000 V/cm? The voltage across a resistor is 1 V, and the length

of the resistor is 2 µm. What is the electric field in the resistor?

Answers: 5.00 × 103cm/s; 1.35 × 10

6cm/s; 5.00 × 10

3V/cm.

Resistivity of intrinsic silicon

We are now in a position to calculate the electron and hole drift current densities jdriftn and jdriftp:

in which Qn = (−qn) and Qp= (+qp) represent the charge densities (C/cm

3) of electrons and

holes, respectively. The total drift current density is then given by:

jdriftT= jn+ jp= q(nμn+ pμp)E = σ E

This equation defines σ, the electrical conductivity: σ = q(nμn + pμp)(Ω· cm) −1

Resistivity ρ is the reciprocal of conductivity: ρ =1/σ(Ω · cm). The resistivity unit, the Ohm.cm,

is therefore:

ρ = E/ jdriftT and V/cm/ A/cm2 = Ω · cm

Example: Find the resistivity of intrinsic silicon at room temperature (300oK) and classify it as an

insulator, semiconductor or conductor. Suppose μn=1350cm2/V·s, μp=500 cm

2/V·s and ni= 10

10/cm

3

in intrinsic Si.

Analysis: : σ = q(nμn + pμp)

σ = (1.60 × 10−19

)[(1010

)(1350) + (1010

)(500)] (C)(cm-3

)(cm2/V·s) = 2.96 × 10

−6 (Ω · cm)

−1

The resistivity ρ is equal to the reciprocal of the conductivity, so for intrinsic silicon

ρ = 1/σ = 3.38 × 105 Ω · cm

we see that intrinsic silicon can be characterized as an insulator, albeit near the

16

low end of the insulator resistivity range.

Exercise: Find the resistivity of intrinsic silicon at 400oK and classify it as an insulator,

semiconductor, or conductor. Use the mobility values μn=1350cm2/V·s, μp=500 cm

2/V·s.

Answer: 1450 Ω· cm, semiconductor

Exercise: Calculate the resistivity of intrinsic silicon at 50oK if the electron mobility is

6500 cm2/V·s and the hole mobility is 2000 cm

2/V· s. Classify the material.

Answer: 1.69 × 1053

Ω·cm, insulator

Example: Calculate the electron concentration a piece of rod of length 3 mm and 1 x 4 µm cross

section area, if a current 5 mA produces 0.1 V drop. Suppose the mobility of electrons is 1000

cm2/V.s. and classify it as an insulator, semiconductor or conductor.

Solution: Jn = σn * E or ( I/A) = σn *( V/L)

Thus: σn = (I*L)/ (V*A) = (5*10-3

*3*10-1

)/( 0.1* 1*10-4

*4*10-4

) = 3.75*105 Ω. cm

σn = qnμn thus n = σn / qμn = 3.75*105/( 1.6*10

-19*10

3) = 2.3 * 10

21 /cm

3

The rode is conductor.

Note: The materials can be classified according to electron concentration number as follow:

Conductor: n > 1021

/ cm3

Semiconductor 10 < n > 1021

/cm3

Insulator: n< 10 /cm3

Impurities in semiconductors

The real advantages of semiconductors emerge when impurities are added to the material in minute

but well-controlled amounts. This process is called impurity doping, or just doping, and the

material that results is termed a doped semiconductor. Impurity doping enables us to change the

resistivity over a very wide range and to determine whether the electron or hole population controls

the resistivity of the material. The following discussion focuses on silicon, although the concepts of

impurity doping apply equally well to other materials. The impurities that we use with silicon are

from columns III and V of the periodic table.

Donor impurities in silicon

Donor impurities in silicon are from column V, having five valence electrons in the outer shell.

The most commonly used elements are phosphorus, arsenic, and antimony. When a donor atom

replaces a silicon atom in the crystal lattice, as shown in Fig. 1.10, four of the five outer shell

electrons fill the covalent bond structure; it then takes very little thermal energy to free the extra

electron for conduction. At room temperature, essentially every donor atom contributes (donates) an

electron for conduction. Each donor atom that becomes ionized by giving up an electron will have a

net charge of +q and represents an immobile fixed charge in the crystal lattice.

Acceptor impurities in silicon

Acceptor impurities in silicon are from column III and have one less electron than silicon in the

outer shell. The primary acceptor impurity is boron, which is shown in place of a silicon atom in the

lattice in Fig. 1.10 (b). Because boron has only three electrons in its outer shell, a vacancy exists in

the bond structure, and it is easy for a nearby electron to move into this vacancy, creating another

vacancy in the bond structure. This mobile vacancy represents a hole that can move through the

lattice, and the hole may simply be visualized as a particle with a charge of +q. Each impurity atom

that becomes ionized by accepting an electron has a net charge of −q and is immobile in the lattice.

17

(a) (b)

Figure 1.10 (a) An extra electron is available from a phosphorus donor atom. (b) A hole is created

after boron atom accepts an electron.

Electron and hole concentrations in doped Semiconductors

We now discover how to calculate the electron and hole concentrations in a semiconductor

containing donor and acceptor impurities. In doped material, the electron and hole concentrations

are no longer equal. If n > p, the material is called n-type, and if p > n, the material is referred to as

p-type. The carrier with the larger population is called the majority carrier, and the carrier with

the smaller population is termed the minority carrier. To make detailed calculations of electron

and hole densities, we need to keep track of the donor and acceptor impurity concentrations:

ND= donor impurity concentration atoms/cm3

NA= acceptor impurity concentration atoms/cm3

Two additional pieces of information are needed. First, the semiconductor material must remain

charge neutral, which requires that the sum of the total positive charge and negative charge be zero.

Ionized donors and holes represent positive charge, whereas ionized acceptors and electrons carry

negative charge. Thus charge neutrality requires: q(ND+ p − NA− n) = 0

Second, the product of the electron and hole concentrations in intrinsic material was given before.

as pn = n2i, the relation is true even for doped semiconductors in thermal equilibrium, and is valid

for a very wide range of doping concentrations.

If the concentration of donor atoms is ND, where ND is usually much greater than ni, the

concentration of free electrons in the n-type silicon will be: nn = ND

where the subscript n denotes n-type silicon. Thus nn is determined by the doping concentration and

not by temperature. This is not the case, however, for the hole concentration. All the holes in the n-

type silicon are those generated by thermal ionization. Their concentration can be found by noting

that the relationship in pn = n2

i applies equally well for doped silicon, provided thermal

equilibrium is achieved. Thus for n-type silicon: pnnn = ni2

Substituting for nn we obtain for Pn = ni2/ND

Finally, we note that in n-type silicon the concentration of free electrons will be much larger than

that of holes. Hence electrons are said to be the majority charge carriers and holes the minority

charge carriers in n-type silicon.

Similarly for p-type material, If the acceptor doping concentration is NA, where the hole

concentration becomes NA>> ni; pp = NA and np =ni2/ NA

18

Example: A silicon bar is 3 mm long and has a rectangular cross section area 0.05 x 0.1 mm and

ni = 1010

/cm3

at room temperature (300 oK). Calculate the voltage applied across the bar to produce

a steady current of 1 µA, if the semiconductor is: (1) intrinsic, (2) if it is doped 5*1014

/ cm3

of donor

atoms (n-type), suppose the electron mobility is 1500 cm2/V.s.

Solution:

(1) When the semiconductor is intrinsic n = ni = 1010

/ cm3

σn = qnμn = 1.6* 10-19

* 1010

* 1500 = 2.4* 10-6

(Ω. cm)

Jn = σn * E or ( I/A) = σn *( V/L)

Thus: σn = (I*L)/ (V*A) = (1*10-6

*3*10-1

)/(V*0.05*0.1*10-2

) = 2.4* 10-6

(Ω. cm)

V = 2500 volt (2) When the semiconductor doped with donor n = ND = 5* 10

14 /cm

3.

σn = qnμn = 1.6* 10-19

*5* 1014

* 1500 = 1.2* 10-1

(Ω. cm)

σn = (I*L)/ (V*A) = (1*10-6

*3*10-1

)/(V*0.05*0.1*10-2

) = 1.2* 10-1

(Ω. cm)

V = 0. 05 volt

Note: the voltage required to produce the current when the semiconductor is doped is much lower

than the voltage required when the semiconductor is intrinsic. In other words the material is

changed from seminsulator to semiconductor

Diffusion currents

As already described, the electron and hole populations in a semiconductor are controlled by the

impurity doping concentrations NA and ND. Up to this point we have tacitly assumed that the doping

is uniform in the semiconductor, but this need not be the case. Changes in doping are encountered

often in semiconductors, and there will be gradients in the electron and hole concentrations.

Gradients in these free carrier densities give

rise to a second current flow mechanism,

called diffusion. The free carriers tend to

move (diffuse) from regions of high

concentration to regions of low concentration

in much the same way as a puff of smoke in

one corner of a room rapidly spreads

throughout the entire room. A simple one-

dimensional gradient in the electron or hole

density is shown in Fig. 1.11.

Figure 1.11 Carrier diffusion in the presence

of a concentration gradient.

The gradient in this figure is positive in the +x

direction, but the carriers diffuse in the −x

direction, from high to low concentration.

Thus the diffusion current densities are

proportional to the negative of the carrier

gradient:

The proportionality constants Dp and Dn are

the hole and electron diffusivities, with units

(cm Diffusivity and mobility are related by

Einstein’s relationship:

19

The quantity (kT/q = VT) is called the thermal voltage VT, and its value is approximately 0.025 Vat

room temperature. Typical values of the diffusivities (also referred to as the diffusion coefficients)

in silicon are in the range 2 to 35 cm2/s for electrons and 1 to 15 cm

2/s for holes at room

temperature.

Exercise: Calculate the value of the thermal voltage VT for T = 50 K, 300 K, and 400 K

Answers: 4.3 mV; 25.8 mV; 34.5 mV

Exercise: What are the maximum values of the room temperature values (300 K) of the diffusion

coefficients for electrons and holes in silicon, let the mobility’s μn=1350cm2/V·s, μp=500 cm

2/V·s.

Answers: Using VT = 25.8 mV; 35.1 cm2/s, 12.8 cm

2/s

Exercise: An electron gradient of +1016

/cm3.µm exists in a semiconductor. What is the diffusion

current density at room temperature if the electron diffusivity = 20 cm2/s? Repeat for a hole gradient

of +1020

cm4 with Dp= 4cm

2/s.

Answer: +320 A/cm2; −64 A/cm

2

Total current

Generally, currents in a semiconductor have both drift and diffusion components. The total electron

and hole current densities jTn and jTp can be found by adding the corresponding drift and diffusion

components:

combined with Gauss’ law: ∇·(ε*E) = Q

where ε = permittivity (F/cm), E = electric field (V/cm), and Q = charge density (C/cm3)

Energy band model

This section discusses the energy band model for a semiconductor, which provides a useful

alternative view of the electron–hole creation process and the control of carrier concentrations by

impurities. Quantum mechanics predicts that the highly regular crystalline structure of a

semiconductor produces periodic quantized ranges of allowed and disallowed energy states for the

electrons surrounding the atoms in the crystal. Figure 1.12 (a) is a conceptual picture of this band

structure in the semiconductor, in which the regions labeled conduction band and valence band

represent allowed energy states for electrons. Energy EV corresponds to the top edge of the valence

and represents the highest permissible energy for a valence electron. Energy EC corresponds to the

bottom edge of the conduction band and represents the lowest available energy level in the

conduction band. Although these bands are shown as continuums in Fig. 1.12 (a), they actually

consist of a very large number of closely spaced, discrete energy levels. Electrons are not permitted

to assume values of energy lying between EC and EV. The difference between EC and EV is called the

bandgap energy EG= EC− EV

Electron–hole pair generation in an intrinsic semiconductor

In silicon at very low temperatures (≈ 0 K), the valence band states are completely filled with

electrons, and the conduction band states are completely empty, as shown in Fig. 1.12 (b). The

semiconductor and in this situation does not conduct current when an electric field is applied. There

are no free electrons in the conduction band, and no holes exist in the completely filled valence

band to support current flow. The band model of Fig. 1.12(b) corresponds directly to the completely

20

filled bond model of Fig. 1.10 (a). As temperature rises above 0 K, thermal energy is added to the

crystal. A few electrons gain the energy required to surmount the energy bandgap and jump from

the valence band into the conduction band, as shown in Fig. 1.12 (c). Each electron that jumps the

bandgap creates an electron– hole pair. This electron–hole pair generation situation corresponds

directly to that presented in Fig. 2.31.10 (b).

(a) (b) (c)

Figure 1.12 (a0 Energy band model for a semiconductor with bandgap EG. (b) Semiconductor at 0

K with filled valence band and empty conduction band. (c) Creation of electron–hole pair by

thermal excitation across the energy bandgap.

Energy band model for a doped semiconductor

Figures 1.13 (a, b, and c) present the band model for extrinsic material containing donor and/or

acceptor atoms. In Fig. 1.13 (a), a concentration N of donor atoms has been added to the

semiconductor. The donor atoms introduce new localized energy levels within the bandgap at a

donor energy level ED near the conduction band edge. The value of (EC− ED) for phosphorus is

approximately 0.045 eV, so it takes very little thermal energy to promote the extra electrons from

the donor sites into the conduction band. The density of conduction-band states is so high that the

probability of finding an electron in a donor state is practically zero, except for heavily doped

material (large ND) or at very low temperature. Thus at room temperature, essentially all the

available donor electrons are free for conduction. In Fig. 1.13 (b), a concentration NA of acceptor

atoms has been added to the semiconductor. The acceptor atoms introduce energy levels within the

bandgap at the acceptor energy level EA near the valence band edge. The value of (EA- EV ) for

boron is approximately 0.044 eV, and it takes very little thermal energy to promote electrons from

the valence band into the acceptor energy levels. At room temperature, essentially all the available

acceptor sites are filled, and each promoted electron creates a hole that is free for conduction.

Compensated semiconductors

The situation for a compensated semiconductor, one containing both acceptor and donor

impurities, is depicted in Fig. 1.13 (c) for the case in which there are more donor atoms than

acceptor atoms. Electrons seek the lowest energy states available, and they fall from donor sites,

filling all the available acceptor sites. The remaining free electron population is given by n = (ND−

NA). The energy band model just discussed represents a conceptual model that is complementary to

the covalent bond model. Together they help us visualize the processes involved in creating holes

and electrons in doped semiconductors.

21

Figure 1.13 (a) Donor level with activation energy (EC− ED). (b) Acceptor level with activation

energy (E A− EV). (c) Compensated semiconductor containing both donor and acceptor atoms with

ND> NA

.