Notes on Semiconductor Physics for Electronic Devices

Notes On Semiconductor Physics for

Electronic Devices

Kenneth H� Carpenter

Department of Electrical and Computer Engineering

Kansas State University

September ��

Introduction

The following notes are a supplement on semiconductor physics for use in a �rst course in electronics�where the principal textbook used may have only a brief section on device physics� For more completepresentations of semiconductor physics the reader should consult one of the works given in the bibliog�raphy below� The small volumes in the series by Pierret and Neudeck are especially recommended fora student wanting more detail on the topics presented here� �These are new enough to be available forpurchase� The other works would need to be found in a library��

Bibliography

�Pierret� Robert F� Pierret� Semiconductor Fundamentals� �nd� ed�� Vol� of the Modular

Series on Solid State Devices� Robert F� Pierret and Gerold W� Neudeck� editors�Addison�Wesley� ��

�Neudeck� Gerold W� Neudeck� The PN Junction Diode� �nd� ed�� Vol� � of the Modular Series

on Solid State Devices� Robert F� Pierret and Gerold W� Neudeck� editors� Addison�Wesley� ��

�Az�aro � Leonid V� Az�aro � Introduction to Solids� McGraw�Hill� ��

�Millman�Halkias� Jacob Millman and Christos C� Halkias� Electronic Devices and Circuits� McGraw�Hill� ��

�Ferry�Fannin� David K� Ferry and D� Ronald Fannin� Physical Electronics� Addison�Wesley� ��

�Kittel� Charles Kittel� Introduction to Solid State Physics� fourth edition� John Wiley �Sons� Inc��

�� Semiconductors in Equilibrium


�� Introduction to solids

Solids come in three classes�

Amorphous� no long�range order

Polycrystalline� order within grains

Single crystal� regular� repeated pattern

We will devote our attention to the ideal single crystal at �rst� It can be represented as a repetitionof a unit cell� The array of corners of unit cells makes up the crystal lattice� These three�dimensionallattices have symmetry and the particular class of symmetry is used to classify the crystal� While manytypes of symmetry are possible� we will only consider cubic symmetry� as it is the type found in theprincipal semiconductors�

There are three subclasses of cubic crystals�

Simple� the sites for atoms are at the corners of the cubes

bcc� body�centered�cubic with sites at the corners and at the center of the cube

fcc� face�centered�cubic with the sites at the corners and at the centers of each face of the cube

These types are shown in Fig� �� In addition� these arrangements can be complicated by having morethan one atom associated with each site in an orderly way� giving rise to the di erent crystal structures�The two structures we will study are the diamond structure and the zincblende structure� which are thestructures of Si and GaAs� respectively� These structures have atoms on fcc sites and on a second set ofidentical sites displaced �� of the cube diagonal along a diagonal from the �rst sites� They are shownin Fig� ��

With the structure of a crystal known� one can calculate the length of the side of the unit cell andthe distance between centers of nearest neighbor atoms if one knows the chemical composition of thecrystal and its density�

�� Miller indices

In a crystalline solid� certain directions and certain planes have physical properties di erent than others�Thus when making an integrated circuit� for example� one wants to cut the crystal along a particularplane relative to the crystal structure� These planes and directions can be determined in a crystal bymeans of x�ray di raction� The speci�cation of particular planes and directions is made through theuse of Miller indices�

To �nd the Miller indices of a plane in a crystal� choose coordinate axes along the principal directionsin the crystal �along the unit cell edges�� Take the three numbers that result as the intersection of aplane with the three axes� take their reciprocals� and multiply the results by the smallest value thatwill give three integers� These three integers are the Miller indices of the plane�

To �nd the Miller indices of a direction in a crystal� take the three components of a vector in thatdirection along the three axes� Multiply these components by whatever is needed to reduce them to thesmallest set of integers� and these integers are the Miller indices of the direction�

Miller indices of a plane are placed in parentheses� If one of them is negative� the minus sign isplaced over the integer instead of in front of it� �� A set of equivalent planes has the Miller indicesplaced in braces� f��g� Directions are placed in square brackets� �� while equivalent directions areplaced in angle brackets� h��i�

�� Semiconductors in Equilibrium �

Face Centered Cubic

Body Centered CubicSimple Cubic

Fig� �� The three types of cubic crystalline symmetry�

Diamond Structure

Zincblende Structure

Fig� �� The diamond and zincblende crystal structures�

�� Problems

� Given that Si has atomic weight �� and density �� gm�cc� �nd the spacing between centers ofnearest neighbor atoms in the Si crystal and �nd the length of the side of the unit cell� �Avogadro�snumber is ��


�� Salt �NaCl� forms crystals of sodium and chlorine ions alternating at the corners of simple cubes�Find the Miller indices of a plane in NaCl that contains only Na atoms� but has the highestpossible density of them in the plane�

�� Purity

Crystals can have two types of imperfections� defects and impurity atoms� The latter are occasionalatoms of types other than the ones that make up the de�ned crystal�s unit cell� The former are placeswhere the ideal unit cell repetition is interrupted in some way� Both can have undesirable e ects in asemiconductor� unless they are intentionally introduced� Pure semiconductor material can be preparedso that� e�g�� impurity atom exits for �� Si atoms��Pierret� page �� Since this is a three�dimensionalcrystal� to help visualize the concentration of impurities� think of a piece of material in the shape of acube having �� atoms along each side� Within this cube there would be one impurity�

Intentionally adding impurities is called doping the semiconductor� and the added material is calledthe dopant�

�� Energy levels

�� Single atoms

A single� isolated atom has its electrons occupying states that are determined by quantum mechanics�Only one electron can occupy a given state� This is called the Pauli exclusion principle� The di erentstates have di erent energies associated with them� In the ground state an atom has all of its electronsin the lowest possible energy states� When one or more electron moves to a state of higher energy theatom is said to be in an excited state� When an electron makes a transition to a lower energy state�the energy lost is given o as a photon� The photon has a frequency� f � that is related to the energylost� W � by W � hf � where h is Planck�s constant� �There are many rules that determine when sucha transition is possible � we will not go into these�� The electrons form shells about the nucleus ofthe atom� The electrons in a shell have similar energies� The number in each shell is determined bythe rules of quantum mechanics� These rules are most easily found by studying a single electron atom�either hydrogen or another with all the electrons but one removed or one in which the nucleus plus theelectrons in closed shells are treated as if a new nucleus of lower charge�� Consider the H atom�If one uses the formula for the electrostatic potential energy of the electron�proton combination� one

�nds the value� P�E� � �q�

��r� The force attracting the electron to the proton can be found from

Coulomb�s law and set equal to the mass times the centripetal acceleration� according to Newton�s law�

to obtain mv�

r � q�

��r�� Since the kinetic energy is K�E� � �

�mv� one �nds that K�E� � � ��P�E��

Thus the total energy is just one�half the potential energy�The contribution of quantum theory is to require that an electron have an angular momentum that

is an integer times Planck�s constant divided by two pi� or mvr � n�h where n is the integer� called theprincipal quantum number� ��h � �� eV�sec��

It takes quite a bit of algebra to do it� but these formulas can be combined to yield a formula for thetotal energy of an electron in terms of the quantum number n as the only variable� W � K�E�� P�E�

W � � q�m��h�

�n� �

The derivation above gives the energy of an electron in terms of a quantum number n� However�quantum mechanics assigns two more quantum numbers to each electron in addition to a spin value�For each n� there is a second quantum number l which can have any value from � to n� � For each lthere is a third quantum number m that can have the values �� l� For each set of the possiblevalues of n� l� and m there can be two electrons� one for each possible spin� Thus for n � there canbe � electrons� for n � � there can be � electrons� and so on� This is the basis in quantum theory forthe periodic table in chemistry�


�� Problems

� Find the frequency of a photon emitted when the electron in a hydrogen atom changes from then � � to the n � state� The constants in the energy level formula for hydrogen evaluate toW � ��eV�n�� Note that eV is a unit of energy equal to that gained by an electron beingmoved through a potential di erence of one volt� It is equal to �� Joules��

�� Valence

The normal situation in a single atom is for all of its electrons to be in the lowest possible energy levels�Thus if the number of electrons is such that the highest energy level occupied is entirely �lled� thereis no way for another electron to be associated with the atom without going to a signi�cantly higherenergy� Such atoms do not form compounds easily� and make up the last column in the periodic table��For atoms with principal quantum number greater than one� the energy is split signi�cantly betweenthe l quantum numbers � hence the complications of the periodic table�� When the highest energylevel is not �lled when all the electrons in an atom are at the lowest level� these levels can be used byelectrons from di erent atoms� The number of the electrons in the highest energy state in the singleatom alone is called the valence of the element� When this number is greater than four� the valence isoften stated as the number of electrons in the highest level minus eight �the number needed to �ll thelevel�� The fact that up to eight electrons can occupy the highest energy level allows a lower energystate to be formed overall if electrons are shared �or transfered in ionic compounds� between these levelsin neighboring atoms to bring the total number associated with the highest level to eight for each� Thislowering of overall energy is responsible for chemical bonding and also responsible for the formation ofcrystals�

�� Crystals

When the atoms of an element or a compound arrange themselves in the regular pattern of a crystal�there is a chemical bonding between neighboring atoms� The regularity of the pattern causes thequantum mechanical energy levels of the single atoms to be changed into continuous bands of allowedenergies� The lower energy levels merge into what are called the valence bands while the highest levelpartially �lled �or not �lled at all� is called the conduction band� Metallic conductors have electrons inthe conduction band when in the lowest energy state� Since these electrons can move to other allowedquantum states without changing their energies by a signi�cant amount �since the band is not full�they can move under the in�uence of an applied electric �eld� Hence the material is a conductor�Semiconductors and insulators have their highest energy band completely �lled when in the lowestenergy state� An amount of energy equal to the separation of the top of the valence band from thebottom of the conduction band must be added to allow an electron into the un�lled band where it canmove in response to an electric �eld� Thus these materials are perfect insulators in their lowest energystate� However� �nite temperatures in the material means that there is energy of thermal motionavailable which can cause electrons to be raised to the conduction band� How good an insulator amaterial is depends on the relative size of the thermal energy and the energy di erence between thebands� Such materials that are not good insulators are termed semiconductors� A qualitative sketch ofthe energy bands of a solid is shown in Fig� ��

The concept of energy bands is the one used to make quantitative calculations regarding the electronsin a crystal� An alternate visualization can be made for qualitative discussions� This alternate view iscalled the bonding model� and is illustrated on the right in Fig� �� While the actual shared electronsbetween atoms in a crystal of semiconductor are arranged symmetrically in three dimensions� the bondsare shown in a two dimensional array in the bonding model� Each circle in the model represents thecore of the atom while each line represents one shared electron� Breaking of bonds can be modeled byremoving the line of the removed electron and showing a circle with a minus in it to represent it as onefree to move� The missing bond line models the hole left behind�


.

.

.

.

.

.

Bonding Model

-

+(hole)

0

-(lowest state)

(free electron)

Energy BandsFig� �� A qualitative description of atomic bonding in solids� On the left� the energy levels of an atom

are expanded into the band structure of the solid� On the right� the bonding model for holesand electrons is given�

�� Materials and carriers

The discussion that closed the last section noted that when thermal energy is added to a lattice ofa crystal that has no electrons in the conduction band when in its lowest energy state� some of theelectrons involved in the chemical bonding between the atoms will leave their places in the valence bandand move into the conduction band� Any electrons in the conduction band are called carriers sincethey can carry electrical charge from place to place� In the same way� the holes they leave behind in thevalence band can move from atom to atom� and so they too serve as carriers� The concept of holes ascarriers may seem strange at �rst� but this is a common way of looking at things in quantum mechanics�The positron was �rst postulated to exist as a �hole� in a sea of negative energy electrons� From thestandpoint of the mathematical model� a hole is just as good a particle as an electron� Thus we willbe talking about two oppositely charged entities which carry currents in semiconductors � holes andelectrons�

Materials deemed good conductors have such an atomic number that in their crystalline form thereare more electrons than required to �ll the valence band� and hence one or more electrons per atom inthe conduction band� Thus there are many negative carriers in a conductor at any temperature�

Materials deemed good insulators have their valence band full and their conduction band emptyof electrons at the lowest energy state� Further� the energy gap� or band gap between the top of thevalence band �on an energy level diagram� and the bottom of the conduction band� EG� has a valuethat is large compared to the thermal energy associated with lattice vibrations at room temperature�


Thus few bonds will be broken at room temperature� and the number of conduction electrons will bemuch less than one per atom� �Less by many powers of ten� in fact��

Materials deemed semiconductors are similar to insulators except that the band gap energy� EG� isnearer to the thermal energy associated with lattice vibrations� Thus there will be more broken bonds�and although the number of conduction electrons and holes will be much less than one per atom� therewill still be enough of them to conduct a signi�cant current when an electric �eld is applied�

The concept of thermal energy of the lattice has been mention several times above� The kinetictheory of heat shows that each degree of freedom in a many body system in thermal equilibrium has�on the average� the same amount of kinetic energy� This kinetic energy has the value �

�kT where T isthe absolute temperature �degrees Kelvin� and k is the Boltzmann constant� The numerical value ofk is �� eV�K� Another conversion factor which is worth remembering is that �� degreesKelvin is equal to one electron volt� i�e�� k � � ��K � �� J� This yields the value of kin Joules per degree Kelvin as �� At room temperature� which is near ��K� we �nd theenergy characteristic of lattice vibrations is kT � ��eV� By contrast� the value of EG in typicalsemiconductors is near eV� while for insulators the value of the band gap is likely to be greater than�eV� While this may not appear to be much of a di erence between insulators and semiconductors� thisvalue appears in an exponent in the theory of thermal excitation �to be discussed below� � hence itschange is greatly magni�ed in the number of broken bonds�

�� Semiconductor doping

When a pure semiconductor is cold enough there are no carriers� At room temperature there will beequal numbers of holes and electrons that serve as carriers but their density will be much less than oneper atom� Such a material is called an intrinsic semiconductor� For example� in Si there are about�� atoms per cubic centimeter �cc�� There are � valence electrons per atom and � bonds per atomin the crystal� or a total of � � �� bonds per cc� At room temperature there will be approximately�� broken bonds per cc �and an equal number of electrons and holes�� This is only one broken bondper �� atoms��Pierret� page ��

Pure intrinsic silicon is not a very good conductor� It can be made to have a larger number ofcharge carriers by intentionally adding impurity atoms� This is called doping and the impurity addedis called the dopant� When impurities from the column in the periodic table to the right of Si �valence�� replace some of the Si atoms in the lattice then the extra electron in each impurity atom cannot�t in the valence band� but occupies an energy level just slightly below the bottom of the conductionband� Thus at room temperature nearly all the extra electrons are in the conduction band� and thusthe concentration of carriers can be controlled by the concentration of dopant atoms� �Note that thepositive ion left behind when an electron from such an impurity atom moves away from it cannot moveitself� Thus there is no hole to correspond to the electron� Doping with such impurities that raise thenumber of electron carriers but not the number of holes yields what is called n�type semiconductor�The n stands for �negative� since the electrons carry a negative charge� The symbol n by itself is usedfor the density of electrons in the conduction band �for both doped and undoped semiconductors�� Asemiconductor that has been doped is called an extrinsic semiconductor� A dopant that causes thenumber of electrons in the conduction band to increase is called a donor since it �donates� an electron�

If the impurity added to a semiconductor to make it extrinsic comes from the column of the periodictable to the left of Si �valence �� then where such an atom replaces Si in the lattice there is one twofew electrons to complete the bond� This is represented in the energy level diagram as a �hole� in thevalence band� near its top� That is� there is an un�lled electron level present� This un�lled level� atroom temperature� acts just like a hole created by the breaking of a bond� There is no correspondingelectron carrier� however� Thus the hole concentration� given by the symbol p is increased by the densityof the acceptor dopant� A semiconductor having acceptor doping is called p�type�

Obviously the density of dopants must be much less than the density of the Si �or other intrinsicelements� in the lattice or else the crystal�s basic structure would be changed and the energy levelswould all be di erent and one would no longer have a semiconductor� However� dopant concentrationscan be as high as �� per cc in Si� which is about one in �fty thousand atoms� before the doping


reaches what is called the �degenerate� state� Even higher levels of doping can be done while retainingthe basic structure of the crystal��Pierret� page ��

There are two more terms to de�ne related to carriers� The carrier �hole or electron� having thegreater density in the particular semiconductor is called the majority carrier� The other is called theminority carrier� Thus in p�type� extrinsic Si� the majority carriers are holes and the minority carriersare electrons�

�� Carrier properties

The charge carriers in a semiconductor have properties that are used in quantitative calculations� The�rst property is the density� The symbol for the density of holes is p� The symbol for the density ofelectrons is n� The densities in an intrinsic semiconductor of the same type as the extrinsic one beingconsidered are represented by the same symbols but with an i subscript� ni and pi� One notes thatni � pi is always true�

The second property is the electric charge� If q stands for the magnitude of the charge on an electron�q � ��C� then the charge of a hole is q and that of an electron is �q� �q is used here instead ofthe more familiar e� since e will be used in exponential relationships as the base of natural logarithms��

The mass of a carrier is another important property� A free electron has a massm � ��kg�Due to the interaction of a moving electron with the lattice� in a solid the conduction electrons act likethey would in a gas except the free electron mass is replaced by an e ective mass� The holes also act likefree particles in a gas but with an e ective mass��Kittel� page �� The e ective masses are representedby placing an asterisk on the m� In Si the values for electrons and holes at ��K are� respectively�m�

n � ��m and m�

p � ��m��Pierret� page ��

�� Problems

� Write a sentence to de�ne each of the following terms�

�a� Dopant

�b� Donor

�c� Acceptor

�d� n�type material

�e� p�type material

�f� Intrinsic semiconductor

�g� Extrinsic semiconductor

�h� Conduction band

�i� Valence band

�j� Majority carrier

�k� Minority carrier

�� Distributions and densities

The previous discussion has established the existence of charge carriers on a qualitative basis� Nextwe turn to the number of carriers and their distributions in energy quantitatively� Recall that isolatedatoms can have only one electron occupy a quantum mechanical state that is de�ned by a set of quantumnumbers �including the electron spin�� When the atoms are packed into the regular relationship of thecrystal lattice the same is still true � there can be only one electron for each distinct quantum mechanicalstate� The number of such states becomes too large to be accounted for individually� and so a density ofstates is used instead� The density is given as a function of electron energy� g�E�� The meaning of thedensity function is that within the range of energies E to E � dE there are g�E�dE distinct quantumstates for electrons to occupy per unit volume of the semiconductor� Calculation of the formulas forthe energy densities in the di erent energy bands is the job for an expert in quantum mechanics� Wecan only quote the formulas here��Az�aro � page ��Millman�Halkias� page �� The density of states inthe valence band�

gv�E� �m�

p

q�m�

p�Ev �E�

��h�� E � Ev


and the density of states in the conduction band�

gc�E� �m�

n

p�m�

n�E �Ec�

��h�� Ec � E�

In the formulas above� Ec is the energy at the bottom of the conduction band while Ev is the energyat the top of the valence band� �Since there can be several energy bands associated with the electronsin the core of the atom� the one we are refering to as the valence band is the band with the highestenergies that is completely �lled in the ground state of the system��

Note that from these formulas the number of states per unit energy starts at zero at the edge of theband and then increases along a parabolic curve as the energy moves away from the band edge� Thebehavior of the density of states is not modeled by these formulas when one approaches the oppositeedges of the bands� We are only interested in states near the edges next to the band gap of thesemiconductor�

The existence of states does not say anything about how many actually have electrons in them ata given temperature� For that information we turn from quantum mechanics to statistical mechanics�This is the discipline that makes statements about the probability of a given state being occupied by aparticle at a given temperature� Again� the derivation of the formula is beyond the scope of this course�However the formula is a famous one� called the Fermi distribution��Kittel� page �� This distributionis described by the Fermi function�

f�E� �

� e�E�EF ��kT

where EF is the Fermi level or Fermi energy and where the function is interpreted to mean that theprobability that states with energies between E and E � dE are �lled is f�E�� This probability appliesonly to the case of thermal equilibrium conditions when the absolute temperature is T � �k is theBoltzmann constant��

Note that the Fermi function approaches the limiting form of a step function when T approacheszero� Note that for non�zero temperature the Fermi function has the value one�half at E � EF � anddecays exponentially as energy increases above the Fermi energy�

With the density of states known and the probability that a state contains an electron known� wecan calculate the density of electrons and holes per unit energy �and per unit volume� as the productof the two� gc�E� f�E� is the density of electrons in the conduction band� and gv�E� � � f�E�� is thedensity of holes in the valence band�

�� Problems

For each of the following� label axes with appropriate units and show speci�c values along the axes�

� Make a sketch of f�E� versus E�kT for the case EF � �kT �

�� Make a qualitative sketch of gc�E� f�E� vs E for EF � Ec�

�� Carrier charge densities

While the densities per unit energy and per unit volume of electrons and holes are important to know�we also need to know the total volume charge density in the semiconductor� To �nd this we multiplythe total density of electrons n times the charge �q and add it to the total density of holes p times thecharge �q� The values of n and p are found by integrating the distributions in energy over the allowedenergies�

p �

Z Ev

Ebot

gv�E�� f�E��dE

n �

Z Etop

Ec

gc�E�f�E�dE�


Because of the complicated nature of the Fermi function� these integrals cannot be carried outanalytically in terms of elementary functions� Instead they can be expressed in terms of a specialfunction called the Fermi�Dirac integral� In the most general case of possible values for the Fermienergy this special function would have to be used� However� for the special case where the Fermienergy satis�es the restriction�

Ev � �kT � EF � Ec � �kT� ��

the integrals can be closely approximated by elementary functions� leading to the results

p � NV e�Ev�EF ��kT ��

n � NCe�EF�Ec��kT ��

where

NV � �

��m�

pkT

h�

��

NC � �

��m�

nkT

h�

��

The condition of eq�� is the degeneracy condition� When it is satis�ed� the semiconductor is saidto be nondegenerate and when it is not satis�ed� the semiconductor is said to be degenerate� We willrestrict our formulas to ones that apply only to nondegenerate semiconductors� �The degenerate case isalso of interest� but due to the advanced mathematics required� it cannot be considered in these notes��

�� Problems

� Find the numerical values of NC and NV in Si at T � ��K�

�� Relationships between n� p� and the Fermi level

The formulas found above require knowledge of the Fermi energy before actual numbers can be cal�culated for n and p� In the following we will try to �nd ways to use material properties� such asconcentration of dopants� to obtain the desired values�

�� Intrinsic semiconductors

In an intrinsic semiconductor the values of n and p must be equal� For any semiconductor� let us de�nethe corresponding carrier densities in the same material at the same temperature but without anydopants present as ni and pi� These are the intrinsic electron and hole densities� Since these must beequal� we can see from eq� �� and eq� �� that if NC � NV then EF must lie half�way between the loweredge of the conduction band and the upper edge of the valence band� EF � �Ev �Ec�� However� thee ective masses are not quite equal� so this relationship does not hold� Instead� if the e ective mass ofelectrons is higher than that of holes� the Fermi level must be slightly below the average of the bandedge energies� The value of the Fermi energy in the intrinsic material is given the symbol Ei� Ei � EF

in an intrinsic semiconductor�By using eqs� �� and �� along with the same equations with EF replaced by Ei� one can substitute

for NV and NC to obtainp � nie

�Ei�EF ��kT ��

andn � nie

�EF�Ei��kT � ��

Alternately� one can eliminate the Fermi energy from the formula for ni to obtain

ni �pNVNCe

�EG��kT ��


where the band gap width is EG � Ec �Ev�From eqs� �� and �� one has immediately �for the nondegenerate case� the relationship

np � n�i � ��

�� Problems

� For Si� EG � ��eV� Using the other properties of Si� given above� �nd ni in Si at ��K�

�� Repeat the previous problem for GaAs� given that EG � ��eV andm�

n � ��m�mp � ��m�

�� Relationships between n� p� doping� and temperature

De�ne ND to be the density of donor atoms in the semiconductor and NA to be the density of acceptoratoms in the semiconductor� Then for room temperature these donors and acceptors will be totallyionized� creating an equal number of electrons and holes� respectively� Unless stated otherwise� we will

always assume total ionization of the donors and acceptors in the following discussion� The requirementof conservation of charge will then yield

p� n�ND �NA � ��

If one uses eq� � to eliminate p from eq� �� one obtains a quadratic equation for n in terms of ni�ND� and NA� Solving this quadratic yields

n �ND �NA

��

��ND �NA

�

��

� n�i

��

which then� with eq� � gives a similar result for p with NA and ND interchanged�

p �NA �ND

��

��NA �ND

�

��

� n�i

��

From these last two equations we can identify several special cases of interest�

Intrinsic semiconductor � When NA � ND � � the material is intrinsic� and n � p � ni�

Compensated semiconductor � When NA � ND �� then the values are again n � p � ni� Thispoints out that it is not the concentration of donors or acceptors that determines the semiconductorproperties but only the di erence between their concentrations�

Doped� p�type semiconductor at normal temperature � When NA �ND � ni� p � NA �ND

and n � n�i �p�

Doped� n�type semiconductor at normal temperature � When ND �NA � ni� n � ND �NA

and p � n�i �n�

Doped semiconductor at elevated temperature� As the temperature is increased from the levelwhere the dopants are all ionized and the dopant concentration far exceeds the intrinsic carrierconcentration� to a higher values� the ni term in eqs� � and �� begins to in�uence the valuesof p and n� When a high enough temperature is reached that ni � jNA � NDj then� as in theintrinsic case� n � p � ni�

Note� If the temperature is high enough that the material becomes degenerate� then eqs� �and �� no longer apply� However� the result is still the same� n � p � ni at su�ciently hightemperature� The wider the band gap the higher the temperature must become for degeneracyto be reached and the higher the temperature must become for n � p � ni� When n � p � ni adevice depending on doping �such as a diode or transistor� ceases to operate�


�� Fermi levels in doped and intrinsic materials

The Fermi level in intrinsic material is de�ned as Ei� Its value is obtained in terms of the e ectivemasses for temperatures for which the nondegenerate restriction holds by combining equations �� and �� to obtain

Ei �Ec �Ev

��

�

�kT ln

�m�

p

m�

n

��

Next one takes logarithms of eq� �� to obtain EF � Ei � kT ln�n�ni�� This� along with eq� �� willyield the Fermi energy for any condition of temperature and doping� Note that the e ect of temperatureon the total number of carriers means that the Fermi level itself becomes a function of temperature�These formulas depend on the solution for the Fermi level being such that the nondegenercy conditionis ful�lled� One could work out explicit formulas for the Fermi level for each of the special cases of thecarrier densities given above� but this will be left as an excercise for the student�

�� Problems

� For a p�type semiconductor� �nd the maximum value of NA that can be used and still have thesemiconductor be nondegenerate at normal temperatures� Give the answer in terms of kT � EG�Plank�s constant� and the e ective mass for holes� assuming that the e ective mass for holes andelectrons are equal�

�� Find a numerical answer for the formula of the previous problem� assuming that the e ective massequals the electron rest mass� temperature ��K� and EG � eV�

�� Find the special case formula for the Fermi energy for n�type material that is nondegenerate andat normal temperature� in terms of ND� NC � Ec� and the temperature�

�� Dynamic Processes in Semiconductors

�� Currents

The previous discussion on semiconductors has established the concentrations of carriers in a semicon�ductor under conditions of static� thermal equilibrium� We now turn our attention the dynamics ofthe carriers � how they move when subjected to electric �elds and�or density gradients� Furthermore�the establishment of thermal equilibrium involves the breaking of bonds to give hole�electron pairs andalso the re�establishment of bonds with the recombination of hole�electron pairs� When the carriers arenot in thermal equilibrium these e ects are not necessarily balanced� Thus there are three dynamicfeatures of carrier behavior to study� drift� di�usion� and generation�recombination� The term drift

current is used for the electric current produced due to the presence of an electric �eld intensity� Theterm di�usion current is used for the electric current produced due to a density gradient � a change of nor p with position in the semiconductor� Recombination�generation does not produce a current directly�but a ects the other current processes by changing the carrier densities� Each of these e ects will beconsidered separately in the following sections� and then they will be combined to give the overall rulesfor carrier dynamics�

�� Drift

The application of an electric �eld to a semiconductor means that a Coulomb force �beyond thoseassociated with the establishment of the energy bands and the concept of e ective mass� will act on thecharged carriers� �The energy band model� along with the concept of e ective mass� has allowed thesecarriers to be considered as if they were gases of free particles acted on by any electric �eld in additionto the �elds responsible for the model itself� The additional �eld may be produced by net charge in the�gas� due to unequal concentrations of holes and electrons within a region of the semiconductor as wellas due to charges outside the semiconductor material itself��

The charge carriers are not stationary when in thermal equilibrium but are in random motionwith average kinetic energy proportional to kT � When an electric �eld is applied� the random motioncontinues� but its average velocity changes from zero to a �nite value directed along the electric �eld�When a dynamic equilibrium is reached� the average velocity of each carrier type will be a functionof electric �eld intensity� As long as the magnitude of the electric �eld intensity does not exceed alimit determined by the material� the average velocity will be directly proportional to the electric �eldintensity� This can be expressed as an equation�

vdp � �pE and vdn � ��nE � ��

where vdp is the average� or drift� velocity of holes� and vdn is the drift velocity of electrons� Thesymbol E has been used for electric �eld intensity �since E has been used for energy�� The constants ofproportionality are called the mobilities with �p being the mobility of holes and �n being the mobilityof electrons� The minus sign is explicity placed with the equation for the electrons� drift velocity sothat the mobility will be a positive number�

The current density due to a distribution of moving charges is given by the average over the velocitiesof the charges� of their charge times their individual velocities� This averaging process produces theconcept of drift velocity� Thus the current density is just the product of charge times drift velocitytimes carrier density�

Jn�drift � ��q�n��n�E Jp�drift � qp�pEJdrift � Jn�drift � Jp�drift � q�n�n � p�p�E ��

�

�� Dynamic Processes in Semiconductors �

Since the electrical conductivity � is de�ned by J � �E we have for a semiconductor

� � q�n�n � p�p��

�� Temperature and doping dependence of conductivity

The conductivity depends on both the carrier densities and mobilities� In a doped semiconductor�for normal temperatures� the density is nearly constant at the value determined by the density ofdopants� For elevated temperatures the densities increase exponentially� For intrinsic semiconductorsthe densities increase exponentially with temperature� In any case� the densities either remain �xed orincrease with increasing temperature� This tends to increase conductivity with temperature�

The mobility represents the ease with which electrons or holes can move through the lattice� Whathinders the motion is collisions� These collisions can be with lattice atoms due to their thermal vibrationsand with the ions at the sites of dopants� The higher the temperature the more the motion is hindered�The higher the doping level the more the motion is hindered� But with higher doping levels the mobilityis less temperature sensitive� Thus on the basis of mobility alone the conductivity would decrease withincreasing temperature�

The overall results can be stated as follows� First� in intrinsic material there is a net increase ofconductivity with temperature� Second� for extrinsic materials there is a net increase of conductivitywith concentration of majority carriers� Third� the temperature behavior of conductivity is complicatedfor extrinsic �or compensated� semiconductors� but tends to be toward increasing with temperature�

�� Problems and numerical values

Typical values for mobilities��Pierret� page ��For n�type Si at ��K having ND � ��cm�� n � ��cm��V�s�For p�type Si at ��K having NA � ��cm�� p � ��cm��V�s�Note that the mobility of holes is less than that of electrons for the same doping level� This is typicalof semiconductors�

In GaAs material the mobility of electrons is signi�cantly higher than in Si�

� Find the conductivity of n�type Si having the values given above�

�� Find the conductivity of p�type Si having the values given above�

�� Diusion

In semiconductors� in addition to ohmic conductivity� represented by �� there is also a current dueto di usion if the density of carriers is not uniform throughout the material� The thermal motioncauses carriers to move away from their present locations� Thus� on average� the regions with greaterdensities gradually have their densities reduced while the regions with lower densities gradually havetheir densities increased� This process is called di�usion and since it implies a net �ow of electrons andholes from regions where their densities are higher to regions where their densities are lower� it alsoimplies the existence of a current� called the di usion current� which is not proportional to the electric�eld intensity�

�� Di�usion equation

Di usion is expressed quantitatively by equating the �ux density of the particle �ow to a constanttimes the gradient of the particle density� When the particles are charged� the current density is the�ux density of the particle �ow multiplied by their charge� Thus

Jp�diff � �qDprp and Jn�diff � qDnrn ��


Here Dn and Dp are the di usion constants for electrons and holes� respectively� Both are positivenumbers� Both are constants for reasonably wide ranges of density gradients� Both will depend ontemperature and doping levels �as will be seen below��

Eq� �� is a vector equation� Current density is a vector �as is electric �eld intensity�� and the deloperator� r � x��x� y��y � z��z� converts a spatial dependence of charge density into a vector�When one wishes to simplify calculations for illustrative purposes� or when the geometry of a problem issimple enough that a linear symmetry is valid� the di usion equation can be reduced to one dimensionby assuming that the density varies only along one direction� Suppose this direction is the x direction�Then the del operator in eqs� �� r� will be replaced by the derivative with respect to x� ��x� andthe equation is reduced from a vector one to a scalar one for the x component of the current density�

�� Net current density

When both electric �elds and density gradients are present in a semiconductor� the net current densityis given by

Jp � qp�pE � qDprp Jn � qn�nE � qDnrnJ � Jp � Jn � q��p�p � n�n�E �Dprp�Dnrn��

This expression can be simpli�ed for the various special cases where certain terms are dominate� Forexample� in n�type material at normal temperatures� the �rst term in eq� �� can be omitted and thevalue of n in the second term replaced by ND �NA� Both di usion terms must be retained� however�since the gradient of minority carriers may be signi�cant� even if the minority carrier density is not�

�� Current density contribution to changes in carrier concentration

The formulas for current density depend on carrier densities� but do not directly show how currentdensity itself a ects the densities� This can be shown by means of the electrical continuity equation�r � J � ��

�t � Since the charge density for each species is just the charge per particle times the particledensity� one obtains

�p

�t

��drift

�r � Jp�drift

�q � ��

along with similar equations for �n��t due to drift and for both p and n�s derivatives due to di usion�

�� Relationship between di�usion constant and mobility

The two phenomena of drift and di usion are related by an equation named the Einstein relationship�We will consider a simpli�ed derivation of this relation in a later section� The equation that we will�nd is

Dn

�n�

Dp

�p�

kT

q��

�� Problems

For the problems of section �� above� calculate the corresponding values of the di usion coe�cients�

�� Recombinationgeneration

Generation refers to the process by which bonds are broken� creating hole�electron pairs� It requiresenergy to raise a valence band electron to the conduction band� This energy must come from thermalvibrations or from another source� such as a photon of radiation� Recombination is the process of anelectron returning to the hole in the valence band from the conduction band� This is accomplished


by the release of the potential energy di erence between the two states� These processes are occuringsimultaneous at all times throughout a semiconductor� When the rate of recombination equals that ofgeneration the carrier density is in equilibrium� When the densities are not at equilibrium values� thenthe rate of either generation or recombination exceeds the other until equilibrium is established�

Recombination�generation �abbreviated as R�G� can be enhanced by the presence of certain typesof impurities �such as metal atoms in Si� or by lattice defects� by introducing energy levels for electronsnear the center of the band gap� By having these extra levels at the locations of the impurities ordefects� either generation or recombination can occur in two steps� each of which takes only aboutone�half the usual energy� These locations are known as R�G centers� or traps� and the energy levelassociated with each is given the symbol ET � �In semiconductor materials used for electronic devices�the density of traps� which are detrimental to most devices� will be small compared to the density ofdopants��

�� Dynamics of RG

When carrier concentrations are not at their equilibrium values� the R�G processes work to restore thatequilibrium� The theory gets quite complicated� but the results can be stated simply for special caseswhere the departure from equilibrium is not too great� Usually the thermal R�G process is dominatedby the traps� so the rate of return to equilibrium should be proportional to the density of the traps �R�G centers�� NT � When departure from equilibrium is slight� there will be only a small percent changeof the majority carrier density from the equilibrium value� There may� however� be a large change inminority carrier density from its equilbrium value� As long as the minority carrier density remains muchless than the majority carrier density we say there is a low level injection of minority carriers� This isthe only case we will consider� Let the non�equilibrium densities be n and p while the correspondingequilibrium values are n and p� The changes from equilibrium will be !n � n� n and !p � p� p�

Consider a n�type semiconductor for de�niteness� Then !p � n� But since np � n�i � !p canbe much greater than p� A small change in the majority carrier will have little e ect on electricalproperties� but a large change in the minority carrier can have a major e ect on electrical properties�Thus it is the dynamics of the return of the minority carrier to equilibrium that is of most interest�This is expressed mathematically by the equation

�p

�t

��therm�R�G

� �!p

�p��

which applies to holes in n�type material� or by the equation

�n

�t

��therm�R�G

� �!n

�n��

which applies to electrons in p�type material�The constants �p and �n are called the minority carrier lifetimes since they characterize the average

time a minority carrier in excess of the equilibrium lasts in the material before recombining� The � �s�in turn� are proportional to NT � We will forgo more details of this process� �See �Pierret� section ��for more details on R�G��

�� Generation by other processes

Excess hole�electron pairs can be generated by other than thermal means� One example is the productionof them by photons of incident light on the semiconductor� This is the basis of the photo diode andalso of the erasing of the EPROM integrated circuit� The rate of generation will be proportional to theintensity of the light and the area illuminated� It will also depend on the match of the wavelength ofthe light to the energy of the band gap� For our purposes we will merely assume these e ects combineto create a constant rate of generation�

�n

�t

��light

��p

�t

��light

� GL� ��


�� Combined equations mathematical description of carrier

dynamics

All of the processes of carrier dynamics discussed so far are all happening simultaneously in a semicon�ductor� They can be combined into a single equation which represents the total time rate of change incarrier concentration at a point in the material� The total time rate of change has contributions fromdrift� di usion� thermal R�G� and other R�G �such as due to light�� By combining equations for eachprocess we can obtain the single composite equation desired� Speci�cally� eqs� �� and ��along with the corresponding ones for other currents�� used with eqs� �� and �� in the following basicde�nitions�

�p

�t

��total

��p

�t

��drift

��p

�t

��diff

��p

�t

��therm�R�G

��p

�t

��light

��p

�t

��otherproc�

��

�n

�t

��total

��n

�t

��drift

��n

�t

��diff

��n

�t

��therm�R�G

��n

�t

��light

��n

�t

��otherproc�

� ��

give the results�

�p

�t

��total

� ��pr � �pE� �Dpr�p� !p

�p�GL �

�p

�t

��otherproc�

��

�n

�t

��total

� �nr � �nE� �Dnr�n� !n

�n�GL �

�n

�t

��otherproc�

� ��

�� Minority carrier di�usion equations

The general equations just derived for time rate of change of carrier densities need simpli�cation beforethey can be applied to solve for the results useful in device modeling� These simpli�cations will yieldwhat are called the minority carrier di�usion equations� The assumptions are�

� The equations apply to the rate of change of minority carriers only�

� The system is one�dimensional so that r becomes ��x�

� The equilibrium concentrations are not functions of x�

� The low level injection restriction is satis�ed�

� The �other proc�� term is zero�

� The electric �eld is zero�

While these may seem to be too many restrictions� the regions of interest for solving the resultingequations will satisfy them� The results are to reduce eqs� �� and �� to

�!pn�t

� Dp��!pn�x�

� !pn�p

�GL ��

�!np�t

� Dn��!np�x�

� !np�n

�GL� ��

�� Di�usion length

Consider the following example� A bar of n�type Si extends along the x axis� The end at x � � has lightfalling on it so that there are excess holes of the amount !pn at x � �� For x � eq� �� applies withGL � �� In equilibrium� the left side of eq� �� equals zero� Thus we have Dpd

�!pn�dx� � !pn��p�


subject to boundary conditions� !pn � !pn at x � � and !pn � � as x � The solution to thisdi erential equation which �ts the boundary conditions is

!pn�x� � !pne�x�

pDp�p � ��

From the nature of the exponential function� the constant dividing x in the exponent in eq� ��is the average distance that minority holes di use before being annihilated� Thus this constant� Lp �pDp�p� is called the minority carrier di�usion length for holes in n�type semiconductors� The analogous

problem for p�type yields Ln �pDn�n as the minority carrier di usion length for electrons in p�type

semiconductors�

�� Problems

� Using values given in section �� including T � ��K�� nd the minority carrier di usion lengthfor p�type Si having �n � �sec�

�� Repeat for n�type having �p � ��sec�

�� Energy levels revisited

There are three aspects of the energy band model that need further consideration before our study ofelementary semiconductor theory is complete� These are the e ect of electric �elds on the levels� theinterpretation of the Fermi level when not in equilibrium� and the derivation of the Einstein relation�Each of these will now be presented in a simpli�ed manner�

�� Band bending

The energy band model for electrons in semiconductors provides formulas for densities of carriers interms of characteristic energies� e�g�� Ei� Ev � etc� However� one should note that all the formulas fordensities depend on the di erences between two electron energy levels� not on the absolute value of any��EG is the di erence between two levels and so does not invalidate this observation�� This is due to thefact that the choice of the zero level for potential energy is arbitrary � it is the di erences in potentialthat have physical meaning�

The usual calculation of the band edge energies in quantum mechanics takes the zero level as thatof a single electron removed a great distance from the rest of its atom� The calculations give the bandmodel picture of constant energy levels as one moves throughout the material� But when electric �eldsexist within the material� the energy levels must be shifted by the work done in moving an electronthrough the �eld � this is a potential energy in addition to the quantum mechanical one� Thus when a�eld is present� the picture of the energy levels versus distance through the material is changed� Theelectron energy bands are �bent� in the opposite way from the value of the electric potential versusdistance� since by the de�nition of energy and potential� energy is charge times potential� Likewise� theelectric �eld is in the direction of force on a positive charge� so electric �eld is the negative gradient ofpotential� For example� suppose the electric �eld is in the �x direction� If the potential is taken to bezero at x � � it will increase as x increases" thus the electron energy levels will decrease as x increases�This is illustrated in Fig� ��

�� Fermi level in a semiconductor

The Fermi level is a property of equilibrium statistical mechanics� It is a constant that appears in theequation for the probability that an electron occupies an allowed state� This constant is independentof the electric potential and must have the same value throughout a semiconductor whose carriers arein equilibrium� Thus when band bending occurs due to electric �elds� the Fermi level will change itsdistance from the conduction band with position in the material� �Our equations will be valid only ifthe Fermi level stays far enough away from the band edges for non�degenerate conditions to remaintrue��


EF

electric field

x

Ec

iE

Ene

rgy

leve

ls

vE

Fig� �� Band bending in the presence of an electric �eld�

�� Nonuniformly doped semiconductors

Suppose the doping of a semiconductor depends on position� Then the distance of the Fermi level fromthe band edge must also vary with position when equilbrium is established� But the Fermi level mustbe constant� This means that the band edges must change level with position� But this in turn meansthat there must be an electric �eld present� Thus in equilibrium� non�uniformly doped semiconductorscontain electric �elds�

�� Einstein relationship

Consider a non�uniformly doped semiconductor in equilibrium �with no external electrical conductorsattached�� Let us suppose it is n�type for de�niteness� Then eq� �� must have the total electron currentbe zero for equilibrium� This gives �assuming a one�dimentional situation�

�nnE � Dn�dn�dx� � ��

The �band bending� relationship between electric �eld and energy levels at the edge of the conductionband is

qE � �dEc�dx��

We recall eq�� which is the fundamental relationship between band edge energy� Fermi energy� andelectron density� n � NCe

�EF�Ec��kT � Since EF and NC are constant with x� one can evaluate dn�dxfrom this latter equation to obtain

dn�dx � ��n�kT ��dEc�dx��

Substituting the value of E in eq�� and the value of dEc�dx in eq�� into eq�� and thencanceling the common factors� yields the Einstein relationship�

�� QuasiFermi levels

The concept of the Fermi energy implies equilbrium conditions� When a semiconductor is not inequilibrium it is still convenient to represent the density of carriers by a simple exponential formula�This is done by de�ning a quasi�Fermi level for electrons� Fn and another for holes� Fp� With these�somewhat arbitrary� de�nitions we can write

n � nie�Fn�Ei��kT ��

p � pie�Fp�Ei��kT ��

These de�nitions allow simpli�cation of other equations applied to semiconductors not in equilibrium�and allow visualization of non�equilbrium processes on energy band diagrams�

�� PN Junctions

�� Introduction to PN junctions

When the doping in a single piece of semiconductor is changed with position so that the piece is dividedinto two parts� one part being n�type and the other part being p�type� the region where the transitionbetween the two parts occurs is called a PN junction� The junction may be abrupt� with uniform ND inthe n�type up to the boundary between the parts and with uniform NA in the p�type up to the boundary�Such an abrupt change yields a step junction� Or the junction may be made more gradually� with thedensity of donors and acceptors gradually changing as the border between the parts is approached�This latter case is called a graded junction� The graded junction may be more reasonable physically�and from a manufacturing viewpoint� but the step junction is easier to analyze� Thus we will directattention to the step junction�

The physical semiconductor we will be considering in the following will consist of a bar of material�with p�type material in the left half and n�type material in the right half� Within each half the dopingis uniform� All changes will occur along the dimension that is the length of the bar� We will place thex axis along the length of the bar with the zero at the step junction and positive values to the right inthe n�type�

Our analysis to obtain the current�voltage characteristic of a PN junction diode constructed fromthis bar of material will require a sequence of simplifying assumptions� These assumptions will notalways be consistent with each other� but their combined e ect will yield a mathematical theory for thediode which corresponds well to the experimental case�

�� Step junction in thermal equilibrium no ohmic connections

When a piece of semiconductor containing a step junction is isolated� that is to say there are no electricalcontacts �ohmic contacts� made to it� then the electrons and holes will distribute themselves in sucha manner that equilbrium exits � no properties will be changing with time� As we have seen for asemiconductor with only a single doping throughout� the carrier densities depend on the Fermi level�srelative value compared to that of the electron energy levels at the edges of the conduction and valencebands� These relations are given in eqs� �� and �� The alternate equations for the carrier densities�eqs� �� and �� show how the Fermi level in the intrinsic material� Ei� and the Fermi level in the dopedmaterial together �x the densities� Now for the bar of material we are considereing� there must be twodi erent sets of eqs� �� and �� for the two halves of the bar� But the Fermi energy� EF � must be thesame throughout the material when it is in equilibrium� Also the intrinsic density must be the same forboth halves� Thus the energy levels of the band edges� and hence Ei �which is approximately half�waybetween the edges of the conduction and valence bands� must be di erent in the two halves� This isonly possible if there is an electric �eld present to yield a di erence in electrical potential energy to addto the energy levels of the crystal obtained from quantum mechanics� These relationships are shown inFig� ��

In the left half of the bar� the p�type material will have p nearly equal to NA� which will be largecompared to ni� Thus in the left half� the Fermi level must be less than Ei� In the right half� the n�typematerial will have n nearly equal to ND� which will be large compared to ni� Here the Fermi level mustbe greater than Ei� With the Fermi level the same in both halves� we see that the bands bend towardslower energies as one goes from left to right� passing through the junction� This� in turn� implies thatthe left half is electrically more negative than the right� since the change in energy levels for electronsdue to traversing an electric potential di erence V is !E � �qV �

�� PN Junctions ��

�Note� In the paragraph above� and throughout this discussion of the PN junction� we are assumingthat the p�type material has only acceptor dopants and that the n�type material has only donor dopants�If this is not the case then NA would have to be replaced by NAp � NDp and ND would have to bereplaced by NDn �NAn in all of the formulas��

There is a simple explanation for why the potential is negative on the p�type half of the bar andpositive on the n�type half of the bar during thermal equilibrium� The reason is that near the junctionholes will di use into the n�type and electrons into the p�type until a great enough electric �eld isestablished to counter the di usion current with an equal drift current� However� with the energy levelsthen displaced through the junction toward lower values on the right� the Fermi level near the junctionwill be near Ei� which means that the equilibrium value for n and p will be near ni� There must bean anhilation of the holes and electrons that were visualized as crossing into the opposite material inorder for the equilibrium to be created�

Once the picture of the equilibrium situation is understood� we can obtain a numerical value for thedi erence in electrical potential between the two halves of the bar� The potential di erence must bethe change in the value of Ei from one end to the other� divided by the charge on the electron� �q�Replacing p in eq� �� by NA and n in eq� �� by ND and solving for the value of Ei in each region�one obtains Eip � kT ln�NA�ni� � EF and Ein � kT ln�ND�ni� � EF � Thus the electrical potentialdi erence across the junction in thermal equilibrium conditions �taken as positive on the p�type side� is

Vpni �Eip �Ein

�q � �kT

qln

�NAND

n�i

��

�� Problems

� Find the numerical value for Vpni in Si at ��K having ND � NA � ��cm��

�� Step junction in thermal equilibrium with ohmic connections

Suppose the semiconductor bar� considered above� now has ohmic conductors attached to the extremeleft and right ends� The junction between the conductor on the left and the p�type semiconductor willconstitute another type of junction � one the details of which we cannot consider here� �There will bea potential di erence across this junction due to the di erence in carrier densities on the two sides� justas for the PN�junction�� The junction between the conductor on the right and the n�type semicondutorwill be a third type of junction� Suppose the conductors are looped around and connected together toform a closed circuit� Under thermal equilibrium� no current can �ow in this circuit �due to the �rstand second laws of thermodynamics�� Thus there must be potential drops across the junctions betweenthe conductors and the semiconductors� called contact potentials� which exactly cancel the potentialdi erence across the PN junction� These contact potentials must be taken into account when the PNjunction diode terminal characteristics are determined�

�� The depletion model

When the PN junction is in equilibrium� as described above� the density of carriers near the junction issmall � on the same order of magnitude as the intrinsic value� Since this is much smaller than the densityof dopants� the net e ect is that there is a charge density near the junction that is approximately equalto the dopant density times the charge on the ionized dopant atoms� This charge density will graduallydecrease towards the bulk semiconductor region which will have no net charge� For this reason� theregion near the junction is called the depletion region�

The depletion model simpli�es calculation of the electric �eld intensity due to the charge densityby making the assumption that within the depletion region the charge density is equal to the dopantdensity times charge � without any gradual reduction as one moves away from the junction� The chargedensity is taken to be zero in the bulk region� The width of the depletion region on each side of thestep junction is determined by the di erence in potential between the bulk regions on either side of the

�� PN Junctions �

depletion region� and by the requirement that the amount of charge in the depletion region on one sideof the junction is equal and opposite to the amount on the other side� These assumptions are not quitetrue� but they allow analytic solution for the depletion region width and the electric �eld� The result isa good enough approximation to use in the succeeding calculations for the PN diode characteristics� Abetter solution would require self consistent numerical calculations of potential and charge distributions�

�� Solution for the electric �eld in the depletion model

Gauss�s law in electrostatics� when applied to a one�dimensional system and reduced to a di erentialequation� is dE�dx � �� where is the charge density and � is the permittivity of the medium� Thisequation can be solved for the electric �eld intensity by integration for the case of constant chargedensity as E � x�� K� where K is the constant of integration which must be determined from theboundary conditions�

First condsider the bulk regions far from the junction� There the �eld must approach zero� Thus�since is zero there� K must also be zero�

Second� consider the depletion region within the p�type material� Here the value of x will be negative�The boundary between this region and the bulk region in the p�type material will be at x � �b� whereb is a positive distance� At x � �b� E � � is required� since E must be a continuous function of x� Alsoin this region� in the depletion model� � �qNA� Combining these facts yields

E � ��q��NA�x� b� ��

Identical arguments apply for the depletion region within the n�type material� except that this regionis for � � x � a� Here

E � �q��ND�x� a��

The third step is to require that eq� �� and eq� �� give the same value for E at x � �� This yieldsthe relationship between a and b corresponding to equal but opposite total charges in the positive andnegative halves of the depletion region�

NAb � NDa� ��

Eqs� �� and �� along with the fact that E is zero for x � �b and x a completely speci�esthe electric �eld intensity for the depletion model� From this knowledge of the electric �eld intensityone can obtain the potential di erence between the p�type bulk region and n�type bulk region by thefundamental de�nition of potential as the integral of electric �eld�

Vpn �

Z a

�b

Edx � � q

��NAb

� �NDa��

Eq� �� along with eq� �� allows expressing the widths of the two halves of the depletion region in termsof the potential drop across the junction�

a �

r��

q��Vpn�

rNA

ND

r

NA �ND

b �

r��

q��Vpn�

rND

NA

r

NA �ND��

Eqs� �� and �� can be solved simultaneously to obtain the width of the depletion region in thermalequilibrium� when no current is �owing through the junction�

�� Junction capacitance

�� Depletion capacitance

The PN junction� with a depletion region� acts in a manner similar to a capacitor� When the valueof Vpn is changed� the width of the depletion region is changed� When this is because of an external


voltage applied through conductors connected to the left and right ends of the bulk regions� and whenthe voltage is applied so that the net external voltage is negative on the p�type end� then the chargescovered or uncovered in the depletion region must move through the external circuit� since few chargecarriers move across the depletion region �in this case�� This is the same relationship as would be foundin a capacitor� A change in voltage produces a change in equal but opposite charges on the plates of anordinary capacitor� Here the change in charge is given �in magnitude� by AqNAjdbj � AqNDjdaj whereA is the cross sectional area of the semiconductor� One can evaluate db�dVpn by using eqs� ��

db

dVpn�

��ND

bqNA�NA �ND��

The capacitance due to this e ect is a function of the potential� and so is non�linear� However� theincremental capacitance for small variations in potential about a bias value is given by C � jdQ�dV j�where Q is the charge on one plate� V is the voltage� and C the incremental capacitance� From eq� ��and this de�nition� we �nd

CJ ��A

b� �NA�ND��

�A

a� b� ��

Note that this result is the same as if we had used the formula for a parallel plate capacitor having aspacing between plates as given by the depletion width of the PN junction� This phenomenon is usedto make electrically variable capacitors �called variactor diodes� for use in electronic circuits�

�� Di�usion capacitance

The junction capacitance� discussed in the last section� depends on the change in width of the depletionregion with change in junction potential di erence� However� when the diode is carrying a forwardcurrent there is another capacitance e ect due to the presence of minority carriers in the depletionregion which must be swept out before conduction can cease when the applied potential is reversed inpolarity� This e ect produces a capacitance called the di�usion capacitance which is proportional tothe value of the forward current� and can be much larger than the junction capacitance� We will omitthe derivation of this capacitance� but with the simplifying assumption of one of the dopants being atmuch higher concentration than the other it can be found to be�Millman�Halkias� page��

CD � �Iq

kT� ��

where � is the lifetime of the minority carrier assumed to be carrying nearly all the current I � Whendoping is more nearly equal for both p and n regions the derivation and the result are much morecomplicated��Neudeck� pages ��

When the current is not zero� then the e ective capacitance the diode presents to an external circuitis the sum of CD and CJ �

�� Problems

� Find the width of the depletion region for Si at ��K if NA � ND � ��cm�� Assume nocurrent �owing��

�� Find the capacitance of the junction in the previous problem if the cross sectional area of thejunction is mm�� and given that � for Si is �� where � � �� farads�meter��

�� Find the di usion capacitance for a diode that is forward biased and carries mA of current�assuming the n�type material is much more heavily doped than the p�type� and that the resultinglifetime of minority carriers in p�type is �S�


�� The I�V characteristic of a PN junction

�� Qualitative considerations

When an external voltage is applied to ohmic leads connected to the two sides of a PN junction� thejunction acts as a �non�ideal� diode� When the negative side of the applied voltage is connected to thep�type material� then the result is that the depletion region widens until the junction potential �alongwith the contact potentials� just balances the external voltage� and so no current �ows �ideally�� Iftoo large an external voltage is applied� then the high electric �eld may break down the material� emi�e�� more carriers will be generated� and reverse conduction will occur� In any case there will be comesmall reverse current as will be seen in the quantative formulas below��

When the positive side of the applied voltage is connected to the p�type material� then the depletionregion will narrow slightly and the junction potential decrease� However� this narrowing will changethe equilibrium so that a large current will �ow across the junction� The current will be seen toincrease exponentially with di erence between the magnitude of the junction voltage without the appliedpotential and that with it present� Thus the ohmic losses will be come su�ciently large that the actualvoltage across the junction will always be in the reverse direction� �If the voltage were so large that thiswere not true� the diode would be destroyed#� That the current will increase exponentially with changein junction potential can be see from all the formulas for carrier densities depending on potential in anexponent�

�� Assumptions needed for quantative derivation

The current�voltage characteristic for a PN junction can be calculated based on the information wehave obtained in the preceding sections� along with a few additional assumptions� We now remove theassumption used in the depletion model that the density of carriers is zero throughout the depletionregion� but still keep the value for the electric �eld found using that assumption� The self consistentelectric �eld intensity that would be found with this assumption removed� will be qualitatively the sameas the one derived using the depletion model� but the details will be di erent� The electric �eld muststill vanish when one is su�ciently far removed from the step junction� and this region where the �eldis zero is what will be called the bulk region�

The assumptions needed to complete the derivation of the I�V characteristic are��Neudeck� page ��

� E is zero in the bulk regions�

� There is no recombination of holes and electrons in the depletion region�

� The minority carrier densities in the bulk regions remain at low level injection values�

� Dynamic equilbrium is established so that all time derivatives are zero�

� The total hole and electron currents within the depletion region are much less than the individualdrift and di usion currents �which are in opposite directions��

The last assumption seems to contradict the assumptions of the depletion model� However� witha current �ow assumed through the depletion region� where drift and di usion currents will be inopposite directions� the net current �ow must be of a lower magnitude than either the drift or di usioncomponent� This assumption only requires that the net current be not just less than the componentcurrents� but much less� The equation that results from this assumption is crucial to the theory� andyields the accepted form of the I�V characteristic�

�� Relation of potential across junction to minority carrier injection

The last assumption above can be expressed in terms of eqs� �� by setting Jp � � and Jn � �� Thusfrom these equations we have

E �Dp

�p

dp

pdx� �Dn

�n

dn

ndx�


or� with the use of the Einstein relationship�

Edx �kT

q

dp

p� �kT

q

dn

n� ��

These latter equations can be integrated across the depletion region to obtain the relationshipbetween the density of majority carriers� the potential drop across the junction� and the density ofminority carriers at the edge of the depletion region� The results are�

npjx��b � nneqVpn�kT

pnjx�a � ppeqVpn�kT � ��

�� Relation of potential across junction to potential applied to diode

When an external potential is speci�ed for a PN junction diode� it is not the same as the potential acrossthe junction� Recall that when the externally applied potential is zero� the potential across the diode isVpni� The contact potential where the wire leads join the ends of the bulk region of the semiconductorsum to the negative of the equilibrium junction potential� When the externally applied potential isnot zero the contact potentials do not change� only the junction potential �assuming no resistive dropthrough the bulk material��Millman�Halkias� page � Thus the V in the I�V characteristic is relatedto the junction potential by Vpn � V� Vpni� Note that with the sign convention we have chosen� withthe positive reference for voltage being on the p side of the junction� Vpni is a negative number� In thesame way� Vpn must remain negative if the assumption of no �eld in the bulk region is to remain valid�Thus the value of V must be less than �Vpni for the I�V characteristic we will derive to be valid� Withthese considerations� we can now replace Vpn in eqs� �� with V to obtain

npjx��b � npeqV�kT

pnjx�a � pneqV�kT � ��

The Vpni in the exponents in eqs� �� was eliminated by using eqs� �� with V�� for then there mustbe no injection of minority carriers into either bulk region at the depletion region boundary�

Finally� we can write the formula for the excess minority carrier densities at the boundaries betweenthe depletion region and the bulk regions as

!npjx��b � npjx��b � np � np�eqV�kT � �

!pnjx�a � pnjx�a � pn � pn�eqV�kT � ��

�� Drift in the bulk regions

Having established the minority injection levels at the boundaries of the depletion region with thebulk regions� we next use the minority carrier di usion equations� eqs� �� and �� to obtain thedistribution of minority carriers and the current they carry in the bulk region� Since by assumption alltime derivatives are zero� if there is no light incident on the material the di usion equations reduce tothe second order di erential equations in x�

d�!pndx�

�!pnL�p

andd�!npdx�

�!npL�n

�

These equations have exponential solutions� The boundary conditions are� far from the depletion regionthe excess minority carriers must vanish� due to recombination� while at the boundary with the depletion


region the values must be as found in eqs� �� Thus the solutions to the di usion equations whichmeet the boundary conditions are�

!pn � !pnjx�a e��x�a��Lp and !np � !npjx��b e�x b��Ln � ��

The di usion currents that result from these minority carrier densities in the bulk regions are� fromeqns� �� in the p�type bulk region�

Jp�diff � �qDpd

dx�!p� � �

qDp

Lp�!p��

and in the n�type bulk region�

Jn�diff � qDnd

dx�!n� � �

qDn

Ln�!n��

�� The �nal formula for the IV characteristic

One last assumption must be involked� It is the one that says there is no recombination or generationwithin the depletion region� Thus whatever currents enter at the boundaries must continue completelythrough the depletion region� At the boundary where x � �b the current density due to holes must bejust that of eq� �� when evaluated at that position� since there is no electric �eld intensity in the bulkregion and hence no drift current component� In the same way the current density due to electrons mustbe just that of eq� �� when evaluated at x � a� The sum of these two current densities must be thetotal current density within the depletion region� and since dynamic equilibrium �DC� conditions areassumed� this must be the total current density throughout the entire diode� Multiplying this currentdensity by the cross sectional area A then yields the current� I� for the I�V characteristic� Combiningeqs� �� with eqs� �� and �� gives the desired result�

I � qA

�DppnLp

�DnnpLn

��eqV�kT �

��

The multiplier of the exponential term in eq� �� is called the reverse saturation current since it isthe value of I when V is large compared to kT and negative� We �nally write the I�V characteristic as

I � I

�eqV�kT �

��

where the reverse saturation current is

I � qA

�DppnLp

�DnnpLn

�� qAn�i

�Dp

LpND�

Dn

LnNA

��

�� Problems

� Find the value of reverse saturation current for a PN junction diode made from Si having thevalues for the various constants and properties that have been used in previous problems� �T ��K� NA � ND � ��cm�� etc��

�� Make a sketch of the I�V characteristic for a PN junction diode using I�I for the vertical axisand qV�kT for the horizontal axis� Show numerical values along the axes�

�� Modi�cations required for real diodes

The derivation of eq�� was possible due to the many simplifying assumptions used� some of whichseemed self contradictory� Thus it is not surprising that a real diode�s I�V characteristic is somewhatdi erent and more complicated� The formula of eq�� may be used� however� to a considerable degreeof accuracy� if empirically obtained values are used for I� and if the value of kT�q in the exponent is


multiplied by a constant �� which lies between and � in value� and represents e ects that were notconsidered in the derivation above��Millman�Halkias� page �� Other modi�cations are also requiredin the formulas to take into account other phenomena��Neudeck� pages �� Finally� we note that areal diode will have an ohmic series resistance due to the conduction in the bulk region� and will haveadditional reverse current due to leakage� When these additional factors are important� it will probablybe necessary to use a computer�aided circuit simulation� such as Spice� to take them into account�

Notes on Semiconductor Physics for Electronic Devices

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