Semiconductor Devices Module 1 Notes

8/10/2019 Semiconductor Devices Module 1 Notes

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109. Semiconductor Devices Module 1

Department of ECE, VKCET Page 1

Syllabus

Texts and References:

1. Streetman & Banerjee : Solid State Electronic Devices, 6/e, PHI

2. S.M.Sze: Physics of Semiconductor Devices, 3/e, Wiley Eastern

3.

V. Suresh Babu: Solid State Devices & Technology, Sanguine, Bangalore, 2005.


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Crystal structure:

The structures of solid materials are described with three main categories. These

categories are:

1. Crystalline

2. Amorphous

3.

Polycrystalline

Crystals are arranged in a periodic fashion. That is, there is some basic arrangement

of atoms that is repeated throughout the entire solid.

Amorphous have no periodic structure.

Polycrystalline are composed of many small regions of single-crystal material.

The crystal structure is known as the latticeor lattice structure.

The locations of each of the atoms in the lattice are known as the lattice points.

A unit cell is a volume-enclosing group of atoms that can be used to describe the

lattice by repeated translations.

A primitive cellis the smallest possible unit cell. Often primitive cells are not easy to

work with and thus we often use slightly larger unit cells to describe the crystals.

Cubic Lattices

There are three of very simple basic unit cells that are

often seen in crystalline structures. They are:

1. Simple cubic (sc) structure

2. Body-centered cubic (bcc) structure

3. Face-centered cubic (fcc) structure

The sc structure has an atom located at each corner of the unit cell. The bcc lattice has

an additional atom at the center of the cube, and fcc unit cell has atoms at the eight

Question:

Part B

Define lattice. Explain about

different lattice structure.


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corners and centered on the six faces. All three structures have different primitive

cells, but the same cubic unit cell.

Maximum packing fraction of sc lattice structure:

Consider the sc structure shown below:

The number of atoms per unit cell =1

8 8 = 1atom

Let a is inter-atomic spacing and r is the radius of an atom. Then relationship between

a and r is = 2. ---(1) Consider unit cell as a sphere, then volume ofsphere = 1 43 3. ---(2)

The volume of unit cell, = 3. ---(3) Packing fraction is given as =

. From (2) and (3) = 4333 and substituting (1), = 43 383 = 6 = 0.524 = 52.4%Maximum packing fraction of bcc lattice structure:

Consider the bcc structures shown below:

The number of atoms per unit cell =

1

8 8 + 1 = 2atoms.

Question:

Part A

Give the relationshipbetween radius of atom

and lattice constant of

different lattice structures.


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Consider the face diagonal, by right angled triangle ABC, = ()2 + ()2 = 2 + 2 = 2.In right angled triangle ACD, = ()2 + ()2 =

2

2 +

2 =

3

.

Also from bogy diagonal, = 4.Then equating above equations, 4 = 3 --- (1) Consider spheres in unit cell, then volume of

sphere = 2 43 3. ---(2) The volume of unit cell, = 3. ---(3) Packing fraction is given as =

. From (2) and (3) = 83 33 andsubstituting (1), = 83 333

433 = 38 = 0.6801 = 68.01%Maximum packing fraction of fcc lattice structure:

Consider the fcc structures shown below:

The number of atoms per unit cell =

1

8 8 +

1

2 6 = 4atoms.

Consider the right angled triangle ADC, = ()2 + ()2 = 2 + 2 =2.Also, = 4.Then equating above equations, 4 = 2 --- (1)

Consider spheres in unit cell, then volume of sphere = 4 43 3. ---(2) The volume of unit cell, = 3. ---(3) Packing fraction is given as

=

. From (2) and (3)

=

16

3 3

3

and

substituting (1), = 163 322433 = 32 = 0.7404 = 74.04%

Question:

Part A

How many atoms are found

inside a unit cell of a sc, a bcand an fcc crystal? How far

apart in terms of lattice

constant a are the nearestneighbor atoms in each case,

measured from center to

center?

Question:

Part B

Find the fraction of the fccunit cell volume filled with

hard spheres.

Question:

Part A

Compare packing fraction of

different lattice structure.


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Problem:

Solution:

For bcc, = 34 . Given that = 5,then = 34 5 = 2.165 The volume of sphere = 2 43 3 = 2 43 2.1653 3 = 85.01 3 The volume of unit cell, = 3 = 125 3 Packing fraction, =

= 85.01125 = 68.01%Planes and Directions:

Planes

It is necessary to specify directions and planes in crystals.

Many material properties and processes vary with directionin the crystal.

Directions and planes are described using three integers

called Miller Indices.

General rules for lattice directions, planes & Miller Indices: x, y, z are the axes (on

arbitrarily positioned origin). a, b, c are lattice parameters (length of unit cell along a

side) and h, k, l are the Miller indices for planes and directions expressed as planes -

(hkl) and directions - [hkl].

Conventions for naming: There are no commas between numbers. Negative values

are expressed with a bar over the number.

Steps to find Miller indices for plane are:

1. Find the intercepts of the plane with the crystal

axes and express these intercepts as integral

multiples of the basis vectors (the plane can be

moved in and out from the origin, retaining its

orientation).

2. Take the reciprocals of the three integers found in step1 and reduce these

to the smallest set of integers h, k, and l, which have the same relationship

to each other as the three reciprocals.

3. Label the plane (hkl).

Question:

Part B

How planes anddirections arerepresenting?

Question:

Part A

Give the steps to obtain

Miller indices.


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Many planes in a lattice are equivalent. The indices of such equivalent planes are

enclosed in braces { } instead of parentheses. Representation of cube faces by {100}

is shown below:

Problem 1:

Solution:

a)x y z

Intercepts 3 3 3

Reciprocal 1/3 1/3 1/3

Smallest integers (h, k, l) 1 1 1

Plane: (111)

b)

x y z

Intercepts 3 2 2

Reciprocal 1/3 1/2 1/2


Plane: (233)


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Problem 2

Solution:

a)

x y z

Intercepts a aReciprocal 1/ 1/a 1/a


Plane: (011)

b)

x y z

Intercepts a a

Reciprocal 1/a 1/a 1/


Plane: (110)

Directions Direction in a lattice is expressed as a set of three integers with the same relationship

as the components of a vector in that direction.

Miller indices are also used for the representation of direction, but the three integers

are the components of the vector in that direction.

The direction is labeled with in [ ] and the equivalency is labeled in < >.

Examples:


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Energy bands and bonding forces in solids:

Electrons in solids also are restricted to certain energies and are not allowed at other

energies.

In the solid, the electron has a range (or band) of available energies.

The discrete energy levels of the isolated atom spread into bands of energies in the

solid because i) in the solid, the wave functions of electrons in neighboring atoms

overlap, thus, it affects the potential energy term and the boundary conditions in the

equation, and different energies are obtained in the solution, and ii) an electron is not

necessarily localized at a particular atom.

Potential energy and wave function are represented by Schrodinger wave equation.

Bonding forces

It is due to the interaction of electrons in neighboring atom.

Different types:

1.Ionic bonding

Consider NaCl, each Na atom is surrounded by 6 Clatoms. The crystal is made up of ions, where Na has

+ve ion and Cl haveve ion.

Thus, an electrostatic attractive force is established between atoms, and the

balance is reached when this equals the net repulsive force.

In this bonding all the electrons are tightly bound to the atom.

Since there are no loosely bound electrons to participate in current flow NaCl

is a good insulator.

2.Metallic bonding

In metals, the whole solid made up of ions with closed shells immersed in asea of free electrons.

Coulomb attraction force between the ions and the electrons hold the lattice

together.

Free electrons are easy to move under the influence of an electric field, hence

the metal have high electrical conductivity.

3.Covalent bonding

Bonding occurs in diamond structure and compound semiconductor.

Each atom surrounded by four nearest neighbors, each having four electrons

in the outermost orbit.

Each atom shares its valence electrons with its four nearest neighbors.

The bonding forces arise from a quantum mechanical interaction (repulsive

force) between the shared electrons.

No free electron at 0K (insulator), but by thermal or optical excitation

electrons can excited from this bond and crystal behaves as conductor.

Question:

Part ACompare different bonding

forces.


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Energy bands in solids:

As isolated atoms are brought together to form a solid, the electron wave functions

begin to overlap.

Various interactions occur, and, at the proper interatomic spacing for the crystal, the

forces of attraction and repulsion find a balance.

Due to Pauli Exclusion Principle, the discrete energy levels of individual atoms split

into bands belonging to the pair instead of to individual atoms.

In a solid, due to large number of atoms, the split energy levels for essentially

continuous bands of energy.

The splitting of individual energy levels to energy bands as Si atoms are brought

closer together is shown below:

Si atom have an energy structure 1s22s

22p

63s

23p

2in the ground state. Each atom

has available two 1s states, two 2s states, six 2p states, two 3s states, six 3p states, and

higher states.

For N atoms, there will be 2N, 2N, 6N, 2N, and 6N states of type 1s, 2s, 2p, 3s, and

3p, respectively.

With a reduction in the interatomic spacing, these energy levels split into bands, and

the 2s and 2p bands merge into a single band having 8N available states.

As the interatomic spacing approaches the equilibrium spacing of diamond crystal,

this band splits into two bands separated by an energy gap, where no allowed energy

states for electrons exist forbidden gap.

Question:

Part B

Explain about the energybands of solid.


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The upper band (called the conduction band) and the lower band (called the valence

band) contain 4N states each.

Metals, semiconductors and insulators

For electrons to move under an applied electric field there

must be states available to them.

A completely filled band cannot contribute to current

transport; neither can a completely empty band.

Thus, semiconductors at 0 K are perfect insulators.

With thermal or optical excitation, some of these electrons

can be excited from the valence band to the conduction band, and then they can

contribute to the current transport process.

At temperatures other than 0 K, the magnitude of the band gap separates an insulator

from a semiconductor, e.g., at 300 K, (diamond) = 5eV (insulator), and (Silicon) =

1.12eV (semiconductor).

Number of electrons available for conduction can be increased greatly insemiconductors by reasonable amount of thermal or optical energy.

In metals, the bands are either partially filled or they overlap thus, electrons and

empty states coexist great electrical conductivity.

The energy band diagram of metal, semiconductor and insulator at 0K are shown

below:

Question:

Part BCompare metal,

semiconductor and insulator

with the help of energy banddiagram.


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Direct and indirect band gap semiconductors

The wave function of a single electron traveling through a perfectly periodic lattice is

assumed to be in the form of a plane wave moving in the x-direction with propagation

constant k, also called a wave vector.

In quantum mechanics, the electron momentum can be given by

=

, where

is

Plancks constant and = 2 The space dependent wave function for the electron is

Where the function U(kx, x)modulates the wave function according to the periodicity

of the lattice.

The periodicity of different crystals may vary, so E-k relationship will also vary.

E-k diagram for GaAs and Si are shown below:

Based on E-k relationship there are two types of semiconductors: direct band gap and

indirect band gap semiconductors.

For direct band gap semiconductor, the minima of the conduction band and the

maxima of the valence band occur at the same value of k (k = 0) and an electron

making the smallest energy transition from the conduction band to the valence band

can do so without a change in k.

For indirect band gap semiconductor, the minima of the conduction band and the

maxima of the valence band occur for different values of k, thus, the smallest energy

transition for an electron requires a change in momentum. Electron falling from conduction band to an empty state in valence band

recombination. Recombination probability for direct band gap semiconductors is

much higher than that for indirect band gap semiconductors.

Direct band gap semiconductors give up the energy (Eg) released during

recombination in the form of light used for optoelectronic applications (e.g., LEDs

and LASERs).

Question:Part B

Discuss about direct and

indirect band gapsemiconductors.


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For 0 x 0.38, valley has minimum energy with direct band gap. For x = 0,

material is GaAs and for x = 0.38, material is Al0.38Ga0.62As. Both have direct band

gap, but radiate different colour of light.

For x > 0.38, X valley has minimum energy with indirect band gap. For x = 1,

material is AlAs and energy release in the form of heat.

Charge carriers in semiconductors:

Electrons and Holes

For T > 0 K, there would be some electrons in the otherwise empty conduction band,

and some empty states in the otherwise filled valence band.

The empty states in the valence band are referred to as holes.

If the conduction band electron and the valence band hole are

created by thermal excitation of a valence band electron to

the conduction band, then they are called electron-hole pair

(EHP).

After excitation to the conduction band, an electron is surrounded by a large number

of empty states, e.g., the equilibrium number of EHPs at 300 K in Si is 1010

/cm3,

whereas the Si atom density is 1022

/cm3.

For every electron moving with a given velocity, there is an

equal and opposite electron motion somewhere else in the

band.

Under an applied electric field, the net current is zero, since

for every electron j moving with a velocity vj, there is a corresponding electron j

moving with a velocity -vj.

In a unit volume, the current density J can be given by = = 0=1 , whereN is the number of electrons/cm3in the band, and q is the electronic charge. Now, if the j

thelectron is removed and a hole is created in the valence band, then the

net current density = = (+) =1 Thus, the current contribution of the empty state (hole), obtained by removing the j

th

electron, is equivalent to that of a positively charged particle with velocity vj.

For simplicity, therefore, the empty states in the valence band are called holes, and

they are assigned positive charge and positive mass.

Question:

Part A

How free electrons and holes

formed in crystals?

Question:

Part A

What is EHP? Explain.


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The electron energy increases as one move up the conduction band, and electrons

gravitate downward towards the bottom of the conduction band.

On the other hand, hole energy increases as one moves down the valence band (since

holes have positive charges), and holes gravitate upwards towards the top of the

valence band.

Effective Mass

The "wave-particle" motion of electrons in a lattice is not the same as that for a free

electron, because of the interaction with the periodic potential of the lattice.

To still be able to treat these particles as "free", the rest mass has to be altered to take

into account the influence of the lattice.

The calculation of effective mass takes into account the shape of the energy bands in

three-dimensional k-space, taking appropriate averages over the various energy

bands.

The effective mass of an electron in a band with a given (E,k) relation is given by = 222 Problem:

Find the dispersion relation (E-k relation) for a free electron, and,

thus, observe the relation between its rest mass and effective mass.

Solution:

For free electron, the electron momentum p is given by = 0---(1)Where m0is rest mass of electron and v is velocity.

Also = --- (2),Where = 2, h is Plancks constant and k is wave vector.

We have energy = 122, using (1) = 0 and = 12 20. Then using (2), = 1

2

20 2---(3) This is E-k relation and is parabolic with wave vector.

Obtaining second derivative of (3) w.r.t kgives the relation of effective mass and rest

mass. Then = 20 and 2 = 20and = 22 2 = 0

Question:

Part B

Obtain the E-k relation, rest

mass and effective massrelation. Explain.


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This states that for a free electron, the rest mass and the effective mass are one and

the same, which is due to the parabolic band structure. Most materials have non-

parabolic E-k relation, and, thus, they have quite different rest mass and effective

mass for electrons.

Thus, the effective mass is an inverse function of the curvature of the E-k diagram:

weak curvature gives large mass, and strong curvature gives small mass.

The curvature22is positive at the conduction band minima; however, it is negative

at the valence band maxima. Thus, the electrons near the top of the valence band have

negative effective mass.

The electron and hole effective masses are denoted by and respectively:

Effective

mass /m0Ge Si GaAs

mn* 0.55 1.1 0.067

mp* 0.37 0.56 0.48

Intrinsic Semiconductors

A perfect semiconductor crystal with no impurities or lattice defects.

No carriers at 0 K, since the valence band is completely full

and the conduction band is completely empty.

For T > 0 K, electrons are thermally excited from the valence

band to the conduction band causes EHP generation. EHP

generation takes place due to breaking of covalent bonds and

the required energy isEg.

The EHP in the covalent bonding model of the Si crystal is shown below:

The excited electron becomes free and leaves behind an empty state (hole).

Since these carriers are created in pairs, the electron

concentration n/cm3is always equal to the hole concentration

p/cm3, and each of these is commonly referred to as the

intrinsic carrier concentration (ni). Thus, for intrinsic

material n = p = ni.

Question:

Part B

Write short notes onintrinsic

semiconductors.

Question:

Part A

Define intrinsic carrierconcentration.


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These carriers are not localized in the lattice; instead they spread out over several

lattice spacings, and are given by quantum mechanical probability distributions.

To maintain a steady-state carrier concentration, the carriers must also recombine at

the same rate at which they are generated. Recombination occurs when an electron

from the conduction band makes a transition (direct or indirect) to an empty state in

the valence band, thus annihilating (nullifying) the pair.

Let the generation rate of EHPs is g i(EHP/cm3-s) and

there combination rate is ri, then equilibrium requires

that ri = gi. Each of these rates is temperature

dependent. For example gi(T) increases with rise in

temperature, a new carrier concentration niestablished

such that the higher temperature recombination rate

ri(T) balances generation.

At any temperature, the rate of recombination is proportional to the equilibrium

concentration of electrons and holes, and can be given by = 00 = 2 = ,where proportionality is constant depends on the mechanism by which therecombination takes place, 0and 0are equilibrium concentration of electrons andholes respectively.

Question:

Part A

Explain the dependence of

temperature on intrinsic

carrier concentration.

(April 2014)


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Extrinsic semiconductor

Instead of thermally generated carriers, it is possible to create

carriers in the semiconductor by purposely introducing

impurities into the crystal. This process is called doping.

Doping is a most common technique for varying the

conductivity of semiconductors. By doping, the crystal can

be made to have predominantly electrons or holes. Then

there are two types of extrinsic semiconductors: n-type

semiconductor (predominant in electrons) and p-type semiconductors (predominant in

holes).

When a crystal is doped such that the equilibrium concentrations of electrons (n0) and

holes (p0) are different from the intrinsic carrier concentration (ni), the material is said

to be extrinsic.

Doping creates additional levels within the band gap.n-type semiconductor:

In Si or Ge, V group elements (e.g., P, As, Sb) doping introduce energy levels very

near (typically 0.03 - 0.06eV) the conduction band.

At 0K, these levels are filled with electrons, and very little thermal energy (50K to

100K) is required for these electrons to get excited to the conduction band. Since

these levels donate electrons to the conduction band, they are referred to as the donor

levels and the impurities are referred as donor impurities.

The covalent model of n-type Si is shown below:

The energy band model of n-type semiconductor at 0K and 50K are shown below:

Question:

Part B

Explain about extrinsic

semiconductors.

Question:

Part A

Define doping.

Question:

Part B

Why electrons are referred amajority carriers in n-type

semiconductors? Explain.


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Thus, Si doped with donor impurities can have a significant number of electrons in

the conduction band even when the temperature is not sufficiently high enough for

the intrinsic carriers to dominate, i.e., n0>> (ni, p0) and the material is referred as n-

type material, with electrons as majority carriers and holes as minority carriers.

p-type semiconductor

In Si or Ge, doping III group elements (e.g., B, Al, Ga, In) introduce energy levels

very near (typically 0.03-0.06 eV) the valence band.

At 0K, these levels are empty, and very little thermal energy

(50K to 100K) is required for electrons in the valence band

to get excited to these levels, and leave behind holes in the

valence band. Since these levels accept electrons from the

valence band, they are referred to as the acceptor levels and

the impurities are referred as acceptor impurities.

The covalent bond model of p-type Si is shown below:

The energy band model of p-type semiconductor at 0K and 50K are shown below:

Thus, Si doped with acceptor impurities can have a significant number of holes in thevalence band even at a very low temperature, i.e., p0>> (ni, n0), and the material is

referred as p-type material, with holes as majority carriers and electrons as minority

carriers.

The extra electron for column V elements is loosely bound and it can be liberated

very easily ionization; thus, it is free to participate in current conduction. Similarly,

column III elements create holes in the valence band, and they can also participate in

current conduction.

Question:

Part A

Give the examples of donors

and acceptors. Why is it

called so?


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Carrier concentration

Obtaining semiconductor electrical properties and analyzing device behavior, it is

necessary to know the concentration of charge carriers free electrons and holes.

To obtain equations for the carrier concentrations an investigation of the distribution

of carriers over the available energy states is required. This is not difficult to

calculate, but the derivation requires some back ground in statistical methods.

Fermi-Dirac distribution

Fermi-Dirac distribution function is a function in Fermi-Dirac statistics.

Electrons in semiconductor obey Fermi-Dirac statics. By this the distribution of

electrons over a range of allowed energy levels at thermal equilibrium is given by = 11+ ()

where k is Boltzmann's constant (k = 8.62 x 10-5

eV/K = 1.38 x 10-23

J/K),

T is absolute temperature,

f(E) is Fermi-Dirac distribution function which gives the probability of the available

energy state at E will occupied by an electron,

EF is Fermi level and it represents an important quantity in the analysis of

semiconductor behavior.

If energy level E = EF, the occupation probability is = 11+0 = 12

A closer examination of f(E) indicates that at 0K the distribution takes the simple

rectangular form as shown below:

For T = 0K, distribution takes place in rectangular form. At T = 0K and E < EF,

=

1

1+

= 1. At T = 0K and E > EF,

=

1

1+

= 0.

This rectangular distribution shows that at 0K every available energy state up to EFis

filled with electrons, and all states above EFare empty.

If T > 0K, some probability exists for states above the Fermi level to be filled.

The Fermi function is symmetrical about EFfor all temperatures.


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Proof:

Show thatprobability of a state E above EFis filled equals the probability that a state E below

EFis empty.

Solution:

Probability that state E above EFis filled

E + E

F=

1

1+ (E +EFE F )/kT=

1

1+E/kT

Probability that a state E below EFis filled EF EProbability that a state E below EFis vacant 1- EF E = 1 1

1+EFEEFkT = 1 1

1 + E/kT = E/kT1 + E/kT = 11 + E/kT ie, E + EF =1- EF EFermi-Dirac distribution for intrinsic semiconductors

Here the probability of occupancy in conduction band equals to probability of vacancy in

valance band. ie, EC =1- EV Then Fermi level is at the middle and is also referred as intrinsic level. Relative plot of Fermi-Dirac distribution and energy-band diagram of intrinsic

semiconductor is shown below:

Fermi-Dirac distribution for extrinsic semiconductors

For n-type semiconductors, probability of occupancy in the conduction band is much

greater than the probability of vacancy in valance band. ie 1 (). To satisfy this, Fermi level EFmoves closer to EC and (ECEF) gives the measure of

carrier concentration n.


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Relative plot of Fermi-Dirac distribution and energy-band diagram of n-type


In case of p-type semiconductor, probability of vacancy in the conduction band is much

greater than the probability of occupancy in valance band. ie 1 () . Here the Fermi level EF is closer to EV and (EF - EV) gives the measure of carrier

concentration p.

Relative plot of Fermi-Dirac distribution and energy-band diagram of p-type


From the above plots it is clear that EF is an important parameter to indicate the

probability of energy level distribution.


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Electron-Hole concentration at equilibrium

The concentration of electrons in the conduction band is given by0 = ---(1)whereN(E) dErepresents the density of states in the energy range dE.

The number of electrons per unit volume in the energy range dE is the product of thedensity of states and the probability of occupancy f(E). Thus the total electron

concentration n0 is the integral of product of density of states and probability of

occupancy over the entire conduction band.

N(E) can be evaluated by quantum mechanics and Paulis exclusion principle and it is

shown that 1/2. This shows that density of state in conduction band increaseswith increase in temperature.

Fermi-Dirac distribution function becomes extremely small at large energies. Therefore,

f(E)N(E) rapidly decrease above ECand the electron concentration above the conduction

band is very low. And most of the electrons occupy the lowest available energy levels in

the conduction band.

The concentration of holes in the valance band is given by0 = (1 ) --- (2) In this (1- f(EV)) is the probability of vacancy of electron (or probability of occupancy of

hole) at valance band and it decrease rapidly below EV. Therefore holes occupy top most

available states in the valance band.

Graphical representation of carrier concentration in semiconductors

The carrier concentration in intrinsic semiconductor is as shown below:

For intrinsic semiconductors N(E)f(E) and N(E)[1-f(E)] are concentration of energy at

conduction band and valance band respectively. And are equal, so most of the carriers are

concentrating in the conduction band and valance band.


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The carrier concentration in n-type semiconductor is as shown below:

The area underN(E)f(E)in the conduction band is the measure of electron concentration

n0. Similarly the area under N(E)[1-f(E)] in the valance band is the measure of hole

concentration p0.

For n-type semiconductorsf(EC) >> [1-f(EV)]. ThenN(E)f(E)is larger thanN(E)[1-f(E)]

and n0>> p0.

The carrier concentration in p-type semiconductor is as shown below:

For p-type semiconductors [1-f(EV)] >> f(EC). ThenN(E)[1-f(E)]is larger thanN(E)f(E)

and p0>> n0.

Question:

Part A

Plot the energy band

diagram along with Fermi-Dirac distribution function

for n-type semiconductor.

(April 2014)


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Analytical representation of carrier concentration in semiconductors

By simple notation, electron concentration n0 in the conduction band can represent by

product between density of states of energy in conduction band and Fermi-Dirac

distribution function.

Then

0 =

(

)---(1),

where NCis the effective density of states in the conduction band and is given by

= 2 22 3 2 ---(2),where is effective mass of electron.

We have

= 11 + ()

Then = 11+ ( ) .If

, ie the electron concentration is not very

high.Then () 1and we can approximate 1 ( ) = () ---(3).

We can write electron concentration in conduction band at equilibrium is (by substituting

(3) to (1)), = () ---(4) Similarly hole concentration p0in the valance band0 = [1 ]---(5),

where NVis the effective density of states in the valance band and is given by

= 2 2 2 3 2 ---(6),where is effective mass of hole.

Then 1 = 1 11+ ( ) = 1+ ( ) 11+ ( ) = ( ) 1+ ( ) .If , ie the hole concentration is not very high.

Then () 1and we can approximate

1 ( ) ---(7). We can write hole concentration in valance band at equilibrium is (by substituting (7) to

(5)),

= () ---(8) Equations (4) and (8) are the carrier concentration for any semiconductors. And also bytaking product of the two equations00 = =

= ---(9),where = and is called band gap energy difference.

Question:

Part B

Derive the relation n0p0=

ni2. (April 2014)


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At room temperature:

Si Ge GaAs

Eg(in eV) 1.12 0.66 1.424

For intrinsic semiconductors, carrier concentration are equal to intrinsic carrier

concentration ni and Fermi-level EF is same as intrinsic level Ei. Therefore, (4) and (8) can

be written as 0 = = () ---(10)0 = = ( ) ---(11) Taking product of (10) and (11)2 = () ( ) = ( )

= () ---(12) Comparing (9) and (12), we can relate

=

and is

called mass action law. This law states that the product ofequilibrium electron and hole concentration in a

semiconductor is a constant for any doping and is equal to

the square of the intrinsic carrier concentration at that temperature.

In intrinsic semiconductor NC= NV, therefore from(12)2 = 2() = () (2 ) ---(13).This is the relationship between intrinsic concentration and energy band difference

At room temperature:

Si Ge GaAsni(in cm

-) 1.45 1010 2.4 1013 1.75 106

Take (4) and (10), and findNCfrom (10) gives: = ()

Put this into (4), 0 = () () = () ---(14) Similarly using (8) and (11):

= () 0 = ( ) () = () ---(15) Equations (14) and (15) gives the relationship between equilibrium carrier concentration

and intrinsic carrier concentration. Also taking the product of (14) and (15) gives another

proof of mass action law, = .

Question:Part A

State mass action law.

Question:

Part B

Analytically explain about

equilibrium carrierconcentration in

semiconductors.


26/51



From (14), if EFmoves away Eitowards conduction band, n0increases exponentially over

ni. At the same time p0 decreases exponentially over ni. This will happen when donor

impurities are doped into the semiconductor and the material behaves like n-type

semiconductor.

Similarly from (15), if EF moves away Ei towards valance band, p0 increases

exponentially over ni. At the same time n0 decreases exponentially over ni. This will

happen when acceptor impurities are doped into the semiconductor and the material

behaves like p-type semiconductor.

Problem 1:

Solution:

Given As is as dopant, then donor concentration Nd >> ni and we can approximateequilibrium carrier concentration 0 = 10173 .

By mass action law, 00 = 2,then 0 = 20 = 1.5101021017 = 2.25 1033

We have 0 = () , from this () = 0 . Apply ln on both sides, = ln 0 = 0.026 ln 10171.51010 = 0.407 The resulting band diagram is:


27/51



Problem 2:

Solution:

Given that NC= NV, Eg= 1.1eV, therefore EC- EF= 1.1eV

Donor concentration Nd = 1015

/cm-3

n0, electron concentration

kT = 0.026eV

EC- Ed= 0.2eV, EC - EF= 0.25eV

Then EF - EV= 0.85eV

We have 0 = () Then

=

0

() = 1015 e0.25 0.026 = 1.56 1019cm3 =

Also 0 = ( ) = 1.56 1019 e0.85 0.026 = 8.71 104cm3 We have 00 = 2,then = 00 = 1.56 1019 8.71 104 = 9.35 109cm3

Problem 3:

Solution:

Give that In acceptor atom concentration, then Na= 10

16

/cm

3

. Ea- EV= 0.16 eV, EF- EV= 0.26eV.

Fraction of Eastates filled = 11+ ( ) = 0.16 0.26 = 0.1 = 11+ (0.1) 0.026 = 0.979 Unionized In atoms 1 = 1 0.979 1016 = 2.1 10143

Problem 4:

Solution:

We have 0 = = () and = () (2) Equation above equations, () = () (2)


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() + () (2) = =

3 4

Taking natural log on both sides,

2 =

3

4

= 0.026

3

4

0.56

1.1 =

0.013

So = 2 0.013

Temperature dependence of carrier concentration

The intrinsic carrier concentration = () (2) . This shows that intrinsic

carrier concentration is a function of temperature.

Also = ln 0 , then Fermi level is also a function of temperature and henceextrinsic carrier concentration is a function of temperature.

From the above equations, variation in temperature strongly dependent on nibecause it is

dependent on the exponential quantity of temperature.

We can write = (

) (2

)

--- (1)and put = 2 2 2 3 2 , = 2 2 2 3 2 .Then = 2 22 3 2 3 4 () (2) = (1)3 2 () (2) --- (2)Where 1 = 2 22 1 2 , this is not an exponential function of temperature.

Band gap energy of semiconductors decreases monotonically with increase in

temperature. Then

=

0

1

, where

0is band gap energy at 0K, b1is rate of

decrease in band gap energy with increase in temperature. Put this equation to (2) = (1)3 2 (01) (2) = (1)3 2 (0) (2) 1 2 = 2(0) (2) Where 2 = (1)3 2 1 2 and is less temperature dependent quantity.

Question:

Part A

Explain about the temperaturedependency of intrinsic

semiconductor.


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The plot of niverses T for different semiconductor is as shown below:

For extrinsic semiconductor, at 0K, the carrier concentration is zero. If temperature

increases carrier concentration increases due to the excitation of electron/hole in

donor/acceptor level. ie, n0 Nd+

(for n-type) and p0 Na-

(for p-type), where Nd+

andNa

-are concentration of donor and acceptor ions respectively.

At low temperature, carrier concentration is equal to

ionized dopant concentration and ni is negligible. This

operating region of extrinsic semiconductor is called

ionized region.

When temperature is increased to medium, all impurities

get ionized and niincreases, so that electron concentration.

ie n0= Nd + ni. Comparing to Nd, ni is very less, then n0 Ndand is independent of

temperature. This operating region is called extrinsic region. (For p-type p0= Na+ niNa)

When temperature increases more than 500K, nibecome more comparable to or greater

than the doping concentration. Then n0 ni and the minority carrier p0 also increases.

This region is referred as intrinsic region.

The carrier concentration of Si and Ge w.r.t T is shown below:

Question:

Part B

Explain about the temperaturedependency of extrinsic

semiconductor.


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Compensation and space charge neutrality:

Under equilibrium semiconductor remains neutral in charge.

ie, total +ve charge = total ve charge. Let Nddonors/cm3

and Naacceptors/cm3, then 0 + + = 0 + ---(1)

where

+ represents ionized donor concentration and

represents ionized acceptor concentration.

The above equation is called charge neutrality equation.

If temperature > 150K, almost all impurities get ionized in a uniformly doped

semiconductors, then + = and = Therefore, 0 0 = ---(2)

Using mass action law and the above equation can be solved to get the carrier

concentration n0and p0in a semiconductor under thermal equilibrium.

ProblemA Ge sample is doped with 5x10

13As/cm

3. Determine the carrier concentration and Fermi level

position at room temperature.

Solution:

Given that Nd= 5x1013

/cm3, for Ge ni= 2.5x10

13/cm

-3at room temperature.

ie Nd ni.

Use mass action law and charge neutrality condition for obtaining carrier concentration.0 0 = 0 + 0 = 0 +

2

0 = 002 0 2 = 00 = 2 + 422

0 = 5 1 013 5 1 0132 + 42.5 101322

0 = 6.035 10133(Neglectve value)Then

0=

2

0=

2.51013

2

6.0351013= 1.035 1013

3

Fermi level position

= ln0 = 0.026 ln6.035 1013

2.5 1013 = 0.023

Question:

Part A

Explain about the chargeneutrality in semiconductor.


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Carrier transport in semiconductors:

Carrier transport in semiconductors is mainly by two mechanisms: 1) drift 2) diffusion.

The current due to drift is by the movement of electrons/holes under an electric field and is

similar to current flow in a conductor. The current due to diffusion motion is due to the gradients

in carrier concentrations.

Drift velocity, current density and mobility of electrons Under thermal equilibrium, carrier transport is in a constant motion with a mean thermal velocity

of electron = 3 At room temperature, the thermal motion of electron is in random scattering and due to this there

is no net motion along any direction for a group of n electrons/cm3. This is as shown below:

If an electric field is applied in x axis, each electron experiences a force from the field.Thus the group of electrons has a net average velocity in x direction and this mean velocity is

called drift velocity.

If px is the total momentum of the group of n electrons/cm3, then the accelerating force due to

electric field is given by = ---(1)

The accelerating force due to electric field is balanced in steady state by decelerating force of the

collision process by electrons.

Let N0 be the group of electrons at time t = 0 and N(t) be the number of electrons have to

undergone collision at time t. Then the rate of decrease in N(t) at any time is proportional to N(t).


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ie,

=

Solving this equation gives = , where mean time between collisions is ormean free time. The probability of collision of an electron in any time interval dt is

. Therefore the differentialchange in momentum due to collision is =

Then decelerating force due to collision = ---(2)

At steady state the sum of accelerating and decelerating effect must be zero, ie by (1) and (2)

=

= and < > = ---(3)

where < is constant average momentum. This indicates that the electrons have constant average velocity

inve x direction given by

< >= = ---(4) This represents the drift velocity of an average electron in

response to electric field. This usually much less than . The current density due to electron drift is the number of electrons crossing unit area per unit timemultiplied by the charge of an electron. Its dimension is

Substituting (4), = ---(5) By Ohms law,

=

---(6)

Where = is called conductivity of electron in (-cm)-1and can be written as = ---(7)Where = ---(8) is called mobility of electron and is the representation of electron drift in amaterial.

Question:

Part A

Define drift velocity of

carriers.

Question:Part A

Define conductivity and

mobility of semiconductors.


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Mobility can also defined as average drift velocity per unit electric field, = cm2/V-s ---(9)(Dimension cm-s-1/ V-cm gives cm2/V-s)

Then current density in terms of mobility is

= ---(10) Similarly for hole, = + --- (11)or = ---(12) = ---(12)

Then current density due to holes is = ---(13) If both carriers participate the flow of current, then overall current density is sum of (10) and (13)

= + = ( + )---(14) = ---(15)Where = ( + ) ---(16) , conductivity due to carriers and resistivity = -cm

Problem 1:

Solution:a)

We have = ( + ) At equilibrium state, = ( + ) Substituting = ,

= ( + ) For minimum conductivity at electron concentration n0min

=

= ( )( ) = =


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= Then = , hence proved.

Solution: b) Minimum conductivity is due to the minimum concentration of carriers.

We have

= ( + )Then

= ( + )From above problem,

= = +

= + = Solution: c)

For Si ni= 1.5 x 1010cm-3, and q = 1.6 x 10-19C, n= 1350 cm

2/V-s p = 480 cm2/V-s = = . . = .

Intrinsic conductivity

=

(

+

)

= . . + = .

This shows that .


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Problem 2:

An intrinsic Ge sample at room temperature has resistivity of 50 -cm. The sample is uniformly

doped to the extent of 6 x 1013Arsenic atoms/cm3and 1014Boron atoms/cm3. Find the conduction

current density if an electric field of 4 V/cm is applied across the sample. n= 3800 cm2/V-s; p=

1800 cm2/V-s.

(April 2014)Solution:

Given doping concentrations are almost close to ni of Ge. So use charge neutrality and mass

action law to fine carrier concentratio.

Given intrinsic resistivity i= 50 -cm, n= 3800 cm2/V-s p= 1800 cm

2/V-s, = 4 / = 6 1 0133 = 10143 We have intrinsic conductivity = ( + ) = = ( + )

= ( + ) =1

1.6 1019 50(3800 + 1800) = 2.23 1013

3

We have 0 0 = 0 = 20

0 20 = 02 ( )0 2 = 0

02

1014

6 1 013

0

2.23 1013

2 = 0

0 = 4 1 013 ( 4 1013)2 (4 2.23 10132)2 =

4 1 013 1.6 1027 + 1.99 10272

=4 1 013 5.99 1013

2

= 4.99 10133 (ignoreve result)0 = 2.23 10132

4.99 1013= 9.97 10123

= ( + )= 1.6 1019(9.97 1012 3800 + 4.99 1013 1800) 4= 1.6 10193.79 1016 + 8.98 1016 4= 81.72 103 = 0.0817 A/2


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Problem 3

The intrinsic carrier concentration of Si at 300K is 1.5 x 1010 cm-3, n = 1500 cm2/V-s and p = 450

cm2/V-s. The Si sample is doped with donors of concentration 5 x 1016cm-3 and then compensated by

acceptor doping. The compensated Si sample has resistivity 1-cm. Determine acceptor concentration,

assuming that the mobility remains unchanged.

Solution: Give Nd= 5 x 10

16cm-3 >> nithen n0 Nd= 5 x 1016cm-3

Then resistivity due to n0is = = . = . After compensation,

Case 1: Nd> Na = = , where n0 is new carrier concentration. = = . = . /

=

= = . = . /Case 2: Na> Nd = = , where p0 is new carrier concentration. = = . = . / = = + = + . = . /Effects of temperature and doping on mobility:

There are different mechanisms which limits the mobility of charge carriers in semiconductors.

Some mechanism are:

i) Lattice scattering

ii) Ionized impurity scattering

iii) Neutral impurity scattering

iv) Carrier-carrier scattering

Lattice scattering is due to the vibration of lattice. As

temperature increases, lattice vibrates at a faster rate. Therefore

vibration due to this scattering increases with increase in temperature. Then mobility of carriers

decrease with rise in temperature, such that mobility due to lattice scattering L T-3/2.

Ionized impurity scattering is due to scattering of charge carriers with ionized impurities. Thistype scattering increases with increase in doping, but decrease with increase in temperature. Then

mobility due to ionized impurity scattering is give by I T3/2

.

Neutral impurity scattering is due to the difference in size of neutral impurity atom and host

lattice. This causes a strain to carriers in lattice and is negligible, because the number of neutral

impurities is very small.

Carrier-carrier scattering is due to the deflection of electron and hole during the transport. This is

negligible, because in extrinsic semiconductors any one of the carrier is major than other.

Question:

Part A

What are the scattering

mechanisms in asemiconductor?


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Variation of mobility with temperature:

At low temperature, mobility of carriers depends on ionized

impurity scattering, which increases with increase in

temperature.

At high temperature, mobility of carriers depends on lattice

scattering, which decreases with increase in temperature. The plot of these situation is shown below:

Variation of mobility with doping:

The concentration of impurity increases also increase the ionized impurity scattering and causes

decrease in mobility.

An appropriate plot of mobility as function of doping is shown below:

Question:

Part B

Explain temperature and

doping dependence ofmobility.


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Constancy (invariance) of Fermi-level at equilibrium:

This concept states that there is no discontinuity or gradient can

arise in the equilibrium Fermi level EF.

Consider two materials in contact like Ge and Si, n-type Si and

p-type Si, a metal and semiconductor, heavily doped Si and

lightly doped Si, etc. as shown below:

The transfer of electrons from material 1 to 2 must be exactly balanced by the opposite transfer of

electrons from 2 to 1.

Let N1(E) is desity of state in material 1, N2(E) is density of states in material 2, f1(E) is

probability of state being filled at E in material 1, f2(E) is probability of state being filled at E in

material 2, EF1 is Fermi level of material 1 and EF2is Fermi level of material 2.

At energy E, the rate of transfer of electrons from 1 to 2 is proportional to the number of filled

states at E in material 1 times the number of empty states at E in material 2.

ie rate from 1 to 2

11.

2(

)[1

2]---(1)

Similarly rate from 2 to 1 22.1()[1 1]---(2) At equilibrium these rate must be equal,11.21 2 = 22.1()[1 1]1. 1 2 = 2. [1 1]1 12 = [2 21]1 = 2---(3)ie

1

1+ (1) = 11+ (2) This shows that EF1 = EF2and no gradient exists in Fermi level at equilibrium or

Question:

Part B

Show that the gradient in

Fermi-level is zero underthermal equilibrium.

(April 2014)


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High field effect:

Drift current is proportional to electric field and the proportionality constant is conductivity. But

this assumption is valid only for a range of field intensity . For large (usually over 103V/cm), there is a limitation to electron drift velocity vdand is same

as thermal velocity vth. This is due to hot carrier effect.

The limitation of carrier lift velocity to a constant at high electric field in Si and Ge is shownbelow.

Hall effect:

This is mainly used to determine majority carrier concentration,

conductivity and mobility of semiconductor bar.

Consider a p-type bar with current Ixflowing in x-direction due

to the known applied voltage VCDacross side C and D. There is

also electric field

in x-direction and holes flow from C to D.

The length of the bar is L, width w and thickness t.

When a magnetic field with intensity Bz is applied in z-direction, the net force experienced in

hole due to the combined electric and magnetic field is given by = + ( ) ---(1)where =

Question:

Part B

Explain Hall Effect. Explain

the procedure to measure

majority carrier concentrationand mobility of a

semiconductor specimen.


40/51



We know that = = = = 0, then = 0 0+ 0And = ()

The net force experienced by the hole is in y-direction and = ()---(2)This shows that an hole acceleration in y-direction is due to force

and

(

).

Under steady state condition = = 0 = ---(3) Physically this electric field is set up when the holes are slightly shifted to ve y-direction

(towards A) due to magnetic field. The force due to this motion balances the force due to

magnetic field and the holes drift along the bar.

The establishment of electric field is known as Hall effect, the field is known as Hall field andthe voltage developing across A and B VABis called Hall voltage. = ---(4)

We have = +0 = 0Then Hall field = 0 ---(5) = ---(6)This shows that Hall field is proportional to product of current density and magnetic field

intensity. The proportionality constant is called Hall coefficient and = 10. For p-type bar Hall voltage and Hall coefficient are +ve, but for n-type bar both areve.

From (5) and (4),

0 = 0 = ( . ) ( ) = ---(7)

Since all quantities are measurable and the majority carrier concentration in semiconductor bar

can be measured by Hall Effect.

We know that the resistance of the bar

= . = .. = ---(8)Here also all quantities are measurable and the resistivity of the semiconductor is measurable by

Hall Effect.


41/51



For p-type semiconductor, conductivity = +0 . Then = 0 = From (7) and (8)

= 1 ( . ) ( ) = ---(9)Here also all quantities are measurable and the mobility of the semiconductor is also measurable

by Hall Effect.

Problem:

Solution:

Given = 5 = 5 1 05/2, Ix = 4mA, w = 0.25mm = 0.25x10-1cm, t = 50m =50x10-4cm, L = 2.5mm = 2.5x10-1cm, VAB= -2.5mV and VCD= 170mV.

VABisve, then the material is n-type.

=

=41035105

1.610

19 5010

4(

2.510

3)

= 1 1 017

3

= = 2.5103.5105. = . 1052/


42/51



Excess carriers in semiconductors:

Excess carriers are charge carriers in semiconductors generated by optical or electron

bombardment in addition to thermally generated carriers.

When a photon of energy greater than the band gap energy of semiconductor falls on it, the

energy of photon is absorbed by an electron in the valance band and is excited to the conduction

band. This results generation of an EHP and is the concentration of such carriers are equal. The optically generated carrier concentration are in excess of the thermal equilibrium value and = 0 + = 0 + =

where and are excess electron and hole concentration respectively. If the light of intensity I0 falls on the sample of thickness l, then the amount of light transmitted

through it is given by = 0 ---(1)where is called the absorption coefficient and it depends on the the incident wave length or

energy = The absorption coefficient as a function of incident photon energy is shown by the following

figure.

If the incident energy is equal or greater than band gap energy, the velocity of excited electron

will be greater than mean thermal velocity.

The photo-generation and recombination process is illustrated below. During the recombination

of carriers energy is released from the semiconductor and is Eg.


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The energy of radiation depends on the band gap energy as well as the recombination mechanism,

absorption coefficient etc.

The band gaps of some common semiconductors relative to the optical spectrum are shown

below.

Luminescence:

When an EHP is generated in semiconductor and they fall into its equilibrium states emits light

according to the band gap energy Eg. This property is called luminescence. Different types of luminescence are:

1) Photoluminescence The EHPs are generated by

photon absorption and the radiation results from the

recombination of the excited carriers. E.g. Fluorescent

lamps.

2) Cathodoluminescence The EHPs are generated by

high-energy electron bombardment of the material and the radiation results from the

recombination of the excited carriers. E.g. LASERs

3) ElectroluminescenceThe EHPs are generated by the applied current into the sample

and the radiation results from the recombination of the excited carriers. E.g. LEDs

Excess carrier and carrier life time:

When excess holes and electrons are created in a semiconductor,

there is an increase in conductivity.

If the excess carriers are generated by optical energy, the

increase in conductivity is called photoconductivity.

Consider a semiconductor with direct recombination mechanism,

the rate of decay of electrons at any time t is proportional to the

number of electrons remaining at t and the number of holes, with

some constant of proportionality for recombination r.

Then the net rate of change in the conduction band electron concentration is the thermal

generation rate minus the recombination rate () = 2 ()---(1) Let assume that excess carriers are created at t = 0 are n and p (initial values). Then the

instantaneous concentration of carriers is also equal, i.e. n(t) = p(t).

We can write instantaneous values of carriers = 0 + () = 0 + () Then using equation (1), we can write

Question:

Part A

What is luminescence? Explaindifferent types in brief.

Question:

Part B

Explain the transit decay ofexcess carriers in a

semiconductor with direct

recombination. What isminority carrier life time?


44/51



() = 2 0 + 0 + ()the rate of change of excess carrier concentration

(

) =

2

0 +

0 +

(

)

= 2 00 0+ 0 + We know that 00 = 2 = ()Then () = (0 + 0)+ 2---(2)

Consider a low level injection of excess carrier

2

(

0 +

0)

And for p-type semiconductors p0>> n0

Then () = 0---(3) Solution of this first order differential equation gives() = 0 () = ---(4)

where = 10is called recombination life time of excess electron in p-type semiconductorand also called minority carrier life time.

Similarly for excess hole recombination life time is =1

0.Steady state carrier generation Under thermal equilibrium, the generation rate is balanced by

recombination rate, so that the equilibrium values of n0 and p0

are maintained. Then = 2 = 00---(1) Consider a steady state optical generation rate gopand adding it

to thermal equilibrium, carrier concentration will increase to new steady state values. Then the

balance between generation and recombination is maintained by

+

=

---(2)

We know that = 0 + = 0 + Then (2) becomes + = (0 + )(0 + )+ = 00 + 0 + 0 + ---(3)

For direct recombination, =

Question:

Part B

Derive = = .


45/51



We have00 = 2, and using (1), (3) becomes = [(0 + 0) + 2]Neglecting 2,

=

(

0 +

0)

= where = 1(0+0)---(4) is called minority carrier life time in general.

Then excess carrier concentration = = ---(5)Quasi Fermi-levels

If excess carriers are present in semiconductors, mass-action law

is not valid.

The Fermi-level EF is the representation of thermal equilibriumcarrier concentration n0and p0. When excess carriers are present,

the carrier concentration cannot be represented by the

equilibrium Fermi-level EF.

The separate Fermi-level to represent electron and hole concentration including excess carrier are

called Quasi Fermi-levels.

They are not stable levels; it changes to EFaccording to the concentration of excess carriers.

Quasi Fermi-level for electron is represented as Fnand for hole represented as Fp. Then the carrier

concentrations can be represented as

=

()

= (

)

Recombination mechanisms Excess carrier generation is also the function on

recombination/generation rate as shown in (5).

Different recombination mechanisms are:

1. Direct band to band recombination: An electron at

conduction band minimum directly falls into hole at

valance band maximum, releasing the difference energy as a photon. This type of

mechanism is found in GaAs, InP, etc.

2. Indirect recombination: Due to the presence of impurity atoms, defects, etc. there is a

localized energy level deep in the band gap, away from the band edges. These levels act

as intermediate levels between conduction band and valance band, and this cause an

enhancement in the recombination process. The following figure shows the

recombination of EHP through a deep level.

Question:

Part A

Define quasi Fermi-level?

When do they exist?

Question:

Part A

What are the different

recombination mechanisms?


46/51



Here the energy releases according to Et. This type mechanism is found in Si, Ge etc.

3. Auger recombination: This takes place in three particle process and is possible in either

direct band to band or indirect recombination. In direct band recombination, an electron

in conduction band makes a transition to valance band and the energy of EHP is

transferred to nearby electron and is excited. The excited electron gets kinetic energy and

comes back to conduction band minimum.

4.

Trap aided Auger recombination: In this the trapped electron makes a transition to the

valance band by giving its energy to an electron in the conduction band. This process

happens only in heavily doped n-type semiconductor.

Diffusion of charge carriers:

Due to the presence of excess carrier concentration in

semiconductor, the carrier concentrations vary with position in

the sample. Then charge carriers move from high concentration

to low concentration. This type of motion is called diffusion and

the current due to this motion is called diffusion current.

The spreading of pulse of electrons by diffusion is shown below:

If a pulse of electron injected at t = 0 and x = 0, it will spread out in time. At x = 0 excess

electrons are concentrated and time elapses these excess electrons spread out and finally the

carrier concentration becomes uniform throughout and n(x) become constant.

Derivation for diffusion current:

Consider one dimensional distribution of electrons n(x) as shown below:

Question:

Part B

Explain diffusion process.

Derive expression for diffusion

current.


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Divide x into segments of width , which is referred as mean free path with n(x) evaluated at thecentre of each segment. The electrons in any segment have equal chances of movement towards

left or right.

Consider two segments with electron concentration n1and n2respectively.

Total number of electrons in segment 1 = 1Total number of electrons in segment 2 = 2where A is area of cross section of segment.

The net number of electrons crossing x0 from left to rightis one mean free time

=1

2(1 2)

The rate of electron flow in +ve x-direction per unit are (electron flux density) is given by

() =

=

12

(1 2) =

1

2(12) ---(1)

is small differential length, then (1 2)can be written as


48/51



where x is the centre of segment 1 and x = In the limit of small x we can write

= () ---(2)where = 22---(3) is called electron diffusion constant in cm2/s.

Similarly = () ---(4)where Dpis called hole diffusion constant Then the diffusion current density due to the flux density of the particle can be written as

=

Using (2) = + () ---(5)Similarly = + = () ---(6)

The total current in semiconductor is the sum of diffusion and drift current due to electron and

hole and is

---(7)

The total current density

---(8)

The direction of current flow and carrier flow in a given electric field and carrier gradient is

shown below.


49/51



Energy band diagram in a semiconductor with electric field

The energy bands of a uniformly doped n-type

semiconductor under thermal equilibrium condition

with an electric field along +ve x-direction bends

upward. It is due to the gradient in the electron energy.

Energy bands are flat when the applied field is zero.

Einsteins relations:

Consider semiconductor under electric field with

intensity () in x-direction, then the electrostaticpotential is given by

=

---(1)

Then the potential due to electron is = () ---(2) By definition the electric field intensity

=1

+ () ---(3)We know that all energies in the energy band bend equally in the given field, then

(

)

= = = Then we can write = 1 ---(4) Under thermal equilibrium, Jn = Jp = 0, but there is a diffusion of carrier due to

gradient and setup an electric field results a drift of carrier. Then we can write (in

case of electron) = +

Question:

Part A

How do the energy band

diagram change withapplication of electric field?

Question:

Part B

State and derive Einsteinsrelation.


50/51



= 0+ 0 = 00 = 0

=

0 0

---(5)

Using (4)

1 = 0 0 ---(6) We know that 0 = ()

Then 0() = () 0 = () 1 = 0() 1

---(7)

Substituting (7) into (6)1 = 0 0() 1

= ---(8)We know that EFhas no gradient and

= 0 Then (8) becomes

=

= ---(9)Similarly = ---(10)In general = ---(11)This is shows that ratio between diffusion coefficient and mobility of carriers under

thermal equilibrium is a constant and equivalent to thermal voltage kT/q in V

This relation is called Einsteins relations.


51/51


Continuity equation:

(Assignment No. 1)

Semiconductor Devices Module 1 Notes

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Transcript of Semiconductor Devices Module 1 Notes