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MIT 6 012 - Semiconductor Physics

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6.012 Lecture 2Electronic Devices and Circuits - S2007 1Lecture 2Semiconductor Physics (I)Outline• Intrinsic bond model : electrons and holes• Generation and recombination• Intrinsic semiconductor• Doping: Extrinsic semiconductor• Charge NeutralityReading Assignment:Howe and Sodini; Chapter 2. Sect. 2.1-2.36.012 Lecture 2Electronic Devices and Circuits - S2007 21. Silicon bond model:electrons and holes• Electronic structure of silicon atom:– 10 core electrons (tightly bound)– 4 valence electrons (loosely bound, responsible for most of the chemical properties• Other semiconductors:– Ge, C (diamond form)– GaAs, InP, InGaAs, InGaAsP, ZnSe, CdTe (on the average, 4 valence electrons per atom)Si is Column IV of the periodic table:IIIAIVAVAVIAIIBBCNOAl SiPSZnGaGeAsSeCdInSnSbTe567813 141530 31333448 4950 511632526.012 Lecture 2Electronic Devices and Circuits - S2007 3Silicon crystal structure• Diamond lattice: atoms tetrahedrally bonded by sharing valence electrons– covalent bonding• Each atom shares 8 electrons– low energy situation• Si atomic density : 5 x 1022cm-35.43 A°A°3sp tetrahedral bond2.356.012 Lecture 2Electronic Devices and Circuits - S2007 4Simple “flattened” model of Si crystal• All bonds are satisfied – ⇒ all valence electrons engaged in bonding• No “free” electronsAt 0K:two electrons in bond4 valence electrons (–4 q),contributed by each ion silicon ion (+ 4 q) border of bulk silicon region6.012 Lecture 2Electronic Devices and Circuits - S2007 5At finite temperature• Finite thermal energy• Some bonds are broken• “free” electrons – Mobile negative charge, -1.6 x 10-19C• “free” holes – Mobile positive charge, +1.6 x 10-19CCaution: picture is misleading!Electrons and holes in semiconductors are “fuzzier”: they span many atomic sitesborder of bulk silicon region+incomplete bond (mobile hole)–mobile electron6.012 Lecture 2Electronic Devices and Circuits - S2007 6A few definitions:• In 6.012, “electron’ means free electron• Not concerned with bonding electrons or core electrons•Define:–n ≡ (free) electron concentration [cm-3]–p ≡ hole concentration [cm-3]6.012 Lecture 2Electronic Devices and Circuits - S2007 72. Generation and Recombination• Requires energy from thermal or optical sources (or external sources)• Generation rate: • In general, atomic density >> n, p ⇒– supply of breakable bonds virtually inexhaustibleG = G(th)+Gopt+....[cm−3•s−1]G≠f(n,p)GENERATION=break-up of covalent bond to form electron and hole pairsRECOMBINATION=formation of covalent bond by bringing together electron and hole• Releases energy in thermal or optical form• Recombination rate: • 1 recombination event requires 1 electron + 1 hole ⇒R=[cm−3•s−1]R∝ n•pGeneration and recombination most likely at surfaces where periodic crystalline structure is broken6.012 Lecture 2Electronic Devices and Circuits - S2007 83. Intrinsic semiconductorTHERMAL EQUILIBRIUMSteady state + absence of external energy sourcesGeneration rate in thermal equilibrium:Recombination rate in thermal equilibrium:Go= f(T)Ro∝ no•poIn thermal equilibrium:Every process and its inverse must be EQUALGo(T) =Ro⇒ nopo=koGo(T)nopo= ni2(T)ni≡ intrinsic carrier concentration [cm−3]In Si at 300 K (“room temperature”): ni≈ 1x1010cm-3 In a sufficiently pure Si wafer at 300K (“intrinsic semiconductor):no= po= ni≈ 1 ×1010cm−3niis a very strong function of temperatureT ↑⇒ ni↑Only function of T6.012 Lecture 2Electronic Devices and Circuits - S2007 94. DopingDoping = engineered introduction of foreign atoms to modify semiconductor electrical propertiesA. DONORS: • Introduce electrons to semiconductors (but not holes)• For Si, group V elements with 5 valence electrons (As, P, Sb)IIIAIVAVAVIAIIBBCNOAl SiPSZnGaGeAsSeCdInSnSbTe567813 141530 31333448 4950 511632526.012 Lecture 2Electronic Devices and Circuits - S2007 10Doping: Donors Cont’d...• 4 electrons participate in bonding• 5th electron easy to release⇒– at room temperature, each donor releases 1 electron that is available for conduction• Donor site become positively charged (fixed charge)• If Nd<< ni, doping is irrelevant– Intrinsic semiconductor→no=po=niDefine:Nd≡ donor concentration [cm-3]border of bulk silicon regionAs+immobile ionized donor–mobile electron6.012 Lecture 2Electronic Devices and Circuits - S2007 11Doping: Donors Cont’d...• If Nd>> ni, doping controls carrier concentration– Extrinsic semiconductor⇒n0= Ndpo=ni2NdNote: no>> po: n-type semiconductorExample:Nd=1017cm-3→ no=1017cm-3, po=103cm-3In general: Nd≈ 1015-1020cm-3• Electrons = majority carriers• Holes = minority carrierslg Ndlg nolg ponoponiniintrinsicextrinsic6.012 Lecture 2Electronic Devices and Circuits - S2007 12Doping : AcceptorsA. ACCEPTORS: • Introduce holes to semiconductors (but not electrons)• For Si, group III elements with 3 valence electrons (B)IIIAIVAVAVIAIIBBCNOAl SiPSZnGaGeAsSeCdInSnSbTe567813 141530 31333448 4950 511632526.012 Lecture 2Electronic Devices and Circuits - S2007 13Doping: Acceptors Cont’d...• 3 electrons participate in bonding• 1 bonding site “unsatisfied” making it easy to “accept” neighboring bonding electron to complete all bonds⇒– at room temperature, each acceptor “releases” 1 hole that is available for conduction• Acceptor site become negatively charged (fixed charge)• If Na<< ni, doping is irrelevant– Intrinsic semiconductor→no=po=niDefine:Na≡ acceptor concentration [cm-3]B–+immobile negatively ionized acceptormobile hole and later trajectory6.012 Lecture 2Electronic Devices and Circuits - S2007 14Doping: Acceptors Cont’d...• If Na>> ni, doping controls carrier conc.– Extrinsic semiconductor ⇒po= Nano=ni2NaNote: po>> no: p-type semiconductorExample:Na=1017cm-3→ po=1017cm-3, no=103cm-3In general: Na ≈ 1015-1020cm-3• Holes = majority carriers• Electrons = minority carrierslg Nalg nolg ponoponiniintrinsicextrinsic6.012 Lecture 2Electronic Devices and Circuits - S2007 155. Charge Neutrality• The semiconductor remains charge neutral even when it has been doped– ⇒ Overall charge neutrality must be satisfied• In general: ρ= qpo−no+Nd−Na()Let us examine this for Nd= 1017cm-3, Na= 0We solved this in an earlier example:no= Nd= 1017cm−3,po=ni2Nd= 103cm−3Hence:ρ≠ 0!!What is wrong??border of bulk silicon regionAs+immobile ionized


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MIT 6 012 - Semiconductor Physics

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