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Physical Electronics

by: Manuela Boyer PhD

Physical Electronics EEE 166

Manuela Boyer PhD

GPA 3.66

Russell Tatro

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Russell Tatro
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This 73 page Class Notes was uploaded by Manuela Boyer PhD on Monday October 5, 2015. The Class Notes belongs to EEE 166 at California State University - Sacramento taught by Russell Tatro in Fall. Since its upload, it has received 22 views. For similar materials see /class/218827/eee-166-california-state-university-sacramento in Electrical Engineering at California State University - Sacramento.

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Date Created: 10/05/15
Prevlew Inthe ree Chapter 9 p edlng two ehapters we eonsrolereolthe pnjun t on and s urneolthe sernreonduetorrnatenal as the sarne throughout the entare strueture The type ofjunetlon ls referredto as ahamylmctmn 1n thrs ha ter we eonslder the metalrsemleonduetorjunetlon andthe sernreonoluetor heterojuneuon m whleh rnatenal on eaeh slde ls not the MetalSemiconductor and same Semiconductor Heterojunctions ernreonoluetordeyrees mustmak eontaetwrth the outsroleworlol Thls rough nonreetlfylng metalrsemleonduetor tuneu ons or Ohm canmcts e eontae An ohrn r alowrreslstanee junctlon proyrdrng eurrent eonoluetaon m both dlreetlons Seetron 9 l The Sehottky Barrrer Dlode lreular ohmle eontaetrn ahlgh power devlee f rorn y y Teehnlsehe Unlversltelt Elndhoven use ofmetal touehrng a sernreonoluetorto proolu enrsteol slnee the early 1900s Slnee th e a dlode has e 1950s there has been rnueh effort wth sern oluetor anol yaeuurn teehnology to fabrleate reproduelble anol rellable metalrsemleonduetor eontaets lnlnmannncllmuulwtrwnl numan eumutout I The Sehotthy barner dlode ls ametalrsemleonduetorreeufylng eontaet otuulc os eases the reetlfylng eontaets are rnaole on nrtype WWquot sernreonoluetors Drawmg by V Mterostruetur 39439 Matenals Engrn quot quot39 eerlng M0 a laboratory Kyoto Japan Seeuon 9 1 1 Qualltatlve charactensues Table 9 1 Work Funetrons ofsome elernents ldeal energyrband dlagram for a partaeular rnetal anol netype sernreonoluetorbefore maklng eontaetrs shown below tnlnnunt Ehgkcmmm v ls the metal warrunoanhnvults A 1 T trsunsennonuntorwnlrnlnn Us I v l t Table 9 2 Electron afflmty of some semleonduetors 7 un nu u The vacuum level ls used as a reference level laA stuntquot mutt uu luntt t Qualrtahve Charaetenstres Qualrtahve Charaetenshes The ldeal thermal equlllbnum energyrband dlagram Under thermal equrlrbnum the Ferml level mustbe eonstant through the system hauls the ldeal barner herght ofthe semreonduetor eontaet Eleetrons from the semreonduetor must ow rnto the lower energy states tr mm t 39 t m ofthe metal for the Ferml level to aehreve rts eonstant state tn Posrhvely eharged donor atoms remam m the semreonduetor ereatrng a d Before eontaet the Ferml level m the semleonduetorwas above that m eplenon reglon the metal Qualrtahve Charaetenstres Qualrtahve Charaetenshes The bamer hauls known as the Schat cy ddmet andls grven rdeally by Ifwe apply aposrtrve voltage to the metal wrth respeetto the semreonduetor forward bras the semlconductorrtormetal barnerherght 1 Vb deereases whlle Euremams eonstantrn ths ldeallzed ease u mrx The next sllde shows the energy band dlagrams for the reverse and forward blas eases on the semreonduetor slde Vhls the bulltrln potentral barrrer Srnee the energy band dlagrams are srmrlarto the pn Junctlon dlagrams we enpeet slmllarbehavlor The eurrent meehamsm here however ls Vol au 39 n due to the ow of majority earner eleetrons Ifwe apply aposmve voltage to the semreonduetorwrth respeet to the metal reverse bras the semlconductorrtormetal barrrer herghtrnereases whlle on remams eonstant m thrs ldeallzed ease Flgure 9 2 Sectlon 912ldeal Junction Properties Reverse Blas he analysls ofthe eleetrostahe propemes of th5 metal to semlconductorjuncnon shows that the E eldls allnearfunctlon of drstanee from the junction the depletron reglon wrdth w ls rdentreal to that ofthe one sldejuncuon 0quot 1 E Fleld 5 Deplehon reglon wrdth C 3925 Junetron Capaertanee 1 Vr Forward Blas Exercise on page 331 Ideal ehrorniurntontype silieon Schottky diode at T 300 K Nd 3 x 10 5 cm393 Determine the ideal Schottky barrier height BD457401049V fromTableQ 1 and92 Determine the builtin potential barrier xlo gcm 0 2368V N 28 Vln 002591n lb Nd 3x10 5m named 049702370253V Determine the peak electric eld for VR 5V i A a hawk Zll7885 10 gt102534r5 ngwama 9N1 16x10 C3x10 cm 15054x10 cm Example 92 on page 331 nu silieon diode experimental data shown is gure 9 3 at T 300 K eapaeitanee Note that vb 0 4v the datais plotted as VP zaNa iiz mm mm 0 am The slope ofthe line ofthe experimental data can he given by la la d1C 7 my R1 A1C 7 2 M 7 n 25 AVn 725 Now we can nd the value ofthe slope from the gure Example 92 Solve for Nd 2 2 N N4 25 4 4X10 1 6 x10quot911 78 85 xlo 4 4 x10 2 744x10 7cm 3 This is the donor doping in the ntype sernieonduetor Now let us calculate the barrier height Exercise on page 331 Ideal ehrorniurntontype silieon Schottky diode at T 300 K Nd x 10 5 cm393 Continue nding the peak eleetne eld u u a 16X10 0mm m gt15xlo 70 69535 ll7885xlo g ap 7 x 3 Determine the junction eapaeitanee per unit area for vR 5v C 2511311 16x10quotyC1178 85x10quot gt3X10 cm 3 2Vm VR 20 2535 4 731x10quot7 i Figure 9 3 Finding the slope twill Example 92 The energy qquot EC 7 Ep and is 2 8x10 95m 002591n 27x10 7cm 3 0 02594 542 0 120V So the aetual Schottky barrier at the metal ntype junetion is oh which is approximately the ideal barrier height om db Vh 4 040012 052V Section 9 1 3 Nohtdeal Etfeets oh the BamerHetght The aetttal Sehottky bantet hetghtvattes tom the toleal sltghtly tom several effects Read thts seettoh fot the Schonky effect the effect ofthe 313cm eld on the bathet The bamethetght ts also aftthettoh of the surface states th the semteoholuetot Thts seettoh eoheluoles wtth the statement that sthee surface state olehstty ofthe semteoholuetot ts hotpte etable wtth ahy aeeutaey the bantet hetght must be found expettmehtally CurrentrVoltage Relataohshtp Fot a oheedtmehstohal bathet wtth forward blas voltage V applteol Thete are two euneht olehstttes 9mts the electron eutteht olehstty oltte to the ow of electrons tom the semteoholuetot thto the metal JM ts the electron ow tom the metal thto the semteoholttetot CurrentrVoltage Relataohshtp k Haw Let all ofthe electron ehetgy above Ec be klnenc ehetgy m E7Emw11 2mgquotij and K v Pluggmg thts tesult th equattoh 9 20 4 2 quot 39 e E7 it an quotJim V ap E h 2 kT expttm The veloetty ts foundby settahg the kthette ehetgy equal to the potehttal aetoss the nrtype tegtoh m39vel 2 t 24 Seettoh 9 14 OlmentrVoltage Relataohshtp The eutteht transport th ametalrsemlconductor tuhettoh ts oltte mathly to amets The basl ptoeess th the teettfythg eohtaet wtth ah nrtype semteoholuetot ts by transport of eleettohs ovet the potehtaal bantet whteh ts olesettbeol b the thetmtohte emtsstoh theoty The thetmtohte emtsstoh theoty asstt es the electrons have an ehetgy latget than the top ofthe bantet hetght whteh ts mueh latget than kT so LhatMaxweleroltzmann apptoxtmattoh appltes Thus the ptoeess ts uholet thetmal equtltbttum Thts ts hottttte ahdts only arough approxlm at on See wotk ofBethe ctowell and 523 CunentrVoltage Relattohshtp The eutteht olehstty 139mm a fuhettoh othe eoheehttattoh of electrons whteh xrdlrected veloettaes suf clentto overcome the bathet 1W EIvdn Where E ts the mthtmttm ehetgy tequtteol totthetmtohte emtsston thto the metal vxls the eametveloetty tn the otteetton ofh anspon The thetemehtal electron eoheehttatton dn gcEfE 01E a oor 42 Equations 3 72 and 3 80 CunentrVoltage Relattohshtp he thtegtal mustbe solved tot all thtee dtmehstohs dx oly dz The sultxs Where Aquot ts ealled the etfeetzve Rtehardsoh eohstant New 11 km thts ease ts Boltzmann39s eohstant Exermse 9 5 on page 341 The Schultkybamerhaght era 5111cm Schultkyjunmums 1 n 59 v IK T 311 4 MK2 m Crussrsamuml araA 1nt m Ca1eh1ate mama re ase E mun amt 1 1 59 1312 11411 J Artex e114 3111 z x l r p K J STEM rm ltzmzy l 39eWXUUmZIZ e 121mm Ca1ematethe dmde eurrmz term 1 KW 131x1nquot1n725n 141x1uquot1141m1 gure 9 1n 1 139 The afeetwe mman 1m1tage er 1 the Sehnttky mnde1s1essthan b Schuuh I that uflhepnjuncnun dmde 1111119 1 T 1 A 4 1 C 1 1 391 1 1 1 r 1 1 t 391 1 0 1 04 1m vuunns gt conprnsonomr SchnnkyEamszmdz andthz pnhmcnan Dad The seeono1 major drfferenee between a Sehottky barner dwde and apn junction dwde 1s 1n the frequency response or swrtehrng eharaeterrstaes The Scho ky devree 1s a majority earner devree Thus there 15 no dAqusmn eapae1tanee assoe1ated w1th a forward b1ased Schonky 61ch The ehrnrnataon ofthe drffusron eapaertanee makes the Schonky dwde a h1gherrfrequency devree wrth swrtehrng tarnes 1n the preoseeondrange versus nanoseeono1 range for the pn juncuon Seaman 915 conprnsonomr SchnnkyEamexDmdz andthz pn1nne1nn Dmdz A1though the 1dea1 cunentrvoltage re1atronshrp of the Sehottky barrrer 61001315 ofthe sarne orrn as that ofthe pnjuncuon dwde there are two rrnportant drtrerenees The rnagnrtudes ofthe reverse saturatron eunent densrtres are many orders of rnagnrtude dAfferent Schonky 1s brggere see example 9 5 As s1de effect ofth1s faet 15 that the tumron voltage forthe Sehottky dev1ee1s1ower Example 9 s on page 342 Censmer dmdes erthnsem Example 9 s T J 1Schuuky s 98 x 1115 Annl J pnjc1 xl uAEml Ca1eu1atethe furward has 1m1tages required gven11n Annz V J 1n JJ ex 771 VVIn7 1125917 P11 1 1 lt MKS nn25912nzn3115V FortheSehottky dwde Furthepnjunmun V J 1n JJex7712V h7nnzsu17 P11 1 1 lt Mhrorru 11 11259125 331 682V For the pn junction dwde Seet1on 9 2 MetalrSemmonductor Ohrme Contaets o taets rnust be made between any sernreonduetor devree andthe outsrde wor1o1 These eontaets are made mauhmx contacts 0hrnre eontaets are rneta11osernreonduetor eontaets but 1n thrs ease are not re cufymg eontaets An ohrnre eontaetrs alowrres1stance Juncuon promdmg eonduetaon 1n a1 both dArecuons between the met and the sermeonduetor There are two genera1 types of ohrnre eontaets the 1dea1 nonrecufymg barner andthe tunnehng barner Seeuoh 9 2 1 Ideal Nonrecnfylng Bamers In the Sehottky barrrer we eohsrderedthe ease where om gt Q In the ease ofthe ohrrue eohtaet we wan lt hi Whether forward or reverse blased electrons can ow easlly m elther dlrectlon Seeuoh 9 2 2 Tunnelmg Barner The tuhhehhg eurreht has the form mfgj were em Thus the tuhhehhg eurreht lncrease expohehually wrth doplng eoh eehtrau oh Seeuoh 9 2 3 Speel e Contact Reslstanee For a metalrtorsemlcon duetor Junctlon wrth hlgh rmpuhty doplng eoheehtrauohs the tuhhehhg proeess wlll dorhrhate 1h th5 ease Rc wlll be The theory offermlng ohrrue eohtaets ls stratghttorward A good ohrrue eohtaet needs a low barrrer and uses ahlghly doped serhreohduetor at the surface However rh praeuee rt ls not easy Mueh experrrhehtal workls requrredto successfully ereate agoodlowresrstauee eohtaet Seetloh 9 2 2 Tunnelmg Bamer epleu oh reglon wrdth m a rectlfylng metal rsemlconductor eohtaet ls lnversely proporuohal to the square root ofthe serhreohduetor doplng The wrdth ofthe depleuoh reglon deereases as the doplng eoheehtratroh m the serhreohduetorrhereases Loeally heavy doplng lncrease the probabrhty oftunnellng through the barrrer Seetroh 9 2 3 Speclfl Cohtaet Resrstauee A gure of merrt of ohrrue eohtaets ls the speclflc eohtaetresrstauee Rc Thls parameter ls de ned as the reclprocal ofthe dehvatrve ofeurreht denslty wrth respeet to voltage evaluated at zero blas a a M h m1 r w tlzc e as small as posslble for art ohmlc eohtaet Forlowto moderate serhreohduetor doplng therrmorue emlsslon eurrehtr aut ah ls glven by dorhrh d R 3 J T Seetloh 9 3 l Heterojunctlon Materlals The type ofsermeohduetorrhaterral dlscussed so far ls ealled a h mtgumth The rhthhsre eryst rst e same or eaeh slde ofthe Juheuoh wth doplng eohtrolhhg the behavlor ofthe devlce 1t ls posslble to bulld up from a substrate seed many layers that are dlsnncdy dlfferentwlth dlfferent rhatehals ahd doplng levels Thls type ofstxucture ls ealled aheterqlmctmn devlce more eorhpheated strueture ear gulde both the eamers electxo and holes and photons m more emerehtways Mueh to flndmatenals where the two erystal latuees rhateh Any latuee rmsrhat results rh straths and defects whlch degrade the performance f the devlce us work has oeeurred Flgure 7 example of more eomplreated heterojunctlon devlce eaten miU1 sham mmufmuh mumm IMhhtmmmm39me manusumu mme RuAanemmxLPmml Seetaon 9 3 3 Tworddmenslonal Eleetron Gas Please read thrs seetron but the seetron wlll be sklppedfornow Seetron 9 3 4 Heterojunctlon equlllbnum eleetrostatres The prevlous ean be alteredto ndLh e bulltrln voltage m terms of the valenee band and eonduetaon band shrlts V M Hm 51 Valenee band andhole 39 pnn N eoneentxauon equatlon n N V eeAEgztrln 4 WVquot N Conduetdon band and eleetron eoneentxauon equatlon Example 9 a mfageKSS Cunsdanrlype Ge dupedwthNd In A dFrGaAs dupedwlthN 1EI em m Em3andleleGaAs143 ev n 3nnllt Gen Axlulznnzand letlzwme 672V We sull needtu deternnevm me Table 3 4 NGe a x ln x tmzarld WGaAs7xll xtm3and Fustweneedthe nununty mmerhul euneentratlunlnthe ntype n 24xln smquot1 n quot7 757am quot N l cmquot m p l N VNAEVkT1n A m 1 6x10 u 7Ule U259aVnWW n7naVn U259 V12I U7UaVEI3thl Ille Seetron 9 3 2 EnergyrBand Dlagrams The gure below shows the energyrband d separated nrtype andPrtype he m agram oflsolated physlcally whlch atenal has the larger bandg matenal T eaprtalpmdleates aP The vaeuum level ls used as a referenee The eleetron afflmty of the wl er bandgap matenal exglsless than 39 494 th t the narrow bandgap matenal each I e a It 71 39 39 39 Mr AE Elw39v Exam Mm The last equataon ls the statement ofthe eleetron afflmty rule base on the vaeuum level belng a eonstant between the two materlals Exampleg a unpagezss CunsldernrtypeGedupedwthNdrl mmrzandlelEqAGzAsr 43ev AndFerAsdapedwthN lU Em3 T3nnllt Gen24xln13un3andlet5wce a7ev DewmnneAEvAEvandefanheaneluFrGaAshelaqunmmn Frumeduauun 9 34 aandtahle 9 2 AEKlzlner41374 U7le n my the afflmty ofamatenal ls 1 the wrdth ofthe eonduetron e Energyrequlredto lrberate an electro om an free spaee ethe orh s t dthe electxo afflmty s olts as ls eommon Hels attemptang to elue m the reader by the notataon ex and henee my use e eabove equataon AEKAEV em 33 emmemz l43ena7weu my I7EIle Semun94 Summary A metal un a llgntly duped semcunducmr En pmduee a renlrylng euntad that ls lmuwn as a Schuuky burner dlude The ldml burner haght between the metal and semeundudur ls the Menace between the metal wmk runetlun and the semeundueter eleetrun ardnlty When a pusltlve wltage ls applled tu an nrlype serneundletur wlth remeet m the metal reverse has the harner between the serneundletur and metal mmses su that there ls essmually nu uw ur eharged Emers Whena pusltlve Wltagels applledtu themetal wlthremeette annrtype sermeundleturaurward has the harner between the semleundletur and demses su that deetmns En uw mslly mm the semleundleturtu the metal by thennune emlsslun Section 9 4 Summary M r junctions providing conduction m both directions Section 9 4 Summary The ideal I 39 L 5Na C it EField 5 k1 4 5XP f Lowmoderate doping Where RF 2 thermionic emission dominates W m Depletion region Width A T eN C ZQNJ Junction Capacitance R 2 m ah High doping Where tunneling process 2Vm VR exp 1 W dominates J ATzexp7quotexpil A E Airzrz ez h Section 9 4 Summary L A r M 2m e 1 End of Chapter 9 AE AEV AEW Chapter 2 Introduction to Quantum Mechanics In order to understand the currentvoltage characteristics of a semiconductor We need some knowledge of the electron behavior in a crystal When subjected to various potential functions Work ruheu on The minimum energy to remove an electron from a material is called the Work function This is the energy to liberate an electron from a bound state in an atom or molecule and move it to free space The electron does not have excess energy in the form of momentum at this min value This minimum energy is supplied by a photon offrequency vu Thus the maximum kinetic energy T in this text of the freed electron is Tmax 2 mass velocity2 2 mv2 hv r hvu Where v 2 VB Exercise 2 1 Ehv 011 fmqv b A 10 A 1 nm a thousand times shorter A This is in the xray Wavelengths 6 625 x10 Jsee3x1051 he 15 Ehv WS 1988x10 J x m 1 988 10quot5J In terms of 2Vgt1 988x10quot5J X i9 1241 eV 1 502 x10 6 Seem 2 1 Principles of Quantum Mechanics Energy Quanta Experiments With the photoelectric effect and dark body radiation show that the energy comes in discrete packets E hv Where v frequency h Planck s constant 66262 x 103934 J sec The text uses h 6625 x 103934 J sec instead ofthe more exact value E is in units of Joules and can be expressed in electron volts eV 1 eV1602x1039lg I If an electron is accelerated through apotential of 1 volt then the energy i T kinetic energy 1 e V 1602 x 103919 J Thus We frequently interchange volts With electron volts to mean the same energy See appendix F on page 721 Exercise 2 1 page 29 Determine the energy ofa photon at a particular wavelength 5 Ehv 011 fruit1 a A 10000 A where 1 A 1039m m 0 1 x 10399 m 10398 em This is in the far infrared wavelengths he 6 625x10 3 Jsee3x1051 Lem 7 10 000 104D S 1 988x10quot9J x m 49 In terms of 2Vgt1 988x1049J 1 241 eV gtlt Light Waxermine Duality Waves exhibit particlelike behavior photoelectric effect Light Waves at only a certain minimum Wavelength Will liberate electrons in the above dev1ce Light WaverPamcle Duality Particles exhibit wavelike behavior diffraction interference F holugwmz mquot A stream of particles photons in a light beam can create an interference pattern that depends on the energy per photon de Broglie Wavelength In terms ofthe duality oflight de Broglie hypothesized that the wavelength of aparticle can be expressed by h de Broglie Wavelength 7 Note that these relationships are for VERY small particles smaller than neutrons and protons The Uncertainty Principle Heisenberg 1927 showed that we cannot describe with absolute accuracy the behavior of subatomic particles Thus we will describe the probabilities of particle behavior Iti 39 r quot pieci n the pu iliuu and momentum of a particle Ap is the uncertainty in the momentum Ax is the uncertainty in the position 39Ihen Ap Ax 2 h where h hZn 1054 x 1039 J sec Photon Momentum The momentum ofa photon is stated in terms of energy since the photon is massless The momentum is given by E p2 Exercise 2 2 page 29 a Find the momentum and energy of a particle with mass 5 x 103931 kg de Broglie A 180 A h 7 6625x103934Js m 36806x1039 21 180x10 m p 26kgquot 35805x10 Ns The energy of the particle is all kinetic in this case so 1 E thekmetic energy Ema212 where p mvel gt vel 1 2 m 2 2 3 68gtlt103916 kg 39m E71 p 7p 3 e2 4 737quot z 7E W 135x10 J 846x10 eV The Uncertainty Principle It also is impossible to simultaneously know the exact energy of a particle at some speci c instant of time t AB is the uncertainty in the energy At is the uncertainty in the time Then AEAt 2 h whereh hZn The physical meaning ofthis is shown by the exact position of an electron We cannot know the exact position ofthe electron at time t What we can develop is a probability density inction of its position Exercise 2 3page 30 The uncertainty in position ofan electron is Ax 12 A T 39 39 39 39 uuLcuaiut in corresponding uncertainty in kinetic energy it 6625x10 34Js 26 mph 3 8787gtlt10 AP AP Ax 2n12x10quot m r 2 Mtgm 2 8 787x1039 Wu 2m 2911x10 3 kg 4 237x10 2 J 0 02632V Schrodinger s Wave Equation 7 me independent form Read the derivation of the time independent form of the wave equation on page 31 on your own For the reading recall that v freq and on 27w or v 0321 Time Independent Wave Equation also called the position dependent equation TE7Vxyx 0 Where E the total energy of the particle Boundary Conditions The probability of nding the particle somewhere is certain thus Ely3de 1 this is over all space The following conditions are postulates but physical arguments d and them wx must be nite singevalued and continuous awx ac must be nite singevalued and continuous kgm Section 2 2 Schrodinger s Wave Equation The Wave Equation in one dimension and nonrelativistic VOWx I J39h iaw 0x2 6 6 We I 2m I Where ixt is de ned as the wave inction and is a complex Vx is de ned as the potential inction and is independent of time m is the mass of the particle which is still avery very small particle Physical Meaning ofthe Wave Equaaon We have a position dependent and a time dependent inction We 0 wx t wooe j jt A complex inction cannot by itself represent a real physically quantity But the magnitude lyxtl2 is the probability of nding the particle somewhere between x and x Ax at time t lyxtl2 is called the probability density inction of position since it is independent of time Section 2 3 Applications ofthe Wave Equation Electron in free space If no force is acting on the electron then the potential inction And E gt Vx or the electron would not be in free space For simplicity let Vx 0 which is a constant 821100 2mE 2 2 it thus Vx 0 Tnmlmg Wave Sahman The solutron to the wave equatron forthe eleetron m free spaee ls Llrmen Wm Ae Amen 4 82 quot Whleh ls m the form A 3quotquot H3 e39IX whleh ls atravellng wave CoetflelentA m the flrstterm ls for a wave txavelrng m the x ddrectlon Coemelent B m the seeondterm ls for awave txavelrng m the e x ddrectlon The lnrrnue Potenuel Well A partrele m the rn nrte potenual well ls a elassre example ofa e bound pamcl 41 H For an rn nrtely hlgh barrrer and a paruele wrth a nrte energy E then W zero m reglons 1 anle Thls means no paruele m reglon 1 and 111 The lnrrnue Potenuel Well 7 BoundaryCondmons Boundary condluons mustnow be applled Wx0 Ixa0 I Atx 0 may 0 A eos Kx0 A2 sln Kx0 thus 0 A 1 A2 0 50 A zero At a 11xeaeoA2srn Ka where KaZnnfornr SoKnma The parametern ls referredto as a quantum number pmnletnyehnmnreetnnonly Let a partrele travel m the n dreetron only thus B 0 turn yew A2 where the wave numberir 27 Thewayelengthrsgryenby a M h ZmE A free partrele wth a wellrdeflned energy wlll also have awellrdeflned wavelength d momentum The probabrlrty denslty funeuon wxt went AAquot whlch ls a eonstant lndepend nt of posruon Thls says that afree partrele wrth a wellrde e momentum ean be found anywhere wth equal probabrlrtyl The lnrrmte Pnlenlul WellrSahmnnm Dl exemal 1n reglon JIMx zero thus the Wave equauon beeomes ZmE h 1004 figx or The lnrrmte Pntenlul Well 7 Namahmonn Candmans We must use the normahzatron boundary eondtron to nd A2 Normalrzauon Condmon Wx I39xdx1 FF errts oflntegrauon are zero to aquot srnee h box quot So zslnzhdx n eparuelers ma Solution A The lnrrnue PotenoalWelle sundrng Wave Frnally the umerlndependentwave solutron ls glyen by we 3sh where nl23 a a hz 2 2 where energy E E quot7 Zma The energy of abound paruele m an rn nrte potentral well ls quanuzed A bound eleetron m the rn nrte potenual waye ls a standlng wave as shown by the next sllde gure 2 s on page 38 Erersrse 2 5 on prge 2 r Eleetron m an rn nrte potentral well y F The wdth uttheputenual well a 1EI A Caleulate the rst three energy levelsln terrn ufeV 2 2 e 1 1 E 1 72 5M J S M o 022x104 1h2 2m 29 Hxl tgouxl m J2 UmtsJNmkg quot m 30 1 2 kgm lg J El s 022x10 ml12 6 022x10quotquot 0 3743M E2 7 6 022x10quot 12 2 409mm 497w 53 s 022x10 132 5 4198x104 3 3682V Th suppotentulpunsoon Srnee the re eetron coef clentR e 1 then all the parueles are eventually re eetedbaek quotquot rnto reglon 1 Further analysls shows that the probabrlrty densrty funeuon ls not equal to zero m reglon Ill 50 a paruele rnay penetrate rnto reglon H at least for a whlle untrl rt returns to reglon 1 as are eetron Pameles are nerther absorbed nor transrnrttedrnto reglon 11 Frgure 25 V a b E four lowest Corresp ondmg Correspondlng dlscrete energy waye funcn ons probabrlrty leyels funetrons Th steppntentul Funcnan A ux ofpamcles ls rnerdent on the potenual barrrer trayelrng m the r drreeuon Let the energy ofthe parueles be less than th5 step potenual barner erght Read the text for the denyauon pages 3941 ofthe results that follow The analysls shows 1 An rnerdenttrayelrng waye ofveloclty y 2 Are eetedtrayelrng waye ofveloclty y 3 y yyeloerty oflncldent and re eeted waves are equal 4 The re eeuon eoefflelentR l Exerc15227anpug242 The pmhabullty uf ndmgapamde ata dlstanee dln A regun ll eurnparedtu thatatx n ls gyenhy mid Cunslder an eleurun un regun l unyellng ngnt at a yeluutyy 1EI5 ms Thepu39enual bumerhelght 3 E hneue energy urthe eleurun Flnd the prubuhlhty ur ndlng the eleetrun at a dlslanee d 1EI A lntu regun ll 1 m w Dllxlquot kgl mi 456x1 39 Exererse 2 7 t The Potenm anrner Zm A Gwen K2 2Wn E A ux ofparheles 15 rnerdent on the potenhal r on banner travehng m the rr ohreeh Set the potentra1 banter herghtvuto 315 Let the energy ofthe pamcles be less than U 1 146575 272nm thebamerherght E ltVu The 3 regron39s solnhons are we 42quot 32 for Regionl 9 20 7 A12 325 1 for Regionl 14 x A121 332 for Region3 zuzzrmymdxmxmmm 00872 872 Th mm Emquot um e Aw E rtr or Regrnnx The Potenonmnrner M WrltxAra Eraquot rm Ewan The wave funehons thru the potentral bamer nehwrwgpn ror Raglan r r n lt Vuv 13 501mm 2mg 2m shows that there 15 afrnrte Whm K t T and K2 72 5 7 probabrhty ofapamcle m regron 111 The phenomenon rs eanedtunnehng Smce there rs nothrng to cause a re eetron m regron 111 the eoemerent B3 0 The raho ofrnerdent ux from regron Ithatreaehes regron 111 15 grven by E E m where a barner wrath and T A3 T 16711 7 T he tunnehng probabrhty A A n n nouercoreterVerene anntnnn Tnnnzhng Exams 2 2 PE 45 n Esurrale the tunnehng pmbatnhty eran eteetmntunnehngunrua remangularbamer wtth Eama39hagulufvn 1 ev v dlhuf 15 W Roller eoasterreleased from A eannotreaeh E m the Newtonran world 1911X1U39nkg 4e 7 Hum 1mm 1 1 ustxlm X 4 595mmquot However there is a chance of tunnehng to the other state m quantum 2 mmquot IDEIZ64 theory mane Burnth pnrnrnxee af gchnm Mata 1rd Demes Semen 2 4 Section 24 Extensions of the Wave theory to atoms Will not be covered and is not testable End of Chapter 2 Introduction to Quantum Mechanics Chapter 10 The Bipolar Transistor Seeuon 10 1 The Blpolar Translstor Aeuon The blpolartranslstorhas three separately dopedreglons anoltvvo pn junctions The blpolartranslstorhas three separately dopedreglons anoltvvo pn junctions The three terrnlnal eonneeuons are ealledthe ernltter base and eolleetor Flgurelnz I unnt lurr t t sum um rrnt rtttumu rquot rntutsrn Prevlew The slngle junctions devlces consldered so far are passlve devlces ln thatltls not eapable ofgam translstorls amulufuncuon sernleonoluetor olevleet a ln h t conjunction Wth other elreult elernents ls eapable ofeurrentgaln voltage gatn anol slgnal powergam The basl translstor aetlon ls the eontrol of eurrent at one terrnlnal by voltage applled aeross tvvo other terrnlnals oth devlce The three basl translstor types are the blpolartranslstor the metalr oxlde s rnleonoluetor eld effect translstor MOSFET anolthe junction eldreffect translstor J39FET The Blpolar Translstor Aeuon The vvlolth ofthe base reglon ls srnall eornpareolto the rnlnonty earner dlffusl on length The H anol notataon ludlcates the relatave rnagnltuoles of the lrnpunty dopmg eoneentrata ons The above gures are the slrnple vlew Wth the nent gure eloser to the physlcal reallty that the dopmg eoneentrataons anol geometry ofthe ernltter anol eolleetor ean be vastly dlfferent Seetlon 10 l l The Basl Pnnclple of Operauon The up d pnp translstors are eornplernentary devlces T at ls lt takes both to olo the eornplete job quot Thls ehaptervvlll develop the npn translstor but the pnp devlces use the sarne e uatlons e baserernltter BrE pn junction ls forward blased The baser Th eolleetor Brc pn junction ls reverse blased ln norrnal operatlon The Basre Pnnerple ofOperatlon The Basre Pnnerple of Operauon Thls eonfrgurauon ls ealled the forwardractxve operatrng rnode We enpeet the eleetron eoneentratron m the base to be st PM The baseremltter 137E Junctlon ls forward blased so eleetrons from the ernrtter are lnjected aeross the BrEjuncnon rnto the base Th large gradlent m the eleetron eoneentrauon rneans that eleetrons e The basercollector Becnunetron ls reverse blased so the rnrnontv lnjected from the ernrtterwrll dlffuse aeross the base reglon m e e ner eleetron eoneentrauon at the edge othe Bectunetron ls ldeally depleuon reglon where the eleetrre eldwlll sweep the eleetrons rnto the The Basre Pnnerple ofOperatlon Seetron 10 l 2 Slmpllfled Transrstor Current Relatrons The B715 and Bectuneuons are elose enough to be ealledrnteraetrng The minority tamer romantramns are shown m the nextfxgure for e mmo e W Jnnetrons forward blas Ideally th anler eleetron eoneentratron m the base ls allnearfuneuon ofddstance l e no reeornblnauon The eleetrons dlffuse aeross the base and are sweptrnto the eolleetor by the eleetrre eld m the BVC depleuon reglon The above gure shows a eross seeuon ofan npn transrstorwrth the lnjectlon ofelectxons from the nrtype ernrtter henee the narne ernrtter and the eolleeuon ofthe eleetrons m the eolleetor henee the narne eolleetor Flgure ID a Mnunty Ema dlmbuums vnth busle mmlsr fm39ward blas collector cmem mu u u t on Assumlng the ldeal llnear eleetron drstnbuuon m the base the quot quot quot v eolleetor eurrent ean be wrrtten as a dlffuslon eurrent urn Dtdrd j Hamp j Where A5515 the crossrsecuonal area othe BrEjunctlon nEn ls the therrnal equlllbnum eleetron eoneentrauon m the base The dllfuslon of electrons ls m the x ddrectlon so that the eonventlonal eurrent ls m the ex ddrectlon Thus the eurrent through the eolleetorterrnrnal depends on the voltage quot applled ae s the base and ernrtterterrnrnals Thls ls the basll transrstor aetron Ernltter Current Ernrtter eurrentrE1 ls due to the ow of eleetrons lnjected from the ernrtter rnto the base Thls eunent r5 ls then equal to the eolleetor eurrent l V4 Vt ere ls another eornponent to the ernrtter t r V 41 eurrent Slnce the baseremltterjuncnon ls 39 forwardblased majority carnerholesln the base are lnjected aeross the BE junctlon rnto s the ernrtter H 1 Exp J Trammde mly Ernltter currouent Irr1rrEXPEVVJ From the last two equauons we have rc arE andthat the ernrtter u rent ls rE an exponenual funeuon ofthe BrE voltage Thus as long as the BC Junctlon ls reverse brased the eolleetor eurrent ls rndependent ofthe basercollector voltage The blpolar transrstor aets llke a eonstant eurrent souree wrth gam B as e Current The rauo ofeolleetor eurrentto base eurrentls a eonstant ealled 13 Where brs ealledthe commonremltter eurrent gam Normally the base eurrentwrll be relatwely small so that m general the commonremltter eurrent gam ls on the order of 100 or greater Ernltter Current I 11Ep V J Trammdeunly The eurrent r52 lnvolves the rnrnonty earner hole pararneters m the ernrtter and not dlscussed yet The total ernrtter eurrent rE ls the surn ofthe two eornponents tie51 V z 1 Mfrs IerXP K The rauo ofthe ernltter and eolleetor eurrent ls the eornrnon base gan Srneerc ltlEthen alt1 Base current The ernrtter eunent eornponent rmrs aBrE junction eurrent As a eunent rnto the base terrnrnal rtrs calledlh From the ernrtter eurrent equatron we saw that tlus eurrent ls proporuonal to expoJV There ls also a eunent due to reeornbrnauon ofrnrnonty canlers wrth majority earners m t the base Thls eurrent ls ealledrm The total base eurrent ls the surn ofthese two ts eurren n1 Seetlon 1013 The Modes of Operauon 1n thrs eornrn onremltter erreurt con guration the transrstor may be blasedln three dlfferent rnodes of operatron l Cutoff 7 V 0 e no rn onty earner eleetrons lnjected rnto the base VCE 20 7 zero orreversed blas All eunents m the transrstor are zero 2 ForwardAetweRegron VEE gt 0 e sumerently to forward blas CE 7 se All eurrents were descnbedln the last seeuon v The Modes ofOperatron s The Modes of Operathn In th5 commonremltter erreurt eonfrguratron The blpolar transrstor commonremltter currentrvoltage eharaetenstres the transrstorrnav be blasedln three dlfferent rnodes of operahon Colleetor Current L Saturahon VEE gtgt 0 the forward blas rnereases the eolleetor eurrent and VR Ulhrnatelv VCE g 0 wlll beeorne forward blased Both VBE and VCE are forward blased and the eolleetor eunent ls no longer eontrolled by the VE voltage Voltage VCE When Mrs large enough the VCE rs gt zero Junctlon ls reversed blased The Modes ofOperatron The Modes of Operathn wrrtrng the KV39L equation around the LE loop nght slde we nd The load lrne supenrnposed on the txanslstor eharaetenstles can be used to vlsuallze the blas Eond tlon and chfVKw 0 VM V iVK operahng rnode ofthe transrstor Va V 39IcRc The eutoff rnode oeeurs when 1C 0 Saturatron oeeurs when there ls no longer a ehange m eolleetor eunent Thls llnear relauonshlp between VCE andIc ls ealled a load 1mg for a ehange m base F orwardractlve rnode oeeurs when the relahon 1c mg ls valld Inverse aetlve nut envaed Seehon 10 14 Arnplrfreatron wrth Blpolar Transrstors Arnplrfreatron wrth Blpolar Transrstors Voltages and eunent ean be arnplrfred by blpolar transrstors m slnusmdal lnputxmltage conjunctlon wrth other elernents Thls wlll qualrtanvelv desenbedrn thrs etron The de voltage sourees VW and VCC are usedto 7 blas the txanslstonn the forwardracuve rnode 39 deand slnusnrdal cunmts i he voltage sourees vrs ailmervarylng rnput thatneeds to be ampllfled Smusm l vulta eatmss resl urR Slnce rc mg arelahvely large slnusoldal eolleetor eurrent ls g supenrnposed o a de value eolleetor eunent Slnce rc ows m the resrstor RC th5 rneans a slnusoldal voltage exrsts m VCE The erreurt has produeed avoltage gam m the hrne varylng slgnal and ls known as a voltage arnplrfrer Seeuoh 10 2 Mnonty CarnerDrstnbuuoh he eurrerrts m the bxpolartransxstor1ust as m the pnjunchon are deterrrurred by rrurrorrty earrrer ddffusxon The rrurrorrty earrrer gradrerrts produee the ddffusxon so the terrtrrow deterrrurres the steady state rrurrorrty earrrer drstrrbuuorr m eaeh ofthe three trarrsrstor regrorrs Seetable101rrerrtshde or the rrotatrorr usedm thrs arra1ysrs Seeuoh 10 21ForwardrAcuve Mode rurur u to tourquot P dwi r r urn t 7 r4 The above gure rs the model used for the arra1ysrs Eaeh regrorr rs uruforrruy doped eaeh regrorr may be ddfferently doped Note the use ofx39 and xquot whreh are the Edges of the dep1etrorr regrorrs 1r forwardracuve rhode BrEJuncuon rs forward brased vEE gt 0 Becrurreuorr rs reverse brased VCE lt 0 Mrrrorrty earrrer drstrrbuuorrs lookhke those m gure 10 14 next shde Base Regrorr The steady state trrrre rate ofchange equals zero excess rrurrorrty rrer 1eetrorr eorreerrtrauorr rs found from the arrrbrpolartrarrsport equauorr LetEorrr base regrorr 625n5x7 5n x BET 15 n 5n x as x 7 ma rE rs the rrurrorrty eamer e1eetrorr concentration rs the base The subserrpt zero agarrr rrrdreates the thermal equrhbrrurrr value DE rs the rrurrorrty earrrer drffusrorr eoeffrererrt m the base eleetrorrs m thrs e s In rs the rrurrorrty earrrer electron hfeurrre m the base Tableml ttrkurrmaruur rueuutwruuuran s in numb rmmmu mlelrmlmn t r u uh Frgure m 14 In Furwardramve made nllcumr LIVIiHL I rue rue rrunurrtyhaxes Base Regrorr The derrvatrorr on page 380 nds the excess rrurrorrty earrrer electron eorreerrtrauorr rs a hyperbohe sme furreuorr um exp L 71 smh 5quot esrhh l V L5 L5 smh l 1 5 r Thrs equatrorr m afauly drffererrtforrhat appears m the terrsrorr of powerhrre eab1es whreh rs a eosh equatrorr The eorrrparrsorr rs that both are founded on the use ofa constant terrsrorrquot T rrt m he gradre r ease or gravrty on the eab1e versus the stretehrrrg force betweerr the powerhrre po1es on Base Regroh Wrth the base wrth XE yery srha11 1h eorhpansoh to drtrusroh1ehgth LE we have s1nhx x for x 1 The excess e1eetroh eoheehtrauoh tn the base ear how be approxrrhated by We s n EXP 1M WV7ts lltrrerex 7 Exemse 1n 1 sureuh rtph EmmerNdr 1nx m3Easeer13 v3 11 mm furwardbms Ca1eu1ate the Excess mmuuty Ema e1eetruh Emcentmuun m the base at x xsz y t W Jquot 39 t L mn2n W 39 L 1 1 1n 3 smKulmu1m2 smnznn2n1z 39 uauu izlsmpmsoxm39 um 7 x A 7T 718947 lD cm Tah1e1n2 These expansmns are used by the text 1h thrs seeuoh Exeruse1n1unyugezb1 SlllEmnpn quot EmllaMl xtrnzEaseN NElU rm3 v32 umnrurwardbuas xE umdtrruauh1mgth111num 4 Calmlaletheexcessmnumymm demmcunca mhunmmebasealx V See gurel 5mm W e n m M XP74 turths eruatuh 1w ensnnm mt t 7139 Mb hugger some shaggy 225nnmquot 1 sszsxrumr es 81x1n cmquot so mm a 3 81X10quotm 3 e 22500sm 81X10quotm 3 Exemse 1n 1 1 sureuhhph EmllaMl xtrnzEaseN rNElU rm3 vRE umnrurwardbuas x pmdlffusunlmglh14l pm Use the lmmr apprumrmum tu a1eu1ate the Excess mmuuty Ema e1eetruh Duncantmum m the base at x M m z p ngmqhme 4mm fi f39psguuev 1w e 3 2184 x1 Wmquot Themhu urtheaetua1 m theappmmauums 11 99511 Su as1uhgas uurrurwardhras 1s sumuent the 11m appmhmauuh 1s uutte guud Erhrtter Reg1 or The steady state excess rhrhohty earnerho1e eoheehtrauoh 1s a1so found from the amb1polar transport equatroh LetE tn the erhrtter regroh 62W K51 ma 2 Agam through bouhdary condmons the steady state excess H rhrhohty earrrer ho1e eoheehtrauoh 1s a1so foundto be Pm 61 J s h o srhh LE 1 5 5nd Enntttet Reglon wtth quallflcanons the last equauon may also be expressed as approxlmahon The entena ts that th L ahnear l eonnpatedto the dttruston length e dtstanee XE nnustbe smal 1 mm o i m exp e lx5 e x E t the dtstanee ngts comparable to the dttruston length LELhen the concentration shows an enponenttal d ependenee on XE notltneat any the above equatton ts not valld Exanse In 3 unfage 384 Stlteun npn htased tn the fm39Ward aettye tegun cuncmtmuun meh 95 unhethmnal enutlthn Atwhat value urtr39 eunparedtu Lg dues themagnttude urthe nununty at umwlue 7 t 7 Mod pox ptn pod Pcn pcx psn7ptn xp Su uurpmblemlslu nd39hexquot at n 95 pm n 95pmpmleexp n 95 x 1 ex eusseuusm feign A KP L nus L tum 3 ThecmcemmhunlsgsVouf39heLhamalequlhbnumvalueal L ahuuzthtee dl usmlenghs Seettun ID 2 2 OLherMudes ufOpemum Th seeuon ts a qulckrevlew ofthe four modes Ihaye sklpped over the mverseractlve mode for now CollectorReglon The excess mmonty camerhole concentration tn the collectorreglon ean be detetnntnedfnonn D Mux own TiT Agatn thtough boundary eondtttons the steady state excess Earner hole concentratlon for thts reverse blased tunetton mtnottty ts Exemse ID 2 Stlteunnpn EmllerNd l xtrn3EaseN Vmn mu rutwardtuas x24 um dtr slmlenghLE4 um eTTE VmSTTlWFl ml 7mlxluquotu52u 7 H75 znsn skMEISCIJEIS2H skMIJl 175 1 dgxln icmquot Seetton 10 3 LowrFrequency Commoanase Cunent Gatn The basl pnnetple of operation ofthe blpolar txanslstor ts the control ofthe collector eunent by the BVE yoltage V55 The collector eunentts afuncuon ofthe numbet of nnatonty carners teaehtng the collector a erbemg mjectedfrom the enntttet aetoss the BVE tuneuon The common base current gum ts deftned as the tatto of eolleetot eunent to emtttet eunent Secuon 10 3 1 Contxlbuhng Factors There are a nurnber of ux components m the npn brpolartransrstor basrc prrncrple of operauon ofthe blpolar transrstor ls the control ofthe collector current by the BF yoltage V55 The current densrues are de ned as follows see gure 10 19 JnE Due to the dlffuslon of rnrnonty carner electrons m the base mn0 Inc Due to the dlffuslon ofrnrnorrty carner electrons m the base at KB I The dlfference between Id and Inc whrch ls due to the recornbrnatron ofexcess rnrnonty carner electrons th majorlty carrrer holes m the base The IRE current ls the ow of holes rnto the base to replace the holes lost by recornbrnauon r hunter Contnbuung Factors 1 1 615 mug The last equatron can be rewntten to ernphasrze relauonshrps 1 Le 14 Arm 7 5 JnEJpE Jar JnEJKJPE 7 The factors are de ned as f ernrtterrruecuun ef an ctur 13pr la 1 hasetranspurtfactur fix J 1 5 quot L xecmnhnahanfaclm mug Contnbuung Factors JFE Due to the dlffuslon ofrnrnorrty carrrer holes m the ernrtter at rr39 0 JR Due to the recornbrnauon ofcamers m the forwardrblased BF Junction JPCD Due to the dlffuslon ofrnrnorrty carnerholes m the collector at xquot 0 JG Due to the generatron ofcarners m the reverserblasedBrCJunctlon The currents IRE JFK and JR are BF Junction currents only and do not contnbute to the collector current The currents 1 JG are BrCJunctlon currents only These ve currents do ot contrrbute to the transrstor acuon orto the current gam Contnbuung Factors The dc commonrbase currentgaln ls de ned as We wlll assume that the acuve crossrsectlonal areas are the same for the collector andthe ernltter Then Srnce we are pnmanly rnterestedrn how the collector current wrll e m change wrth a chang the ernrtter current we wlll wnte the current Jtr W Srnce JG and JPCD are not functlons quot P of the ernrtter current Contnbuung Factors The emlttenmecnon e rcrency factor39ytakes rnto account the rnrnonty carnerhole dlffuslon current m the ernrtter Thls curr nt ls part ofthe rnrtter current but oes not contnbute to the transrstor acuon m that JFE ls not part ofthe collector current e base transport factor aortakes rnto account any recornbrnauon of excess rnrnorrty carrrer electrons m the base Ideally we want no recornbrnauon m the base The recomblnauon factor takes lnto account the recornblnatlon m the forwardrblased BeEJunctlon The current JR contrlbutes to the ernltter current but does not contrlbute to the collector current Section 10 3 2 Mathematical Derivation ofCurrent Gain Factors The results ofthis section will be presented Please read the derivation on your own 1 1 FtthLa quothi nBDDBLE tanhl For the ideal y 1 thus we should have pEEI ltlt nEn assuming the other variables are xed 2 n and EU 4 N For the emitter injection efficiency to be close to unity the emitter doping must be large compared to the base doping Base Transport Factor In order for CLT to be close to unity the neutral base width xB must be much smaller than the minority carrier diffusion length in the base LE Under sufficient forward bias the base transport factor is approximately Jnc N 1 J quoti cosh LE LE For xE ltlt LE we may expand the cosh inction in a Taylor s series so 1 Z 2 1 X5 7 X N 2 Fit J cosh ij Hg LE 5 LE LE 27 Section 1033 Section Summary Although we have considered an npn transistor in all of the derivations e act e analysis applies to the pnp transistor The s e minority distributions will be obtained except that the electron oncentrations will become hole concentrations and vice versa The commonbase current gain already discussed is a 2 It The commonemitter current gain is defined as u The sum of the currents is IEIBIC Section 10 3 2 Mathematical Derivation of Current Gain Factors The last condition allows us to approximate y as Now we will move on to the base transport factor Recombination Factor Ifwe assume that JFE ltlt J E then the recombination factor is JnE is due to diffusion ofmc electrons at x JPE is due to diffusion ofminc holes at x 0 5 me J R J R JMJKJ JMJK 1 J1 J n J in The recombination factor is a inction ofthe BE vo tage As VBE 39n the recombination current becomes less dominant and the recombination factor approaches unity 39 39 factor also mnltt39 charge recombination at the surface of the semiconductor The main effect is to reduce the carrier lifetime and to increase the recombination Section Summary I A k current gains is It I7 1 I I 1 1 1 1 awau jigand u 11 u u an an an an u 17 This relationship holds for both the dc and smallsignal conditions so e can drop the subscript and a i 1 TablelEIK merMr rm u i Seeuun 1u4Nurudm1 E ems Thus r we have euuudued a npn bxpularlmnsmur at T mu K wnh umfmmly duped regunsluw1memmcm anlmn a and base wmlhs an energy bandgzp umfurm ldal cunsmal eurrmz denuues andjuncnunswhmh ureuuun bmkduvm Ifany urmese ldal euuduuus are nut presmL men re transsturpr pa ues wru deviate um re ldal haradmsucs deuved su r Idu nulm39audlu dweu un thesexssues butwul mcrudreememruryuur nnhanly Frgure m 21 ILIW Mm mg gt1mu uhm c 1 wilh IHL39 mu 1 anldgc lucrc x ing mmm ily ghldk m L 0 Example 1n 3 unpage 395 npn Eupulartransimral T mu K Assume39helm luxAmrzandjg mHAmrz Calculate the furward uused vE2 wltage redured u scheme a reeumumauuu aur eduu1 u 5 n 9957 52 1 209967 1 1Jmexpe imam J Jquot 221 1039 Salve fur v32 V 211000expe V W expey VIEZVLn 71 mquot nggm z M12 51 2m n259x12 m n 6536 Jmu Seeuun 1n 4 1 Easedethudulauun Theneutml has uddth x52 uueuur urme arc wltage AsLheErC exarsest vulng de ned as vE m gure m a mmss me a 7c deplenun reguu wm39h mnases whmh reduees xE see gure 1n 3 A ehauge mLhe ueuuu1 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euheehbauuh Th The base urThe SehuTTky trans ur Wuyreduees The sTuage Timer sTuTageTTmes uh The quer un hs nrless are eummuh EEE 166 Physical Electronics Smmnnuunmn I MYSIES 7 AND EVIEES asi Mummies UumM Mrmnzu mm m Instructor Russ Tatro Of ce Riverside 5010 Phone 916 2784878 Course Website httpWWWcsuseduindiVttatror Review of Semiconductor Device Operation I ll start with a quick review of three major devices in EEE 108 Diodes Bipolar Junction Transistors BJ T Metal Oxide Semiconductor Field Effect Transistor MOSFET The material in this review is mainly from Sedra Smith Microelectronic Circuits 6th Edition Oxford University Press 2010 Ideal Diodes A diode is one of the simplest semiconductor devices Anode Cathode K I 5 J v The device has nonlinear characteristics ideal device shown below T k Reverse bias gt Forward bias 0 v I 139 gt gt oo o o o o o o v 7 v vlt0i0 igt0v0 Ideal Diodes The general circuit analysis approach is to assume the diode is either off or on complete the circuit analysis and then check that the answer makes sense The following diode is ideal Find the values of I and V Assume the ideal diode is on 25 k Thus V 0 Volts We may consider this as 0 meaning the voltage is just barely positive to turn on the ideal diode The current I now only depends on the voltage drop across the resistor 5 0 2x10 3A 2mA 2500 Example Ideal Diodes The following diode is ideal Find the values of I and V Assume the ideal diode is oft Thus V 5V This is the reverse bias case The current I is now zero Silicon Diodes The real diode differs from the ideal as shown below Compressed l39zx scale o lt Breakdown Reverse Expanded scale The forward bias for turnon is not zero but approximately 07 volts The reverse biased diode shows a small current called the saturation current IS The forward bias current i is i 15eVLT 1 Silicon Diodes The thermal voltage VT is given by k Boltzmann s constant 862 X 10395 eVK 128 X 103923 JK T the absolute temperature in kelvins 273 temp in 0C q 160 X 103919 coulomb At room temperature of 20 0C gt VT 253 mV We can nd the voltage across the diode by solving for v in the current equation iISe l gt vVTln1il VTlnILf0rigtgtIS Example Silicon Diodes Find the change in diode voltage if the forward current through the diode changes from 01 mA to 10 mA Notice that the problem does not give Is So we need to find a relationship for the voltage that excludes the saturation current variable iISe l We can set up a ratio i1ISe l Z lOmA e l Elooze l i2 13e l O39IMA e l e l I will approximate by assuming eX 1 z eX V1 V e T V142 V2 VT 100 6 V1 V2 VT lnlOO 253mV46l 1165an Silicon Diodes The modeling of the diode circuit analysis can be done by several methods depending on the desired level of accuracy The load line is a graphical analysis under forward bias The following circuit shows the load line parameters I R u VDD VI Diode characteristic Q operating point Load line Silicon Diodes A quick method is to assume the diode has a constant voltage drop regardless of the current through the diode Assume VDD is 5 Volts R 1 k9 and VD 07 V Solve for ID By KVL we have VDD IDRVD 0 510100007 0 D mA Example Here is an example that shows the closeness of the constant voltage approximation Determine R in the following circuit so that the output voltage is 24 Volts Assume the diodes have 07 V drop at a forward current of 1 mA HOV By the constant voltage model we can only get v0 21 Volts regardless of the forward current So R could be any value In real circuits the current through the diode must be limited due to heat dissipation issues Thus R should be large enough to keep the total Wattage below the speci ed value Example We can also solve for R by including the current voltage dependence We will need to determine the saturation current IS from the given conditions lmA 58 0 1 mA9637 X 10715 9637 X10 18A 2 X l mAe The desired drop across each diode is 08V across each diode 3Di0des Thus the current I through the series diodes is 08 I Ix 9637 gtlt10718Ae m53 9637 gtlt 10 18A5403 X1012 521X10 3A Now we have the data we need to nd the resistor R RZlOV ZAV 76V 21469 521mA 521mA Silicon Diodes Diodes enable a wide variety of circuits including Recti ers Clamping and Limiting Voltage Regulators Zener based Varactors voltage variable capacitors Photodiodes Light Emitting Diodes LEDs Later we will see that all these device operations are the result of two or more semiconductor materials in intimate contact to form a pn junction Bipolar Junction Transistors BJ Ts The Bipolar Junction Transistor is a three terminal device Metal mtype Flype Hype contact Fmiu Emitter Base Collector C H l 0 ea or re I0quot re 1011 re ion E g 8 g C Emitter base Collector base junction junction Base EBJ CBJ B The three terminals are the Emitter Base and Collector The basic device operation is the use of the voltage between two terminals to control the current owing in the third terminal Charge carriers of both polarities electrons and holes participate in the current conduction process Hence the term bipolar Bipolar Junction Transistors BJ Ts For example the NPN B T terminal voltages are de ned below Fonvurdrbiused Reverserbiased C 37 0 V IE lt 1 v I t Injected holes 111 Rammme 4 391 electrons In 1 m 1 1 I z I iL I E v I 4 I 39gt gt ulul l 39l39l VRE VCR The three voltages are VBE base to emitter voltage VCB collector to base voltage VCE collector to emitter voltage VCB VBE Bipolar Junction Transistors BJ Ts The device operation depends on the magnitude and polarity of these thre VOltagCS Fortrall dfhirised Reverse biased E 33 quotQ r T r Injected hrnles igu Remmbincd i electrons 1321 i3 i3 III D 1135 O 5 CB 3 B Mode EBJ CBJ Cutoff Reverse Reverse Active Forward Reverse Saturation Forward Forward Notice that the text lists the state of the junction either forward or reverse biased You must relate that state to a speci c voltage 1n the form of the three voltages VCE V B and VBE Bipolar Junction Transistors BJ Ts The current ow in the B T can be rather complex and will be studied in detail in this course For now the main currents are 1C collector 1B base IE emitter 1C Ise C 2813 The parameter 3 is the fraction of the current from the base versus the total collector current ow Bipolar Junction Transistors BJ Ts The Emitter current ow can be de ned in terms of the previous currents iEiCiB IC IB 13 The lower case i implies all current ow including any time varying component iEszi l C Let a 3 3 Example Measurement of an npn BUT in a particular circuit shows iB 1446 uA iE 1460 rnA VBE 07 V Calculate 0c 3 and IS C E IB 21460 mA l446uA 1460 001446mA 2144554 mA Now find CL C 144554 mA 0 09901 1E 1460 mA NOW find 3 IBIC 144554 MA 21000 B 001446 uA Or find 3 from the calculated value of 0c 3 05 09901 21000 l a 1 09901 39 Example Now find IS IC 56 1C 144554 mA 144554 mA 144554 mA Is 2 mi 2 L 27668 2W eVT 600253 6 X l39gtlt1015A The reverse saturation current IS is the current due to the reverse biased collectorbase junction and is not the total current across the base to the collector This points out the complexity of the BJ T versus the simpler diode We must now track system currents due to effects in specific portions of the device Example 10 V Find IE 1B Ic and Vc for this circuit The base terminal is grounded and given VE 07 V This is consistent with the voltage drop across the baseemitter terminal VE 0 07 V 07V 07 10V 93V 093mA 1 E 10k IOkQ GivenB50 Now find 1c OtI C 31CotIE iIE 50 093mA 0912mz4 1E 31 501 Nowfind 1B zl C 2135 M 21824x10 6A 10V B 3 Nowfinch 10V VC 1C W chlOV IC5kQ 10V O912mA5kQ 10 456 Volts 544 Volts Bipolar Junction Transistors BJTs The two types of BJTs are npn and pnp C E B B E C npn pnp Note how the voltages are de ned for each transistor Example 10 v Find all node voltages and branch currents for this npn circuit 11 RC 47 m The emitter is grounded through the 33 k9 resistor With the base at 4V the junction is most likely forward biased I ll assume the base emitter junction voltage drop is 07 Volts VE 4 VBE 4 07V 33 V The current IE can now be found 21m 33kQ With the base terminal at 4V and the collector terminal at 10 V the collectorbase junction is most likely reverse biased The BJT thus appears to be in the active mode Example 10 v Given B 100 11 RC 47 km a 2i 2 100 09901 51 1001 IC 051E 099011mA 09901 mA The voltage VC can now be found VC 2 10V IC 4769 2 10V 4653V 535 V This voltage is consistent with the transistor being in the active mode as we assumed earlier The voltage VB was given by the circuit The only parameter not yet found is the base current IczlE IB gtIBIE IC l 09901mA 001mA 10V Example Or we can find the base current from other know values leE IC IE aIE IE1 a 1E1 1 1EM IE 1 2991A zoormA t l 1 10V 099 x 1 099 mA t 47 kg 10 7 099 X 47 53 v gt 100 099 00I mA 470733v zlmA 33 k Bipolar Junction Transistors BJ Ts There is much much more to the design and operation of B T circuits But our focus in this course is why the device operates the why it does and not the subtle device design issues MetalOxideSemiconductor Field Effect Transistors MOSFET The MetalOxideSemiconductor Field Effect Transistors MOSFET is a four terminal device Source S Gale G Drain D Oxide Si02 T Metal thickness I L A Channel V I n region n lt L ptype substrate Body l Body The four terminals are the Source Gate Drain and Body The basic device operation is to create or modulate a channel under the gate that allows current ow from the source to the drain Often the body and source are tied together so that the MOSFET becomes a three terminal device MOSFET Some common circuit symbols for the nchannel enhancementtype MOSFET are shown below D D D Go IB GH39B Go Il i s S s a b c a Since the drain and source are interchangeable this symbol has no speci c indicator Which is the drain and Which is the source b This symbol indicates device polarity in the case nchannel b The simpli ed circuit symbol When the source is connected to the body or When the effect of the body on device operation is unimportant MOSFET Regions of operation of the Enhancement NMOS Transistor Triode 19 Saturation 05 lt Um I 05 2 av UGS Vm W Cutoff T Slope vGS lt Vm gDS ris 0 VZVWOV UOV vDS Note that VGS and VDS determine the device operation VGS must be suf cient to create a channel in the body of the deVice under the gate There is a threshold voltage Vt Which just creates this channel MOSFET VGS gt Vt enhances the channel and increases the channel conductivity Sedra and Smith de ne this voltage greater than Vt as the overdrive voltage VOV VOV VGS Vt For an nchannel device the voltage VGS must be positive at the gate If the source and body at tied together the drain must be at a positive voltage with respect to the source We are now in a position to look at MOSFET circuits at DC Example VDD 25 V Determine the values of RD and RS under the following conditions I D ID 04 mA l RD VD 05 V VD Vt 07 V unCox 100 uAV2 P L 1 pm W 32 um ID i Rs To establish a dc voltage of 05 V at the drain we must find RD VD 25 05V 04mA 25k V35 25V D Note that VD 05 V and VG 0 V Thus the NMOS transistor is operating in the saturation region and we can use the saturationregion expression for ID 1 W I 1D Eluncmc fVon 3 VOZV D l W C 21 quotng Example VDD 25 V Now we can calculate the overvoltage VOV I l D R I 04mA D V03 D 025 V2 V D nch lt10039 312 um I VOV 05 V 39 I The required gate voltage can be found by the relation VOV VGS V D l RS V V V O5O7 V 12V GS 0V t VSS 25 V The gate is grounded Thus VS 12 V Note that VS is not labeled in the figure Now we can find RS RS z 2325 kg D 04mA This completes the design as requested MOSFET In most of the transistor circuit analysis a mode of operation is assumed Then the results checked for consistency The three modes are cutoff triode and saturation regions Cutoff and triode regions are useful if the MOSFET is being used as a switch If the MOSFET is being used as an amplifier it must be operated in the saturation region Frequently the channellength modulation effect on current is neglected In the next example we Will assume that gate current is zero ignore channel modulation and assume the transistor is biased in the saturation region Example VDD 10 V Analyze this MOSFET circuit and determine the voltages at all nodes and the currents through all R01 2 10 M9 RD 6 k0 branches Vquot 1 V kvm1mA RGZ10MQ RS6kQ n L V2 Where is the transconductance parameter Since we assume the gate current is zero a very good assumptionl VG can be found from the voltage divider IOMQ 1 V V 10V 5V D DDIOMQ10MQ 2 With 5 V at the gate terminal the NMOS transistor will be turned on But in What region Triode or Saturation We will assume saturation solve the circuit and check that the solution is valid Example The circuit is now redrawn with what we know The left voltage divider current is 10V i 2 dtvtder The voltage drop across the source is VS 2 ID in mA6kQ 6ID Volts The voltage at the drain is VD 10V IDm mA6kQ 210 61 Volts The gate to source voltage VGS is VGszVG VS 5 6ID Volts Example Now we invoke the assumed saturation region for the current ID ID 2 VGS Vin2 1 lmA 2 7 V2 5 61D 1 Expand the quadratic 1 1 A 2 1D 7 2 j5 61D 25 61D1 2 2 1 A 1 E39V 361g 481D16 1813 241D8mA We can form the quadratic as 1813 255 820 m mA Example The solution to the quadratic is two roots ID 089 mA and ID 05 mA ID can only be one of these roots Try the 089 mA root rst VD 10 61D 10 6089 466 V VS 61D 6089 534 V This gate to source voltage would mean that the transistor is in cutoff and does not make physical sense Try again with the other root VD210 6ID 10 605 7V VS6ID 605 3V VGszVG VS 5 3V 2V Since VD gt VG Vm the transistor is operating in the saturation region as initially assumed and the solution is valid MOSFET The overdrive voltage in Sedra and Smith terms lvOVl lvGSl thl is the key quantity that governs the operation of the MOSFET For the MOSFET to operate in the saturation region which is the region for amplifier application lvDSl 2 lvOVl With a current of ID nCox W y If lvDSl lt lvOVl the MOSFET operates in the triode region If lvGSl lt thhresholdl the conductive channel cannot form and the transistor is in cutoff Device Review Solid state semiconductor physics covers why the behaviors just reviewed exist The course is not concerned with device applications and extensive circuit analysis But connecting device operation with the physics of the device will give you great insight into the uses of transistors Please review the appropriate device operation as the course discusses that device End of Device Review


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