1EE105 - Fall 2005Microelectronic Devices and CircuitsLecture 5PN JunctionDiode2AnnouncementsHomework 2, due todayHomework 3, due next weekLabs start this weekI am away on Thursday; lecture by Sebastian HoyosReading: Chapter 3 (3.1-3.6), Chapter 623Lecture MaterialLast lectureIC capacitorsPN junctionThis lecturePN junction in reverse and forward bias12Plot of Fields In Depletion RegionE-Field zero outside of depletion regionNote the asymmetrical depletion widthsWhich region has higher doping?Slope of E-Field larger in n-region. Why?Peak E-Field at junction. n-typep-typeNDNA––––––––––––––––––––+ + + + ++ + + + ++ + + + ++ + + + +DepletionRegion)()(00xxqNxEnsd−−=ε)()(0 posaxxqNxE +−=ε313Continuity of E-Field Across JunctionRecall that E-field diverges on charge. For a sheet charge at the interface, the E-field could be discontinuous In our case, the depletion region is only populated by a background density of fixed charges so the E-Field is continuousWhat does this imply?Total fixed charge in n-region equals fixed charge in p-region! Somewhat obvious result.)0()0(00==−=−== xExqNxqNxEpnosdposanεεnodpoaxqNxqN=14Potential Across JunctionFrom our earlier calculation we know that the potential in the n-region is higher than p-regionThe potential has to smoothly transition form high to low when crossing the junctionPhysically, the potential difference is due to the charge transfer that occurs due to the concentration gradientLet’s integrate the field to get the potential:∫−++−=xxposapopdxxxqNxx0')'()()(εφφxxposappxxxqNx0'2')(2−⎟⎟⎠⎞⎜⎜⎝⎛++=εφφ415Potential Across JunctionWe arrive at potential on p-side (parabolic)Do integral on n-sidePotential must be continuous at interface (field finite at interface)20)(2)(psappoxxqNx ++=εφφ20)(2)(nsdnnxxqNx −−=εφφ)0(22)0(2020ppsapnsdnnxqNxqNφεφεφφ=+=−=16Solve for Depletion LengthsWe have two equations and two unknowns. We are finally in a position to solve for the depletion depths202022psapnsdnxqNxqNεφεφ+=−nodpoaxqNxqN=(1)(2)⎟⎟⎠⎞⎜⎜⎝⎛+=daadbisnoNNNqNxφε2⎟⎟⎠⎞⎜⎜⎝⎛+=addabispoNNNqNxφε20>−≡pnbiφφφ517Sanity CheckDoes the above equation make sense?Let’s say we dope one side very highly. Then physically we expect the depletion region width for the heavily doped side to approach zero:Entire depletion width dropped across p-region02lim0=+=∞→adddbisNnNNNqNxdφε;abisaddabisNpqNNNNqNxdφεφε22lim0=⎟⎟⎠⎞⎜⎜⎝⎛+=∞→18Total Depletion WidthThe sum of the depletion widths is the “space charge region”This region is essentially depleted of all mobile chargeDue to high electric field, carriers move across region at velocity saturated speed ⎟⎟⎠⎞⎜⎜⎝⎛+=+=dabisnpdNNqxxX112000φεµ11012150≈⎟⎠⎞⎜⎝⎛=qXbisdφεcmV10µ1V14=≈pnE619Have we invented a battery?Can we harness the PN junction and turn it into a battery?Numerical example:2lnlnlniADthiAiDthpnbinNNVnNnNV =⎟⎟⎠⎞⎜⎜⎝⎛+=−≡φφφmV600101010logmV60lnmV262015152=×==iADbinNNφ?20Contact PotentialThe contact between a PN junction creates a potential differenceLikewise, the contact between two dissimilar metals creates a potential difference (proportional to the difference between the work functions)When a metal semiconductor junction is formed, a contact potential forms as wellIf we short a PN junction, the sum of the voltages around the loop must be zero:mnpmbiφφφ++=0pnmnφpmφ+−biφ)(mnpmbiφφφ+−=721PN Junction CapacitorUnder thermal equilibrium, the PN junction does not draw any currentBut notice that a PN junction stores charge in the space charge region (transition region)Since the device is storing charge, it’s acting like a capacitorPositive charge is stored in the n-region, and negative charge is in the p-region:nodpoaxqNxqN=22Reverse Biased PN JunctionWhat happens if we “reverse-bias” the PN junction?Since no current is flowing, the entire reverse biased potential is dropped across the transition regionTo accommodate the extra potential, the charge in these regions must increaseIf no current is flowing, the only way for the charge to increase is to grow (shrink) the depletion regions+−DbiV+−φDV0<DV823Voltage Dependence of Depletion WidthCan redo the math but in the end we realize that the equations are the same except we replace the built-in potential with the effective reverse bias:⎟⎟⎠⎞⎜⎜⎝⎛+−=+=daDbisDnDpDdNNqVVxVxVX11)(2)()()(φεbiDndaadDbisDnVxNNNqNVVxφφε−=⎟⎟⎠⎞⎜⎜⎝⎛+−= 1)(2)(0biDpdadaDbisDpVxNNNqNVVxφφε−=⎟⎟⎠⎞⎜⎜⎝⎛+−= 1)(2)(0biDdDdVXVXφ−= 1)(024Charge Versus BiasAs we increase the reverse bias, the depletion region grows to accommodate more chargeCharge is not a linear function of voltageThis is a non-linear capacitorWe can define a small signal capacitance for small signals by breaking up the charge into two termsbiDaDpaDJVqNVxqNVQφ−−=−= 1)()()()()(DDJDDJvqVQvVQ+=+925Derivation of Small Signal CapacitanceFrom last lecture we foundNotice thatL++=+DVDDJDDJvdVdQVQvVQD)()(RDVVbipaVVjDjjVxqNdVddVdQVCC==⎥⎥⎦⎤⎟⎟⎠⎞⎜⎜⎝⎛−−===φ1)(0biDjbiDbipajVCVxqNCφφφ−=−=11200dadabisdadabisbiabipajNNNNqNNNqNqNxqNC+=⎟⎟⎠⎞⎜⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛==φεφεφφ22220026Physical Interpretation of Depletion CapNotice that the expression on the right-hand-side is just the depletion width in thermal equilibriumThis looks like a parallel plate capacitor!dadabisjNNNNqC+=φε20010112dsdabissjXNNqCεφεε=⎟⎟⎠⎞⎜⎜⎝⎛+=−)()(DdsDjVXVCε=1027A Variable Capacitor (Varactor)Capacitance varies versus bias:Application: Radio Tuner0jjCC28“Diffusion” ResistorResistor is capacitively isolation from substrate Must Reverse Bias PN Junction!P-type Si SubstrateN-type Diffusion
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