Introduction To Semiconductors - MIT OpenCourseWare

6.012 - Electronic Devices and CircuitsLecture 1 - Introduction to Semiconductors - Outline Introductions/AnnouncementsHandouts:1. General information, reading assignments (4 pages)2. Syllabus3. Student info sheet (for tutorials, do/due in recitation tomorrow!)4. Diagnostic exam (try it on-line)5. Lecture 1Rules and regulations (next foil) Why semiconductors, devices, circuits? Mobile charge carriers in semiconductorsCrystal structures, bondingMobile holes and electronsDopants and doping Silicon in thermal equilibriumGeneration/recombination; nopo productno , po given Nd, Na; n- and p-types DriftMobilityConductivity and resistivityResistors (our first device)Clif Fonstad, 9/10/09Lecture 1 - Slide 1

Comments/Rules and expectationsRecitations: They re-enforce lecture.They present new material.They are very important.Tutorials:They begin Monday, September 14.Assignments will be posted on website.Homework: Very important for learning; do it!!Cheating: What you turn in must be your own work.While it is OK to discuss problems with others, you shouldwork alone when preparing your solution.Reading assignment (Lec. 1)Chapter 1 in text*Chapter 2 in text* "Microelectronic Devices and Circuits" by Clifton if Fonstad, 9/10/09Lecture 1 - Slide 2

SEMICONDUCTORS: Here, there, and everywhere! Computers, PDAs, laptops,Silicon (Si) MOSFETs, Integrated Circuits (ICs),anything “intelligent”CMOS, RAM, DRAM, flash memory cells Cell phones, pagers, WiFiSi ICs, GaAs FETs, BJTs CD players, iPodsAlGaAs and InGaP laser diodes, Si photodiodes TV remotes, mobile terminalsLight emitting diodes Satellite dishesInGaAs MMICs Optical fiber networksInGaAsP laser diodes, pin photodiodes Traffic signals, carGaN LEDs (green, blue)taillights, dashboardsInGaAsP LEDs (red, amber) Air bagsSi MEMs, Si ICsThey are very important, especially to EECS types!!They also provide:a good intellectual framework and foundation,anda good vehicle and contextwith whichto learn about modeling physical processes,andto begin to understand electronic circuit analysis and design.Clif Fonstad, 9/10/09Lecture 1 - Slide 3

Silicon:our default example and our main focusAtomic no. 1414 electrons in three shells: 2 ) 8 ) 4i.e., 4 electrons in the outer "bonding" shellSilicon forms strong covalent bonds with 4 neighborsSi bondingconfigurationSilicon crystal ("diamond" lattice)Figure by MIT OpenCourseWare.Silicon crystal("diamond" lattice)Clif Fonstad, 9/10/09Lecture 1 - Slide 4

Intrinsic silicon - pure, perfect, R.T.:Electronenergy All bonds filled at 0 K, po no 0ConductingstatesSi-Eg At R. T., po no ni 1010 cm-3 Mobile holes ( ) and mobile electrons (-) Compare to 5 x 1022 Si atoms/cm3Clif Fonstad, 9/10/09Eg 1.1 eVBondingstatesDensity of electronenergy statesLecture 1 - Slide 5

Intrinsic Silicon:pure Si, perfect crystalAll bonds are filled at 0 K.At finite T, ni(T) bonds are broken:Filled bond " Conduction electron HoleA very dynamic process, with bonds breaking and holes andelectrons recombining continuously. On average:! Concentration of conduction electrons " n Concentration of conduction electrons " pIn thermal equilibrium : n no p poandn o po n i (T)The intrinsic carrier concentration, ni, is very sensitive totemperature, varying exponentially with 1/T:!n i (T) " T 3 / 2 exp(#E g /2kT)10#3In silicon at room temperature, 300 K: n i (T) " 10 cmClif Fonstad, 9/10/09!In 6.012 we only "do" R.T.!A very importantnumber; learn it!!Lecture 1 - Slide 6

1010 cm-3 is a very small concentration and intrinsic Si is an insulator; we need to do somethingExtrinsic Silicon:carefully chosen impurities (dopants) addedColumn IV elements (C, Si, Ge, lumn V elements (N, P, As, Sb):too many bonding electrons electrons easily freed to conduct (-q charge) fixed ionized donors created ( q charge)Clif Fonstad, 9/10/09Lecture 1 - Slide 7

A column V atom replacing a silicon atom in the lattice:ElectronenergySb Ed One more electron than needed for bonding.Easily freed to conduct at RT.Impurity is an electron "donor."Mobile electron (-) and fixed donor ( ); Nd Nd.Clif Fonstad, 9/10/09Ed 45 meVDensity of electronenergy statesLecture 1 - Slide 8

Extrinsic Silicon, cont.:carefully chosen impurities (dopants) addedColumn IV elements (C, Si, Ge, lumn III elements (B, Al, Ga, In):too few bonding electrons leaves holes that can conduct ( q charge) fixed ionized acceptors created (-q charge)Clif Fonstad, 9/10/09Lecture 1 - Slide 9

A column III atom replacing a silicon atom in the lattice:ElectronenergyBEa 45 mevEa One less electron than needed for bonding.Bond easily filled leaving mobile hole; at RT.Impurity is an electron "acceptor."Mobile hole ( ) and fixed acceptor ( ); Na Na.Clif Fonstad, 9/10/09Density of electronenergy statesLecture 1 - Slide 10

Extrinsic Silicon:carefully chosen impurities (dopants) n IV elements (C, Si, Ge, α-Sn)IIColumn V elements (N, P, As, Sb):too many bonding electrons electrons easily freed to conduct (-q charge) fixed ionized donors created ( q charge)Column III elements (B, Al, Ga, In):too few bonding electrons leaves holes that can conduct ( q charge) fixed ionized acceptors created (-q charge)Clif Fonstad, 9/10/09Lecture 1 - Slide 11

Extrinsic Silicon:What are no and po in "doped" Si?Column V elements (P, As, Sb): "Donors" Concentration of donor atoms " N d [cm-3 ]Column III elements (B, Ga): "Acceptors"! Concentration of acceptor atoms " N a [cm-3 ]At room temperature, all donors and acceptors are ionized: N d " N d!! N a- " N aWe want to know,"Given Nd and Na, what are no and po?"Two unknowns, no and po, so we need two equations.Clif Fonstad, 9/10/09Lecture 1 - Slide 12

Extrinsic Silicon: Given Na and Nd, what are no and po?Equation 1 - Charge conservation (the net charge is zero):q( po " n o N d " N a" ) 0 # q( po " n o N d " N a )First equationEquation 2 - Law of Mass Action (the np product is constant in TE):!n o po n i2 (T)Second equationWhere does this last equation come from?The semiconductor !is in internal turmoil, with bonds being broken andreformed continuously:Completed bond " # Electron HoleWe have generation:Completed bond ""# Electron Hole!occurring at a rate G [pairs/cm3-s]:!Clif Fonstad, 9/10/09Generation rate, G Gext go (T) Gext " gm (T)mLecture 1 - Slide 13!

And we have recombination:Electron Hole ""# Completed bondoccurring at a rate R [pairs/cm3-s]:!Recombination rate, R n o po ro (T) n o po " rm (T)mIn general we have:dn dp G " R Gext # gm (T) " n p# rm (T)dt dtmm!In thermal equilibrium, dn/dt 0, dp/dt 0, n no, p po, and Gext 0,so:0! G " R # gm (T) " n o po # rm (T) mm#gmm(T) n o po # rm (T)mBut, the balance happens on an even finer scale. The Principle ofDetailed Balance tells us that each G-R path is in balance:!gm (T) n o po rm (T) for all mThis can only be true if nopo is constant at fixed temperature, so wemust have:2np n(T)ooi!Clif Fonstad, 9/10/09Lecture 1 - Slide 14

Another way to get this result is to apply the Law of Mass Action fromchemistry relating the concentrations of the reactants and productsin a reaction in thermal equilibrium:Electron Hole " # Completed bond[ Electron][Hole] [Completed bond ] k(T)We know [Electron] no and [Hole] po, and recognizing that most!of the bonds are still completed so [Completed bond] is essentially!a constant*, we haven o po [Completed bond] k(T) " A k(T) n i2 (T)Back to our question: Given Na and Nd, what are no and po?!Equation 1 - Charge conservation (the net charge is zero):q( po " n o N d " N a" ) 0 # q( po " n o N d " N a )First equationEquation 2 - Law of Mass Action (the np product is constant in TE):!n o po n i2 (T) Second equationClif Fonstad, 9/10/09* This requires that no and po be less than about 1019 cm-3.!Lecture 1 - Slide 15

Extrinsic Silicon, cont: Given Na and Nd, what are no and po?Combine the twoequations:Solving for no we find:no (# n i2&% " no N d " N a ( 0 no'n o2 " ( N d " N a ) n o " n i2 0(N d- N a ) (N d - N a ) 4n i2! 2(N d- Na ) " 1 2 #2 ),%2n i2 1 .'2.* (N d - N a ) -'&(N d- Na ) " 1 2 #%4n i2'1 2(N d - N a ) '&Note: Here we have used1 x " 1 x 2 for x 1This expression simplifies nicely in the two cases we commonlyencounter:! ( N d " N a ) n iCase I - n - type : N d N a and!Case II - p - type : N a N d and (N a " N d ) n iClif Fonstad, 9/10/09Fact of life: It is almost impossible to find a situationwhich is not covered by one of these two cases.!Lecture 1 - Slide 16

Extrinsic Silicon, cont.: solutions in Cases I and IICase I - n-type: Nd Na:, (Nd - Na) ni"n-type Si"Define the net donor concentration, ND:We find:N D " (N d # N a )n o " N D , po n i2 (T) /n o " n i2 (T) /N D!In Case I the concentration of electrons is much greater thanthat of holes. Silicon with net donors is called "n-type".!Case II - p-type: Na Nd:, (Na - Nd) ni"p-type Si"Define the net acceptor concentration, NA:N A " (N a # N d )We find:po " N A , n o n i2 (T) / po " n i2 (T) /N A!In Case II the concentration of holes is much greater than thatof electrons. Silicon with net acceptors is called "p-type".!Clif Fonstad, 9/10/09Lecture 1 - Slide 17

Uniform material with uniform excitations(pushing semiconductors out of thermal equilibrium)A. Uniform Electric Field, ExDrift motion:Holes and electrons acquire a constant net velocity, sx,proportional to the electric field:εxNo fieldE-field appliedsex " µ e E x , shx µ h E x xNo fieldE-field appliedFigure by MIT OpenCourseWare.At low and moderate E , the mobility, µ, is constant.At high E the velocity saturates and µ deceases withincreasing E .Clif Fonstad, 9/10/09Lecture 1 - Slide 18

Uniform material with uniform excitations(pushing semiconductors out of thermal equilibrium)A. Uniform Electric Field, Ex , cont.Drift motion:Holes and electrons acquire a constant net velocity, sx,proportional to the electric field:sex " µ e E x , shx µ h E xAt low and moderate E , the mobility, µ, is constant.At high E the velocity saturates and µ deceases.!Drift currents:Net velocities imply net charge flows, which imply currents:J exdr "q n o sex qµ e n o E x!J hxdr q po shx qµ h po E xNote: Even though the semiconductor is no longer in thermalequilibrium the hole and electron populations still have theirthermal equilibrium values.Clif Fonstad, 9/10/09Lecture 1 - Slide 19

Velocity saturationThe breakdown of Ohm'slaw at large electric fields.108SiliconCarrier drift velocity (cm/s)GaAs (electrons)Ge107Above: Velocity vs. field plotat R.T. for holes andelectrons in Si (log-logplot). (Fonstad, Fig. 3.2)106Left: Velocity-field curves forSi, Ge, and GaAs at R.T.(log-log plot). (Neaman, Fig. 5.7)Si105102103T 300KElectronsHoles104Electric field (V/cm)105106Figure by MIT OpenCourseWare.Clif Fonstad, 9/10/09Lecture 1 - Slide 20

Conductivity, σo:Ohm's law on a microscale states that the drift current density islinearly proportional to the electric field:J xdr " o E xThe total drift current is the sum of the hole and electron driftcurrents. Using our early expressions we find:J xdr J exdr J hxdr qµ e n o E x qµ h po E x q (µ e n o µ h po ) E x!From this we see obtain our expression for the conductivity:" o q (µ e n o µ h po )![S/cm]Majority vs. minority carriers:Drift and conductivity are dominated by the most numerous, or"majority," carriers:!n-typen o po " # o qµ e n op-typepo n o " # o qµ h poClif Fonstad, 9/10/09Lecture 1 - Slide 21!

Resistance, R, and resistivity, ρo:Ohm's law on a macroscopic scale says that the current andvoltage are linearly related: v ab R iDw vABThe question is, "What is R?"σοJ xdr "!o E xviwith E x AB and J xdr Dlw#tWe have:Combining these we find:iDv AB #ow"tl!which yields:v AB l iDl1iD R iDw " t #owhere!tR l1l %ow " t #ow"tNote: Resistivity, ρo, is defined as the inverse of the conductivity:Clif Fonstad, 9/10/09!" o # 1 o[Ohm - cm]Lecture 1 - Slide 22