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by: Immanuel Brakus PhD


Immanuel Brakus PhD
GPA 3.66


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Class Notes
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This 17 page Class Notes was uploaded by Immanuel Brakus PhD on Monday October 19, 2015. The Class Notes belongs to ECSE 2210 at Rensselaer Polytechnic Institute taught by Staff in Fall. Since its upload, it has received 26 views. For similar materials see /class/224781/ecse-2210-rensselaer-polytechnic-institute in ELECTRICAL AND COMPUTER ENGINEERING at Rensselaer Polytechnic Institute.





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Date Created: 10/19/15
Chapter 32 Diffusion and band bending We will learn two new topics in this lecture Diffusion 7 a process whereby particles tend to spread out or redistribute as a result of their random thermal motion migrating on a macroscopic scale from regions of high particle concentration to region of low particle concentration Examples of diffusion Perfume in a room Ink drop in a bottle of water Hot point probe measurements Band bending 7 resulting from the presence of electric eld inside a semiconductor No band bending means the electric eld is zero Hotpoint probe measurement This is a commonly used technique for determining whether a semiconductor is ptype or ntype Carriers diffuse more rapidly near the hot probe This leads to a particle current away from the hot probe and an electrical current away ptype or towards ntype the hot probe Energellc hula Energcuc eleclmns ay aw diffuse aw 1 us 4y 3 lb Figure 313 Diffusion current For diffusion to occur there must be a concentration gradient Logically greater the concentration gradient greater the ux of particles diffusing from higher concentration region to lower concentration region IfF is the ux ie the of particles cm2 s crossing a plane perpendicular to the particle flow then F D d n n particle concentration dx where D is called the diffusion coefficient The sign appears because for positive concentration gradient dndx the particles diffuse along the negative x direction Particle diffusion Concentration gradient dndx positive Concentration 11 Particles ow along x direction Diffusion current H 1 Electrons oes O O O O O O C C O O O O C O O O O O O O O O C O O O O O O O O O O O O O x hole ux electron ux hole diffusion current JP diff qudPdx Jnldiffandndx What is the unit of diffusion coefficient D Total currents Jp Jpldn39 Jpldiff quPE liq311 drift diffusion Jn Jnldd Jnldiff qunnfl 421 The total current owing in semiconductor is given by J Jn Jp electron diffusion current Band bending Band diagram represents energies of electrons 7 so far we have drawn it as independent of position When E eld is present EC and EV change with position called bandbending This is a way to represent that an E eld is present No E fzeld x indicates presence of E fzeld 7 Band bending and electrostatic variables Diagram represents total energy of electrons with x KE E EC for electrons PE EC Eref for electrons From elementary physics PE q Vfor electrons V 1W Ec Em g dVdx1qdEcdx Flume 1w Band bending Crudely inverting EC in eV versus x diagram results in electrostatic potential Vin Volts versus x diagram Similar to potential energy V is relative with respect to some arbitrary reference l V 3 EC 33 IfEC Eref is given in eV we use e 16 X 10 19 C to convert from eV to Joules Thus values of Vin Volts are numerically equal to Ec Eref expressed in eV The slope of EC energy in eV versus x diagram gives the E eld versus x plot ldEc idEVidEi qu qu qu E eld expressed in V cm will be numerically equal to dEi dx if E is in eV and x in cm Example 1 Exercise 32 Plot electrostatic potential V and E fzeld E versus x for the case shown below u a o h 2 3 0 O I EC 8 in d 39 E1 LF T O r 1 o E r oles n k Review Resistivz39 ormula qup ann I Jdn39 Jn dd Jp d quinnJerp I Drift currentdensz39ty dn dp Jn diff anE and Jp diff qDPE Dz uszon current denszty d Jp den39 Jpditr 111 0111311317 Total hole and dn electron current Jn Jndn39 Jr1 diff qn E ana density Total current density Logic gate 4 Circuit entity invcner ls I pnri l ln Bit 01 out Bit vicw end invcrtcr Wafer Figure 3 Equivalent descriptions of a digital integrated circuit gt X id lmplantcd donnls a Mask de nition h After anncaiing Figure 33 Patterning sequancc for 1 doped n iinc 5 lt Figure 32 Minimum width and spacing quot 3 w quot01 4410 Flgum 3A Depiction regions due in parallel n lines a Contact Size b Side view Figun 35 chmm39y of a comma cul gt7llt a Mask Design b Registration Tolerance Flgul 36 Canlacl spacing rule Afrlvl NsEIECr pauhurnic 11 Curran musk wing Figun 37 Formalmxl of n39 mgmns m m mahanncl MOSFET NSELECT ACIIVE E psubslralz 1 n lrlmmxl mask mmg Acxive area Figure 38 Gal overhang in MOSFET layout Acli we arca border Flynn Gale sp an n edge un acing from My gm In Nu rwclhnng yawn h Whh miullgnmcnl m Arm Implnm Figure 39 Ellen nrgalc misalignment wimmn overhang a RCSISI Pnucm m Isnlmpic clch b Amsnlmplc clch Figure Ml Poly iliccn L ICh pro les 4 L From View Poly M N w Side View le channel width ECSE 2210 Microelectronics Technology Class Activity 2 Solution A voltage of 10V is applied to point B with respect to point A in the gure below Assume C is in the middle If an electron at rest is placed at point C and released E Fquot 0 3 1 D In which direction will the electron move Towards A or towards B The electron will move towards B Electrons move towards positive terminal since the electron charge is negative Calculate the electron energy in Joules when it reaches B Energy q V 16 x1049 C x 5 v 80 x1019 J Note C X V J A X V W A Cs Make sure that you use consistent units Calculate the electron energy in eV when it reaches B A new unit of energy is electron volt eV It is the energy acquired by an electron when subjected to 1V potential difference Since 1 electron has the charge q 16 X 10719 C 1 eV turns outto be 16 x 10 x 1v ie 16 x 10 J So 1 eV 16 x 10 19J Therefore Energy 80 x 1039 J 5 eV What will be the electron velocity in ms when it reaches B The electron energy at point B is 05 mov2 where v is the velocity at point B and m0 is the rest mass of electron So 12 mov2 80 X 10719 J 05 x 911 x 103931kg x v2 80 x1019 Jresults in v 13 X106ms Calculate the electric eld E that exists between point A and point B if the distance between A and B is 5 cm Indicate the direction of the E i eld by an arrow The electric eld is directed from the positive terminal to the negative terminal ie from B to A If we take the positive xdirection as from A to B then the direction of electric eld is in the negative x direction The electric eld magnitude is potential differencedistance 10 V 5 cm 2 Vcm 2 E 4 So the electric eld is 72 V cm assuming that the positive x direction is from A to B Unit of E is V cm or Vm The direction is from V to 7V Calculate the energies in eV for the rst three allowed orbits of the hydrogen atom Use equation 21 in text and plug in the values Make sure you are consistent with the units Note that h i 27239 Answers should be 7 136 eV for n 1 734 erorn2 7151erorn3 Suppose an electron falls from the n 3 orbit to the n 2 orbit in hydrogen Calculate the wavelength of radiation emitted during this process Express this wavelength in Angstroms Is the emitted light visible When electron falls from n 3 to n 2 orbit the energy of radiation is 7151eV77 34eV 189 eV UseEhc7t and A h cE663 x103934 Js x 3 x 1081ms189 eV x 16 X103919JeV 6577 x 10397m 6577 x 10399 m 6577 nm 6577 A This corresponds to red light Note that the visible spectrum is approximately from 4000 A to 7500 A 4000 A being the violet color and 7000 A being red color a What is the number of Si atoms per cm3 See your class notes There are 8 atoms per unit cell The unit cell volume is a3 with a 542 A 542 X 10398 cm So number of atoms per cm 8542 X 10 8 cm3 5 X 1022 atomscm b What is the total number of electrons present in 1 cm3 of Si Each Si atom has 14 electrons so total number of electrons per cm3 is given by 14 x 5x1022 cm393 What is the total number of electrons present in the valence band of Si at 0 K 0 There are 4 electrons per atom in the valence band or 4 X 5 X 1022 valence electronscm3 3 1 How many electrons are free to conduct electricity at 0 K None since the valence band is full ie every state in the valence band is lled with electrons Note that there are 4 states per atom or 4 X 5 X 1022 statescm3 in the valence band and there are 4 valence electrons per atom 4 X 5 X 1022 valence electronscm3 A full band cannot conduct current because for every electron moving from one state to another state in one direction there must be another electron moving in the opposite direction Is Si an insulator or a conductor at 0 K D Si will be an insulator at 0 K At 0 K the valence band is full and cannot conduct current The conduction band is empty and cannot conduct current either If you think in terms of the bond model each and every electron is used up in making the bond So no current conduction is possible unless you break the bond There is no energy available to break the bond at 0K 5 Suppose you have some electrons to spare and you want to put them in the conduction band of Si What is the total number of electronscm3 that you can put in the conduction band This is the number of statescm3 available in the band You can put up to 4 X 5 X 1022 electrons in the conduction band since there are that many states available in the conduction band Of course it is a very hypothetical question 6 Using the bond model for a semiconductor indicate how one visualizes a a missing atom b an electron c a hole This is problem 22 in the textbook See gure 27 in text Mela Gm m V Gale oxide pvsubsuale Saurce Dram Bulk Bulk a Crosssection b Circuil symbol Figure 23 An nchznnel MOSFET Figure 2 4 Topview of an nchanncl MOSFET Gave Galeuxide nwcll Bulk Bulk a Crosssecxion b Circuit symbol Figure 26 Topvicw o a pchanncl MOSFET pcpitaxial layer t for devices 1 piypc wafcr Figure 28 Epitaxial layer growth Location of nMOSFET Location of pMOSFET lt lt gt ptypc epitaxial layzr Figure 29 MOSFET placement Stressvrelicf oxide Silicon Nitride st3 N4 w Field39 W Implant b FOX Growth Figure 210 Active area dz nitinn and cld pxide growth P39WW b Gal oxide gmwlh Polysilicon gm c Poly deposition and paucming Figure 2quot Gate oxide growth and poly gale formation a Boron p implam Arsenic inns 6 Arm anncaling Figure 212 Drainsnurcc ion implants m 533 3 pMSFET h Afler Melall deposilion and paucming Figure 213 Mom I layering LTO Oxide IlIIIIA 33 km pMOSFET Via 74 911111 IIlI nMOSFET a LTO oxid and VIA de nilions b Melal 2 Dcpasilion Figure 214 LTD and Metal 2 deposition Gm nxidc panth P39WPc 2 Formation of 605 Paraxlut GOS Dindc h Diode Modcl Figure 217 Galeroxide lthons G05 Defect duc lo 1 line break Linc break


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