Microelectronic Circuits ECE 3040
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Lecture 28 Operational Ampli ers Reading Jaeger 111115 and Notes Operational Ampli er Operational Ampli er or OpAmp is a multistage ampli er that is used for general electrical signal manipulation The numbers of applications possible with Opamps are two numerous to list Most everyone agrees OpAmp analysis is signi cantly easier than transistor analysis Though they are often internally complex their use in circuits most often simpli es the overall design The circuit is modeled by an ideal voltage ampli er Circuit Symbol MOdel Georgia Tech ECE 3040 Dr Alan Doolittle Ideal Operational Ampli er 0Rm In nity Voltage Gain Afln nity at all frequencies R 0 out Model Vin Avviquot Vout Georgia Tech ECE 3040 Dr Alan Doolittle Ideal Operatlonal Amph er reeuunm to control the overall gain to a nite value Consider the circuitto the rightwith v5 R1 IfAv gt amhe above equationis onlysatis ed for v 0 39 39 39 inum ground 39RI de2 form 1 Feedback Network Inverting Amph er in u mi c feedb negative summation node 1Vn1merzRiV 0 2 i due to in nite input resistance in R but v 0 due to the virtual ground 3 in Ln R1 Combining 1 2 and3 wig m rL mRi 7v 0 Overall circuit gain is nite negative for this feedback con guration and set by the feedback resistor network Invertmg Amph er Input Resistance Output Resistance VI 1le iZRl but 11 Osincevr 0 and 1112 thus vt00 R Y 0 am NonInverting Ampli er The virtual yound requires mat v v 50 Vmv 0 Same circuit as for Non invening case Unity Gain Buffer or Voltage Follower v Same as Nan r mvemng amplx er except R1 0 and R no R 14 R A R2 R2 Output is a scaled sum of inputs 39Scaling can be controlled by mu39os of resistors R2 R2 Vmi rvmei rvmi 39 Rla 39 R117 39 R15 Summlng Amph er 1 R15 Vin 11 in Run WinkIE 39 R 1i C 1 P vimc I Gem39g39a Te V2 v werposition We can corn R2 m R L R 1 Difference Ampli er v21 vnut R1 47 vin b and Rm Ri HR Vinh0 R vm vm 1 2 R R Vii Vim 2 R inz v 2 bine the results of the Dwelling and Noninverting solutions 0 1 R2 v v on ib n E 1122 7 mi R on R vi 7 i This circuit ampli es die difference oftwo signals FK F UIMIVU I nnnlinln NonIdeal Real World Operational Amph ers Finite OpenLoop Gain Real r r quot in nite open r 39 gain 39Voltage gains are typically large but nite 104105 VV 39Finite gain causes a deviation from ideal ampli er behavior R1 m R R2 M where R is known as the feedback factor R R2 Vm Aww ivAVPWV rm v out A P P where A xsthelaap gam an a m Amway V i W mm m aim If AWN gtgt1 1 R2 AWWW 1R approaches the in nite gm result 1 NonIdeal Real World Operational Amph ers Finite OpenLoop Gain Finite open100p gain means the Virtual Groundis not peifectz AND V szH but finim 7 uffsetbetwem 1 Awpar nwp 1 Anymbwp andr terminals 39The Gain LHUI V 1 Awij I Anm GE 7 r 7 1 Anm NonIdeal Real World Operational Ampli ers Finite Output Impedance Real OpAmps have a small but nite output impedance Ro We want to find the Output impedance of the various circuits we have examined All the con gurations have a common circuit for calculating the output impedance gt4 AA ILA Georgia Tech ECE 3040 Dr Alan Doolittle NonIdeal Real World Operational Ampli ers Finite Output Impedance 1x 10 12 1 vx Av0penloop v v7 R1 v vx vx R1Rz Ram 71 ix 1 Avopenloop 1 Tm R Rout R1 Rz v0penloop R0 is very small so this term is EXT REMELYsmall Georgia Tech ECE 3040 Dr Alan Doolittle NonIdeal Real World Operational Ampli ers Finite Input Impedance NonInverting Case Real OpAmps have a large but finite input resistance RID V V 1X RID Neglecting the currentiX compared to i1 and i2 due to RID gtgt R1 or R2 i V7 i1R1 z l3921121 it R R quot 1 V7 V014 V014 a 27100 V V7 1 R1R2 I 6 I AWPM Amway V V 1 Awenzmp 3 VF Wm V 1A 1X Rm 1 Amalgam Rm Rm 1 AVUpEnlUUp RID V Wenzmp X RID is very large so Rin is wlargg is 3040Dr Alan Doolittle Georgia Tech Lecture 27 Amplifier Configurations Reading CECS Jaeger 136 139 1310 1311 CCCD Jaeger 141 143 CBCG Jaeger 141 144 and Notes Ampli er Con gurations Voltage Amplifier Voltage input and Voltage output at Q U a a m m 43 CE Resmtance Amphfier Inpu Resistance Amphfier Outp Load Resistanc Source Amplifier Load Any signal source has a finite source resistance Rs The amplifier is often asked to drive current into a load of finite impedance RL examples 8 ohm speaker 50 ohm transmission line etc The controlled source is a VoltagecontrolledVoltage Source AVOpen Circuit Voltage Gain can be found by applying a voltage source with R50 and measuring the open circuit output voltageno load or RLinfinity Georgia Tech ECE 3040 Dr Alan Doolittle Georgia Tech Ampli er Con gurations Why is the input and output resistance important RS 91C Only the voltage vi11 is amplified to Avvin Since RS and Rin form a voltage divider that determines Vin you want Rin as large as possible for a voltage amplifier for maximum voltage gain Since RL and Rout form a voltage divider that determines vout you want Rout as small as possible for a voltage amplifier for maximum voltage gain ECE 3040 Dr Alan Doolittle Ampli er Con gurations Current Amplifier Current input and Current output 139 i In 0 gt I lt S gt Rs Rm Allin Rout RL 39 The controlled source is a CurrentcontrolledCurrent Source AiShort Circuit Current Gain can be found by applying a current source with R5 in nity and measuring the short circuit output current No Load or RL0 Only the current iin is ampli ed to Aiiin Since RS and Rin form a current divider that determines iin you want Rin as small as possible for a current ampli er for maximum current gain Since RL and Rout form a current divider that determines iout as possible for a current ampli er for maximum current gain Georgia Tech ECE 3040 Dr Alan Doolittle you want R as large out Wmm A Ampli er Con gurations Transconductance Ampli er Voltage input and Current output R l 5 0 The controlled source is a VoltagecontrolledCurrent Source GmTransconductance Gain can be found by applying a voltage source with RS0 and measuring the short circuit output current No Load or RL0 Only the voltage vin is ampli ed to ioutvain Since RS and Rin form a voltage divider that determines Vin you want Rin as large as possible for maximum transconductance gain Since RL and Rout form a current divider that determines i as possible for maximum transconductance gain Georgia Tech ECE 3040 Dr Alan Doolittle out you want Rout as large Ampli er Con gurations Transresistance Amplifier Current input and Voltage output The controlled source is a CurrentcontrolledVoltage Source RmTransresistance Gain can be found by applying a current source with Rsin nity and measuring the open circuit output voltage RLinfinity Only the current iin is ampli ed to vomRmiin Since RS and Rin form a current divider that determines iin you want Rin as small as possible for maximum transresistance gain Since RL and Rout form a voltage divider that determines V as possible for maximum transresistance gain out you want Rout as small Georgia Tech ECE 3040 Dr Alan Doolittle Amph er Con gurations Input Resistance With the load resistance attached Apply a test input Voltage and measure the input current RmHli Or Apply a test input current and measure the input Voltage Rm Vlil Output Resistanc e With all input Voltage sources shorted and all input current sources opened Apply a test Voltage to the output and measure the output current REL Vi1 Or Apply a test current to the output and measure the output Voltage Rum Vti Flnal Summary of Tranmstor Amph er Analy51s I a r r I 39 39 39 Do not a BJTand b calculate small signal parameters gm r In etc 2 Convert to the AC only model Dc Voltage sources are replaced with shorts to ground DC Current so es are replaced with open circuits Large capacitors are replaced with short circuits Large inductors are replaced with open circuits 3 Use a Thevenin circuit where necessary on each leg oftransistor 4 Replace transistor with small signal model 5 Simplify the circuit as much as necessary and solve for gain 5 Solve for Input Resistance with the loadresistance attached a Apply a test input measure the input voltage Kn vi 7 o e lo 21 H r current meit or b Apply a test current to the output and measure the output voltage Rm Input R istance Transistor Ampli er Configurations Common Emitter and Common Source VmAC u Resistanr ltput esistancw R v u V Can be modeled as a current ampli er Modeled as transconductance lc31B or a transconductance amplifier amplifier iDSKvGS Overall Ampli er Con guration IEmitterSource is neither an input nor an output IInput is between baseemitter or gatesource IOutput is between collectoremitter and drainsource IIs a transconductance am li er see small signal models We have used in prewous examples ECE 3040 7 Dr Alan Dnnlittle Trans1stor Amph er Con gurations Common Emitter and Common Source Portion due to Portion due Portion due to Portion due to gt bias circuitry to transistor tmnsistoi bias circuitry 1 H r7K is replaced with an open circuit for the MOSFET case Previously we have analyzed voltage gain Now let us look at the ampli er input and output resistance these are small signal parameters Rin R2 H Rl H r for the BJT or Rin R2 H R1 for the MOSFET Rout rg R5 for the BJT or MOSFET P g Summary of Common Emitter and Common Source Characteristics Very Large Voltage Gain These propemes Inverting Voltage Gain due to igmrt g xgiSiy High Input Impedance g d for high gain stages of ampli ers High Output Impedance Now let us consider the other two con gurations of transistor ampli ers Common GateCommon Base Common DrainCommon Collector Transistor Ampli er Con gurations Common Collector and Common Drain DC Circuit Collector or Drain is neither an input or output Input is Base or Gate Output is Emitter or Source Georgia Tech ECE 3040 Dr Alan Doolittle Transistor Ampli er Con gurations Common Collector DC Circuit converted to AC Equivalent reduced Note the extra ground due to C2 f fv AC mi Circuit 300K 150K 395 u w 1 o S 4 X Georgia Tech ECE 3040 Dr Alan Doolittle Transistor Ampli er Con gurations Common Collector DC Circuit converted to AC Equivalent reduced RS AC f i j g C 9 g R7 R2 E R1 r1 1311 1 0 Jam Vs 5 Clrcult i 300K 150K 1 E AC Circuit m reduced v0 o 1z39bR4II 1 R7 vth 1 1 n o 1R4 1 R7 Georgia Tech ECE 3040 Dr Alan Doolittle Transistor Ampli er Con gurations Common Collector AC Voltage Gain 7 nwmm W W 42 1R4llnllk7 R2 H R1 W 7 RZHRI Aux me 1 y AquotRzuxnkImyw 1 h RL mllnllm multiplyingnumemlurand dennmmatnr by g 2 1 RZHRI ng A R2R1K K m m gquot m m Transistor Ampli er Con gurations Common Drain Conversion from DC to AC Equivalent Circuit DC 22 22M 322141 C2 21 I l C C1rcu1t quot AC L 392K vs R2 3 C1rcu1t 5 2 Georgia Tech ECE 3040 Dr Alan Doolittle Transistor Ampli er Con gurations Common Emitter and Common Source DC Circuit converted to AC Equivalent reduced I G D AC l vs R2 K y u C1rcu1t l R7 8 l AC Circuit G D reduced gmquot g v m vth V m 25de i 3 A v Qa 1 R7 7 Georgia Tech ECE 3040 Dr Alan Doolittle Translstor Amph er Con guratlons Common Drain AC Voltage Gain v0 gvaSUMH r0 H M v vGS gvaSR4Hro H117 vss1gmR4H r0 H 117 V RZHRI S RZH R1Rx A v0 ELME RZHRI gmR4HroHR7 V VGS v v R2HR1RS 1gmR4Hro HR7 R AV amp g39quot WhereRL R4Hro H117 mummy 1ngL But for R2 H R1 gtgt Rs andngL gtgt1 AVE1VV a Transistor Ampli er Con gurations Common CollectorDrain Input Resistance i 0 130 1i Input Resistance With the load resistance attached apply a test input voltage and measure the input current RmZVXl where iix V RinBJT 2 r7 u 1RL Resistance in the emitter circuit is multiplied by transistor to increase the input resistance R Vx vx oo inMOSFET l 0 Georgia Tech ECE 3040 Dr Alan Doolittle Transistor Ampli er Con gurations Common Collector Output Resistance Output Resistance With all input voltage sources shorted and all input current sources opened apply a test voltage to the output and measure the output current R VXlx out V v v v v x 3i x x quot r R 0 r r R 0 r R R I h o I h I h L where RL 2 r0 HR4 i i i V 1 l 801 r0 WABJT 1 1 rir 4 th E m1 Two resistors in parallel RL and Resistance in the base circuit is multiplied by transistor to decrease the output resistance Georgia Tech ECE 3040 Dr Alan Doolittle Transistor Ampli er Con gurations Common Drain Output Resistance NA v th ngGS Output Resistance With all input voltage sources shorted and all input current sources opened apply a test voltage to the output and measure the output current R VXl out vx vx lx gvaS gmvx R 2 L R vx l ouzMOSFET Z 1 Where RL 2 2 llR4 r gm H l Two resistors in parallel RL and inverse transconductance Georgia Tech ECE 3040 Dr Alan Doolittle Transistor Ampli er Con gurations Summary of Common Collector and Common Drain Characteristics These properties make the Unity Voltage Gain CCCD con guration very goo INonInverting Voltage Gain transformation IE Very High Input Impedance imp edmces to 10w Low Output Impedance ampli ers due to their very low output impedance IE very little voltage drop in the output resistance ofthe amp Transistor Ampli er Con gurations Common CollectorDrain Current Gain AiMOSFET 2 00 th Note since R7 was originally defined as the load the current gain should actually be EH R4Hr0R4Hr0R7 using a current divider Georgia Tech ECE 3040 Dr Alan Doolittle Transistor Ampli er Con gurations Common Base and Common Gate DC Circuit VDD12V 22 m 3 300 kn lt 160 m Base or Gate is neither an input or output Input is Emitter or Source Output is Collector or Drain Georgia Tech ECE 3040 Dr Alan Doolittle Transistor Ampli er Con gurations Common Base DC Circuit converted to AC Equivalent reduced DC Circuit AC Note Jaeger let s ro go to in nity which makes the math dramatically easier B C1rcu1t 4 A M 1 7 gmv u 9 E R3 ka R4 2 Georgia Tech ECE 3040 Dr Alan Doolittle Transistor Ampli er Con gurations Common Base AC Equivalent reduced B C J I Vquot r 3i gmvan E Yquot i C1rcu1t E R N 5 AC Circuit i reduced W R4 Vth VI mi Vm v 08667K R4RS RrhRiiR4 73K Transistor Ampli er Con gurations Common Base Voltage Gain 1 6C v i a V ngnC39o 39Eici3gmvrii3 r0 lt RL 18K L 39 quotW Fifth 173K 13 i5 gmv v v v rt rir Vth iERLh 713 7r i rquot um fl m vm V1 r v va v v rm gmv RLH r Vhv R4 08667V rm rir rir R4R also Rm R HR4173K l v v gm r va7iCR 7 gmv RL so vu7vr RL Va Fri r11 1 gm v iv Rr RL l L r Vm gmvir G eorgia ech ECE 3040 Dr Alan Doolittle Common Base Volmge Gain g 7 a A a ngL AV 1Rgn Large A 11 g Common Base Volmge Gain Translstor Amph er Con gurations Common Base Input Resistance iC I RL LN i 18K i 39z m Vx Jv Input Resistance With the load resistance attached apply a test input Voltage and measure the input current qulgi my 272 Hz v V 1 z ng39 y y Transistor Ampli er Con gurations Common Base Input Resistance Lettzrgg r gt no Tran51stor 1 1er Con 1gurat10ns P Common Base Output Resistance V l ignvuy rquot 3 Rim Replace RL by avoltage l source vX m lt1 73w 1 Result follows exactly a er discussion in Jaeger pages 668670 and 683684 Tran51stor Amphfler Conflguratlons Common Base Omput Resistance v vyv I mom v but rR R n h V m M an fhux Very Large Transistor Amplifier Configurations Common Base Current Gain Rth ith i1 Um RL va J i I A 0 N 1 lth Georgia Tech ECE 3040 Dr Alan Doolittle Trans1stor Amph er Con gurations Common Gate Solution The Common Gate solution can be found by recognizing that the following translations can be made in our small signal model iaoo 2 EN aw ng ENE Anwzil e Avmm R il igmk 39 l jl ll M 39 1 Ram 1 J g i 39 R R R r RI r RI r RI r RI a R ng a 1 g R1 1 Aw my 1 9 Aw Transistor Ampli er Con gurations Summary of Common Base and Common Gate Characteristics The input and output High Voltage Gain imp edances are the opposite of what is NonInverting Voltage Gain typically t u 39Very LOW Input Impedance Common EmitterSource Ivery ngh Ontput Impedance used instead of Common Bas eGate Multistage Ampli er Con gurations You can combine or Cascade con gurations to produce High Performance ampli ers with High input impedance low output impedance and huge voltage gains V vthan V1 v v v s s mmpm cr ampli er v1 v2 05 ampli er lt R 20 k1 1 k5 l 39 h CE prov1des ng Input Impedance CS provides High Input Impedance Moderately high negative gain high gain and corrects the negative gain from previous stage CC provides Low output Impedanc no gain but maintains positive gain from previous stage P1 393 P1 be o in Georgia Tech 0 Dr Alan Doolittle Multistage Amplifier Configurations ForACCoupled ampli ers capacitors between stages the DC solution reduces to three parallel and independent circuits l5 V 91 k9 Q3 RG R53 120 kg 9 33 k Georgia Tech ECE 3040 Dr Alan Doolittle Multistage Ampli er Con gurations For ACCoupled ampli ers capacitors between stages the AC solution reduces to three circuits each of which has a load dependent on the input resistance of the next stage Continued I CS ampli er CE ampli er l CC ampli er l Georgia Tech ECE 3040 Dr Alan Doolittle Multistage Ampli er Con gurations Continued ForACCoupled ampli ers capacitors between stages the AC solution reduces to three circuits each of Which has a load dependent on the input resistance of the next stage CS ampli er C E ampli er CC ampli er 99 kg lt lt gmlvl lt gt R ngUZ lt ltgt C 339 I 5 V l A V A A A V V V 2 f0 l A A 1 J Georgia Tech ECE 3040 Dr Alan Doolittle Continued Multistage Am 99 k Georgia Tech pli er Con gurations U 0 v lt V gm 4 04 gt gt gt gt U U gt ltgt gmquot lt r0 RH nz 5quotquotsz 702 R12 RG 1 V 2 lt HR H gt 39 R R gml 73901 11 772 s s G vth V1 v3 gm2r0211R1211einQ4 V3 v3 v0 v4 1 Va V4r gm3r03 Rm IIS 1 Va 2 v3 v0r gm3r0311RL3 IIS 1 gm3r03 RL3 v0 rr3 v 1 3 1r gm3r0311RL3 Ir3 AW vovm V1 v2vsv V V vth V1 V2 V3 gm2r02 11R12 773 1RL3 0 1 gm1r01 HRH 11rz2 gmzr02 11R12 773 1RL3 1 gmsjo os Rm rr3 1 1 gm3r0311RL3 rr3 ECE 3040 Dr Alan Doolittle Lecture 24 MOSFET Basics Understanding with no math Reading Pierret 171172 and Jaeger 41410 and Notes MOS Transistor Qualitative Description Flow of current from Source to Drain is controlled by the Gate voltage S G D VD VD 2 0 Control by the Gate voltage is achieved by modulating the conductivity of the semiconductor region just below the gate This region is known as the channel Georgia Tech ECE 3040 Dr Alan Doolittle MOS Transistor Qualitative Desm39pu39on nchannel MOS p channel MOS Transistor Transistor D D VDS G G B VBS VGS VSG S Note All voltages are shown in their positive direction obviously VYfVXY for any voltage CFGaLe DDram SSource BBody substrate butto avotd eonfuston wtth substrate B lsused MOS Transistor Qualitative Description Assume an n channel receives it s name from the type of channel present when current is owing device with its source and substrate grounded i e VSVBO V For any value of VDS when VGS lt0 accumulation the source to drain path consists of two back to back diodes One of these diodes is always reverse biased regardless of the drain voltage polarity D G vG VD V1320 1 fzmzszz km A Sio2 MKJ m w J Ptype if when VGS ltVT depletion thgre is a deficit of electrons and holes making the channel very highly resistive gt No Drain current can ow S GVG DVD V1320 l 4 n J High p due to i Depletion 3 Georgia Tech ECE 3040 Dr Alan Doolittle MOS Transistor Qualitative Description Consider now the Inversion case F1rst VDS 0 when VGS gt VT an induced n type region an inversion layer forms in the channel and electrically connects the source and drain Inversion layer ntype Genrgjz Tech ECE 3040 7 Dr Alan Dnnlittle MOS Transistor Qualitative Description Inversion case VGS gt VTcontinued When VDS gt0 the induced n type region allows current to ow between the source and drain The induced channel ast like a simple resistor Thus this current ID depends linearly on the Drain voltage VD This mode of operation is called the linear or triode region S Inversion layer n type D I I I P type Triode is a historical term from vacuum tube technology Georga Tech ECE 3040 7 Dr Alsm Doolittle MOS Transistor Qualitative Description Inversion case VGS gt VTcontinuedI Drain current verses drain voltage when in the linear or tri0de region VDsal Georgia Tech ECE 3040 Dr Alan Doolittle MOS Transistor Qualitative Description lnversmn case VGS gt VTcontinuedC When VDS increases a few tenths of a volt gt0 The depletion region near the drain Widens N drain is positively biased le reverse biased with respect to the substrate The electron concentration in the inversion layer near the drain decreases as they are sucked out by the Drain voltage Channel conductance decreases resulting in a drop in the SlOpe Of the ID39VD cuI39Ve Reduced electron concentration in the Inversion layer near the drain S D N Georgia Tech ECE 3040 Dr Alan Doolittle MOS Transistor Qualitative Description Inversion case VGS gt VTcontinuedI Drain current verses drain voltage for increasing VDS still in the linear or triode region VDsal Georgia Tech ECE 3040 Dr Alan Doolittle MOS Transistor Qualitative Description Inversion case VGS gt VTcontinuedI The inversion layer eventually vanishes near the drain end of the channel This is called PinchOff and results in a Flat IDVDS curve Georgia Tech ECE 3040 7 Dr Alan Doolittle MOS Transistor Qualitative Description Inversion case VGS gt VTcontinuedI IDVDS curve for the Saturation Region The drainsource voltage VDS at which this occurs is called the saturation voltage Vsalt While the current is called the saturation current IDsat 1D B IDsat A J 0 VD VDsal Georgia Tech ECE 3040 Dr Alan Doolittle MOS Transistor Qualitative Description Inversion case VGS gt VTcontinuedI For VDSgtV value AL the channel length L effectively changes by a sat The region of the channel AL is depleted and thus is high resistivity Accordingly almost all voltage increases in VDSgtV are dropped across this portion of the channel sat High electric elds in this region act similarly to the collectorbase junction in a BJT in active mode stripping or collecting carriers from the channel Georgia Tech ECE 3040 7 Dr Alan Doolittle MOS Transistor Qualitative Description Inversion case VGS gt VTcontinued2 If ALltltL the voltage at the end of the channel will be constant V for all VDSgtV ID will be constant sat sat39 If ALL the voltage dropped across the the channel VSAT varies greatly with VDS due to large modulations in the electric eld across the pinched off region EVDS VSATAL In this case ID increases slightly with VDS b ALL ALltltL Georgia Tech ECE 3040 Dr Alan Doolittle Lecture 21 Bipolar Junction Transistors BJT Part 5 Hand Example SPICE Example and Limits on output swing imposed by having to stay Within Forward Active Mode Reading Note s An Example that puts it all together Consider an Integrated Circuit npn BJT with Emitter Doping NDE75e18 cm393 Collector Doping NDcl5e16cm393 Substrate doping Ngub 5e15 cm393 Minority Carrier Diffusion coef cient in the emitter DpErmules 5 chs Minority Carrier Diffusion coef cient in the base Dn me mns 10 chs Base Quasineutral width W 300 nm Minority Carrier Diffusion length in the emitter L 250 nm Minority Carrier Diffusion length in the base LE 100 um Areas Aemitterrbase 25 m2 w Acuiiemupaase 100 m2 w AsubsuaterCullectnr 500 m2 Resistances and Early Voltage rb 250 E2 rE 200 E2 reX 5 E2 VA 35 V Find the complete small signal model for Ic 100 mA VCE ilVy an cf V and Vbe0395Vbn Furthe mama An Example that puts it all together First recognize that WltltLE so we can use the simplest form ofthe transport paramet s Next recognize that our choice ofgeneric labeling allows us to use the m r r 1 r 1 given a npn transistor D L N IBM 1 I H 5 am 125 DEWNE 1 W DEWNE 1410c 7 i DELENE 2 LE 1 1 BBC 1 0992 DC 1 DEWNE1 W 2 w 1 oc DELENE 2 LE DELENE An Example that puts it all together Transconductance I C 386 mS gm y 21 VT Input Resistance V r T 324kQ Output Resistance y V C V gm ra A GE 370kQ y22 IC or using the approximate solution r L 350k9 s llalargevalue y22 IC 1 An Example that puts it all together Forward Base Transport Time 2 1F 45 pS 1 Watctheunits Mm39mum bare mmqmrt limited operational equemy N 35 GHz Ba se Emitter Diffusion Capacitance CB gmrF 173fF c CECJE An Example that puts it all together General Zero Bias Depletion Capacitance cm 22m 17m deplztmn mpm tan 2 where Vb mmmw 2 1714112 In voltagz for Ma E r B jummm e m my hump NN Ddalas NAND VblkTh1 A20 4 2 NANDV Cm C lmym q 739 Junction VB Area cmz Cm fF Applied Voltage C fF 0947 250E07 23335 0 73 33000 BC 0786 100E06 37248 2000 19782 CSubsh39ate 0768 500E06 177201 2000 93332 Thus C CBCJE 206fF C 198 fF COS 33fF SPICE BJT Modeling Consider for example the BaseCollector capacitance c 2 22m 171m depletion capacx tan 2 whm WC 2 1714112 m voltage for the B r c junctmn MJC E B r Cexponzntml factor related 20 the dapmg profile Note the negative sign here SPICE BJT Modeling PSPICE Name Units IS A Most Common Model Parameters Ilransport saturation current B B Ideal maximum forward bias beta pp BF Forward Early volta e VAF V Ideal maximum forward bias beta pp BR Base resistance rb RE 9 mitter resistance In RE 9 Collector resistance rc RC 9 B E zero bias depletion capacitance QED CJ39E F E built in potential Vb or on VJE V E 39unction exponential factor MJ39E B c zero bias depletion capacitance cw CJC F B c built in potenti Vb or om VIC V B c junction exponential factor MIC Substrate z o de letion capacitance cm CJS F Substrate built in potential Vb or dag VJS V Su stra e jun 39on exponential factor MJS Ideal forward transit time rF TF seconds Thls can optionally be replaced with the Eb aerull parameters Basereulleeturleakzge saturation current Imam lsc A B saturation current 1mm ISE A akzge saturation current lss A quots i c i ir MShe i g voltage gain via a transient analysis and an AC analysis What happens to the gain when C3 is removed Why when VinAcl V Why is there a difference In class example Simulate this circuit using Vsl mV and determine the What happens when you do a transient analysis using VinACl V and VinAC Rs Genrgjz Tech ECE 3040 7 Dr Alan Dnnlittle iWhat sets the Maxmiutiiiiiints ofigfieratlon of the B T ireuit Forward active mode lies between saturation and cutoff Thus the maximum voltage extremes that one can operate an amplifier over can easily be found by examining the boundaries between forward active and cutoff and R1 R1 V i Cmax Sup ly 2 15V Q2N3904 A Vbase A VE min Supp yGnd Genrgjz Tech ECE 3040 7 Dr Alan Dnnlittle R1 2 A Vbase A Genrgz Tech iWhaE sets the gaxuniutgiLiithlt ofig egglon of thei iralait 15V QZN3904 and the boundaries between orward active and saturation 07V Cmin ECE 3040 7 Dr Alan Dnnlittle Lecture 16 PN Junction Diodes Part 6 Photodiodes Solar Cells Photovoltaic devices and Multiple Diode Analysis Reading Pierret 92 and Notes Photodiode 433353 ram mm mums O 0 0 E i o Reversed Bias Diode with no light Reversed Bias Diode WITH light illumillatiOn illumination results in extra drift current due to photogenerated ehp s that can reach the junction Georgia Tech ECE 3040 Dr Alan Doolittle Photodlode E El II n uul mmm hr 1m mum m1 5sz 5 I IIIIE VD duplrnml me I 2J1IDW 1 VD H r m L fiil L 7IMLI mummm a 39 39 Y My 9 n FHIP Light N oiLight I NmLight V Bias Paint Light Solar Cell 7 Unb1ased Photodlode since npgtna and pngtpa at the junction e es the voltage must be positive due to the 1 law ofthe junction VA7 7 HX XFWXXW 1n 2 a 8 Solar Cell acts as a battery producing power l I Current is negative or out V l V ofthe ptype side anode cine i I Anode Multiple Dlode Analys1s Consider the 2 diode circuit below Solution Possible states for the 2 diodes Make a guess as to one ofthe possible states ofthe circuit a o is assume on ven e cunent D1 D2 If di de 39 d 39 th calculated ows in the conect direction Off Off consistent with the diode being on Off on Lfa diode is assumed off venfy the voltage On Off consistent with the diode being off Only one solution exists for a given model on on ideal oi CVD IWe ll select the ideal diode model Let s first assume both diodes are on then 15 7 0 1A V 15 M 1069 I 0 10V DZ 5159 but 1A 10 Im 21D 705mA Possible states for the 2 diodes 20mA 39This contradicts the assumption that D1 is on IWe ll select the ideal diode model Let s first assume both diodes are on then 15 7 0 1A V 15 M 1069 I 0 10V DZ 5159 but 1A 10 Im 21D 705mA Possible states for the 2 diodes 20mA 39This contradicts the assumption that D1 is on Let s rst assume diode 1 is on diode 2 is off I 7 lsrvay 7 1570V 715m A 1059 1059 39 1m 0 MA Possible states for the 2 diodes vb 710v V094 Va w 010V IOV 39This contradicts the assumption that D2 is om Multiple Diode Analysis IA Va Im vwv 15V I 1016 m D2 516 710V D1 Let s now assume D1 is on and D2 is on A 7 Dz Possible states for the 2 diodes 15V 7100001 7100001 7 7 10V 7 0 1A 717 mA v 7 0 7 VM 715 7100001 7 717V Assumntions D1 0 and D2 on Veri ed It is le as an exercise to di ove the last combination SPICE Computer s1mulat10ns of Circuits SPICE Simulation Program with Integrated Circuit Emphasis 39A general purpose program that simulates electronic circuits PSPICE PC Version of SPICE from MicroSim Corporation which was recently purchased by Cadence Corporation 39References 39Program Version 9 is available in a limited use small number of c omp onents from Also available Version 8 with detailed reference and instructions in the equired text for ECE3041 Schematic Capture Using MicroSim Pspice for Windows 9598NT by Hemiter This is the Version I will use in class one good may deems usm m device modelquot descripth are m P T a m m s Reidel and Ndssm lntxanvctamta Pprce Menla me CA Addison Wesley 1997 SPICE Computer Slmulatlons of ClIcults Types of analysis supported by SPICE Quiescent operating point determination 0 Dc sweeps of current voltage sources DC 39llrnedornain transient response TRAN Srnallsignal frequency response AC Fourier analysis FOUR Noise analysis NOISE Sensitivity analysis SENS 39Thevenin equivalents TF others SPICE Computer s1mulat10ns of Circults List of most Common SPICE diode model parameters Paramemr Svmbul SPICE Name Unis Default Satum un current In Dr IS IS A 10e14 Emissiun memoian n Dr I N r 1 Series resismnce RS RS 0 0 Rune in vnlmge vi Dr 9 VJ v 1 JuncLiun Capacitance cin cm F 0 Grading cn 39 cimt m M r 05 Transit time 1 TT 5 o Breakdnwn vn age v 13v v m I 113v A 10e1o Lecture 29 Operational Ampli er frequency Response Reading Jaeger 121 and Notes Ideal Op Amps Used to Control Frequency Response Low Pass Filter Previously 39 Vout Q Vm R1 R1 Vin Now put a capacitor in parallel with R2 If S j m V Vout R2 P Q 7i Vm R1 R1 vm R2 L Vour 1 sz R2 C2 7 i V 1 m R1 R2 7 25 1 1ch2 E Ideal Op Amps Used to Control Frequency Response Low Pass Filter 39 V out C2 39At DC 50 the gain remains the same as before RzRl 39At high frequency R2C25gtgt1 the gain dies off with increasing frequency V 3 VM R 1 Zir a R1 CZ f H 39At high frequencies more negative feedback reduces the overall gain Ideal Op Amps Used to Control Frequency Response Low Pass Filter I IAV DE 20 Log is the gain expressedindB R m 20 Lo 2 g R IAV IDB 1 3dB drop at fH vuu has dropped in hal Slope 720 dBDecade f 1 H 2n RZCZ f L0gf H 12 asbefme VRZRl A Dc m a gmmms a d mphmmy my a games ammmmgmm 1mp1 ements a Low Pass Filter Lower frequencies are allowed to pass the lter Ideal Op Amps Used to Control Frequency Response High Pass Filter 39 VOL Km 7 R2 Vm R1 At DC 50 the gain is zero At high frequency R1C15gtgtL the gain retums to it s full value RZRl Ideal Op Amps Used to Control Frequency Response High Pass Filter IAVIDE R MADE 20 Lug SSdB drop at fH Van 7 RZCIS Vm 1R1C15 20 dBDecade 1 2n RICI ELog At DC FD the gem 15 zem Athxgh sequency Rcsgtgt1 the gmnretumstmt39s mu valua rm11 39Implements a High Pass Filter Higher frequencies are allowed to pass the lter Ideal Op Amps Used to Control Frequency Response Band Pass Filter combination or high and low pass lter Pass Pass N 7 1 R20 R i 1R2C392 1Rcs s 1 VDE P P Band Pass Filter combination or high and low pass lter 1ch2 1Rcs R MADE 20 Lug f 4 27r39 1Cl 1 f 7 27r RZCZ fL 1 cysl Slopes if 20 dB Decade 3dB drop at fH fL ltlt fa Log Ideal Op Amps Used to Control Frequency Response Band Pass Filter combination or high and low pass lter 1 Ric 1 R202 1 Rcs Slopes if 20 dB Decade R MADE 20 Lug More than a 3dB drop 21th and fH fLltfH and fL gtfH General Frequency Response of a Circuit es and Zeros mum mLiu nfpulynnmials VA A Tus1fzs1fzs I M Tuwwli nwf ll 131w2 v 11 11 11 V y mm m Tum m Tum z emsquot denuminztnr pnlynumizl are called pales untm w my gain mm u mm gun n IEDead 0 Typically FM 390 nanDena Real Op Amp Frequency Response To this point we have assumed the open loop gain AOpen Loop of the op amp is constant at all frequencies Real Op amps have a frequency dependant open loop gain AOpenLaap s 2 S ZUB S ZUB where szjw A0 2 Open loop gain at DC ZUB 2 Open loop bandwidth A0 zUB MT 1 A an 3 dB 2010glAJ l t 20 dBdecade 40 I I I I 103 104 105 106 107 Radian frequency log scale ZD39T 2 Unity gain frequency frequency where Aopenmp 1 Georgia Tech ECE 3040 Dr Alan Doolittle Real Op Amp Frequency Response AdB 3dB 20 log IAJ A0 zUB lAOpenLoop Z 2 2 1339 03 20 IEdecade lAOpenLoop I l I l 103 m4 105 106 107 Radian frequency log scale At Low Frequencies IA 2 A OpenL 00p 0 A0333 wr ZU ZU A At High Frequencies OpenLoop For most frequencies of interest 03gtgt03B the product of the gain and frequency is a constant T w fT z T E Gain Bandwidth Product GBW 7r Georgia Tech ECE 3040 Dr Alan Doolittle WW Real Op Amp Frequency Response For the quot741 Op Amp For the quotOp 07 Op Amp An N 200000 106 dB An N 12000000 141 dB m5 N27z5 Hz m5 N 27005 Hz m N27z1 MHz m N27z06 MHz GBW GBW If the open loop bandwidth is so small how can the op amp be useful The answer to this is found by consideIing the closed loop gain Real Op Amp Frequency Response Previously we found that the closed loop gain for the Non inverting con guration was for nite open loop gain V AOptmLoop R1 Ayvaom mp 7 Where 7R R m p12 00p 1 2 Using the frequency dependent open loop gain A Vi 0 qu quotWW V 1 AW AOWB A 7 s 175 AGE395 LOW Karma 7 7 quotquot AGE395 sn751 Aa Pass 3 1275 AOWB A0 A mam10gt WAG 1 A 7 1 7 1 1 WHOM10 mum my where pper Curef Frequency Closed Loop Bandthh wa1 Aa Real Op Amp Frequency Response IAI dB 80 A The closed Loop 7 Open LOOPGam Ampli er has a lower gain than 60 quot Closed LoopGain the Open Loop Ampli er Mp 1 40 IA 390 20 VJ 0 103 104 105 106 107 Closed Loop Bandw1dth l 1 Radian fr uency 0g sca e wHZZDB1IBAO ZUT eq VClosedL00p DC The closed Loop Ampli er has a higher Closed Loop D C Gam bandwidth than the Open Loop Ampli er AOpenLoop VClosedL00p Georgia Tech 1 3 AOpenLoop ECE 3040 Dr Alan Doolittle Real 0p Amp Frequency Response IAI d5 80 A0 Closed Loop Gain set by feedback network below 03H 60 40 Closed Loop Gain set Open Loop Gain above coH 20 103 104 105 105 107 Radian frequency log scale Gain x Bandwidth Gain x Bandwidth 1 Open Luup Clused Loop Example 741 Op Amp is used as a low pass filter with fL10kHz What is the maximum voltage gain possible for this circuit From before we can write 200000 x 5 Gain x10000 Open Luup Clused Luup Gain Clmd WP 100 Maximum Georgia Tech ECE 3040 7 Dr Alan Doolittle Real Op Amp Frequency Response For the Inverting Con guration By superpusznun V VOL V7 Von R R R Vm R RZR v 1 2 R 1 2 V VmlJ VmJ 2 R1 R1 Vn b141 Von iviAV0panloop 0 R LVMI5VM 2 Avppmoop R1 Von AVvW OOP V 1 AKQunloop AVGosadLoop Real Op Amp Frequency Response Inserting the frequency dependent open loop gain A AyppmpJ i mama 1AW LW6 R AUWE A 7 swag R2 Agmg R2 VC104dlo 7 7 0P ADWE J R 5WEA0mE R swag Anna3 A AWE3 amp WE1Ag3 R2 WM smE1A0 R swag 1A0 R mama6 i e A HAM R VC104dloap 5 71 Wg1Ag3 Lecture 22 Bipolar Junction Transistors BJT Part 6 Understanding BJT Circuits Clearing up some Confusion Reading Note s Understanding a BJT Circuit Why is the base current so much smaller than the emitter and collector currents in forward active mode If the collector of an npn transistor was open ctrcutted tt wouldlook like a diode Hole Current g en forward btased the current tn 0 e has re ttt J c ould g constst ofholes lnjected tnto the rnttter rorn e electron injected tnto the base fro the m rnttter But stnce there more ele s tn th e e mltta than holes tn the base thevastmajorlty of anmge ctmnt the current th1 be due to electrons Why is the base current so much smaller than the emitter and collector currents in forward active mode When the reverse biased collector is added it sucks the electrons out of the base Thus the aseemitter current is due predominantly to hole current the smaller current component While the collectoremitter current is due to electrons larger current component due to more electrons from the n emitter doping Acuve or F arward Achve Hole Current Electron Current Energy Band Diagram Voltage Current Genrgjz Tech ECE 3040 7 Dr Alan Dnnlittle When to use Wthh Model EbersMoll model Always valid Cutoff saturation forward active active and reverse active inverse Simpli ed Ebers Moll Forward active only for DC solution Requires iteration 3 Analysis assume a turn on voltage when given 3 Forward active DC solution only Note 3 only has meaning in forward active mode Small Signal Models yparameter hybrid7t etc Forward active mode solving for the small signal AC solution 0 y Model SubClass1 catlons Ebers Moll model Always Applies Simpli ed EbersMoll model Assume FA mode and neglect small terms Used for DC or transient soluti ans Simpli ed EbersMoll CVD model usmg model adjusted for 3 analYSIS Base Width modulation ssume 1 mm on Add 1VCEV A terms Used voltage for the Base ernmer juncti on and solve the DC solution small Signal Models y based on p or 7 parameter vccs and cccs versions ofthe Hybridpi Used or AC small signals only form or transient soluu39ons Term Confusion VA is the applied voltage across a Diode VA is also the Early voltage for a BJT VT is the thermal voltage kTq but VT Will be used later for the Threshold voltage of a MOSFET BBDC BFO and SON BF neglecting variations in BF With ic Symb 01 Key Increasing Voltage f Increasing Current Decreasing Voltage Decreasing Current Direction of Current Genrgjz Tech Understanding BJT Circuit An increase in base Voltage will 1 produce an increase in base current which will do two things a produce an increase in emitter current which will i develop a larger Voltage across R2 raising Voltage VE b produce an increase in collector current which will i develop a larger Voltage across R1 lowering Voltage Vc ECE 3040 7 Dr Alan Doolittle Understanding BJT Circuit Symbol Key Conclusion Increasing Voltage VE will follow VB as VB increases so does V19 Increasing Current Vc will take the inverse action of VE as VB increases Vc will decrease Decreasing quot33 Use PSPICE and circuit D Cr asmg Cme Qualitative understanding of a BJT Circuit1 without eminer resistor 9 Direction of Current to simulate this to you e Genrg39z Tech ECE 3040 7 Dr Alan Dnnlittle Understanding a BJT Circuit Use PSPICE and circuits Qualitative understanding of a BJT Circuit1 Without emitter resistorcir and Qualitative understanding of a BJT Circuit2 with emitter resistorcir to simulate the DC bias points ofthis circuit by varying the value of Vbase R1 k R1 1 k 15V 7 15V Vbase Q1 Q2N3904 o1 QZN3904 08 7 R2 E1 k lt7 0 0 Use VbaSFODW7V 08V Use Vbase00 0 V 50V Cutoff Active Saturation Cutoff Active gtgtgt Very Sensitive ltltlt gtgtgt Less Sensitive ltltlt Geurgiz Tech ECE Lecture 13 PN Junction Diodes Part 3 Current Flowing through a Diode Reading Pierret 61 Pn Junction IV Characteristics In Equilibrium the Total current balances due to the sum of the individual components Hunoquot rm Llorlmn 39fmlon 39urrHIl urrenl n V5 E gt o o o J O O gt E V o e o 2322quot Male liffuion urrrnl a VA Georgia Tech ECE 3040 7 Dr Alsm Dnnlilllo P n Junction IV Characteristics Electron Diffusion I ElectronNDrift current nvs E Current ow Current is proportional to ewwref due to the exponential decay of carriers into M E the majority o 0 4 0 carrier bands Hf Hole Dril39t Hole D uSIdlgl Current Current ow is Current dominated by majority carriers owing across the junction and I becoming minority I carriers VA gt 0 QuickTim e Mam b Forward bias VA gt 0 Georgia Tech ECE 3040 Dr Alan Doolittle Pn Junction IV Characteristics Electron Drift Current ow Current IN is constant due to thermally Electron Diffusion Current negligible dur generated 39 0 large encrg burner carriers swept c 0 out by E 0 eldsinthe O O 0 depletion O O 393 C E region 0 quot En Hole Diffusion Current negligible E due to large energ barrier 1 Current ow is dominated by minority O H I D f EV carriers owing across IP 1 ml I the junction and becoming majority carriers c Reverse bias VA lt O QuickTimB Movie Georgia Tech ECE 3040 Dr Alan Doolittle P n Junction IV Characteristics Where does the reverse bias current come from Generation near the depletion region edges replenishes the current source Georgia Tech ECE 3040 Dr Alan Doolittle Lecture 12 PN Junction Diodes Part 2 How do they work A little bit of math Reading Pierret 61 Movement of electrons and holes when forming the junction Bo Be Be Be 9 ea em ea Circles are charges free to move E1 EKB Bo Be 9 ea em em 39 Bo Bo Jae ea em 9 ea demons and 110195 Bo Be as EI e 6 ea 9 39 Squares are charges NOT free to A move ionized donor or acceptor t a 01115 Electrgediffusion SE 913 9 BE 9 136 El Highhole Be Concentration High electron E9 E19 Concentration Local region of g positive charge charge due to p due to 1mbalance 1n imbalance in holeacceptor electrondonor concentrations l concentrations x Space Charge or Depletion Region eorgia Tech ECE 3040 Dr Alan Doolittle Movement of electrons and holes When forming the junction E dVdx dedV x m 7 de m dV Va way Vb but 0quot No net current ow Jquot q quotEqDquotE0 in equiliban E E DN dx kT dx de W q WW 7 q kT w Egg H mm quot7 Movement of electrons and holes When forming the junction VhkTh1M kT1n q HGXPU q 739 For N AND1 015cm393 in silicon at room temperature vbg 6 V For a nondegenerate semiconductor tqu K Eg uveresnmate fur the vultage that can be applied m a dwde befnre xtbums xtselfup Movement of electrons and holes When forming the junction Depletion Region Approximation 39 39 o o o o 30 330 If 0 o O 06 0636 06 39 l O O DenletinnRe inn Annm imminn 39 L no L known as the quasineutral region Thus dE p q i 7 n N 7 N 0 wtthm the 1mm 7 neutral re on l K550 K550 p D A q 5391 1 becomes 01E q ND 7 NA within the space chalge region dx K550 Movement of electrons and holes when forming the junction Depletion Region Approximation Step Junction Solution qNA for xprSO 0 qND forOSxan O forxS xp andexn thus N L for xprSO K580 d E forOSxan dx K580 O forxS xp andexn Where we have used dE dx KSED ND NDrNA t1ND Movement of electrons and holes when forming the junction Depletion Region Approximation Step Junction Solution i E t i H E E X x N E dE39zj de for x 3x30 0 exp 80 p N ExLxxp for xp 3x30 KSSO x x and P O n x E06 Ex Georgia Tech ND xn x forOSxan 80 x N IO dE39zj q Ddx forOSxan x KSSO i E E Since EXO39EXO NAXp DXn ECE 3040 7 Dr Alan Doolittle I Movement of electrons and holes when fonning the junction Depletion Region Approximation Step Junction Solution E dx V VVBi 13 xp x for xp SxSO dx qN Dxn x forOSxan KSgo V0 or I I x Vx x N 7 xp xn J0 dV 39Lp1S0 xp x ch for xp SxSO VB v x v v JV dV L KS xn xdx forOSxan Vx Georgia Tech qNA 2 x x or x SxSO 2ngop f p qND 2 x x 0r0SxSx b1 2nggn f n ECE 3040 7 Dr Alan Doolittle Movement of electrons and holes When forming the junction Depletion Region Approximation Step Junction Solution At X0 qNA x ZKSSO Movement of electrons and holes When forming the junction Depletion Region Approximation Step Junction Solution VA dropped here Negligible voltage drop lowlevel injection I Negligible voltage drop lowlevel injection Negligible voltage drop ohmic contact Negligible voltage drop ohmic contact V A0 No Bias V Alt0 Reverse Bias V Agt0 Forward Bias 39VA39 39 Vbi x Georgia Tech ECE 3040 Dr Alan Doolittle Movement of electrons and holes When forming the junction Depletion Region Approximation Step Junction Solution Thus only the boundary conditions change resulting in direct replacement of Vb With VinVJ 2sz NA V 7V d a 1 q NDNAND5 A an a ZKSSO q Movement of electrons and holes When forming the junction Step Junction Solution What does it mean Consider a p n junction heavily doped pside normal or lightly doped n side Georgia Tech T L KLVAV 0 VAgt0 ECE 3040 Dr Alan Doolittle Lecture 31 What makes up an OpAmp 741 example Reading Jaeger 168 and Notes What Makes up an Op Amp The 741 Example C merit Mirror Used as an Active Load 0 V DC Analysis If We can insure that the mirrored omens are equal I I 1I4Icz 2 Genrgz Tech ECE 3040 7 Dr Alan Dnnlmle What Makes up an Op Amp The 741 Example C merit Mirror Used as an Active Load AC Analysis 0 w i c m cA v 7a 01c4 c2 V V 7 v gm2 gm2 gmni Vow gmvdro 2 r04 Rt 2 gmvaL By using an active load transistors we obtain the Full gain of a CE stage even though we have a r singleended output L Remember that for a Gmgiz Tech passive load resistor the single ended gain was half the CE gain What Makes up an Op Amp The 741 Example Output Stages Class A Output Stage V out Maximum occurs when Q1 is saturated vm lt VCC IEIDCiOut Minimum occurs when Q1 is cutoff vm IDCR1 Only at this minimum voltage value would Q1 no longer conduct current So far the Common Collector output stages again chosen for Very loW output impedance has consisted of a single transistor This transistor must carry the output current delivered to R1 and the bias current IDQ This results in Wasted power in the output stage and possible overheating Genrgjz Tech ECE 3040 7 Dr Alan Doolittle What Makes up an Op Amp The 741 Example Output Stages Class B Output Stage When vin drops below 07V Q2 cuts off and vm0 Once vinlt07V Q3 turns on and output follows the input A Common Collector output stage using 2 complementary 1 npn 1pnp transistors where each transistor conducts only for N12 the cycle is known as a class B output stage During the positive half cycle Q2 conducts and Q3 is off All the current through Q2 is delivered directly to the load During the negative half cycle Q3 conducts and Q2 is off All the current through Q3 is delivered directly to the load ECE 3040 7 Dr Alan Doolittle Genrg39z Tech What Makes up an Op Amp The 741 Example Output Stages Class AB Output Stage nut Maximum occurs when Q4 15 saturated vm VCC IE4lOut Minimum occurs when Q5 15 saturated vm VEE IE5lOnt A partially biased Common Collector output stage using 2 complementary 1 npn 1pnp transistors Where each transistor conducts only for N12 the cycle is known as a class B output stage During the positive half cycle Q4 conducw and Q5 is Weakly on Almost all the current through Q4 is delivered directly to the load During the negative half cycle Q5 conducts and Q4 is Weakly on Almost all the current through Q5 is delivered directly to the load ECE 3040 7 Dr Alan Dnnlittle Genrgjz Tech What Makes up an Op Amp The 741 Example MiniDIP N Package OFFSET NULL 1 U Elm INVERTING INPUT 2 1 v NONINVERTING INPUT E 5 DUYPUT ME TO VIEW 5 orrsn NULL Reading Jaeger155 165 167 Georgia Tech R 3 50 m l m x 15 V A A Rm V V V 39 39 Input stage Second stage 39 Output stage 39 ECE 3040 Dr Alan Doolittle What Makes up an Op Amp The 741 Example Bias Stage Q8 27 I10129 uu us gc Differential stage is biased by a current mirror Georgia Tech ECE 3040 Dr Alan Doolittle What Makes up an Op Amp The 741 Example Input Stage Q15 Qa E 27 n p o a D 39 i 1 III Q Qs L39 m 94 Main input is a Inpmgc Differential stage Georgia Tech ECE 3040 Dr Alan Doolittle What Makes up an Op Amp The 741 Example Input Stage Q1 and Q2 form a Common collector difference amp CC used for it s high input impedance Q3 and Q4 form a Common base stage that only increases the input s resilience to overvoltage damage The worst case situation would be cc an VEE or viceversa This would result in 2 reverse biased junctions VB and VBEZ instead ofjust one as is found in the standard differential amplifier input circuit This extra junction provides an extra amount of safety to insure that large break down currents do not ow and damage the input stage Genrg39z Tech ECE 3040 7 Dr Alan Dnnlittle What Makes up an Op Amp The 741 Example Differential Input Load Stage uses an Active Load in the form of a current mirror Input stage I VSccond stage I Output stage Georgia Tech ECE 3040 Dr Alan Doolittle What Makes up an Op Amp The 741 Example Differential Input Load Stage uses an Active Load in the form of a current mirror V IE7 1B5 IB6 Jr I I V I B5 B6 B I EV W1 W1 150Kltlt Thus Q7 insures that we can neglect IB7 But IE5 m REF and R1 R2 so NEE Since VB IE6 R2 VBE6 IE5R1 VBES and Q5 is matchedwith Q6 IREF IE6 Ics Genrgz Tech ECE 3040 7 Dr Alan Dnnlmle What Makes up an Op Amp The 741 Example Differential Input Load Stage uses an Active Load in the form ofa current mirror The 741 allows the user to adjust the current in the active load to insure proper current balance insuring zero offset volmge at the output Current balance is achieve y an external potentiometer 39 cts to replace R1 and R with R a R H KRWWY and R a R H 1XRpomm where 0 S x 1 What Makes up an Op Amp The 741 Example Gain Stage 0 V 670 M Q15 27 a CC CC 39 gt Av1 Av1 229 D 916 R T V I I CE nag 0nd stage I Output stage I AV39gm Q16139o lRianz Georgia T ECE 3040 Dr Alan Doolittle What Makes up an Op Amp The 741 Example Output Stage Complementary npnpnp Common Collector Georgia Tech ECE 3040 Dr Alan Doolittle What Makes up an Op Amp The 741 Example Output Stage Complementary npnpnp Common Collector Georgia Tech ECE 3040 7 Dr Alan Dnnlinlp What Makes up an Op Amp The 741 Example Output Stage Current Limiting the output 279 p gt 220 F QM nd stage Output stage Georgia Tech ECE 3040 Dr Alan Doolittle What Makes up an Op Amp The 741 Example Output Stage Current Limiting the output During the positive Voltage half cycle Once the Voltage across R7 builds up due to large output current IEIS Ql7 turns on and begins to suck current away from the base on15 This prevenm further increases in the output current IEIS During the positive Voltage half cycle Q18 operates in an analogous fashion PIERRET CHAPTER 3 CARRIER ACTION Jeff Davis ECE304O REFERENCES Prof Alan Doolittle s Notes usersecegatecheduaanindexfilesECE3040indexhtm Prof Farrokh Ayazi s Notes usersecegatecheduayaziece3040 Figures for Require Textbooks Pierret and Jaeger DIFFERENT TYPES OF CARRIER ACTION Drift Currents applied electric field Diffusion Currents concentration gradients Recombination annihilation of EHP Generation creation of EHP DRIFT CURRENTS Holes move in direction of electric field Wm I 0 Electrons move in direction opposite of electric field 1 w Q QUALITATIVE MOTION UNDER APPLIED ELECTRIC FIELD Not ballistic Many collisions occur the lattice vibrations 00000 o o 0000 Direction of net motion gt Themal motion of electrons Other scattering mechanism Imperfection in lacttice surface interface and doping atoms Definition of DRIFT VELOCITY average velocity of carriers CURRENT AND CURRENT DENSITY DEFINITIONS charges passing through surface A q time it takes to move all charges out of AAX volume pAAx At Iq Ax vd quotaverage dnft veloc1tyquot At I quvd Amp u I a u Amp quot JPIdrift Z qpvdm Current Dens1ty m2 Also for electron carriers gt Jnldrift andn DRIFT VELOCITY AND ELECTRIC FIELD Low FIELD MOBILITY Mn and up uoE MOE when E a 0 v whenano sat 1 for holes 2 for electrons MOBILITY LL is the mobility of the semiconductor and measures the ease With which carriers can move through the crystal LL cmZVSecond unl360 cmZVSecond for Silicon 300K up460 cmZVSecond for Silicon 300K un8000 cmZVSecond for GaAs 300K up400 cmZVSecond for GaAs 300K ltTgt average time between particle collisions in the semiconductor Mobility of electron or hole cmA2Vsec IMPACT OF DOPING 1600 H H H O N b O O O O O O 800 f 600 f 400 200 O 1E14 1E15 1E16 1E17 Na or Nd cmA3 I mmemxAcmm UmanUmZOm On 259 DRIFT CURRENT AND ELECTRIC FIELD Jpldrift qp jdm qpaupE Jnldriftqnvdm qnlunE gt gt Jdrift Jnd1ift Jpldrift Jdrift Low field approximations RESISTIVITY AND CONDUCTIVITY resistivity conductivity 0E 1 q q 0 We already have an expression relating J and E 3 D11 7er qpup 11 1 qpu nun RESISTIVITY EXAMPLE What is the resistivity of ntype material that with doping ND 1x1017 cm393 nzND 1 1 1 mi 1 1 p 3 2 cm qND LLn 166 19Clel7cm 1350cm V SD DON T CONFUSE RESISTANCE AND RESISTIVITYI Area A 0mg my Resistance to current ow along length L Ie the electric eld is applied along this samples length RpLA or in units ohmcmcmcm20hms WORK AND POTENTIAL ENERGY OF POSITIVE CHARGES Work The transfer of energy from one object to another by a force From the perspective of an EXTERNAL FORCE performing positive or negative work Work is positive if energy is added to the object and negative otherwise W F d E E quotAn applied exteral forcequot d E quotdifferient displacement vectorquot Lerner RG and Rigg GL Encyclopedia ofPhyscs 2nd Edition VCH Publishers In New York 1991 WORK AND POTENTIAL ENERGY OF POSITIVE CHARGES Fhole gt E Negative displacement by an external agent Dr D requires POSitiVe displacement by an positive work The PE ofthe quot 39 39 39 39 D39 39 39 39 39gt eXterrial agent Dr D reqUWGS particle therefore increases negatlve work The PE of the particle therefore decreases W F dr F E quotAn applied exteral forcequot d E quotdiffen39ent displacement vectorquot Change PE is equal to the amount of work done by an external agent WORK AND POTENTIAL ENERGY OF POSITIVE CHARGES a d a E Vxx dx Work done on a postive charge by external agent Dr D is equal to the negative of the work done by the electric field on a particle moving from 0 to X39 Why Because Dr D is opposing the electric field to move this charge in this direction for this example xx39 xx39 gt XX39 d q q WDrDWelecuic eld J FEifield 39dx 2 J qE dX J gal3035 dX x0 x0 x0 xx39 d do EVxdx Wm qVx x39 m 0 Assume that x 0 that the potential is defined as zero In general X39 can be any position X W ARE 2 qVx Assuming that the potential energy at X20 is the referenced as zero then the potential energy of this positive charge is PE qVx WORK AND POTENTIAL ENERGY OF NEGATIVE CHARGES Felectron qE gt E Negative displacement of electron by an external agent Dr D Positive displacement by an requires negative work The PE quot 39 39 39 39039 39 39 39 39gt eXt inal agent Dr D reqUWGS of the particle therefore decreases DOSItlve work The PE ofthe particle therefore increases W F dr F E quotAn applied exteral forcequot d E quotdiffen39ent displacement vectorquot Change PE is equal to the amount of work done by an external agent WORK AND POTENTIAL ENERGY OF NEGATIVE CHARGES dVx a 1 1 quotA dx SK f d E E Vxx dx Work done on a postive charge by external agent Dr D is equal to the negative of the work done by the electric field on a particle moving from 0 to X39 Why Because Dr D is opposing the electric field to move this charge in this direction for this example xx39 X I gt X gt xx39 d q WDr DWelect1ic eld I FEifield 39dx 2 J qE dX J gal3035 dX 0 x0 x0 X PEqVx xx39d V x dx dx gt V WmD qvxx39Vx0 Assume that x 0 that the potential is defined as zero In general X39 can be any position X W APE qVx Assuming that the potential energy at X20 is the referenced as zero then the potential energy of this positive charge is PE qVx V POTENTIAL ENERGY CHANGES FOR POSTIVE AND NEGATIVE CHARGES ll96 y x x X PEqVx x Band diagrams from electron energy perspective POTENTIAL ENERGY CHANGES FOR POSITIVE AND NEGATIVE CHARGES PEqVx For Band Diagrams the Efield points in direction of increasing electron potential energy PE AND BAND DIAGRAMS A 0 T Excess energy is KE E L A C Total ener of electron gy Relative PE EV Eref E KE PE total PE EC Eref HOLE KE AND BAND DIAGRAMS O T Excess energy is KE of electron E C 1 Ev Excess energy is KE of hole 0 BAND BENDING UNDER APPLICATION OF E FIELD EC Energy Ener T gy E T EC gt gt EV position position Electric field makes electron energies a function of position BAND BENDING MEANS PRESENCE OF EF1ELD This is negative because these are electron ener ies PE EC Eref K 9 Ec Eref qV E EC eEi dE dV dx qE EV 1dEC lt E7 dx E Because EC E and E1 have the same relative position then ldE ldE 1dE E c v z qu qu EE Band Bending E Total I E electron energy 7 gt EV x a EC 7 Ev KE b II A A KE Ec E EV Ere J c V A x d 48 a Sample energy band diagram b Carrier energies kinetic 0 Electronic potential energy EC Eref qV d Potenital J ldEc 1dEv 4 e Electrlc eld q dx q dx q dx DIFFUSION CURRENTS Particles flow from greater concentration to lesser concentration ltquot39 Plcnrf JN ldirr Diffuse Diffuse 9 69 Q9 9 9 9 G G G 9 G G G G 7 x 9 9 G G gt x Figure 312 Visualization of electron and hole diffusion on a macroscopic scale FICK S LAW Fick s law describes diffusion as the ux F of particles in our case is proportional to the gradient in concentration F DVn where 17 is the concentration and D is the di usion coefficient Flux F particlesAreasec Jndiff 39an Jpdiff qu FICK S LAW AND DIFFUSION CURRENT For electrons and holes the diffusion current density ux of particles times q can thus be written as Jp lDi usion 0r Jn lDi usion Note in this case the opposite sign for electrons and holes 1 D EXAMPLES HOLES Px POO A a A P p gt 0 8x hole fl J lt 0 JP IDij cMsiongt O 9 Dl usmn X a P JP IDl CItsion CID pax 1 D EXAMPLESI ELECTRONS Jn IDi usion N A X l a ngt0 8x Electron f w J nDi gt0 X L V an Jn IDi CMsion a V TOTAL CURRENT 3 D Since Jp Jp lDrift Jp lDi usiorz qupEquVp and Jn Jn lDrift Jn lDi usiorz QHnnE anV and J Jp J TOTAL CURRENT 1 D Since 8 1 1 IDrift 1 IDi uSiorz qu and 8 Jn Jn IDrift Jn IDi usiorz 8x and J Jp J Don t forget the negative sign here EQUILIBRIUM CONCEPT Steadystate with no external electrical forces or temperature gradients across a given material The Fermi level inside a material or a group of materials in intimate contact is invariant as a function of position The Fermi level is FLAT aEf aEf aEf 0 dx dy dz WHY WHY IS THE FERMI LEVEL FLAT Fermi Function fE gives the probability that an energy state in a material is occupied 1 fE39 f51 f 16 kT Energy probabilities are the same throughout the materiall ie fE is position independent Orgtmm5gtr gtzgtrOQlt ZOZmOCIwNClt mgt ZO uOZ Orgtmm0gtu wgt u wgtum ltltI wgtum OZ Im umn ltltI gt 151mm u 0wult On IgtltZO LPZOmZ 3m OOV O O O O O 00 o o o o o o o o o o o o yOMAUV OO O O O o o o o o o vgtlt CLASSICAL ANALOGY WE LET THIS SYSTEM EVOLVE INTO EQUILIBRIUM WHICH IS A STEADY STATE CONDITION THE PROBABILITY OF A BALL HAVING A CERTAIN ENERGY BECOMES POSITION INDEPENDENT 8 QQQQOQEQIS g oo Q pbgb OPOQOOOxOoO Vgtlt EQUILIBRIUM EXAMPLE WITH A TWIST NON UNIFORMLY DOPED SEMICONDUCTOR LnND X Fewer electrons but higher energy Ec MWCI energy E Energy EQUILIBRIUM EXAMPLE WITH A TWIST NON UNIFORMLY DOPED SEMICONDUCTOR No net current can ow otherwise we have a perpetual motion machine But dEcdX is nonzero so we have a drift current component The drift current component MUST be balanced by a diffusion current component H JnlDrift JnlDiffusion EC O E LnND f EQUILIBRIUM EXAMPLE WITH A TWIST NON UNIFORMLY DOPED SEMICONDUCTOR JnlDiffusion JnlDrift0 and JplDiffusion JplDriftZ0 Thus for nonuniform doping in equilibrium we have Ef is constant No net current Carrier Concentration gradients that result in a diffusion current component give rise to a Built in electric field that produces a drift current BOTH electron and hole components must sum to zero LE J J 0 n p EINSTEIN RELATIONSHIPS dn Jn Dri Jn Di itsion an a 0 dE dE but E l and n mew By and f 0 q dx dx Thus taking the derivative of n dx kT dx 2 inE kT Thus becomes eEf EikT dEf EE kT M M anDn i 0 dEl dn eEf EikT dx kT or D D Likewise for holes p k T ElnSt m Mn up q Relatlonshlp mmoozgzaaz gt20 sz mw552 Om mTu 3 RECOMBINATION AND 3 GENERATION MECHANISMS BandtoBand gt EC W Photon Light gtltlt o Ev Recombination RG Center quotquot g 5quot y Recombination Auger H EC Recombination Thermal EC gt energy or Light W o Ev Generation Generation Generation aka impact ionization BANDTOBAND RECOMBINATION Ea W Photon Light E XltO V Band to Band or direct directly across the band recombination Some semiconductors are more likely to produce a photon are called direct bandgap materials eg GaAs Some semiconductors are less likely to produce a photon and are called indirect bandgap materia s Basis for light emission devices such as semiconductor LASERS LEDs etc DEEP STATE IMPURITIES IN SEMICONDUCTORS Es Ev Au Cu Mn Cr Fe Gold Copper Magnesium Chromium Iron R G CENTER RE COMBINATION In the form of E multiple phonons C or Thermal 9lt ET gt 1 energy 0 8 Ev I I r39 39 RGCeuLt1 I Also known as ShockleyRead Hall SRH recombination Two steps 1 1st carrier is trapped localized at an unintentional or intentional defectimpurity 2 2 01 carrier opposite type is attracted to the RG center and annihilates the 1st carrier AUGER RECOMBINATION Collision lt 3 Thermalization occurs Ec 2 EHP annihilation energy given to 2nd carrier Ev Xi BAND TO BAND GENERATION Thermal EC 39gt energy or Light W o Ev Band to Band or direct directly across the band generation direct bandgap material absorb photons more efficiently Thermal energy is the mechanism that resulm in intrinsic concentration Basis for light absorption devices such as semiconductor photodetectors solar cells etc RG CENTER GENERATION 2 Trapped e39 promoted to conduction band EC Thermal energy ET 1 Valence e39 is trapped 0 EV IMPACT IONIZATION MECHANISMS Ev High energy electron collides with bonding electron to produce EHP With high electric field this can produce AVALANCHE EFFECT momentum Not quite What we have in a Xtal but a starting conceptual Classical model REAL ENERGY BAND DIAGRAMS DIRECT VERSES INDIRECT BANDGAPS a Direct semiconductor b Indirect semiconductor Figure 317 General forms of E k plots for direct and indirect semiconductors REAL ENERGY BAND DIAGRAMs DIRECT VERSES INDIRECT BANDGAPS Phonons lVLgt Photon W k k a Direct semiconductor b Indirect semiconductor Figure 318 E k plot Visualizations of recombination in direct and indirect semiconductors REAL ENERGY BAND DIAGRAMs DIRECT VERSES INDIRECT BANDGAPS Direct Bandgap Conduction band AK E 14 eV P Valence band GaAs a FIGURE 1 12 Indirect Bandgap E A Conduction band IQ 11 eV band Vaknce 5i 7 Energyband diagram with energy vs momentum for rm GaAs direct and b Si indirect PHOTOGENERATION Light with photon energy hv lt Eg is not easily absorbed A A convenient expression for the ihv lt Eg energy of light is El247t Where 7 is the wavelength of the light in um and E is in eV hVgtEg Light with energy hv gt Eg is absorbed with the unabsorbed light intensity as a function of depth into the semiconductor is Light 01quot 1X loe390 X where 10 is the initial light 0 intensity X is distance and 0c is Semiconductor J the absorption coef cient lcm 1X 106quot X x a cm1 ABSORPTION COEFFICIENT AS A FUNCTION OF WAVELENGTH 105 I I l a 104 E E E E 124 103 photon A 3 ch 102 I I e z 3 0 3 Ge 10 r E Photon energies are 2 too small I I 104 06 08 10 12 14 16 18 A um PHOTOGENERATION Each Photon with energy greater than Eg can result in one electron hole pair Thus we can say 9 9P ax EHP ElLightzglLigsz xyl Where GLxlGLOe X cm3 Sec GL x A quotphotogeneration ratequot If oc is small near bandgap light the generation pro le can be approximately constant If oc is large light with energygtgt bandgap the generation pro le can be approximated as at the surface IMPORTANT NOMENCLATURE no p0 carrier concentrations in the material under analysis when equilib rium conditions prevail n p carrier concentrations in the material under arbitrary conditions An E n n0 deviations in the carrier concentrations from their equilibrium values Ap E p p0 An and Ap can be both positive and negative Where a positive devia tion corresponds to a carrier excess and a negative deviation corre sponds to a carrier de cit NT number of R G centerscm n AnnO and pAppO In nonequilibrium np does not equal niz Low Level Injection ApAn ltltnO and nnO in n type material ApAn ltltpO and ppO in ptype material CARRIER CONCENTRATIONS AFTER A PERTURBATION Steadysttate if perturbation has been applied for a long time NonSteady State Equilibrium Perturbation Nonequilibrium Apt0gt Aptgt po 01 gt Apt0ltAptltp 39gt l E C 11 Ap can be gtgt p0 T 0000 O O G E If Apgtgtp0 pAp 00 r V pOO Ap ltlt 710 After the carrier concentrations are perturbed by some stimulus leftmost case and the stimulus is removed center case the material relaxes back toward it s equilibrium carrier concentrations INDIRECT THERMAL RECOMBINATION RG CENTER RECOMBINATION RATE Rate proportional to RG center concentration 0C N T dt R Rate proportional to number of holes X p dt R Recombination decrease holes so it is negative c N p dt p T R CIo capture coefficient for holes INDIRECT THERMAL GENERATION RG CENTER GENERATION RATE EC Thermal ET energy 1 0 Rate proportional to the number of EMPTY RG center states It is unaffected by an increase in the number of holes therefore it can be approximated as its value in equilibrium Ev dP dP er dt G equilibrium INDIRECT THERMAL GENERATION RG CENTER GENERATION RATE EC Thermal ET energy 1 O In equilibrium the recombination and generation rate are in steady state such that they are equal and opposite Ev dp dp CN dtG d2 Tpquot G equilibrium R equilibrium INDIRECT THERMAL RG ELECTRON AND HOLE RATES For holes in an ntype material dP dt dP d19 dl R dt CpNT p p0 CpNTAp G indirect thermal R G The more holes the faster the rate of RG center recombination For electrons in an ptype material dn dt The more electrons the faster the rate of RG center recombination anTAn indirect thermal R G INDIRECT THERMAL RG ELECTRON AND HOLE RATES For holes in an ntype material d A 1 p pwhere L39p dt indirect thermal R G Tp CpN T IP minority carrier lifetime For electrons in an ptype material dn An 1 Where Tn dt indirect thermal R G Tn CnN T In minority carrier lifetime MATERIAL RESPONSE TO NONEQUILIBRIUM RELAXATION CONCEPT Consider a case when the hole concentration in an ntype sample is not in equilibrium ie pn does NOT equal ni2 a A 1 p p where IF 2 TP CPNT at thermal R G where Tpis the min orily carrier lifetime cp IS a proportionality constant N T is the trap concentration The minority carrier lifetime is the average time a minority carrier can survive in a large ensemble of majority carriers olf Ap is negativeGeneration or an increase in carriers with time If Ap is positiveRecombination or a decrease in carriers with time Either way the system tries to reach equilibrium The rate of relaxation depends on how far away from equilibrium we are BEYOND THE LOW LEVEL INJECTION ASSUMPTION Previous expression are for LOW LEVEL INJECTION the more general expression for arbitrary levels of injection are an 8p n np at thermal R G at TpnnlTnppl thermal R G where ET Ei Ei ET n1 2 nie H and p1 E nie kT EXAMPLE AFTER A LONG TIME ON A LIGHT IS SWITCHED OFF 8n An at In in An at In aAn An a r t Ant Ange Where Ann initial excess electron concentration nt no Ant t A nt mg Ange Anon A A tL Q AnO v 110 Time 1D CURRENT CONTINUITY EQUATION BEAN COUNTEREQUAHON AX J0 39 Jx AX t p gt Qxt A of charges IN of charges OUT Build up of charges in AX time time time Acharge from t to tAt At Qxt AtAx QxtAx At Ixt 1x Axt Qxt At Qxt Ax m Ixt IxAxt Ixt IxAxt mmmagn0 ax at Qx t 2 quotlinear charge densityquot 310602 BMW pXt quotVolume charge densityquot ax at 3D CURRENT CONTINUITY EQUATION BEAN COUNTER EQUATION JW 39 m a Ja Ja Jm ax 3y az a a xt 3 Definition of Divergence jx t quotCurrent Densityquot pXt quotVolume charge densityquot CONTINUITY EQUATIONS Drift Current out Diffusion Current out Drift Current in Diffusion Current in processes an an an Drift Diffusion 5 an an Re combination Generation All other processes at such as light etc 9pap 111 11 1P Drift Diffusion Re combination Generation All other processes at a such as light etc There must be spatial and time continuity in the carrier concentrations INCLUDE CURRENT CONTINUITY EQUATIONS 1 M 1 8 Drift Di usi0n aJNx Ny aJNZ V JN a a 6 8x By 82 q 1 M 1 8 Drift a pDi usi0n aJPx Py aJPZ V JP 8 a 6 8x By 82 q Divergence in the current an 1 an an V J at q N at Re combination Generation at 51262 gli ceetiriets 8p 1 8p 8p V J at q P at Re combination Generation at 5115 gphfceetirets CONTINUITY EQUATIONS SPECIAL CASE KNOWN AS MINORITY CARRIER DIFFUSION EQUATION Simplifying Assumptions 1 One dimensional case We will use 2 We will only consider minority carriers 3 Electric eld is approximately zero in regions subject to analysis 4 The minority carrier concentrations IN EQUILIBRIUM are not a function of position 5 Lowlevel injection conditions apply 6 RG center is the main recombinationgeneration mechanism 7 The only other mechanism is photogeneration E at ASSUMPTION 12 1D CONTINUITY EQUATION 4 an l 8Jnldri Jnldi nsion an 8x 5 ap 8Jpldrl Jpldi nsion ap ax Re combination Generation at Re combination Generation at 19 at All other processes such as light etc All other processes such as light etc ASSUMPTION 22 APPLY TO MINORITY CARRIERS For ptype materials anp l aJnldrift Jnldi nsion an 39 39 39 an at q ax at Re combination Generation at ZZZ12 ggzcjgses For ntype materials 81 Re combination Generation 8 at ax at 8Jpldri Jpdi cusion All other processes such as light etc ASSUMPTION 3 ELECTRIC FIELD APPROXIMATELY ZERO Jn Jndrift n di usion Jn z Jnldi usion quotax Jp z Jnldi usion For ptype materials I l an 1 aJ11ldz usi0n amp an P P at q ax at Re combination Generation at ZZZ1S6 ggzceetscsis For ntype materiags Pn 9pquot Z aup39di qDP ax 9191 9191 at ax at Re combmatlon Genemtlon a Ztahseg39igtoycgg ASSUMPTION 3 ELECTRIC FIELD APPROXIMATELY ZERO For ptype materials an 8211 an an 17 217 P a ax a All other processes such as light etc p Re combination Generation at For ntype materials 81 Re combination Generation at 8p 82p 81 at q P 8x2 at All other processes such as light etc ASSUMPTION 42 MINORITY CARRIER CONCENTRATIONS ARE NOT A FUNCTION OF POSITION IN EQUILIBRIUM n An n0 19 A19 190 For ptype materials amp 1 azAnp amp D at qqnaxz at aAn P a Re combination Generation All other processes such as light etc For ntype materials A 2A A apnDa 19 819 aApn at q P 8x2 at at Re combination Generation All other processes such as light etc ASSUMPTION 5 8c 6 Low LEVEL INJECTION AND RG CENTER THERMAL RECOMBINATIONGENERATION amp Amp Re combination Generation at Tn 819 Apn Re combination Generation at 139 7 For ptype materials amp 161D azAnp amp 6 quot 8x2 T at n All other processes such as light etc a For ntype materials aApn azApn Apn BApn q p 2 a ax T a P All other processes such as light etc ASSUMPTION7 PHOTOGENERATION an 11 2G at iffii i ft at j iffii i cf L For ptype materials BAH 1 82A An an 2 p GL 8 61 8x In For ntype materials EA WA Ap D G at q P 8x2 T L CONTINUITY EQUATIONS SPECIAL CASE KNOWN AS MINORITY CARRIER DIFFUSION EQUATION an 8n an an a a W 3 Wm a gt 3min 82Anp AI1p 4 TZD an Recombination Generation at All other processes such as light etc GL Minority N 2 8x 1quot Carrier 0r Diffusion Equations Mp D mp ltApgt G P 2 L at 8x I P OTHER SIMPLIFICATIONS WE MIGHT ENCOUNTER Steady State a A at at No minority carrier diffusion gradient 82 An 32 Ap p a 0 and D a 0 N 8x2 P 8x2 N0 SRH recombinationgeneration A np O and O Tn 139 p N0 Light GL a0 EXAMPLE PROBLEM 1 Problem A uniformly donor doped wafer maintained at room temperature is suddenly illuminated with light at time t0 Assuming Nd1075cm393 Tp 10396 sec and a light induced creation of 1017 electrons and holes per cm3sec THROUGHOUT the semiconductor determine Apnt for tgt0 FIRST WHAT EXPRESSION DO WE USE 92APn G at p 8x2 139 L I7 EXAMPLE PROBLEM 1 Problem A uniformly donor doped wafer maintained at room temperature is suddenly illuminated with light at time t0 Assuming Nd1075cm393 1p 106 sec and a light induced creation of 1017 electrons and holes per cm3sec THROUGHOUT the semiconductor determine Apnt for tgt0 8A WA A iqu2Pn GL at 8x Tp GL 1017 82A qu 2p 2 0 because uniform illumination ax 8A A pquot GL at 139 I7 EXAMPLE PROBLEM 1 Problem A uniformly donor doped wafer maintained at room temperature is suddenly illuminated with light at time t0 Assuming Nd1075cm393 1p 106 sec and a light induced creation of 1017 electrons and holes per cm3sec THROUGHOUT the semiconductor determine Apnt for tgt0 8A A VPquot GL Physically we would expect the P light generation and thermal t recombination to balance in Apn GL7 Aer steady state Assume at time t0 Apn 0 A GLTp t t 0 1 GL ApnzGLTp 1 6 T I7 EXAMPLE PROBLEM 2 Problem Consider a semiin nite ptype silicon sample with N A1015 cm393 constantly illuminated by light absorbed in a very thin region of the material creating a steady state excess of 1013 cm393 minority carriers What is the minority carrier distribution in the region XgtO 7 A Light absorbed in a thin skin 0 2 0 am p D 8 Amp Amp V21 at N 8x2 In L 82Anp Amp N 8x2 139 n EXAMPLE PROBLEM 2 General Solution Amp x AeXLN BeltXLN where LN E anTn LN is the diffusion length the average distance a minority carrier can move before recombining With a majority carrier Boundary Condition 13 3 Anpx 0 10 cm AB Anp x co 0 A0 BanLiv gt B 0 Anpx Anpx 21013 eHLN cm 3 EXAMPLE PROBLEM 3 Consider a ptype silicon sample With N A1015 cm393 and minority carrier lifetime Tl uS constantly illuminated by light absorbed uniformly throughout the material creating an excess 1013 cm393 minority carriers 1m second The light has been on for a very long time At time t0 the light is shut off What is the minority carrier distribution in for tlt0 X 8A 0 82 O A n 7amp21 7ampp G at N 8x2 1quot L Amp all xt lt 0 GLTn 2107 cm 3 EXAMPLE PROBLEM 3 In the previous example What is the minority carrier distribution in for tgt0 O 0 amp azgp Amp 1 DN 2 L 3 at 8x 1quot mpg b14170 0 g Amp t 107 ewe5 QUASI FERMI LEVELS EqUilibrium Ef no 2 nie and p0 nie Ez Nonequilibrium FN Ei H n nie and p nie 1 Ntype Low level injection Ntype High level injection EC EC Era FN Er Ei Ei FP FP EV E Lecture 8 Equations of State Equilibrium and Einstein Relationships and GenerationRecombination Reading Pierret 3334 Equilibrium Concept Consider a nonunifonnly doped semiconductor ECEf varies with position Since the electrons or holes are free to move anywhere in the material the average energy of the electrons can not change X If the average energy did change from one position to another there would LnND FTer electrons but higher energy More electrons but lower energy be a net mom 0f E C v electrons from high E1 E energy toward low f energy j V Ef must be constant when X no current ows 39 Energy Equilibrium Concept Remember No net current can ow otherwise we have a perpetual motion mac 39ne But dEcdx is nonzero so we have a dri current component The dri current component MUST be balanced by a diffusion current component E J JniDiffusion EC 0 Dr LnND Equilibrium Concept Additionally since electrons and holes operate independently of each other JniDrffusron JniDn 0 and inDrffusron inDn 0 Thus for nonuniform doping in equilibrium we have Efis constant No net current Carrier Concentration gradients that result in a diffusion current component A Built in electric eld that result in a drift current component OTH electron and hole components must sum to zero lE J J 0 Equilibrium Concept Consider the case for electrons Lin Jimw Jimmie WV 90 ix iii air but E i and n mew 5 and quot 0 Thus ak mg mi denvatzve if n gtxmw i dl gttxxtwr ix IrT ix ix ix IrT ix 7iVLE IrT Thus w becames q E7 ED 0 Imam an quotH D Lzrewzse fm holes y T Einstein 9 Relationship Equilibrium Concept Other need to knows kT Energy thermal energy 8617X10395 eVK T in K eV 8617x10395 eVK 16x103919 JeVT in K J kTq Voltage thermal voltage Jcoulomb JJW olts Dn Diffusion coef cient cmZsecond Example For Si ptan 358 27 C gt Dn 00259 V 1358 cmZVsecond 352 cmZSecond Real Energy band Diagrams Direct verses Indirect Bandgaps m bum M an lnrlmcl wmicuudutlur m Dmxl mmlcondwmr m lndlrccl mmmndu m The energy required to liberate an electron from the atom the energy bandgap is the same in all escape directions directions that an electron can leave the atom Example Electrons directed toward a neighboring atom would have a high escape energy while electrons directed toward a channel in the crystal a hole through the crystal would have a lower escape energy Thus the energy band diagram is actually a function of momentum Additionally both energy and momentum directed mass motion must be conserved during any transition Georgia Tech ECE 3040 Dr Alan Doolittle Real Energy band Diagrams Direct verses Indirect Bandgaps Direct Bandgap Indirect Bandgap Pmbab ity Of a dire Conduction transition from valence band E Aconduwoquot band to conduction band zigzirmty Of a band is low but if the valence electron is on an atom transmon vibrating in a direction 23 Iglence P Ie h39as momentum conduction that lowers the energv band is high requlred the probablhty increases GaAs i In h FIGURE 1 12 r r r Energyband diagram with energy vs momentum for m GaAs direct and h S undirecn Georgia Tech ECE 3040 Dr Alan Doolittle Real Energy band Diagrams Direct Verses Indirect Bandgap Variations in Light Absorption Polar materials lightdirect gap Electric eld resonates elds are additive with the atomic dipole and thus is absorbed strongly Magnitude of 7 Electric Field ofa Photon of light Direct Verses Indirect Bandgap Variations in Light Absorption A er the atoms move apart from their equilibrium positions the core is displaced from the electron cloud The photon s electric eldthen resonates elds are additive with the atom core electron cloud dipole and thus is abso e Covalent materials like Si Ge etc tend to 0 be poor light needed to induce a Electron cloud cry stal before the li t can be ab sorb edlndlrect M agnimde of 7 Electric Field ofa Photon of light 3 Recombination and 3 Generation Mechanisms g o E Thermal E 4 W Pliulon Energy Light or Light W gtltltgt Ev O E a Bandtoband recombination d Bandtubnnd generation 0 O E O E E or Thermal Tliemml 2 A7777 77775 X 1 ET energy energy I r 6 E Ev O 5 b R G center recombination c R G center genernnon j EC k E E Ev E c Auger reccmbinarion n Camer generation tin impacl ionization Figure 315 Georgia Tech ECE 3040 Dr Alan Doolittle Recombination and Generation Mechanisms Deep State Impurities Es Figure 316 Near Inidgap energy levels introduced by some common impurities in Si When an impurity introduces multiple levels one of the levels tends to dominate in a given semiconductor sample PERIODIC TABLE OF THE ELEBENTS Georgia Tech ECE 3040 Dr Alan Doolittle Recombination Mechanisms EC Photon 1s1ngle aiticle of li ht or W multiple phonons single quantum of lattice vibration W Ev equivalent to saying theimal Band to Band or direct directly across the band energy recombination Does not have to be a direct bandgap material but is typically very slow in indirect bandgap materials Basis for light emission devices such as semiconductor LASERS LEDs etc Genrgz Tech ECE 3040 7 Dr Alan Dnnlmle Recombination Mechanisms Energy loss can result in a Photon single EC particle of light but is more often multiple 9lt g ET phononsgsingle Quantum of lattice l Vibration eguivalent E to saying thermal O 8 V energy I I F 39 RG CUULUI I Also known as ShockleyRead Hall SRH recombination Two steps 1 1st carrier is trapped localized at an unintentional or intentional defectimpurity 2 2 1 carrier opposite type is attracted to the RG center and annihilates the 1st carrier Useful for creating fast switching devices by quickly killing off ehp s ECE 3040 7 Dr Alan Dnnlittle Genrgjz Tech Auger 7 pronounced O Requires 3 particles Two steps l 151 carrie Recombination Mechanisms 81 j ay recombination r and 2 d carrier of same type collide instantly annihilating the electron hole pair 151 and 3rd carrier The energy lost in the annihilation process is gi phonons until it s energy known as thermalization Georgia Tech yen to the 2 d carrier 2 2 d carrier gives off a series of returns to equilibrium energy EEc This process is ECE 3040 7 Dr Alsm Dnnlinlo Generation Mechanisms Thermal EC gt energy or Light W quot o Ev Band to Band or direct directly across the band generation Does not have to be a direct bandgap material Mechanism that resulm in n Basis for light absorption devices such as semiconductor photodetectors solar cells etc Genrgz Tech ECE 3040 7 Dr Alan Dnnlmle Generation Mechanisms Thermal energy ET 0 Ev Recombination Generation RG Center generation Two steps 1 A bonding electron is trapped localized at an unintentional defectimpurity generating a hole in the valence band 2 This trapped electron is then promoted to the conduction band resulting in a new ehp Almost always detrimental to electronic devices AVOID IF POSSIBLE Genrgz Tech ECE 3040 7 Dr Alan Dnnlmle MOS Transistor IV Derivation With um 39 39 39 and the fact that S 2 F we can determine the Value of the threshold Voltage VT 21 8 QIDF forn channel devices Cox V 5 VT 21D 7 Zi e 2m forp channel devices ox where S C X is the oxide capacitance per unit area x Where we have made use ofthe use of the expression 55 KSSD Genxg39zTech r rmquot n n r MOS Transistor IV Derivation Coordinate Definitions for our NMOS Transistor Xdepth into the semiconductor from the oxide interface ylength along the channel from the source contact FWldLh of the channel Xcy channel depth varies along the length of the channel nxy electron concentration at point Xy unxythe mobility of the carriers at point Xy Genrgjz Tech Device width is Z Channel Length is L Assume a Long Channel device for now do not worry about the channel length modulation effect ECE 3040 7 Dr Alan Dnnlittle MOS Transistor IV Derivation Concept of Effective mobility The mobility of carriers near the interface is significantly lower than carriers in the 5095 semiconductor bulk due to interface A scattering Dram A A JnxeLmjlyz Since the electron concentration also varies with position the average mobility of electrons in the channel known as the effective mobility can be calculated by a weighted average FM i 7 L0 y unx ynx ydx Empirically n e dx 0 Jpn nx y Mn 1 9 Va 7 VT UV de nm where lug and 9 are constants F50 QNy qLU nxy c charge 0m2l i 7 1 PM y 7 dx Q y F0 nxynxy n Gang ech ECE 30407Dr Alan Dnnlittle MOS Transistor IV Derivation Drain CurrentVoltage Relationship In the Linear Region VGSgtVT and 0ltVDSltV015m JN qynnE qDNVn Neglecting the diffusion current and recognizing the current is only in the ydirection 61 J 2 J 2 nE 2 7 n 7 N M Qn y Qn dy Geuxg39zTech r rmquot n n r MOS Transistor IV Derivation Drain CurrentVoltage Relationship In the Linear Region VGSgtVT and 0ltVDSltV015m Io Hdexdy 215 ij 7 Z iii 42W 1 Wm M Q dy z IDdyezm QNd yE 7 4H IDL 72M was zEQN 72 424m IDTLD was We need an expression relating d and QN Genxg39zTech r rmquot n n r MOS Transistor IV Derivation CapacitorLike Model for QN Neglect all but the mobile inversion charge The charge in the semiconductor is a linear function of position along the semiconductor side of the plate VG VG Since Cm Q dV SK 4 MOS Capacitor MOS Transistor QN E CMVGS VT for VGS 2 VT QN E CmVGs VT for V gt VT Neglect the depletion region charge GS Georgia Tech ECE 3040 7 Dr Alan Dmli lo MOS Transistor IV Derivation CapacitorLike Model for QN Neglect all but the mobile inversion charge The charge in the semiconductor is a linear function of position along the semiconductor side of the plate ID 7 1 gm 10 7 jfj 7 cox VG 7 V7 7 W This is known as the square law Geuxg39z Tech MOS Transistor IV Derivation CapacitorLike Model for QN For VDSgtVDsat ZjCox Vin Genxg39zTech r rmquot n n r MOS Transistor IV Derivation ZiC VZ ID VT WA 75 0 S VDS S VDW and V gt V GS vGlt vT VD V Dmt Genrgjz Tech ECE 3040 7 Dr Alan Dnnlittle MOS Transistor Applications Voltage variable Resistor An nchannel MOSFET has a gate width to length ration of ZL1 00 35200 cmZVsec Cox30166 uFcm2 and VEIV We want to develop a resistor that has a resistance that is controlled b an external voltage Such a device In variable gain ampli ers automatic gain control devices compressors and manyother quotquot U 39 39 achieve proper voltage variable resistance operation Find the Onresistance VDsID of the transistor from 15VltVGSlt4Vfor small VDS S a a 7 o E o a First to achieve vuha e 39 39 r 39 39 the linear region Otherwise the current is either a constant regardless of drain 39 39 39 39 quot and depletion Thus VGS VTgtVDS Given the values above 0ltVDSlt05V Continued Genxg39zTech r rmquot n n r MOS Transistor Applications Voltage variable Resistor Using the linear region ID equation ZITWCOX ZECquot Io TVG VTVD T st 7 V Vm for small vDS L Rm Va 7 VD i Z I o ZMECquot st 7V7Vm MWCOXKVG VT 7 001 7 200 015 rsFcm2 1st Thus 1009 3RDr 6009 Geuxg39z Tech MOS Transistor Applications Current Source 39 i u ed for a drainsource Voltage that can be used to achieve a xed current of50 uA Fora 9n 39 n 7 saturation region 2E0 2 2 50uA100200cm VSec0166uFcm Vail 2 VGS1173V This source will operate over a VDSgtVGSVT or VDSgt0173 V Genxg39zTech r rmquot n n r MOS Transistor Deviations From Ideal Channel Length Modulation Effect Above pinchoff when VDSgtVDSBLVGSVT the channel length reduces by a value AL 5 l Thus the expression for drain current l M l l i 7 Z quotCa 2 x i I D IDMI ikVos 7 VT VDmt g VDS r 2L Becomes i Z nCax ID IDMI mkVGS 7 VT 2 VDmt g VDS or since ALltltL 1 2 l 1 L 7 AL L L Zu7nCax AL ID IDMI 2L VGS 7 VVTZ1 Vow g VDS Genrgjz Tech ECE 3040 7 Dr Alan Dnnlittle MOS Transistor Deviations From Ideal Channel Length Modulation Effect But the fraction of the channel that is pinched off depends linearly on VDS so Channel Length Modulation AL causes the dependence of drain A V 1 cunent on the drain voltage in DS Saturn on where 7 is known as the ChannelLength Modulation parameter and is typically 0001 V391lt 7elt0l V 1 i ZiC IDIDMIVGS7VTZ1AVDS VD mt g VDS Genrgjz Tech ECE 3040 7 Dr Alan Dnnlittle MOS Transistor Deviations From Ideal Body Effect Substrate Biasing Until now We have only considered but the substrate Body is o en the case Where the substrate Body intentionally biased such that the SourceBody and DrainBody junctions are reversed biased S G has been grounded The body bias VBS is known as the backgatz bias and can be used to modify the threshold Voltage Note that noW our channel potential has an offset equal to VBS Genrgjz Tech ECE 3040 7 Dr Alan Dnnlittle MOS Transistor Deviations From Ideal Body Effect Substrate Biasing Thus our threshold potential With the body groun VT 2 4r ZZAOdF for n channel devices Ox 5 V7 2 7i LIND 2 for p Channel devices Cox 85 Becomes Surface Potential tbs s ZqN VGE MW 2435 7V55 Ci5 at rVES forn channel dev1ces s Vealmuw 2 V5 Ci quD 7 2 VHS for p channel devices 95 But We Would like to have this in terms ofVGS instead ofVGB Since VGS VGBVBS Vasimmm 2 EAJZZAQPF VES forn channel devices OX 5 mew 2435 7 2i quND 7 2435 VHS for p channel devices Ox 5 ECE 3040 7 Dr Alan Dnnlittle VT Genrgjz Tech MOS Transistor Deviations From Ideal Body Effect Substrate Biasing This can 39 39 voltage to the VESA case VPtenet VWUaeger Vm A 24 7 V7 24 form 7 channel devices VPtenet VJaegey7 V 7 y my 7 V 7 2 ltpr 7 channel devices 1 is known as the may gym parameter Genxg39zTech r rmquot n n r MOS Transistor Enhancement Mode verses Depletion Mode MOSFET We have been studying the enhancement mode MOSFET MetalOXideSemiconductor Field Effect Transistor It is called enhancemen because conduction occurs only a er the channel conductance is improved or enhanced In this case VTNgt0 and Vwlt0 Transistors canbe fabricated such that VTN g 0 and VTP 2 0 These transistors have conduction for VGS0 due to a channel already existing Without the need to invert the near surface region To modulate currents a eld must applied to the gate that depletes the channel Thus transistors of this nature are called Depletion mode MOSFETs Genxg39zTech r rmquot n n r
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