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2.2 EXTRINSIC SEMICONDUCTORS Brief Summary of Previous Lecture & Intro’ Discussed basic ideas of semiconducting behaviour. Energy, E Conduction Band (CB) Energy Gap Eg Valence Band (VB) nc – conc’n of electrons in conduction band pv – conc’n of holes in valence band Discussed how we could calculate nc and pv (E E ) (E E ) nc = Nc exp c F pv = Nv exp F V kT kT 3 3 m kT m kT * 2 * 2 Nc = 2 e Nv = 2 h 2 2 2 2 Intrinsic (pure) semiconductor: nc = pv Typical values of conductivity for intrinsic semiconductors (at room temperature): Si: = 10-4 -1m-1 Ge: = 1 -1m-1 Often measure much higher values than this. Why? small amounts of impurities. Deliberate addition of impurities in a controlled way allows tailoring of charge concentration (and hence ) to desired values. Process is referred to as doping – essential for device fabrication. Doped semiconductors referred to as extrinsic. Most important impurities (dopants) are from: Group V in Periodic Table e.g. P, As Group III in Periodic Table e.g. B, Al [Si, Ge etc. – Group IV] QUALITATIVE PICTURE OF DOPING 1. Intrinsic Semiconductor e.g.Si [2-D representation] each Si atom has 4 nearest neighbours strong, covalent bonds – tetrahedral bonding in 3-D 2. Doping with Group V Impurity e.g. P P atom has 5 available electrons to form bonds with neighbouring atoms e- 4 go to form bonds with neighbouring P Si atoms 5th electron is unpaired and weakly bound to P. Thermal excitation is usually enough to break bond between P and unpaired electron. e- This leaves an ionised P (donor atom) P+ bonded to the Si network, and a free (donor) electron which can carry charge Can represent process on energy level diagram: conduction band c d valence v band At T = 0, all donor electrons are weakly bound to P atoms – represented by level d in energy diagram. Typically c – d = 0.01 – 0.1 eV This is small or comparable with thermal energies (kT = 0.025 eV at room temperature). So at room temperature, most of the donor atoms have given up their excess electron (most excited from E d into the conduction band) large increase in nc (concentration of electrons in conduction band) large increase in conductivity . Semiconductors doped with donors – referred to as n-type. Typical doping levels – 1 part in 10 6. 3. Doping with Group III Impurity e.g. B B has 3 available electrons to form with neighbouring atoms. B adopts tetrahedral bonding but with broken bond. [In fact, broken bond shared B between all 4 bonds to neighbouring Si’s] Can regard as tetrahedrally bonded B binding a hole. B can capture (accept) an electron from a Si-Si bond elsewhere. This leaves a negatively ionised B atom (acceptor), and a broken Si-Si B- bond (mobile or free hole). Can think of B as donating a free hole. Process occurs easily at thermal energies (room temp’). Energy level diagram at T = 0 Ec Ea Ev Energy level diagram at Room Temp’. Ec Ea Ev Thermal energy large compared with E a – Ev. Most holes excited down into valence band. [Alternatively: large numbers of electrons excited from valence band up into levels at Ea – leaves large numbers of holes in valence band]. Large increase in pv. Large increase in . Acceptor–doped semiconductors – referred to as p-type. EXTRINSIC SEMICONDUCTORS - SUMMARY SO FAR n-Type Doped with Group V donor impurities e.g. P, As If ionised, impurities donate extra electron (into conduction band) Nearly all donors (positively) ionised at Room Temperature. Hence: large increase in nc at Room Temp. large increase in at Room Temp. p-Type Doped with Group III acceptor impurities e.g. B If ionised, impurities accept extra electron (from valence band) – leaves extra hole in valence band. Nearly all acceptors (negatively) ionised at Room Temp. Hence: large increase in pv at Room Temp. large increase in at Room Temp. QUASI-BOHR MODEL FOR DONORS Can estimate binding energy for donor electrons in n- type material using a quasi-Bohr model. Recall Bohr model for H atom i.e. electron orbitting positively charged H nucleus (proton). [PA114]. Apply model to n-type semiconductor (doped with P) Weakly bound unpaired electron orbits singly charged P nucleus. r e- + P Electron circles P + under action of central Coulomb force 2 e m r * 2 (1) 4 r 0 2 m* - effective mass of electron - dielectric constant of semiconductor Bohr postulate: angular momentum quantised in units of m r n * 2 [n – integer] (2) Eliminate between eq’ns (1) and (2) to give 4 n 2 2 r 2 0 * (3) em Kinetic energy of orbiting electron T given by 4 * 1 em T m r * 2 24 n 2 2 2 2 0 Potential energy U is given by e e m 2 4 * U 4 r 4 n 0 0 2 2 2 Hence total energy En is given by e m 4 * En = U + T 24 n 2 2 2 0 So orbiting electron can occupy a series of levels just below conduction band edge c. n=1 lowest bound state – ground state n En 0 i.e. impurity atom ionised, electron goes into conduction band So binding energy Eb of donor electron is given by e m 4 * Eb = E1 - E = 24 2 2 0 Example For Si: = 12 Hence (using) m* = m Eb = - 0.09 eV r=6Å i.e. electron is weakly bound [Can do similar calculation for hole orbiting around acceptor in p-type semiconductor] ELECTRON AND HOLE CONCENTRATIONS – NC AND P V - IN DOPED SEMICONDUTORS Doping large increase in nc (or p v) large increase in conductivity Can we calculate nc and pv for a given amount of doping? n-type semiconductor Si sample doped with concentration Nd of P donors. Assumption: complete ionisation of donors Can determine nc and pv using: (1) charge neutrality (2) law of mass action Charge neutrality Positively charged: ionised donors holes in valence band Negatively charged: electrons in conduction band So must have: Nd + pv = nc (1) Law of mass action ncpv = ni2 (2) [true for intrinsic and extrinsic semiconductors] Solve equations (1) and (2) for nc and pv. nc2 – Ndnc – n i2 = 0 N N 4n 2 2 nc = d d i (3) 2 [Take positive root – negative one gives negative nc!] Hence from eq’n (2) pv = n i2/nc So from eq’n (3), if n i << Nd nc Nd and also (from eq’n (1)) pv << nc i.e. can neglect pv p-type semiconductor Similarly analysis for p-type semiconductor with concentration Na of acceptors gives pv Na nc << pv i.e. can neglect nc EXAMPLES! Question 1: A Si sample is doped with 10-4 atomic% of P donors. Assuming complete ionisation of donors at room temperature, calculate the charge carrier concentration and conductivity at room temperature. [For Si: = 2330 kg m-3, atomic weight = 28, e = 0.15 m2V-1s-1] Answer 1 N 10 d 6 N Si where NSi – number of Si atoms per unit volume Obtain NSi from Avagadro’s No. N 6 10 23 28 10 3 Si This gives NSi = 5 × 1028 m-3 So Nd = 5 × 1022 m-3. Complete ionisation, n-type semiconductor: So charge carrier concentration is nc = Nd = 5 × 1022 m-3 (neglect pv) Conductivity = ncee [hole contribution negligible] So = 5 × 1022 1.6 10-19 0.15 -1m-1 = 1200 -1m-1 Question 2: What are the carrier concentrations and Conductivity in intrinsic Si? [For Si: g = 1.1 eV, me* = 0.25me, mh* = 0.5me, e = 0.15 m2V-1s-1, h = 0.05 m2V-1s-1] Answer 2 Intrinsic nc = pv = ni 3 kT ni = 2m m 3 2 exp * * g Obtain n i from: 4 2kT e h 2 For Si: g = 1.1 eV, me * = 0.25me, mh* = 0.5me So at room temperature (T = 300K): ni = 9.8 × 1015 m-3 Conductivity = ncee + pveh = niee + nieh = (2.3 10-4) + (7.8 10-5) -1m-1 = 3.1 10-4 -1m-1 Comparing answers for Q1 and Q2: doping Si with 1 part in 106 (of P) has led to an increase in of a factor > than 106. CONDUCTIVITY OF EXTRINSIC SEMICONDUCTORS AS FUNCTION OF TEMPERATURE T Summary so far Assuming complete ionisation of donor or acceptor impurities: doping semiconductors can lead to large increases in can estimate carrier concentrations and under these conditions Questions How does vary with temperature T in doped semiconductors? Does position of Fermi level F change with doping? Qualitative Picture n-type semiconductor E Ec Ed Eg Ev T = 0K – Insulating! Valence band full. Donor electrons all in bound states – energy Ed. No electrons in conduction band. So E F must lie between Ed and Ec. 1. As T is raised, weakly bound donor electrons are promoted into the conduction band – rises 2. Eventually, nearly all donor electrons have been excited into conduction band. However, T still not high enough for excitation across energy gap E g – levels off. 3. At higher T still, start to get excitation of electrons across Eg – intrinsic behaviour. rises rapidly. Graph of ln vs T-1 for n-type semiconductor ln E slope = [See end Lecture 1] g 2k E E slope = c d 2k 2 1 T-1 Analysis in Tanner (pages 127-130) shows: In region 1 (low T) (E E ) nc = ZN 1 exp 2 c d 2kT d [nc – conc’n of electrons in conduction band] 1 1 N F = (Ec + d) + kTln d 2 2 Z So for an extrinsic n-type semiconductor, F lies between c and d, and falls with increasing T. [Nd << Z] [Nd – donor concentration] [Z – (average) density of states in conduction band] p-type semiconductor Similar results for extrinsic p-type semiconductor. c a v Concentration of acceptors Na For an extrinsic p-type semiconductor, F lies between a and v, and rises linearly with increasing temperature. Hole concentration pv (at low T) given by (E E ) pv = ZN 1 2 exp a v 2kT a

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