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by: Spencer Ondricka
Spencer Ondricka
GPA 3.91

U. Mishra

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U. Mishra
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This 42 page Class Notes was uploaded by Spencer Ondricka on Thursday October 22, 2015. The Class Notes belongs to ECE 132 at University of California Santa Barbara taught by U. Mishra in Fall. Since its upload, it has received 19 views. For similar materials see /class/227035/ece-132-university-of-california-santa-barbara in ELECTRICAL AND COMPUTER ENGINEERING at University of California Santa Barbara.




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Date Created: 10/22/15
ECE 132 Semiconductors are the class of materials that exhibit conductivity between the high values for metals and the very low values for insulators The properties of crystalline materials are a function of the nature of the bonding between the atoms that constitute the lattice Bonding Forces in Solids l IONIC BONDING Ionic bonding between two atoms occurs when metallic elements example Na donate an electron to the more electronegative element example Cl thus creating an Na and Cl39 ion pair The electrostatic attractive force creates the ionic bond between the elements The electrons are in full orbits and tightly bound to both the Na and the C 1 atoms and hence there are no atoms available for current ow NaCl is thus a good insulator 2 METALLIC BONDING In the case of metals the outer shell is only partially lled These electrons are very loosely bound to the atoms and hence are free to move through the metal when subjected to an E eld Thus metals are excellent conductors 3COVALENT BONDING Elemental semiconductors such as Si and Ge exhibit this type of bonding Compound semiconductors such as Galls have an ionic component to a dominantly covalent bond because of the differing electronegativity of the two elements Physical Properties of Semiconductors Wolfe Holonyak amp Stillman HP The important semiconductors today are bonded into a diamond lattice which minimizes the energy of the lattice The bonding is tetrahedral in nature This can be understood by considering say Si The four outer electrons in Si exist in sp3 hybrid orbitals These bonds overlap and the electrons are shared between two atoms fulfilling the required condition of two electrons per orbital Ionic Bondin ex Metallic Bondin Covalent Bondin 5139 5351 St Another Pictorial Representation The energy levels of a hydrogen atom is obtained as e mreAZ2 136eV E n th n2 n2 2 is the atomic number This is obtained by solving the Schrodinger equation for a coulombic potential Energy levels En Ze2 V0 r gt Single atom Solid Next band is empty conduction band W W W Mlt 4 lllllllllllllllll quotIlll llll llllllhvllll lt Last lled band valence band Y7lt1l of Y I4 electrons Note that the energy gaps between the quantized energy levels in a single atom are the origin of the band gap for solids Typical numbers for band gap for various solids Semiconductor E g Si ll eV GaAs l4 eV Ge 07 eV 102 90 eV GaN 34 eV This is the MINIMUM energy required to allow an electron to leave the mostly full valence band and occupy a state in the conduction band An analogy is say a football stadium Step height Band gap Field The step height is analogous to the band gap in that an energy difference less than the step height will not get you to the next step The energy required to move an electron from the valence band VB to the conduction band CB is related to the bond strength and can be provided by one of several means thermal light photons high energy particles etc Thermal energy eXists in each solid at temperatures TgtOK A measure of thermal energy is Q where k8 is the Boltzmann s constant and Q 259 meV at room temperature 300K by convention 1 eV Energy required to move one electron up a potential energy of IV l 6 X 10 19 coulombs IV 16 X 10 19 Joules E 5 E5 E5 E5 lleV l4eV 34eV 90 eV EV s E EV GaN V Galls 5102 At room temperature pure Si has an intrinsic carrier concentration of 15 X 1010 cm393 or 1 electron1012 Si atoms To increase the conductivity of the semiconductors add impurities to make the sc extrinsic Extrinsic Semiconductors Creation of free electrons or holes by the addition of impurities is called doping N Type Case Electrons are majority carriers If a group V atom is introduced into the crystal so that it replaces a Si atom ie SUBSTITUTIONAL IMPURITY then the Si lattice looks like such Si Si Si file SiSlSi gt Si P gtSi Si Si Si The P atom incorporates into the lattice through tetrahedral covalent bonding but has an excess electron that is unpaired and weakly bonded P in a Si lattice now looks hydrogenic The electron at nite temperatures is stripped from the P atom or DONATED to the crystal and is free Hence the name DONOR atom to group V elements like P in semiconductors like Si When the electron is no longer bound to the P atom the P atom has a net positive charge The binding energy of the electron can be calculated as 4 2 AEd E Ed amp n 5 2 2 2n 4727 e h the energy required to make the electron free ie to place it in the conduction band 2 gtk Md 2 116 my 71 e m 136 2 Recall En for a hydrogen atom Intrinsic Semiconductor As stated previously an intrinsic semiconductor has no free carriers at 0K and hence is then an insulator At any nite temperature however the covalent bonds break producing electron hole pairs At steady state 71 E p En Intrinsic carrier concentration Since the electrons can be recaptured by the orbital lacking an electron this is referred to as to RECOMBINE WITH A HOLE At steady State At all tem erature s rag quot Recombination rate Generation rate AsTTgIT3rlT AsEgT nlxl e e E lleV Si 145 X 1010 cm393 Eg V14eV GaAs 179 X 106 cm 3 C C Study the nature of f E 7 2V AtT0K k38o62x1054 w fET 0 1 forEltEF 488x k 0 forE gtEF kBTT 300K 26meV It is impossible for electrons to occupy energy above E F at T OK At niteT A T0 x NAM 1 Ed EF ltI zgt0 H56 kT Na E1 E lZe F 4 T2gtT1 kT EF 39 f E is the probability of occupancy of an AVAILABLE STATE if no state exists and even is f E 1 there will be no electron at that energy In a semiconductor I F for an intrinsic semiconductor Ev EV E 4 f lV 0 At nite T there is a creation of a hole population or a lack of an electron population in the valence band fElt 1 NOTE I f E has a nite value in the forbidden gap but no electron exists because there are no states I f E is symmetrical about E F I f E is the probability of occupancy of a state by an electron l f E is the probability of lack of occupancy of a state g the probability of occupancy by a hole Rewriting l fnE fEW fpE1 fE1 le F eErEFkT 1 ea EFkT NOTE THE SYMMETRY OF ELECTRON AND HOLE PROBABILTY ABOUT E F In Intrinsic Si nt7 R7 Since the concentration is equal the Fermi level should lie close to the center of the gap A EC 0 l3 E E A EF EF EF E V I Ev E 0 l 0 l 0 l o N V 2 Intrinsic n type p type EF is closer to E5 in ntype material and closer to Eguilibrium Electron and Hole Densities for this Class the derivation is not required To calculate the number of electrons in the CB we must know 1 How many states are available 2 What is the probability of occupancy We thus introduce N E dE DENSITY OF STATES cm393 available for occupancy in the energy range dE Total number of electrons in the CB can be written as Top of CB 7 If E N E dE electrons cm 3 Bottom of CB We know that max 4 EC 1 f E W Fermi Diriac Statistics If E EF gt 4kT 1 7E7EFkT l elEiEF e Ecmin Since e5 148 E E Fje455r6406j kT etc At room temperature CT 0026eV We saw that if EC EF 4kT then the probability of occupancy of states at ECmin fEC WENH e4 00183 For higher energies the f E drops sharply Example HHFW e 10 0000045 45 x 106 the contribution to the integral at higher E is reduced to the point that Top of CB nlNEfEdE lNEfdE B och B och Using quantum mechanics and including Pauli s Principle NE EE V5 h h663X10734J S fE N h J 27 414x1015 eVs Here ml is the density of state effective mass no t 5 n 2quot l2 2W 2 EEE eiE7EFdeE E 27 h Multiplying and dividing by kT we get n 2mg OJ 87E7EFk7 E EE dE 272 hi Jkr E De ne EE u kT 3 7 7 1 ZmZkT Aoeergkr Ie o E EE xd E EE 272 hi kT kT Z 00 1 2rka EiEckT u 398 le 0 631 du g From de nite integral tables E EFWVT or n N56 3 2727mjkT JA EFiECkT n 2 oe hi N E Effective density of states in the CB ie Instead of considering the occupancy by electrons of all the states from ECmin to Ecmax NC gives us the number of electrons in the CB by considering all the states in the band to be effectiver at the band edge populated with a probability 7E7 kT function e l C F Likewise proceeding in a similar manner for the holes in the VB we obtain EFrEVyRT 3 27Zmpk7 hi p Nveil where NV 2 I NV Effective density of states in the VB I It is clear now that in n type material EF is close to E5 and as E EF L n ymmwug g PT0R E fMoves towards E for n type Moves towards EV for p type E v AsEFT nTandpr and EFxL niandpl Examine the product n p n p NENVe N N erEcEFrEFEvk7 c v 7557 kT39erEF7EVk7 np NENVe E EN 7557 EkT or np NENVe E5 EVEg E np NENV e A The expressions for n and p are applicable for all types of semiconductors n p or 139 In an intrinsic material the Fermi Level is identi ed as E F nintrinsic r4 Ncelg 7E WT and pintrinsic pl NveEwEvkr m fMM WWH Note that np 712 ZNCNV Er 5 W Now that we can relate E F to the nature of the CB and VB through E m and m we can recalculate EE more accurately Any asymmetry in namp p distribution has to be introduced through differences in N5 amp NV EV l 0 2 n N 8ErEnk7 NN e EgAkT 3 germk7 ergkr NE E EH Eg NV exp o exp kT 2kT NE 3 exp EE EFI E7gJkT 5 Taking ln ofboth sides E N EEER gkT 1n V 2 NE Since Eg EC EV E E N EF1EC 5 VkT 1n V 2 2 N5 E5 E amp 39 EFI Vllen 2 2 c The second turn is the derivation of E E from the middle of the gap So what is the dependence of n vs T A Intrinsic 1017 n E N d l lt Nd 1015 Extrinsic Freeze out region 1013 gt High lOOOT Low temp temp Space Charge Neutrality We can dope a semiconductor with both donors and acceptors Donate electrons J Accept electrons ie remove electrons n E N 01 NA for ntype material effective Peachv EN Nd for ptype These materials are called compensated semiconductors n NE 845 IN andp N eilEF EWT are still valid relations In a bulk semiconductor there must be charge neutrality everywhere because any excess charge will immediately set up an electric eld gtE F qE E22 FqE e The electrons will ow towards the positive charge until the value Q is reduced to 0 Then E gt 0 3 The attractive force F gt 0 or neglecting R7 oooooooooo KEd E conga o o o Conductivity and Mobility 15 At any temperature the individual electrons show randomized motion Scattering with the lattice is the dominant randomizing process If we apply a eld EX then F per electron qEX Net change of momentum of the electron ensemble where n of electronscm3 We want to use this equation to determine the response of an electron to an applied electric eld ie The relationship between its velocity and the applied eld 1 The probability of collision is constant 2 N7 of electons in the group at I 0 3 N t of electrons which have not undergone collisions by time I Let the scattering rate or number of collisionssec be K The rate of decrease of N t the scattered population is given by KN t assumption 1 w Kdt 31nNt Ktc NU Since Nt 0 N clnNU Nt Nge Omi 0 H Let t be the mean time between collisions A Io t Ntdt N0 If te dt t S0 5 Ntdt N0 If and dt The probab111ty that an electron Wlll eXperlence a collision 1n an interval dt is T ThlS t is equivalently the fraction of electrons that undergo a collision Let pX be the momentum in the x direction dt 39 dPx x T t T because momemtum decreases d or rate of decrease of pi due to collls1ons t t dpz E anx rate of 1ncrease of pi due to EX At steady state rate of momentum increase due to EX rate of momentum decrease due to collision csz ls l qtp m mp HFE 3 u lt V gtn is the net drift of the electron distribution under the in uence of Ex The current density resulting from the dri velocity is JpxqpltVX gtp Jm qnn f E Zilex Rewriting as conventional Ohm s Law Tz Note on qnJn ap quF If both electrons and holes are present A Jpx Jm 61nun pup Ex Lattice scattering Impur ity scattering Since the probabilities of scattering add K K1 K2 Since K 06 i i 1 i Matthiesen s Rule I H H Iii Hz 139 Measurement of Semiconductor Parameters 1 Effective mass m can be determined by electron resonance Under a magnetic eld the electron motion is circular 2 mv qu RF energy B V r f Centripetal force T magnetic ux density gt N qBr qB or a Cyclotron resonance frequency If RF energy is incident on the sample it will be transmitted for all frequencies except a when resonance absorption of energy occurs Reduced transmission By measuring a we can calculate m or m 2 MobilitV Hall Effect and Carrier quot Consider a ptype bar which is subject to an electric field E and a magnetic field BZ The force on a single hole F q x j Fy qEy va2 E is the electric field that is set up by the holes that were de ected under the in uence of the force VXBZ Since no current can ow in the ydirection The vx can be obtained from Jr as J v where R7 hole concentration x 0 E J Jr 32 qu Where RH Hall Coef cient Pg p0 J V V V Ey A B amp w w R t V 1 211 where w CD I m an L L wt Excess Carriers in Semiconductors A Recombination hv gt Eg emission of light f direct semiconductor generation The beam is absorbed d x dx 3 06 cm 391 Eg 1x f absorption coef cient a x I 0 e39 0 hv Ez z l24eV 1 MM m is the energy ofthe photon in 2 V Carrier Lifetime amp I 39 quot quot Let us calculate the rate of change in the electron population when I switch the light off We rst assume that the rate of recombination R is proportional to the product of the electron and hole population For example R 06pt nt In the absence of light the thermal generation rate should be equal to the thermal recombination rate or Gm Rt 06 quot227 06 r42 d I Now 1 G R Gm R In the absence oflight dnt emf oc new dt Thermal generation rate Recognizing that on shining light equal numbers of e and hole pairs are created 5nt 5pt and since 067112 ogrzgpm Snt 0th2 06r n 7 5ntpt7 5pt a 070 p 5nt5n2t Low level 1nJectlon 3 ltlt p0 and 672 I lt64 p u Ptype material d 3 5n t a 5n t dt rpg t 3 5nt An 67 An 6 A 06170 1 A lso the minority carrier lifetime THIS ANALYSIS IS VALID ONLY FOR DIRECT RECOMBINATION In Recombination time Not valid for recombination via traps In general traps reduce the a E minority carrier lifetime even y g further mi QuasiFermi level IMREF gt 06110 pa in steady state since generation rate recombination rate We know with light gT g p 0617 06010 5np 5p In steady state with no trapping 5 n 5 p gTg0P an0p0 06 1 270 5n5nzl gt7p 06 p05n Low level injection in a p type material 6n 35nltpgandngltp01 TV 3 In general 511 In gt7p 5p IF gap If In 3 L39p nt7 5n n nieme MW p05ppne E F Electron IMREF EFF Hole IMREF Diffusion of Carriers nx gt n diffusion ma a T The negative sign represents diffusion from higher to lower concentration dpx px p dx n d The diffusion current Jquot q D x In the presence of an electric eld J x Jndrift Jndiffusion dnx dx Jquot x gunnxExan J xqu pxEx qD W F F F dx Jx Jquot x JP x or the total current is the sum of all electron and hole currents Continuity Eguation gt Axlt JP x 14 gt JP xAx J xAx A Area in cm2 and Ax is the length of the box in centimeters Conservation of mass 6 6 1 Increase of hole 7 Decrease of hole 1 concentration concentration T T Rate of Through input Through output hole current current amp net recombination buildup Input current J p x 3 charge X input of carriersunit areasecond Input hole ux 2 Input carrierssecond p x un1t volume JFOC qAx 3 px J p x Ax qAx pxAx 6Pi W151 6t q Ax r F 0pxt 65p 13 5P 6t 6t q 6x I 6571 1quot 5n 6t q 6x In Continuity equation for electrons and holes For purely diffusive currents Z 65nDnd52n 6t 6x I n 65pD 625p5p arFax2 1p Diffusion Length Consider steady state or g 0 625n 5n 6x2 Dnrn L2 n A and 625p 5p 5p 2 2 6x Dprp Lp AP Decay is given by the solution to the 4 above diffusion equation 395px clexLp czei p As x gtooweknow6p gt0 or 010 5px czeizp LP the d1stance where the d1str1but10n drops to X 1n1t1al value e AND LP average distance a hole travels before recombining The hole diffusion current can be calculated at any x as px qux d q FEW qu 5 LP px Currents in relation to E F amp E F dnx dx JKOC 9HnquotxExan Since dnx 1 n1 eEm EF1 quotxdEFn dEF1 dx dx RT RT dx dx AND since Dn LLKRT we get dEm dE Jr x 9UMJCEJC141 nx dx But qEx dE Jn x un nx 61f total electron current 0 E00 9 gt No eld E E5 1 EOC dVx dx But Vx 3 Exi l q dx q q dx pn junctions A pn junction is a metallurgical and electrical junction between p and n materials When the materials are the same the result is a HOMOJUNCTION and if they are dissimilar then it is termed a HETEROJUNCTION Junction Formation e e e e e P N A acceptors II ND donors e B B gtJdiff l Majority carriers diffuse holes from p to n and electrons from n gt p 2 Bare ionized dopants are exposed on either side of the junction Positively charged donors on the nside and negatively charged acceptors on the pside 3 The dopant ions are contained in a region of reduced carrier concentration as the mobile majority charges have diffused as stated in 1 This region is therefore called the depletion region Edpl P n 39 X 39Xn WO P o mefmsion Jdri 4 The process of diffusion continues until the depletion region expands to a width W0 such that the electric field in the depletion region Edepl is large enough to repel the diffusing carriers More precisely Jdiffusion Jdd once equilibrium has been established it FOR EACH CARRIER SEPARATELY 5 The driving force for canier motion is the ELECTROCHEMICAL POTENTIAL DIFFERENCE that exists between the two semiconductors in the bulk prior to junction formation In band diagram terms here are the before and after pictures EVACUUM A A A A qx w W E V v A A V 01 Eg E q P Q g EFn EFp E FP v E V V BEFORE THE TWO MATERIALS ARE SEPARATE De nitions a qX Electron affinity in units of energy use eV or Joules b EFP Fermi Level in the ptype material or electrochemical potential of the ptype material c EFrl Fermi Level in the ntype material or electrochemical potential of the ntype material I NOTE TIHS IS THE ELECTROCHEMICAL POTENTIAL OF ELECTRONS IN BOTH CASES d qQ p and qgn are the work functions of the two materials p and n respectively e Note that the work function difference between the two materials q p Qquot is the difference between the electro chemical potentials of the bulk materials EFrl and EFF Eg 39Xp X 0 Xn AFTER The materials are brought together to form a junction The fermi levels EFn and EFP now equalize or EFn EFP EF IN EQUILIBRIUM a Assume the p material is kept at a constant potential say ground b The p material has to increase its electrochemical potential of electrons upward motion of the bands until the fermi levels line up as shown in the diagram below where the effect is simulated using two beakers of water in equilibrium with different amounts of water in each beaker p n n p POTENTIAL DIFFERENCE Energy increase BEFORE TAP CLOSED 0f the P AFTER TAP OPEN material c The lowering of the electron energy of the ntype semiconductor is accompanied by the creation of the depletion region d The depletion region has net charge and hence the bands have curvature following Gauss a Law 0E E E Electric Field 6x 6 OR 02V 1 0V 6x2 6 6x where V Potential energyof unit positive charge 02E OR 6x2 2 B EC 2 qV Electron energy Joules E or 7V Electron energy eV NOTE NET CHARGE cgt CURVATURE OF THE BANDS NEUTRAL REGIONS 3 NO NET CHARGE cgt BANDS HAVE NO CURVATURE DO NOT CONFUSE SLOPE WITH CURVATURE Neutral regions can have constant slope or equivalently no curvature CALCULATING THE RELEVANT PARAMETERS OF A pn JUNCTION 1 SCHEMATIC OF THE JUNCTION 2 CHARGE PROFILE 0 E 0 0 3 ELECTRIC FIELD ET 3x 6 E Emax the junction X 0 qux 9 i G 4 BAND DIAGRAM NOTE qui qQFp qQFn In the analysis depicted in the foru diagrams above we assume i that the doping density is constant in the pregion at NA and in the nregion at ND and that the change is abrupt at X 0 the junction ii the depletion region has only ionized charges inA Ccm393 in the pregion and qND Ccm393 in the nregion Mobile charges in the depletion region are neglected ie n p ltlt NA39 and ND iii The transition from the depletion region to the neutral region is abrupt at Xp in the pregion and Xn in the nregion Calculation of the built in voltage From the band diagram it is clear that the total band bending is caused by the work function difference of the two materials If you follow the vacuum level which is always re ects the electrostatic potential energy variation and hence follows the conduction band in our homojunction you see that the band bending is the difference of the work function of the ptype material qQP and the ntype material qQH qu 6195 195 The built in potential is therefore the internal potential energy required to cancel the diffusive ow of carriers across the junction and should be exactly equal to the original electro chemical potential which caused the diffusion in the rst place THIS IS REASSURING To calculate Vbi from parameters such as doping let us follow the intrinsic level from the pside E11 to the nside Em Again the total band bending of the intrinsic level is the built in potential or Rip 7 Em qui or de EF 7 Em 7 EF qui De ning Rip 7 EFp as qQFp Note in equilibrium And EFn Ein as qQFn EFn EFF EF We can rewrite qui as qQFn qQFP qui From Fermi Dirac Statistics E EF Exp EFF n e or n e n e Pm 1 RT 1 CT 1 E E s1m11arly nm7 11 e F RT Assume full ionization amp ND n kT kT or qQFp len amp qQFn len N N q Fp qgm kTln Aln D 7 71 NAND 2 or qlE len kT N N K 1n q 71 x 39 39 quot Depletion Region Widths From the electric eld diagram Figure 3 N Emu q Axp e N IE q A W e The magnitude of the area under the electric eld versus distance curve shaded area in gure 3 is by de nition xp 1 E dxl Voltage difference between 7X1 and Xn Vbi or V 1 W E 2 1 EBaseHezght qN A E xp V ltxn x We now invoke charge neutrality Since the original semiconductors were charge neutral the combined system has to also be charge neutral since we have not created charges Now since the regions beyond the depletion region taken as a whole has to be charge neutral Q all the positive charges in the depletion region have to balance all the negative charges If the area of the junction is A cm2 then the number of positive charges within the depletion region from X 0 to X X1 is qND Xn A Coulombs Coulombs cm 3 ltm393 cm2 Similarly all the negative charges contained in the region between X XP and X o is qNA xpA Coulombs charge neutrality therefore requires qNAxp A qNDxn A or IMPORTANT To calculate w X11 and XI we use the above relation in the equation for V171 below 1 V271 Exn xpme p side n side 1 N 1 N Vm3xnxpqxp Vb 3xnxpqEDxn Substituting for Xn l N N N l N N N or Vbl er qAx VI ern qDxn 2 N D p e p 2 N A e 2 N N 2 N N or EVM prz E Vb anz qNl ND qND NA qNA qND NOTE In this analys1s IEmaxl was calculated as xp from the ps1de and x fom the ns1de e e VVJg1xp 2 51 JND qNA l q 1 UNA NAND Np NANDJ EV NAND q 1 NAND NA ND W 5NAN Vm q NAN3 IN EQUILIBRIUM W0 2 6 ii V IMPORTANT 9 NA N Ingeneral 2 E l l Wv 7N AN DV1ilVl V is the applied bias V in reverse bias w expands V in forward bias w shrinks FORWA D BIAS REVERSE BIAS ZERO lg BIAS x39 zom no T 161 V l E Fn x I FqND qNA gt 39 x39 p xn NAx lEmax q p 6 FEATURES 1 Total Band bending is now VbiV lEmax 2 The shaded area dx 2 V2 V FEATURES 6 3 The edges of the depletion region move towards 1 Total Band bending is now Vb IVI the Junctlon or W decreases 2 The Shaded area dxl mi IVI 4 E Fquot EFP VF the electrochemical potentials 3 Depletion region expands separate by an amount equal to the potential difference applied VF 4 E FP EN 2 amount of poMtial difference VR Current ow in pn junctions VERY IMPORTANT A Forward Bias IVb EFn n gt JDIFFUSION lt JDRIFT y JDIFFUSION JDRIFT I Vb V qV p p The depletion width is JDIFFWON JDRIFT 0 reduced which increases the J n J n 0 diffusion current The drift DIFFUSION DRIFT current remains the same The net current is due to the imbalance of drift and diffusion Under forward bias electrons from the nregion and holes from the pregion cross the junction and diffuse as minority carriers in the p and nregions respectively To understand how excess minority carriers are injected and diffuse one has to understand the LAW OF JUNCTION which now follows Consider the two situations shown below one at zero bias and the other under forward bias Note that E is a function of x and is E in the bulk p and ER in the bulk n Note that px is always given by t E x39 EFFX39 x ne P x H Where 131X and EFPX are the intrinsic fermi level and the fermi level for holes at any place X Since we are at zero bias and at equilibrium ElP is not a function of x 1p F quotIv 74v kT kT ne E F EF e E F E xi CT CT or px39pXe gc LAW OF El Ex39 THE where QWOCFT JUNCTION or px decreases exponentially with the local band bending Note that the edge of the depletion region on the nside x x or x W the hole concentration is given by t 7 W x39 W x W e p pm H The total band bending at x Wis Vbi 7 M kT px39 W ppoe We know that px W is the hole concentration in the ntype semiconductor or npo 7 W So npo ppoe k Let us verify that what we derived does not contradict what we learned in the past 2 quotI 7 m n e p0 n0 pp kT 7 qVE 742 e H pponno n V271 k Tln39m ljp0 or with full ionization q 1 CT N N V 1n SAME AS BEFORE q n On application of a forward bias the electron and VI VF hole concentrations continue to follow the relation E1 39 EF IVF E F that p x39 ppoe Wig the law of the junction E The difference from the zero bias case is that at the edge of the junction of x W IIW Vbl VF not Vbi as in the zero bias case 1 quotno rm pquot W Ppoe W quotp xp Pn xn m K 0r pnW Ppoe 396 3 me S 39 n0 Anp xp AquotF xquot pr 0 PnOe k7 Change the Consider THE MINORITY CARRIER CONCENTRATION coordinate system on1y the IS RAISED FROM ITS ZERO BIAS qV VALUE LVF so that xquot in nreglon BY THE FACTOR e M I xquot coordinate is 0 THEN IN THE nregion ll IMPORTANT Apn xquot Apn0equot39 40 Similarly the minority carrier concentrations at the edge of the depletion region on the pside qVF np xp npoe 1 What happens to these excess carriers They diffuse away from the edge of the depletion region to the bulk The pro le that governs the diffusion is set by the recombination rate of the minority carriers in the bulk The situation is analyzed by the continuity equation AP 0 n x l I x 1 At any pomt x the cont1nu1ty equation states V J G R Us1ng J Jp Mm dzpnbc Apnx 2 by neglecting drift currents which will be explained later we get DF 1 Tp 2 2 2 Note that pn xquot pnxquot mu since pn0 0 d2 quot d2 quot FPKOC FAPKX D dimequot Ap ccquot 0 p 2 d 1F We know that far away from the junction the excess hole concentration has to be zero since excess holes have to eventually recombine Apn xquot cle L 026 where Lp JDFTp Diffusion length of the holes which can be proven to be the average distance a hole diffuses before it recombines with an electron Also 01 E 0as Apn gt0as xquot gt00 7 Apn xquot cze L QVF At xquot0 Ap 0 p 017 M7 p 1 VF of A1750 1270 e74 IMPORTANT Ap ccquot Ap0e 2 4l This exponential relationship applies to the minority electrons as well where m x m quotp x j Anp x Anp0e Anp 0 mm 6 H 1 New Coordinate System 42


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