Pn Junctions - High Energy Physics
pn JunctionsMurray ThompsonSept. 13, 1999Contents1 Introduction12 Multiple One Dimensional Potential Wells13 Multiple Three Dimensional Potential Wells24 Fermi Level25 Semiconductors26 p Type and n Type Materials27 Donor Levels in n type Material38 Electrons and Holes49 Acceptor Levels in p type Material510 Adjacent p and n Type material611 p and n Types in Contact912 Now Raise the voltage of the p Type by V1513 Dynamic Resistance of the PN Junction1914 Energy Levels in Junctions211
15 Two pn Junctions2516 An npn Junction Transistor271IntroductionThis is a non-rigorous explanation to give you an understanding of the background for the operation of semiconductor diodes and the Diode Equation.This explanation is intended to fill in part of the missing background inH& H. We advise 623 students to read R.E. Simpson’s book “IntroductoryElectronics for Scientists and Engineers” pages 162 – 185.Since this explanation is an addition to the lectures in which pn junctionsare discussed, it assumes some of the material from the 623 lectures plussome material from the other physics courses and is not intended to be a fullexplanation of pn junctions.2Multiple One Dimensional Potential WellsIn Quantum Mechanics, we learn how the wavefunctions of the electronswhich are held in a potential well such as an atom are eigenfunctions andmust satisfy the boundary conditions. In an artificial but more easily understood one-dimensional situation, these boundaries are on the left and right.Suppose we have found an eigenfunction with an energy eigenvalue E for aone dimensional potential well. Now consider 2 such wells side by side. Wefind that now at energies near E, we have 2 eigenfunctions. Similarly, if 3wells are placed side by side, we have 3 eigenfunctions. If N wells are placedside by side, we have N eigenfunctions with close values of E.3Multiple Three Dimensional Potential WellsAlthough it is not practical to draw the 3D potential wells and eigenfunctions,we find that N wells again give N different solutions or eigenfunctions with Nslightly different energy eigenvalues. These fit within a band of energies withthe lowest being that for the eigenfunction with fewest crossings of Ψ 0and highest being that for the eigenfunction with most crossings of Ψ 0.2
In any object which can be seen, the number N of wells (number of atoms)is extremely high since Avogadro’s Number is NA 6.024 1023 atoms/mole.Thus in any object which can be seen, the energies occupy a band in avirtually continuous way.4Fermi LevelThe “Fermi Level” in a crystal is the highest energy level of electrons whenthe crystal is at a temperature of absolute zero (0 K). In a metal, the conduction band is only partly full and the Fermi LevelLevel lies in the middle of the conduction band. In an insulator, the valence band is full and the conduction band isempty. The Fermi Level lies between the valance band and the conduction bands.Pure Si and Ge crystals act as insulators.5SemiconductorsSemiconductors are materials which would have been insulators except thata very tiny fraction of particular impurities has been added. These impuritiesare called dopants and can have an enormous effect.6p Type and n Type MaterialsA crystal such as silicon can have two types of impurity atoms (dopants)which are infused with very low concentrations which, however, can havevery significant effects. These impurities give two types of doped materials,“p type” and “n type” which we describe below and, for simplicity in mostdiagrams, we will place p type with acceptor dopants on the left and n typewith donor dopants on the right.3
7Donor Levels in n type MaterialConsider the doping of a Silicon (ZSi 14) crystal with a few atoms ofPhosphorus (ZP 15). Phosphorus has 5 valence electrons. (Arsenic ZAs 33 or Antimony ZSb 51 are similar.) Each Phosphorus atom sits amongSi atoms and Si covalent bonds so that the 5th electron of each P atom isvery loosely held. It is said to be in a “donor” level. Its wavefunction hasan energy eigenvalue which is well above those of the bonded electrons andis close to the conduction levels.When the electron moves away from its original donor atom, the donoratom is left with positive charge (it now has 4 electrons instead of 5) (qdonor e 1.602 10 19 C) and so the negative electron is weakly held by thesingle positive charge of the remaining donor nucleus and electrons. Theelectron (-) and its donor atom ( ) form an “atom” with a positive nucleusand a negative electron moving about it. The electron sees a potential welle.eV (r) 4π rwhere relative 0 and relative is the relative permittivity of silicon relative 12. Whereas in the case of Hydrogen, the binding energy of theelectron, in the ground state to the proton is4electron eEB m2(2 13.6 eV ,20 h)the binding energy of the electron to the donor atom is4electron eEB 2(2 mrelative 0.1 eV 0 h)2and has an energy level or “donor level” close below the conduction band.Since there are many of these “donor atoms”, the n type material has manysuch states with the donor energy levels forming a diffuse band. The typicaldonor level is only about 0.05 eV below the conduction band and 1.04 eVabove the valence band and the eigenvalues of the bonded electrons.The donor eigenstate can be thought of as a disturbed valence state. If thetemperature is lowered to 0 K, then the electron in the donor level cannotfall to a lower state since all the other valence states are full. Thus the FermiLevel must lie above the donor level but below the conduction level.The electron carriers have an energy level or “donor level” close underthe conduction band (typically 0.050 eV below the conduction band).4
At normal temperatures (at 300 K, kT 0.026 eV), a large fraction ofthese electrons can lift into the conduction band and move freely about thecrystal. Such freely moving “left over electrons” then exist in a field with mostof the crystal being neutral but the donor atoms being randomly scatteredpositive (qdonor e 1.602 10 19 C) charges and of course the othernegative “left over electrons” making the combination neutral.8Electrons and HolesHoles are found in p type material with “acceptor atoms”. Acceptor atoms,such as Boron (ZB 5), Indium (ZIn 49), Gallium (ZGa 31) andAluminum (ZAl 13), have 3 valence electrons rather than 4 like silicon(ZSi 14) or Germanium (ZGe 32). At each acceptor atom in a Siliconcrystal, 3 electrons are bound in covalent bonds to the 4 adjacent Si atomsbut 1 electron is “missing” and we say this incomplete bond has a “hole”.The hole can be exchanged with adjacent bonds by the electron of thebond being taken and put into the hole to complete its bond. Thus, a holecan be moved about from bond to bond (robbing Peter to pay Paul). (Thiscan be compared with moving a car from one slot in a parking lot to the onlyother available empty slot. The effect is the same as moving the empty slotfrom the second place to the first. The empty slot can be moved all aroundthe parking lot and is similar to the hole being moved all around the crystal.)The electrons are equivalent to the cars and the holes are equivalent to theempty slots.Both the mobile electrons and the mobile holes carry charge ( e and -e )and are referred to as “carriers”. Both p type and n type have both kinds of5
carriers but the p type has a majority of hole carriers while the n type has amajority of electron carriers.9Acceptor Levels in p type MaterialThe quantum mechanics for the “hole” in p type material has an equationlike that of an electron. The hole has an effective mass, which is a littlegreater than the electron mass, and has a positive charge (q e). Whenthe hole moves away from its original acceptor atom, the acceptor atom isleft with negative charge (it now has 4 electrons instead of 3) (qacceptor e 1.602 10 19 C) and so the positive hole is weakly held by the singlenegative charge of the remaining acceptor nucleus and electrons. The hole( ) and its acceptor atom (-) form an “atom” with a negative nucleus and apositive hole moving about it. The hole sees a potential welle.eV (r) 4π rwhere relative 0 and relative is the relative permittivity of silicon relative 12. As before, the binding energy of the hole to the acceptor atomismhole e4EB 2(2 relative 0.1 eV 0 h)2and has an energy level or “acceptor level” close above the valence band.Since there are many of these “acceptor atoms”, the p type material hasmany such states with the acceptor energy levels forming a diffuse band.The acceptor eigenstate can be thought of as a disturbed conduction state.If the temperature is lowered to 0 K, then each electron in the acceptor levelfalls to fill all empty valence states and the acceptor states are ocupied byholes. Thus the Fermi Level must lie below the acceptor level but above thevalence level.6
At normal temperatures (at 300 K, kT 0.026 eV), a fraction of theseholes can fall into the valence band (ie the electrons lift from Valence toConduction bands) and these holes move freely about the crystal. Such freelymoving “holes” then exist in a field with most of the crystal being neutralbut the acceptor atoms being randomly scattered negative (qacceptor e 1.602 10 19 C charges and of course the other positive “holes” making thecombination neutral.10Adjacent p and n Type materialConsider two semiconductors which are p-type on the left and n-type on theright and which are not in contact. First draw the Acceptor atoms in thep-type material on the left and draw the donor atoms in the n-type materialon the right. We ignore the very much more plentiful and more regularlyplaced neutral Silicon atoms.We draw the acceptor atoms such as B, In or Al (which are fixed in place inthe p-type crystal) on the left and draw the donor atoms such as P, As or Sb(which are fixed in place in the n-type crystal) on the right. We draw acceptoratoms as small squaresand donor atoms as small squareswith theirresidual charge after they have accepted the extra electron or donated one oftheir electrons. Although the acceptor and donor atoms are drawn here in afairly regular array for a simple drawing, the acceptor and donor atoms aredopants of the crystals and are really distributed in a fixed but random and7
statistically smooth mannerNow draw the two materials with the surrounding holes h (missing electrons)and extra electrons e. While these h and e have been drawn in particularpositions, they are really diffusing and moving about.Remember that the Silicon atoms, acceptor atoms and donor atoms are fixedin the lattice by their remaining 3 or 4 covalent bonds/atom and cannot move.They provide a relatively neutrally charged medium plus a few positively ornegatively charged stationary centers. The few h’s and e’s move in thismedium and are called the “carriers”.8
The following 2 graphs show, on a vertical logarithmic scale, some typicalhole (p) and electron (e) carrier densities in p-type material on the left andn-type material on the right.Notice that the both of the products of the p n densities on the two sidesis the same pn 1020 acceptor/cm3 .donor/cm3 due to the law of mass-actionand a tiny fraction of e carriers exist even in p-type material and a tinyfraction of hs or (p) carriers exist even in n-type material.[n] type[p] type[p][p] 1016 acceptor/cm3[p] 105 acceptor/cm3[e][e] 1015 donor/cm3[e] 104 donor/cm3The following two sketches, side by side, show the empty conduction and fullvalence bands for the two untouching materials.9
11p and n Types in ContactNOW LET THE TWO MATERIALS BECOME IN CONTACTAgain draw the two materials with the surrounding holes h and electronse. The holes and electrons near the boundary will diffuse into the adjacentmaterial and most of these will recombine. This will leave the boundary withvery few free carriers. We call this region a “Depletion Layer”.10
The charge density (due mostly to the unbalanced charges at Acceptor atomsand Donor atoms) will have a graph of ρ qe (p n Ndonor Nacceptor ): due to these unbalanced chargesThe actual direction of the Electric field Ewill be from Rright to left . The x components will bex1E x 4π ρ(x)dx to distinguish it from the later energyWe write the electric field as a vector EE of each electron. in the x direction, we will have aIf we graph the magnitude of the E negative Ex .We can get V from V Rx dx:E.Thus, the simple joining of the p and n type materials (actually grown together) has caused two surprising effects.1. two charges have appeared near the joining surface – negative in the ptype and positive in the n type.2. A potential barrier V and an associated electric field E x have appearedwith E x having a direction from n type to p type pushing holes from ntype into p type (from the right to the left in our diagrams).11
The Energy of the holes is e (the electric Potential V) where e is theelementary charge e 1.602 10 19 C, and so for our left p type and rightn type appears similar to that of the Electric Potential.Similarly, we can draw the energy of the electrons (with an inversion due tothe electrons having negative charge -e) and, this leads to the energy bandsof the electrons.By placing the 2 materials in contact, they form a pn junction diode withthe symbol matching the left to right p type to n type.)AFTER A SHORT TIME WE WILL HAVE EQUILIBRIUMEven though we have NOT YET applied an external voltage, the existenceof the charged donor and acceptor atoms, with their hs and es, creates anelectric field across the junction which retards and prevents further net carriermovement.12
On both sides of the junction we will have a region which is nearly empty ofcarriers. This is called the Depletion Region.Actually, of course, the carriers still move a little but the net movementbecomes zero.We define two kinds of origins for electron movement.1. Drift movement due to Electric FieldThe movement is not infinite because the electrons are scattered especially when as each electron accelerates, the de Broglie λ hp ofthe electron momentum becomes comparable with the atomic diameter and the electron is likely to be scattered. Thus the x component ofits velocity rises as it accelerates then falls to zero and the accelerationstarts over again. Thus the average speed of the electron is limited.2. Diffusion movement due to a concentration gradient and rangeof energiesThe diffusion flow is proportional to the product of(concentration) (the fraction of carriers which have sufficient energyto overcome any potential barrier E Ef )where Ef is the Fermi Energy and E is greater than Ef .The electrons and holes are fermions and therefore have energy distributions which obey the Fermi-Dirac statistics.f (E) 1(E Ef )/kTe 1where k is Boltzmann’s Constant. If the barrier E is increased, fewerelectrons will be able to diffuse past the barrier and so f (E) is decreased. At normal temperatures, this distribution f (E) cannot bedistinguished from the Boltzmann distribution.f (E) 1(E Ef )/kTeThe distribution of energies does not have the sharp cutoff found atlow temperatures but a few have higher energies.13
For example, considering the electron carriers, if the density is n electron carriers/unit volume, and they encounter an energy barrier withheight E0 eV0 against the electrons moving from n type on the rightto p type on the left, the number of electron carriers/second which candiffuse from n type to p type past the barrier due to the temperatureT will be proportional toRate ne( E Ef )/kTRate ne( eV0 )/kTeV0Rate n e ( kT )where T is the absolute temperature, and k is Boltzmann’s constant.Acceptor dopantslike Boronp typeDonor dopantslike Arsenicn typeNote that in the equation, the first e 2.7182818 while the seconde 1.602192 10 19 C where the electron carries the charge of “ e”.It is hard to avoid this duplication but it seldom causes confusion.14
Consider the combined Drift Diffusion. For the electrons in equilibrium (p type atoms on left, n type atoms onright);With NO applied voltage, we have NO total current and so:Drift of es due to the Electric Field - Diffusion of es due to theConcentration gradient For the holes in equilibrium (p type atoms on left, n type atoms onright);Similarly, with NO applied voltage, we have NO total current and so:Drift of hs due to the Electric Field - Diffusion of hs due to theConcentration gradientThus, in equilibrium,eV0Combined Drif t Dif f usion n e ( kT )15
12Now Raise the voltage of the p Type byVNOW APPLY A VOLTAGE V ACROSS THE TWO MATERIALSWe make the left p type material have a voltage V relative to the right ntype material.1. If V is positive with the p type material on the left having a voltageabove that of the n type material on the right, it is to be expected thatthis voltage will cause the holes on the left to drift to the right and theelectrons to drift to the left. The moving hs and es will intermingle andcancel each other. The movement of the charges will be an electricalcurrent.2. If V is negative, then the holes and electrons will be drawn slightlyfurther apart until the junction field is increased and the movementstops. Except for the brief slight movement, no charges move and sothere is no current.While these effects are easy to understand, we want to get some algebra todescribe the size of the current as a function of the voltage V causing it.The raising of the voltage on the left (p type atoms) will lower the energyof the electrons on the left and will lower the energy barrier E eV0 againstthe electrons moving to the left is changed from eV0 to e(V0 V ). DriftThe combined drift due to the electric field is unchanged because, ifthe barrier is abrupt, the voltage height of the barrier does not affectthe number of carriers which drift over it.16
– While some distance from the abrupt drop and where the potentialslope is gentle, the drift is proportional to the applied ElectricFieldF low eEx .– However, once an electron starts to fall over the potential drop, itwill continue regardless of the potential or height.(An analogy of rate of water flow in a river towards and over a dammay help. In the river, the water flow (in meter3 /second) is dependentupon the slope of the river since the removal of any volume of waterwill make the water behind it flow faster. However the water flow overthe dam is independent of the height of the dam if the fall is abruptsince, once any volume of water has fallen over the edge, the absenceof that volume of water does not cause less water to flow even if thedam height is reduced significantly.) Drift of esAcceptor dopantslike Boronp typeDonor dopantslike Arsenicn type DiffusionThe combined electron and hole diffusion to the left, due to the concentration gradient, past the barrier due depends upon the temperatureand barrier height and becomesdif f usion e (e(V0 V ))kTSince the voltage height of the barrier between the two materials doesnot effect the number of carriers which drift over it, the NET rateof flow of carriers “over” the potential drop is given by the followingequation with the “drift” component unchanged.17
N et F low [dif f usion drif t]N et F low [e (e(V0 V ))kTeV0eV0 e ( kT ) ]eVN et F low e ( kT ) [e( kT ) 1]The resulting current isI e (N ET F low)eV0eVI e e ( kT ) [e( kT ) 1]We combine the proportionality constant, the value of the elementarycharge e 1.602 10 19 C on each electron and hole and the first termeV0eV0e ( kT ) into a constant “saturation current” Is e e kT to get theelectric currenteVI Is [e( kT ) 1] Diode Equation or Ebers Moll Equation. (H&H pg80)Graph of (Voltage of p type on left relative to n type on right) versus INotes1. The voltage at which significant conduction occurs is about 0.7 V forSi and about 0.4 V for Ge. (Of course, the definition of
2(2 0h)2 13:6 eV, the binding energy of the electron to the donor atom is E B m electrone 4 2(2 relative 0h)2 ˇ0:1 eV and has an energy level or \donor level" close below the conduction band. Since there are many of these \donor atoms", the n type material has many such states with t
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