### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Microelectronic Circuits ECE 3040

GPA 3.64

### View Full Document

## 11

## 0

## Popular in Course

## Popular in ELECTRICAL AND COMPUTER ENGINEERING

This 0 page Class Notes was uploaded by Cassidy Effertz on Monday November 2, 2015. The Class Notes belongs to ECE 3040 at Georgia Institute of Technology - Main Campus taught by Staff in Fall. Since its upload, it has received 11 views. For similar materials see /class/233898/ece-3040-georgia-institute-of-technology-main-campus in ELECTRICAL AND COMPUTER ENGINEERING at Georgia Institute of Technology - Main Campus.

## Reviews for Microelectronic Circuits

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 11/02/15

Transfer Functions and Bode Plots Transfer Functions For sinusoidal time variations the input voltage to a lter can be written 111 t Re Viejm where V is the phasor input voltage ie it has an amplitude and a phase and 57m coswt j sinwt A sinusoidal signal is the only signal in nature that is preserved by a linear system Therefore if the lter is linear its output voltage can be written 120 t Re Vgejm where V0 is the phasor output voltage The ratio of V2 to V is called the voltageigain transfer function It is a function of frequency Let us denote T W Slolt We can write T as follows I T A to 67 where A w and go to are real functions of w A w is called the gain function and go to is called the phase function As an example consider the lter input voltage 111 t V1 cos wt 9 Re Vlejeejm The corresponding phasor input and output voltages are WW V0 VleJ39QA w ejww It follows that the time domain output voltage is 110 t Re V1579Aw 97quot 09739 A to V1 cos wt 9 90 This equation illustrates why A w is called the gain function and go to is called the phase function The complex frequency 5 is usually used in place of jw in writing transfer functions In general most transfer functions can be written in the form where K is a gain constant and N s and D s are polynomials in 5 containing no reciprocal powers of s The roots of D s are called the poles of the transfer function The roots of N s are called the zeros As an example consider the function s41 s41 TS45265561 4s21s31 The function has a zero at s 74 and poles at s 72 and s 73 Note that T 0 Because of this some texts would say that T s has a zero at s 00 However this is not correct because N 7E 0 Note that the constant terms in the numerator and denominator of Ts are both unity This is one of two standard ways for writing transfer functions Another way is to make the coefficient of the highest powers of s unity In this case the above transfer function would be written 5 4 s 4 T s 6 6 52556 s2s3 Because it is usually easier to construct Bode plots with the rst form that form is used here Because the complex frequency 5 is the operator which represents ddt in the differential equation for a system the transfer function contains the differential equation Let the transfer function above represent the voltage gain of a circuit ie Ts VgVi where V2 and Vi respectively are the phasor output and input voltages It follows that 52 55 s 1 V 4 1 V 6 6 o 4 2 When the operator 5 is replaced with ddt the following differential equation is obtained 1d2vo Edvo dUI 4 6dt2 6dtvo dtv1 where 110 and 121 respectively are the time domain output and input voltages Note that the poles are related to the derivatives of the output and the zeros are related to the derivatives of the input How to Construct Bode Plots A Bode plot is a plot of either the magnitude or the phase of a transfer function T as a function of w The magnitude plot is the more common plot because it represents the gain of the system Therefore the term Bode plot usually refers to the magnitude plot The rules for making Bode plots can be derived from the following transfer function in s T s K lt gt where n is a positive integer For n as the exponent the function has n zeros at s 0 For in it has n poles at s 0 With 5 jw it follows that Tjw Kji ww0in Kww0in and LTjw in X 90 lf w is increased by a factor of 10 changes by a factor of lot Thus a plot of versus w on logilog scales has a slope of log 10in in decadesdecade There are 20 st in a decade so the slope can also be expressed as iQOn dBdecade As a rst example consider the lowepass transfer function K T S T 1 swl This function has a pole at s iwl and no zeros For 5 jw and wwl ltlt 1we have Tjw 2 K 2 K and mm 2 0 X 90 0 For wwl gtgt 1 Tjw g Kymml lTjwl g Kww1 1and LT 2 71X 90 790 0 On log 7 log scales the magnitude plot for the lowefrequency approximation has a slope of 0 while that for the highefrequency approximation has a slope of 71 The low and highefrequency approximations intersect when K Kw1w or when w am For w wl 1T K 11 jl and LT 7 arctan 1 745 Note that this is the average value of the phase on the two adjoining asymptotes The Bode magnitude and phase plots are shown in Fig 1 Note that the slope of the asymptotic magnitude plot rotates by 71 at w wl Because an is the magnitude of the pole frequency we say that the slope rotates by 71 at a pole A straight line segment that is tangent to the phase plot at w wl would intersect the 0 level at ml481 and the 790 level at 481w1 lTltjwl Alma 1r 1 0 1 w 45 GJ 90 1 a b Figure 1 Bode plots a Magnitude b Phase As a second example consider the transfer function TsK1wil This function has a zero at s iwl For 5 jw and wwl ltlt 1we have Tjw 2 K 2 K and ZTjw 2 0 X 90 0 For wwl gtgt 1 Tjw 2 Kjww112 Kww1 and ZTjw 2 1 gtlt 90 90 On log7 log scales the magnitude plot for the lowefrequency approximation has a slope of 0 while that for the highefrequency approximation has a slope of 1 The low and highefrequency approximations intersect when K K w an or when w am For w wl K and LTjw arctan 1 45 Note that this is the average of the phase on the two adjoining asymptotes The Bode magnitude and phase plots are shown in Fig 2 Note that the slope of the asymptotic magnitude plot rotates by 1 at w wl Because an is the magnitude of the zero frequency we say that the slope rotates by 1 at a zero A straight line segment that is tangent to the phase plot at w wl would intersect the 0 level at an 481 and the 90 level at 481w1 Tjwl Alma 90 K 45 I u I 6quot1 1 a b Figure 2 Bode plots a Magnitude b Phase From the above examples we can summarize the basic rules for making Bode plots as follows 1 In any frequency band where a transfer function can be approximated by K jmm0in the slope of the Bode magnitude plot is in decdec The phase is in X 90 0 Poles cause the asymptotic slope of the magnitude plot to rotate clockwise by one unit at the pole frequency 03 Zeros cause the asymptotic slope of the magnitude plot to rotate countereclockwise by one unit at the zero frequency As a third example consider the transfer function sm1 T S sm1 1 This function has a pole at s 7m1 and a zero at s 0 For 5 jm and mm1 ltlt 1we have 2 Kmm1 and ZTjm 2 90 For mm1 gtgt 1 2 K and ZTjm 2 0 On logilog scales the magnitude plot for the lowefrequency approximation has a slope of 1 while that for the highefrequency approximation has a slope of 0 The low and highefrequency approximations intersect when K mml K or when m m1 For m m1 1T and LT 90 7 arctan 1 45 The Bode magnitude and phase plots are shown in Fig 3 Note that the slope of the asymptotic magnitude plot rotates by 71 at the pole The transfer function is called a highepass function because its gain approaches zero at low frequencies Tjwl Alma K 90 1 45 A I a Q Figure 3 Bode plots a Magnitude b Phase A shelving transfer function has the form 1 T s K 5 1 sm1 The function has a pole at s 7m1 and a zero at s 7mg We will consider the lowepass shelving function for which m1 lt mg For 5 jm and mm1 ltlt 1 we have 2 K and ZTjm 2 0 As m is increased the pole causes the asymptotic slope to rotate from 0 to 71 at ml The zero causes the asymptotic slope to rotate from 71 back to 0 at mg For mmg gtgt 1 2 Km1mg The Bode magnitude plot is shown in Fig 4a If the transfer function did not have the zero the actual gain at m1 would be The zero causes the gain to be between and K Similarly the pole causes the actual gain at mg to be between Km1mg and Km1mg The actual plot intersects the asymptotic plot at the geometric mean frequency The phase plot has a slope that approaches 0 at very low frequencies and at very high frequencies At the geometric mean frequency W the phase is approaching 790 If the function only had a pole7 the phase at an would be 745 approaching 790 at higher frequencies However7 the zero causes the highefrequency phase to approach 0 Thus the phase at an is more positive than 745 At the geometric mean frequency w the slope of the phase function is zero The Bode phase plot is shown in Fig 4b Tjw Alma 1 2 K 0 ca Kw1 quot1 45 72 w 790 6quot1 6quot2 0 b Figure 4 Bode plots a Magnitude b Phase Impedance Transfer Functions RC Network The impedance transfer function for a twoeterminal RC network which contains only one capacitor and is not an open circuit at dc can be written 1 73925 1 TPS ZRda where Rda is the dc resistance of the network7 Tp is the pole time constant7 and 739 is the zero time constant The pole time constant is the time constant of the network with the terminals open circuited The zero time constant is the time constant of the network with the terminals short circuited Figure 5a shows the circuit diagram of an example twoeterminal RC network The impedance transfer function can be written by inspection to obtain 1 R205 1 R1 R2 Cs CR2 L R2 R R 1 1 0 b ZR1 Figure 5 Example RC and RL impedance networks RL Network The impedance transfer function for a twoeterminal RL network which contains only one inductor and is not a short circuit at dc can be written 1 73925 1 TPS where Rda is the dc resistance of the network TP is the pole time constant and TZ is the zero time constant The pole time constant is the time constant of the network with the terminals open circuited The zero time constant is the time constant of the network with the terminals short circuited Figure 5b shows the circuit diagram of an example twoeterminal RL network The impedance transfer function can be written by inspection to obtain zRda 7 1LR 5 RIHRZW Voltage Divider Transfer Functions RC Network The voltageegain transfer function of a BC voltageedivider network containing only one capacitor and having a nonzero gain at dc can be written Va K 1 73925 V do 1 7P5 where Kda is the dc gain C an open circuit TP is the pole time constant and 739 is the zero time constant The pole time constant is the time constant of the network with Vi 0 and V2 open circuited The zero time constant is the time constant of the network with V2 0 and Vi open circuited Figure 6a shows the circuit diagram of an example RC network The voltageegain transfer function can be written by inspection to o tain E X V R1 R2 R3 1 R1 R2 HR3 Cs Figure 6b shows the circuit diagram of a second example RC network The voltageegain transfer function can be written by inspection to obtain E7 R3 X 1R1R2Cs Vi R1R3 1lR1HR3R21 CS HighPass RC Network The voltageegain transfer function of a highepass RC voltageedivider network containing only one capacitor can be written 7 mm where Koo is the in nite frequency gain C a short circuit and TP is the pole time constant The pole time constant is calculated with Vi 0 and V2 open circuited Figure 6c shows the circuit diagram of a third example RC network The voltageegain transfer function can be written by inspection to obtain E 7 R2 X R1 R2 Cs Vi R1R2 1R1R2CS V2 K TPS b c Figure 6 Example RC voltage divider networks RL Network The voltageegain transfer function of a BL voltageedivider network containing only one inductor and having a nonzero gain at dc can be written V2 1 73925 Vi dalfps where Koo is the zero frequency gain L a short circuit7 TP is the pole time constant7 and 739 is the zero time constant The pole time constant is the time constant of the network with Vi 0 and V2 open circuited The zero time constant is the time constant of the network with V2 0 and Vi open circuited Figure 7a shows the circuit diagram of an example RL network The voltageegain transfer function can be written by inspection to obtain E 7 R2 X 1 lLR2HR3l 5 V2 R1 R2 1 LlR1 132 llel 5 Figure 7b shows the circuit diagram of a second example RL network The voltageegain transfer function can be written by inspection to obtain E7 R3 X 1LRlS W7R1R3 1LlR1llR2R3l5 Figure 7 Example RL voltage divider circuits 1 mw JIYF Wq Uv JDP iFek nld gt 0f lt T n ma 3 3 913 ma w e D 3 Cuvo L 3 m w H lt CK ITYF n Marv Moicu05p c m H c MN 0 91 Trp ned 0 9 CPsan V Ho mild gJ VSWCLPP 9 Qtan De J Hmidud Hn U L O I Lunv o 9 dejY Ld r ffp 9n pltm iom V ltnm samf x ow Tlt v mitt V 40W lt0 I ltQ xx 2 Mn EL IAle m W H0 NIHnm IT 7 pou N 90 I MnWT L Qlltr0rdrumka an Smm Trb d 7 b 1 ssopPT PDPCmP SPHVYNlt PSL ob p0 4m 9 MUfnr Jltoci V in Us 3 1T0 OfnLS 05 P xoi fokou PPGMHT Mrnt HltIH30 95 ngo id QU l CoH IFP Nn qo lgtn4 ANA g PP I V rn gtn I amp rh l S ex x 1 PR H gtm l a gt9 A x n 0 20 WV X n A V I I llulul llll c 145 IV CuD l WADMWNV gt41 CVJ M l c Nm Na lt4 Pm u I1 L 9 u 1 Hm A w LP mFP TVFAT F 0c favcgl 0 T TQONIV 833 KOQO M nCSpj 6 Yn AWHMVMJP PXPF WNENFY TYF LICCC HSJVQW ltDJerU4 ng P mhh n Sthpy OCL lt5gy h39h qm MltPS CV 716 1 EXAM YQ CD V4 RC 2 VB OlaBL WK K quotr B l l RE za 4 743 V V4quot S Vquot lgv I QMA QC 75K QB 1001 RE 509 C7 03 39 IEi IE1 VT Rt It ag I Q he Pkg1162 if jig F Wf 7f a npiwt xs M 1 2 H3 lt25 1539 L1 137 Arm 2 4385111quot617 03L VLMA5 Bu 31qu 4 rm 4r 01170 Vv v 0 COMPOMQM5 AMDV1 V w239noc or ah4jS M5 fho a aw 15 0 15 Safar Fog oh o p Qma Commw p m wag Cowfcn m3939 LL39 039 1 414 4 new 413 11G Ar xr39kcw inT Q Erb p ewe G Ll Lwl wwlt sw9 53 9min 3w 9 4m 5 Awe 330 ltomF AS1lt MnUfnL Af lm Qr r Lwxhs WQX l 5 CgtAw CW 54 I I J Cxwvr A A 1T CHrv m V ruhP uS F ltND r yxmof olt dWVIltPr 4 Agriiav IL lnllilulTnlill lt9 TayOP O H 3v 943 c wrni lt4 P Cw mc nv monwp mor abs J vm 05m 3 Tu wofltn hc 810 Co P5P Qlo 136 11 gm o L L Wilb 391 L1 Y kg at Skac CTVCUL AY 5 UI H uvces U a D 1thAaSQ WL39 4 Pov a 0 4390 ARC VLG 3952 Th1 LA 24 FLC k V o o Thus wev objaim va 4cu39vaam Llwcuv 39 a l T Wo t 5 r L Ac 739 3 REI RF v l quot Xe C 43 39 39 3939 1 A7 2M 732 Tkis TS V Sawhe O VSuJ L NON UR fu h via 1 mum A E SWAth M i Gm 5 KCl39 Are 27 O 7 alt2 39 Tkaxz 5 Mo ouAreg foA a5 JUL Ara quotVT cy New U c5 25 v 3 Quasar U a D i5 H2 d wow3rlc THWJVS om 5242 0 He is Smuv kc 1gb oo hw 62 cw Q1 I TV WATS CaSo Ra Cowbimu Th Pcra2 wlkh 2g Quill wk erVL iv mu GZPLRAV A C t H I n 133m H l RV Ad gt9V lWI Fm 7Tw Coe u 6m 4 wa 01 49 m woc ruoru 0rnhxol OeS TrP OVHUCAPw PHltO 4r 003505 fenp qufk oS um Nh o r mcmn rPltp am So lt95me xosx rn 3cm roar T Ult0Csm Hxn Mcnr h Pmbm fc um 4KPWPS y Yn 0mwbrxp 7TP Marxlug Lapw 701 PYFSNUN x1 Tf DOSASAGI SOWP H9C rHOS JV 30 JGSMPlt wf o 17fo 0 TV Cowflemzm4dvj Cowwav Co c uv g TMI 54 5Q 13 Cowvv ohj Usa a l he od u39 4539 Hn am 0 am If axe B as a buil lzv 54062 9 Suf25 69041 0 aux Q1 Ufa793 Ma a3443 10 PM T lt V lt V I bobe Q 0m Q1 Ova c931 942 v L1S Cage U39O 0 now U Igt IK Q 03 OW Gwcl Q1 YLmaTn oPF TVQ OU PQ VONAF CA L 3 Skvew 13 do fUT VBE 7J71 1 WL QSUMD TEE X ComsaL quotAQ 1393 0 v0 Vtvsus U T 3 A zfa F 1 C w 239 TEEquot cows 9 TI W Mow Fov U I lt quot J 07 CU5 01A OLA Q v ma MS 0Q5F IV 1475 Cage Va U 1 WEB Vo 77 lt9 yu39H Ins H2 w6 cuvv 5 oltjz1erj um 513 U39a M Zrl V KT s lfl Vr NW2 New SUfPosa U l TS 0 57H WaJL 4f Q onAub f i Q1 Cov UCA39S rkg oueu foH39asp 6191qu 0 4onowsz 1 ns 09 0256 amp L f G D lt n 40 H5q wwx nix 003Plt Orwqw w 9 97m 04 SACMIV uh PmPL OS rcpr cxHUD gt1 owwWV nqlt PC Chapter 1 Conduction in Semiconductors 1 1 Introduction All solidestate devices eg diodes and transistors are fabricated from materials known as semie conductors In order to understand the operation of these devices the basic mechanism of how currents ow in them must be understood This chapter covers the fundamentals of conduction in semiconductors The chapter is not intended to be an extensive introduction to the area of solidestate physics Instead only those topics which lead to a better understanding of the macro scopic properties of semiconductors are covered The mechanisms of conduction in a metal and in a semiconductor are compared The effects of impurities on conduction in semiconductors are dis cussed The formation of a pen semiconductor junction is described and its conduction properties are discussed 12 Classi cation of Conductors Figure 11 illustrates a twoedimensional view of an atom that is called the Bohr model of the atom It consists of a positively charged nucleus and a system of negatively charged electrons which rotates around the nucleus In a neutral atom the total charge is zero This means that the positive charge on the nucleus is equal to the total negative charge on the electrons The electrons are bound to the nucleus by the forces of attraction between oppositely charged particles They are arranged systematically in layers called shells The closer a shell is to the nucleus the more tightly bound are the electrons in that shell to the atom The shell closest to the nucleus can contain no more than two electrons The outer shell can have no more than eight The number in the shells in between is determined by the laws of quantum mechanics The outermost shell in an atom contains what are called the valence electrons These govern the nature of chemical reactions of the elements In addition they play a large part in determining the electrical behavior of the elements and the crystalline structure of solids The metallic elements tend to have one two or three valence electrons The nonmetals have ve six or seven The inert gases have eight The class of elements which have four valence electrons is called semiconductors If a valence electron escapes its parent atom it becomes free to move about The parent atom then has a net positive charge and is called an ionized atom or an ion If an electric eld is applied to a material the free electrons have forces exerted on them which cause them to move This constitutes the ow of a current in the material that is called a conduction current or a drift current 2 CHAPTER 1 CONDUCTION IN SEMICONDUCTORS Electron t Valence Shell J M g l Figure 11 Twoedimensional Bohr model of an atom showing the nucleus and three shells Depending on the number n of free electrons per unit volume in a solid the material is classi ed as being a good conductor a semiconductor or an insulator For a good conductor n is very large and is independent of temperature A typical value is n z 1028 per m3 For an insulator at ordinary temperatures n is much smaller and has a typical value n z 107 per m3 For a semiconductor it lies between the values for a good conductor and an insulator and is a function of the temperature Silicon is an important semiconductor for which n z 15 X 1016 per m3 at room temperature T 300 K 13 Conduction in Metals Metals are classi ed as good conductors The valence electrons are so loosely bound to the atoms that they are free to move about in the conductor Fig 12 shows a twoedimensional illustration of the atoms in a metal with the free electrons distributed randomly among the immobile ions The free electrons can be visualized as molecules of a gas that permeate the region between the ions Analogous to the random motion of molecules in a gas thermal energy causes the free electrons to be in continuous random motion Observation of an individual electron would reveal that its direction of motion changes randomly after each collision with an ion Because the direction of motion of each electron is random the average number of electrons passing through any area per unit time is zero Thus the average current ow in the metal is zero 131 Drift Velocity If an electric eld E V m is applied to a metal an electrostatic force is exerted on the free electrons which causes a conduction current to ow The arrow indicates a vector quantity The force on an individual electron is given by F iq N where q is the electronic charge q 1602 X 10 19 C The electrostatic forces cause the electrons to be accelerated in a direction opposite to that of the applied eld Fig 13 illustrates the path that an individual electron might take under the in uence of the electric eld If the electron did not collide with the bound ions its velocity would increase inde nitely However energy is lost with each collision so that the average 13 CONDUCTION IN METALS 3 Ionized Atom Q 92 Free Electron Q Figure 12 Twoedimensional view of the atoms in a metal with free electrons distributed randomly among the ions velocity approaches a constant or steadyestate value The average velocity 7 ms is called the drift velocity It is proportional to the applied eld and is given by 7 we 11 where its I112V 1 s71 is the electron mobility The minus sign is required because the negative charge on the electron causes it to move is a direction opposite to the eld The average distance that the electron travels between collisions with the bound ions is called the mean free path As the temperature increases the bound ions vibrate with increasing intensity causing the mean free path between collisions to decrease This effect causes the drift velocity 7 to decrease which is modeled by a decrease in the electron mobility Me with temperature 6 Electric Field 9 Electron Drift Velocity Figure 13 Path taken by a free electron in a metal under the in uence of an applied electric eld 132 Charge Density The charge density p CIma in a conductor is de ned as the free charge per unit volume To relate the charge density in a metal to the density of free electrons let n be the number of electrons 4 CHAPTER 1 CONDUCTION IN SEMICONDUCTORS per m3 Because the charge per electron is iq it follows that the free charge per unit volume in the metal is given by p 7W 12 1 33 Current Density The current density 7 A m2 in a conductor is de ned as the current per unit area owing in a particular direction To relate the current density in a conductor to the drift velocity of the moving charges7 consider a section of wire of length A6 in which a current I is owing This is illustrated in Fig 14 The charge in the section is AQ pAV pSAt where p is the charge density and S is the crossisectional area of the wire Let At be the time required for the charge in the section to move the distance AZ The velocity of the charge is 7 EAZAt where E is a unit vector in the direction of current ow The current I owing in the area S is I AQAt It follows that the current density J can be related to the drift velocity 7 as follows IiAAQiApSAti Ag 7 a E aSAt aSAt paAt 7 1393 AI Current I Figure 14 Section of wire of length A6 in which a current I ows 134 Conductivity Using Eqs 11 through 137 we can relate the current density 3 to the electric eld E in a metal as follows gt gt gt gt J 97 WJ 6513 nweE UE 14 Note that two minus signs have canceled so that j is in the direction of This equation de nes the conductivity 039 Q71 m l of the metal It is given by U mm 15 Because it is independent of temperature in a metal7 it follows that the decrease in electron mobility Me with temperature causes the conductivity 039 to decrease with temperature 14 CONDUCTION IN INTRINSIC SEMICONDUCTORS 5 Example 1 Aluminum has three valence electrons per atom an atomic weight of 002698 kgmol a density of 2700 kgm o and a conductivity of 354 X 107 Sm l Calculate the electron mobility in aluminum Assume that all three valence electrons in each atom are free Solution Recall from introductory chemistry that a mole of any substance is a quantity equal to its atomic weight and contains a number of molecules equal to Avogadro s number which is 602 X 1023 It follows that the number of aluminum atoms per m3 is atoms 1 mol kg 28 atoms gtlt X 2700 6024 X 10 mol 002698 kg m3 m3 602 X 1023 Thus the electron density in the aluminum is n 3 X 6024 X 1028 1807 X 1029 per m3 From Eq 15 the mobility is given by a 354 X 107 36 103 2V 1 1 5 nq 6024 X 1028 X 1602 X 10719 7 X m S 1 35 Resistance Consider the section of wire illustrated in Fig 14 A conduction current I is owing in the wire so that the current density is j aIS where a is a unit vector in the direction of current flow and S is the crossesectional area Let V be the voltage drop across the section of length AZ so that the electric eld in the section is E EVAZ With the aid of Eq 14 we can write 7 aaVAl By equating the two relations for J we obtain a 16 This equation can be solved for the resistance R of the section of wire to obtain V Al R 17 I US Thus the resistance is directly proportional to the length of the wire and inversely proportional to its area Because the conductivity 039 decreases with temperature it follows from this equation that R increases with temperature In most metals the resistance increases linearly with temperature Example 2 The conductivity of copper is 58 X 107Sm If a 1m length of copper wire has a resistance of 1 Q what is the thickness of the wire Assume a circular cross section Solution Let d be the diameter of the wire Using Eq 17 we can write S 7T d22 AZUR 1 58 X 107 X 1 Solution for d yields d 0148 mm 14 Conduction in Intrinsic Semiconductors Semiconductors are the class of elements which have four valence electrons Two important semie conductors are germanium Ge and silicon Si Early solidestate electronic devices were fabricated almost exclusively from germanium whereas modern devices are fabricated almost exclusively from silicon Gallium arsenide GaAs is a semiconductor compound made up of gallium which has three 6 CHAPTER 1 CONDUCTION IN SEMICONDUCTORS valence electrons and arsenic which has ve This semiconductor is making inroads in digital ap plications which require extremely high switching speeds and in extremely highifrequency analog applications However silicon remains the most useful semiconductor material and is expected to dominate for many years to come Semiconductor materials are normally in crystalline form with each valence electron shared by two atoms The semiconductor is said to be intrinsic if it is not contaminated with impurity atoms Fig 15 shows a twoidimensional view of an intrinsic semiconductor crystal Each circle represents both the nucleus of an atom and all electrons in that atom except the valence electrons The links between the circles represent the valence electrons Each valence electron can be assumed to spend half time with each of two atoms so that each atom sees eight halftime electrons Compared to a metal the valence electrons in a semiconductor are tightly bound Q JQxQQQQQ SESESESISESESES QQQQQQ QQ SESESESESESE3EQ Figure 15 Twoidimensional illustration of the crystal lattice of an intrinsic semiconductor The thermal energy stored in a semiconductor crystal lattice causes the atoms to be in constant mechanical vibration At room temperature the vibrations shake loose several valence electrons which then become free electrons ln intrinsic silicon the number of free electrons is approximately one in 1012 of the total number of valence electrons The free electrons behave similarly to those in a metal Under the in uence of an applied electric eld they have a mobility and exhibit a drift velocity which produces a conduction current However because of the small number of free electrons the conductivity of an intrinsic semiconductor is much lower than that of a metal When an electron is shaken loose from an atom an electron vacancy is left which is called a hole The parent atom then becomes an ion The constant mechanical vibration of the lattice can cause the ion to capture a valence electron from a neighboring atom to replace the missing one When such a transfer takes place the position of the hole moves from one atom to another This is equivalent to a positive charge q moving about in the semiconductor The motion of a hole can be likened to the motion of a bubble in water Like free electrons holes have a mobility and exhibit a drift velocity which produces a conduction current under the in uence of an applied electric eld Because of the opposite charge polarity of electrons and holes they drift in opposite directions under the in uence of a eld Figure 16 illustrates the drift of free electrons in an intrinsic semiconductor under the applica tion of an electric eld that is directed from left to right When an electron is shaken loose from its valence shell an electronihole pair is formed The force generated by the electric eld causes the 14 CONDUCTION IN INTRINSIC SEMICONDUCTORS 7 free electrons to drift to the left Fig 17 illustrates the drift of holes In effect a hole drifts to the right when a bound valence electron shifts to the left from one atom to another The arrows in the gure point from the new position of a hole to its former position ie in the direction of movement of the bound electrons in the lattice The movement of holes may be likened to the movement of bubbles of air in water where the water represents the bound electrons and the bubbles represent the holes The movement of a bubble in one direction is really the result of a movement of water in the opposite direction In summary the flow of current in the semiconductor is the result of the flow of two components One component is the flow of free electrons in one direction The other component is the flow of the absence of bound electrons in the other direction Because of the opposite charge polarities the electron current and the hole current add to produce the total conduction current Electric Field gt QQCQJQCQQQ I 3EaaaISESISES EIF MESH1quot QCJQQCQQ QCQ c 8 I c I 3 c 31 S 3 Figure 16 Illustration of the drift of free electrons under the application of an external electric eld 141 Recombinations Because holeielectron pairs are continually created by thermal agitation of a semiconductor lattice it might seem that the number of holes and free electrons would continually increase with time This does not happen because free electrons are continually recombining with holes At any temperature a stable state is reached when the creation rate of holeelectron pairs is equal to the recombination rate The mean lifetime Tn s of a free electron is the average time that the electron exists in the free state before recombination The mean lifetime T10 s for the hole is de ned similarly In the intrinsic semiconductor Tn is equal to T1 because the number of free electrons must be equal to the number of holes However the addition of an impurity to the semiconductor lattice can cause the mean lifetimes to be unequal 1 42 Intrinsic Concentration 3 3 Denote the number of free electrons per m in a semiconductor by n and the number of holes per m by p In an intrinsic semiconductor the hole concentration must equal the electron concentration 8 CHAPTER 1 CONDUCTION IN SEMICONDUCTORS Electric Field V glllili ililg Figure 17 Illustration of the drift of holes under the application of an external electric eld In this case we write n p ni where n is called the intrinsic concentration It can be shown that n can be written n n0T32 exp 18 where no is a constant T is the absolute temperature V0 is the semiconductor bandgap voltage and VT is the thermal voltage The bandgap voltage multiplied by q represents the minimum energy required to cause a bound valence electron to become a free electron For silicon the bandgap voltage at T 300K has the value VG 111 V The thermal voltage is related to the temperature by 7 kT q where k is the Boltzmann constant k 1381 X 10 23JK At T 300 K the thermal voltage has the value VT 00259 V VT 19 Example 3 The initial temperature of a specimen of silicon is T1 300 K By what factor does the intrinsic concentration n increase if the temperature increases by 10 C 18 OF Assume the bandgap voltage at both temperatures is VG 111 V Solution Let T1 300K be the initial temperature and T2 310K the nal temperature At 300 K the thermal voltage is VT1 00259 V At 310 K it is VT2 1380 X 10 23 X 3101602 X 10 19 00267V The factor by which n increases is calculated from Eq 18 as follows 7 T232exp7Vg2VT2 7 E 32 E 71 1 7 300 exp T Hi1 T132 exp VGQVTl 2 00267 00259 143 Conductivity When an electric eld E is applied to an intrinsic semiconductor the free electrons drift with a vee gt gt locity 75 fueE where he is the electron mobility The holes drift with a velocity 7 uhE 15 NiTYPE AND PiTYPE SEMICONDUCTORS 9 where uh is the hole mobility Although the free electrons and holes drift in opposite directions the current densities add because the charge polarities are opposite The electron charge density is p5 iniq and the hole charge density is ph n q where ni is the intrinsic concentration The total conduction current density can be written gt gt gt J pe7eph7hmuemqE UE 110 This equation de nes the conductivity 039 of the intrinsic semiconductor It is given by UmMeMhq 111 Example 4 A rod of intrinsic silicon is 1 cm long and has a diameter of 1mm At room temper ature the intrinsic concentration in the silicon is n 15 X 1016 per m3 The electron and hole mobilities are he 013 m2 V 1 s 1 and uh 005 m2 V 1 s71 Calculate the conductivity 039 of the silicon and the resistance R of the rod Solution The conductivity is calculated from Eq 111 as follows a n pa mg 15 X 1016 X 013 005 gtlt 1602 X 1019 433 X 10 4Sm The resistance is calculated from Eq 15 as follows 03901 2 294MQ US 433 X 10 4 X 7r 05 X 10 3 15 n Type and p Type Semiconductors The preceding example illustrates how poor a conductor intrinsic silicon is at room temperature The conductivity can be increased by adding certain impurities in carefully controlled minute quantities When this is done the semiconductor is called a doped semiconductor There are two classes of impurities that are used These are donor impurities and acceptor impurities Typically one impurity atom is added per 108 semiconductor atoms A semiconductor that is doped with a donor impurity is called an netype semiconductor One that is doped with an acceptor impurity is called a petype semiconductor 151 nType Semiconductor An netype semiconductor is produced by adding a donor impurity such as arsenic antimony or phosphorus to an intrinsic semiconductor Each donor atom has ve valence electrons When a donor atom replaces an atom in the crystal lattice only four valence electrons are shared with the surrounding atoms The fth valence electron becomes a free electron as illustrated in Fig 18 The number of free electrons donated by the donor atoms is much greater than the number of free electrons and holes in the intrinsic semiconductor This makes the conductivity of the n7 type semiconductor much greater that of the intrinsic semiconductor Because the number of free electrons is far greater than the number of holes the free electrons are the majority carriers The semiconductor is called netype because the majority carriers have a negative charge 10 CHAPTER 1 CONDUCTION IN SEMICONDUCTORS Figure 18 Twoidimensional illustration of the crystal lattice of an nitype semiconductor Holeielectron pairs are continually formed by thermal agitation of the lattice in an nitype semiconductor Because of the large number of donor electrons there are many more free electrons available for recombination With the holes This decreases the mean lifetime for the holes Which decreases the number of holes in the nitype semiconductor compared to the intrinsic semiconductor For this reason the current due to the flow of holes in an nitype semiconductor is often neglected in calculations It is important to understand that a donor atom is electrically neutral if its fth valence electron does not become a free electron in the lattice If the fth electron becomes a free electron the number of protons in the atom is greater than the number of electrons by one In this case the donor atom becomes a bound positively charged ion 152 pType Semiconductor A pitype semiconductor is produced by adding an acceptor impurity such as gallium boron or indium to an intrinsic semiconductor Each acceptor atom has three valence electrons When an acceptor atom replaces an atom in the crystal lattice there are only three valence electrons shared With the surrounding atoms This leaves a hole as illustrated in Fig 19 The number of holes created by the acceptor atoms is much greater than the number of free electrons and holes in the intrinsic semiconductor This makes the conductivity of the ptype semiconductor much greater that of the intrinsic semiconductor Because the number of holes is far greater than the number of electrons the holes are the majority carriers The semiconductor is called ptype because the majority carriers have a positive charge Holeielectron pairs are continually formed by thermal agitation ofthe lattice in a pitype semiconi ductor Because of the large number of holes there are many more holes available for recombination With the free electrons This decreases the mean lifetime for the free electrons Which decreases the number of electrons in the pitype semiconductor compared to the intrinsic semiconductor For this reason the current due to the flow of free electrons in a pitype semiconductor is often neglected in calculations It is important to understand that an acceptor atom is electrically neutral if the hole created by the absence of its fourth valence electron is not lled by an electron from an adjacent silicon atom 15 NiTYPE AND PiTYPE SEMICONDUCTORS 11 QQ QQQ gEgagaglgagggag J Q 9 e 8 I c I c I e I 3 c Figure 19 Two7dimensional illustration of the crystal lattice of a p7type semiconductor Once an electron lls the hole the number of electrons in that atom is greater than the number of protons by one In this case the acceptor atom becomes a bound negatively charged ion 153 MassAction Law In an intrinsic semiconductor we have noted that the electron concentration and the hole concen7 tration are both equal to the intrinsic concentration ie n p m If this were not true the material would not be electrically neutral We have seen that adding an n7type impurity to the semiconductor increases n and decreases 13 Similarly adding a p7type impurity increases 13 and decreases n It can be shown that the product of n times 13 is a constant independent of the doping type and the doping level The product is given by my m2 112 where m is given by Eq 18 This relation is called the mass7action law To understand this equation consider an intrinsic semiconductor in which n p m Assume that donors with the density DA m are added to the semiconductor at t 0 This initially doubles the total number of free electrons which causes the recombination rate with the holes to double This causes the hole density to drop from m to mQ The initial free electron density is 2m which drops to 2m 7 mQ after the increase in recombinations Thus the product of the electron and hole concentrations is 2m 7 gtlt mQ 37124 Now suppose that the number of donors is increased by the factor N where N is large In this case the product becomes N71 7 mN gtlt mN 1 7 1N2 Although this is not an exact proof it illustrates the basic mechanism A more detailed proof requires an involved thermodynamic analysis 154 Electrical Neutrality An intrinsic semiconductor is electrically neutral ie there is no net charge stored The addition of n7type or p7type impurities does not change this To state this mathematically let ND be the number of donor atoms per m3 and NA the number of acceptor atoms per m We assume that all donor atoms and all acceptor atoms are ionized so that there are ND bound positive charges per 12 CHAPTER 1 CONDUCTION IN SEMICONDUCTORS m3 and NA bound negative charges per m3 Each donor ion has a charge H1 and each acceptor ion has a charge iq The total number of negative charges per m3 is equal to the number n of free electrons per m3 plus the number NA of bound acceptor atoms per m3 ie n NA Similarly the 3 is equal to the number p of holes per m3 plus the number ND ie p ND Because the semiconductor is electrically neutral the number of positive charges must equal the number of negative charges This gives the condition number of positive charges per m of bound donor atoms per m3 nNApND 113 In an netype semiconductor NA 0 and 13 ltlt n so that the above equation can be solved for n to obtain npNDND 114 The approximation in this equation and Eq 112 can be used to solve for the hole concentration 13 to obtain p z D 115 Similarly in a petype semiconductor we can write pnNANA 116 2 i z 11 n NA lt 7 Example 5 In the silicon rod of Example 4 the number of silicon atoms per m3 is 5 X 1028 A donor impurity is added to the silicon in the concentration of one donor atom per 108 atoms of silicon Calculate the new resistance of the rod Assume that each donor atom contributes one free electron Solution The donor concentration in the silicon is calculated as follows 8atoms 1 donors 5 X 1020 donors N 5 X 102 D m3 1 X 108 atom m3 It follows from Eq 114 that the free electron concentration is n 2 ND 5 X 1020 electrons per m3 From Eq 115 the hole concentration is p z 15 X 101525 X 1020 45 X 109 holes per m3 Because 13 ltlt n we can neglect p in calculating the conductivity Eq 111 gives 039 nque 5 X 1020 X 1602 X 10 19 X 013 1041 Sm The resistance is calculated from Eq 15 as follows At 001 aS 1041 X 7139 05 X 10732 Compared to the intrinsic silicon rod of Example 2 this is smaller by a factor of 24 100 It is possible to add both an acceptor impurity and a donor impurity to an intrinsic semiconduce tor 1f the donor concentration ND is equal to the acceptor concentration NA the semiconductor remains intrinsic because the free electrons of the donors combine with the holes of the acceptors With ND NA Eqs 112 and 113 give n p ni 1f ND gt NA the semiconductor becomes an netype In this case Eqs 114 and 115 become 122kQ npND7NAND7NA 118 15 NiTYPE AND PiTYPE SEMICONDUCTORS 13 n p E 7 ND NA Similarly if NA gt ND the semiconductor becomes a petype and Eqs 116 and 117 hold if NA is replaced by NA 7 ND 119 pnNA7NDNA7ND 120 2 N L n 7 N147 ND 121 155 Conductivity The conduction current density in an intrinsic semiconductor is given by Eq 110 In a doped semiconductor it is given by gt gt gt J W imhME 0E 12 Where n is the electron concentration and p is the hole concentration This equation de nes the conductivity a It is given by U We m q 123 For an intrinsic semiconductor n p n and this equation reduces to Eq 111 1 5 6 Diffusion Current In an netype or a petype semiconductor it is possible to have a component of current that is not a conduction current This current is due to the nonfuniform density of free electrons or holes and is called a diffusion current It is not possible to have a diffusion current in a metal To achieve a nonfuniform density of free electrons or holes the doping concentration in the semiconductor is not constant ie it is a function of position Such a concentration is called a graded doping concentration Figure 110 illustrates a semiconductor in Which the concentration of holes is a function of the coordinate z ie p p In addition 13 is a decreasing function of 2 Consider the plane de ned by z 21 Because the number of holes to the left of the plane is greater than the number to the right it is reasonable to expect that the random motion of the holes due to thermal energy would cause more holes to migrate from left to right than from right to left Thus there is a net current flow across the plane from left to right This current is called a diffusion current In general the hole diffusion current density is given by I ithVp 124 Where Dh is the hole diffusion constant and Vp is the gradient or directional derivative ofp If E a and 3 respectively are unit vectors in the m y and 2 directions Vp is given by 3p A323 A3 z 125 VP x 8m By 82 For the case illustrated in Fig 110 p is a function of 2 only so that Vp Edpdz The total derivative is used because 13 is a function of 2 only Because 13 is a decreasing function of z in the gure it follows that dpdz lt 0 This makes the direction of the diffusion current in the 2 direction 14 CHAPTER 1 CONDUCTION IN SEMICONDUCTORS Z direction gt Figure 110 Illustration of a semiconductor in which the hole concentration is a function of the coordinate 2 In an netype semiconductor that has a nonfuniform density of free electrons the electron diffue sion current is given by a J n qDBVn 125 where n is the electron concentration function and D5 is the electron diffusion constant The diffusion constants are related to the mobility constants by the relation amp amp VT 126 5 Mn where VT is the thermal voltage given by Eq 19 This relation is known as the Einstein equation 1 5 7 Total Current In general the total current in a semiconductor is written as the sum of the electron and hole conduction currents and the electron and hole diffusion currents It is given by gt gt J 0E qDeVn 7 Dth 127 where E is the electric eld intensity and 039 is the conductivity given by Eq 123 In an open 7 circuited semiconductor the equilibrium current density must be zero If we set 0 the above equation predicts an electric eld in the openecircuited semiconductor given by E qDV DV VT V V 128 h P i n P 7 n 39 U 8 7 m M M5 where Eqs 123 and 126 have been used As an application of the preceding results let us calculate the voltage difference between two points in a semiconductor caused by a graded doping concentration Consider a petype semicone ductor in which the hole concentration is a function of the coordinate z ie p p We assume that there are no external sources connected to the semiconductor so that the equilibrium current density is zero The semiconductor is illustrated in Fig 111 Let the voltage at z 21 be V1 and the voltage at z 22 be V2 The hole concentrations are labeled p1 and 132 We assume that p 15 NATYPE AND PATYPE SEMICONDUCTORS 15 gt is a decreasing function of z so that 131 gt 132 This gives an electric eld E that is directed in the negative z direction Because there are more holes at z1 than at 22 it may seem that the electric eld should be directed from left to right To understand why it is directed to the left assume that 131 p2 initially In this case there is no initial electric eld At t 0 let the hole density at z1 be increased by the addition of acceptors As 75 increases some of the holes at z1 diffuse toward 22 leaving bound negative ions at z1 The diffusion of holes toward 22 increases the net positive charge at 212 Thus the voltage at V2 becomes positive with respect to the voltage at V1 and the electric eld is directed from right to left When equilibrium is reached the net current flow is zero because the force generated by the electric eld cancels the diffusion force Electric Field lt V1 V2 1 1 l 1 z D39rect39on 101 p l l 1 l l l Z1 22 Figure 111 pitype semiconductor with nonuniform doping To solve for the electric eld E we set 71 0 and Vp Ezdpdz in Eq 128 to obtain E z 129 Because E is related to the voltage or potential function V by E 7VV and V V z it follows gt gt that E idedz By equating the two expressions for E we obtain AVT dp AdV z 7 7z 7 130 p dz dz when the 2 and the dz are canceled from both sides of the equation we obtain dV VTd p 73 This can be integrated to obtain V2 7 V1 VTln 73 1 131 P2 It follows that the voltage difference depends only on the concentrations at the two points and is independent of the separation of the points A similar equation can be derived for the voltage difference as a function of the free electron concentration 71 in an nitype semiconductor It is given by V2 7 V1 iVTln E 132 n2 16 CHAPTER 1 CONDUCTION IN SEMICONDUCTORS 16 The OpenCircuited p n Junction A pen junction is the junction between an netype semiconductor and a petype semiconductor It is fabricated by introducing donor impurities into one side of an intrinsic semiconductor crystal and acceptor impurities into the other side The transition between the two regions occurs in a very small distance typically 05 pm Fig 112 illustrates the cross section of a pen junction where the donor ions are represented by positive signs and the acceptor ions are represented by negative signs Initially we assume that the only charge carriers in the netype side are free electrons and that the only charge carriers in the petype side are holes Electric Field Bound 9 Bound Acceptor lon Donor lon 0629090 9 OIetel Mobile 6 e e e e e e 9 Mobile Hole gt0 0 o o 0 o o 0lt Electron 9 e e e e e e e 0 O O O O O O 0 e e e e e e e e O O O O O O O O pitype Depletion n type Re ion Figure 112 Diagram of a pen junction with the width of the depletion region greatly exaggerated Because of the unequal electron concentrations and unequal hole concentrations on the two sides of the junction a diffusion current consisting of both holes and free electrons will flow across the junction The diffusion process is similar to the diffusion of different gases between two glass jars joined at the mouths Holes diffuse out of the petype side and into the netype side and free electrons diffuse out of the netype side and into the petype side This causes the ptype side to become negatively charged and the netype side to become positively charged The charges cause an electric eld to build up across the junction which is directed from the netype side to the petype side The polarity of the electric eld is such that the force it exerts on the holes and free electrons opposes the diffusion process Equilibrium is reached when the force exerted on the charge carriers by the electric eld is equal to the diffusion force Let us now consider what happens when thermal agitation of the semiconductor lattice produces a holeelectron pair in the region near the junction The electric eld directed from the netype side to the petype side exerts a force on the free electron and causes it to be swept to the netype side Similarly the hole is swept to the petype side The directions that the charges move are opposite to those due to the diffusion process When equilibrium is reached the net number of both electrons and holes crossing the junction is zero 161 Depletion Region Because no free electrons or holes can exist is the region about the junction there are no mobile charges to neutralize the ions in this region This is illustrated in Fig 112 The ions on the 16 THE OPENiCIRCUITED PiN JUNCTION 17 netype side have a positive charge on them and those on the ptype side have a negative charge These charges are called uncovered charges The region about the junction in which the uncovered charges exist is called the depletion region Other names for this are the spaceicharge region and the transition region Fig 113a illustrates the plot of the net uncovered charge density in the pen junction as a function of distance from the junction The charge distribution is called a dipole distribution because the charge on one side of the junction is the negative of the charge on the other side The uncovered charges on each side of the junction can be thought of as the charges on the plates of a parallel plate capacitor as shown in Fig 113b Charge Density p Distance Q Q a b Figure 113 a Plot of the charge density as a function of distance from the junction b Parallel plate capacitor analog of the charge distribution Because of charge neutrality the total uncovered charge on the netype side of the depletion region must be equal to the negative of the total uncovered charge on the ptype side If the n and p concentrations are equal it follows that the widths of the uncovered charge regions on the two sides of the junction must be equal Now suppose the p concentration is increased while holding the n concentration constant Charge neutrality requires the width of the ptype side of the depletion region to decrease if the total uncovered charge is to remain constant Similarly if the n concentration is increased while holding the p concentration constant the width of the netype side must decrease We conclude in general that increasing either p or n or both decreases the total width w of the depletion region illustrated in Fig 113a This has an important effect on the reversebias breakdown characteristics of a junction This is discussed in the following chapter 162 BuiltIn Potential Because there is an electric eld in the depletion region of a pen junction that is directed from the netype side to the petype side it follows that there is a difference in potential or voltage across the junction This voltage difference is called the builtiin potential or the contact potential It can be calculated from either Eq 131 or Eq 132 Let us use Eq 132 In this equation p1 is the hole concentration in the ptype side and p2 is the hole concentration in the netype side By Eq 116 the hole concentration in the ptype side is pl 2 NA where NA is the acceptor concentration per m3 By Eq 115 the hole concentration in the netype side is pg nZZND where n is the intrinsic concentration per m3 and ND is the donor concentration per m3 It follows from Eq 18 CHAPTER 1 CONDUCTION IN SEMICONDUCTORS 132 that the builtein potential VB is is given by VB VTln NAIZVD 133 7 i The same result is obtained from Eq 131 Example 6 An openicircuited pen junction is fabricated from silicon The acceptor and donor concentrations are NA ND 5 X 1020 per m3 The intrinsic concentration is n 15 X 1016 per m3 Solve for the builtiin potential at room temperature Solution The thermal voltage at room temperature is VT 00259V By Eq 133 the builtein potential has the value 5 1020 2 VB VTln NAIZVD 00259111 0539v i 15 X 10162 17 The ShortCircuited p n Junction Figure 114 shows a pen junction which has metal contacts attached to each end and a short circuiting wire connected between the contacts From the discussion of the openecircuited pen junction it might seem that the builtein potential across the depletion region would cause a current to ow in the external wire If this happened the second law of thermodynamics would be violated This law states that a system in equilibrium with its environment cannot deliver work To see how this law would be violated let us assume that a current ows in the short circuiting wire The current must ow through the semiconductor material which exhibits a nite conductivity 039 given by Eq 123 Because heat is generated when a current ows through a conductor of nite conductivity the system generates energy in the form of heat with no input energy from external sources This is a clear violation of the second law of thermodynamics Metal Contact Metal Contact Figure 114 Shortecircuited pen junction The reason that a current does not ow in the short circuit is because the path through the pen junction and the wire contains two metalesemiconductor junctions as well as the pen junction Like 18 THE BIASED PiN JUNCTION 19 the pen junction both metalesemiconductor junctions exhibit a builtein potential The algebraic sum of the three builtein potentials is zero so that there is no net voltage to cause a current to ow in the wire and the second law of thermodynamics is not violated The characteristics of metalesemiconductor junctions differ primarily from those of pen junctions in two ways First they conduct current well in both directions Second the builtein potential does not change when a current ows through the junction Junctions which have these properties are called ohmic junctions or nonrectifying junctions We will see in the following that the pen junction does not have the properties of an ohmic junction Not all metalesemiconductor junctions are ohmic The Schottkyebarrier diode is an example which is described in the following chapter 18 The Biased p n Junction 181 Reverse Bias Figure 115 shows a pen junction with a dc source connected to it The polarity of the source is chosen so that the positive terminal is connected to the netype side and the negative terminal is connected to the petype side The current that ows across the junction consists of two components a diffusion current caused by unequal carrier concentrations on the two sides of the junction and a conduction current caused by the electric eld across the junction With V5 0 these two currents exactly cancel each other so that the net current is zero Electric Field Bound 9 Bound Acceptor lon lDonor on Mobile 9 9 6 e e B e a are Mobile Hole gt0 0 0 0 0 lt Electron e e e e e e e e e e O O O O O I 9 9 9 9 6 63 63 e 63 63 O O 0 I I I e e e e e e e 63 e e O O O O O I Pitype Depletion H type 1 Region Figure 115 Reverse biased pen junction Now let us examine what happens when V5 gt 0 Because negative charge is attracted by a positive voltage and positive charge is attracted by a negative voltage both the free electrons in the netype side and the holes in the ptype side are pulled away from the junction This causes the width of the depletion region to increase so that there are more uncovered charges on each side of the junction This is illustrated in Fig 115 compared to Fig 112 The potential across the junction which opposes diffusion is increased by the applied bias to the value VB V5 where VB 20 CHAPTER 1 CONDUCTION IN SEMICONDUCTORS is the builtein potential This is greater than VB so that the electric eld across the junction is increased Because the diffusion force on the charge carriers is opposed by the force exerted by this electric eld it follows that the diffusion current is decreased by the applied voltage The diffusion current approaches zero as V5 is increased Although the diffusion current goes to zero the conduction current due to thermally produced holeeelectron pairs in the depletion region continues to flow across the junction When such a holeeelectron pair is generated the electric eld across the junction causes the electron to be swept to the netype side and the hole to be swept to the petype side This causes a very small current to flow in the external circuit in the direction indicated in Fig 115 Because the current is so small the junction is said to be reverse biased 182 Forward Bias Figure 116 shows the pn junction with the polarity of the dc source chosen so that the positive terminal is connected to the petype side and the negative terminal is connected to the netype side Because positive charge is repelled by a positive voltage and negative charge is repelled by a negative voltage both the free electrons in the netype side and the holes in the petype side are forced toward the junction This causes the width of the depletion region to decrease so that there are fewer uncovered charges on each side of the junction This is illustrated in Fig 116 compared to Fig 112 The potential across the junction which opposes diffusion is decreased by the applied bias to the value VB 7 V5 where VB is the builtein potential This is less than VB so that the electric eld across the junction is decreased This decreases the force which opposes diffusion so that the diffusion current increases rapidly as V5 is increased Because a large current ows the junction is said to be forward biased Electric Field Bound Bound Acceptor lon Donor lon Mobile 9 e e e G 6 9V6 Mobile Hole gt0 0 o 0 0 I lt Electron e e e e e e e e O O O O O 0 e 6 9 6 63 G 63 O O O O I I I I e e e e e e e e O 0 O O I Pitype Depletion n type Region lll VS I I Figure 116 Forward biased pen junction It might seem that the potential across the junction which opposes diffusion could be made to go to zero by increasing V5 Should this happen the width of the depletion region would go

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.