1EE105 - Fall 2006Microelectronic Devices and CircuitsProf. Jan M. Rabaey (jan@eecs)Lecture 5: Diode Operation2Overview Last lecture– Capacitance– pn Junction This lecture– pn Junction (cntd)– Diode operation and models23Administrativia Labs starting tomorrow!4PN Junction – Summary so farn-typep-typeNDNA)(0xpaNp =0diNnp20=diffJ0EaiNnn20=Transition RegiondiffJdNn =0––+ +0E0px−0nx The most important device is a junction between a p-type region and an n-type region When the junction is first formed, due to the concentration gradient, mobile charges transfer near junction Electrons leave n-type region and holes leave p-type region These mobile carriers become minority carriers in new region (can’t penetrate far due to recombination) Due to charge transfer, a voltage difference occurs between regions This creates a field at the junction that causes drift currents to oppose the diffusion current In thermal equilibrium, drift current and diffusion must balanceQuestion: width of depletion region35Depletion Approximation Let’s assume that the transition region is completely depleted of free carriers (only immobile dopantsexist) Then the charge density is given by The solution for electric field is now easy)(')'()(000 posaxxsxxqNdxxxEp+−==∫−εερ⎩⎨⎧<<+<<−−≅0000)(ndpoaxxqNxxqNxρ)()(00xxqNxEnsd−−=ε6Plot of Fields In Depletion Region E-Field zero outside of depletion region Note the asymmetrical depletion widths Which region has higher doping? Slope of E-Field larger in n-region. Why? Peak E-Field at junction. Why continuous?n-typep-typeNDNA––––––––––––––––––––+ + + + ++ + + + ++ + + + ++ + + + +DepletionRegion)()(00xxqNxEnsd−−=ε)()(0 posaxxqNxE +−=ε47Continuity of E-Field Across Junction Recall 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 continuous What does this imply? Total fixed charge in n-region equals fixed charge in p-region! Somewhat obvious result.)0()0(00==−=−== xExqNxqNxEpnosdposanεεnodpoaxqNxqN =8Potential Across Junction From our earlier calculation we know that the potential in the n-region is higher than p-region The potential has to smoothly transition form high to low in crossing the junction Physically, the potential difference is due to the charge transfer that occurs due to the concentration gradient Let’s integrate the field to get the potential:∫−++−=xxposapopdxxxqNxx0')'()()(εφφxxposappxxxqNx0'2')(2−⎟⎟⎠⎞⎜⎜⎝⎛++=εφφ59Potential Across Junction We arrive at potential on p-side (parabolic) Do integral on n-side Potential must be continuous at interface (field finite at interface)20)(2)(psappoxxqNx ++=εφφ20)(2)(nsdnnxxqNx −−=εφφ)0(22)0(2020ppsapnsdnnxqNxqNφεφεφφ=+=−=10Solve for Depletion Lengths We have two equations and two unknowns. We are finally in a position to solve for the depletion depths202022psapnsdnxqNxqNεφεφ+=−nodpoaxqNxqN =(1)(2)⎟⎟⎠⎞⎜⎜⎝⎛+=daadbisnoNNNqNxφε2⎟⎟⎠⎞⎜⎜⎝⎛+=addabispoNNNqNxφε20>−≡pnbiφφφ611Sanity Check Does 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=⎟⎟⎠⎞⎜⎜⎝⎛+=∞→12Total Depletion Width The sum of the depletion widths is the “space charge region” This region is essentially depleted of all mobile charge Due to high electric field, carriers move across region at velocity saturated speed ⎟⎟⎠⎞⎜⎜⎝⎛+=+=dabisnpdNNqxxX112000φεμ11012150≈⎟⎠⎞⎜⎝⎛=qXbisdφεcmV10μ1V14=≈pnE713Have we invented a battery? Can we harness the PN junction and turn it into a battery? Numerical example:2lnlnlniADthiAiDthpnbinNNVnNnNV =⎟⎟⎠⎞⎜⎜⎝⎛+=−≡φφφmV600101010logmV60lnmV262015152=×==iADbinNNφ?14Contact Potential The contact between a PN junction creates a potential difference Likewise, 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 well If we short a PN junction, the sum of the voltages around the loop must be zero:mnpmbiφφφ++=0pnmnφpmφ+−biφ)(mnpmbiφφφ+−=815PN Junction Capacitor Under thermal equilibrium, the PN junction does not draw any (much) current But 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 capacitor Positive charge is stored in the n-region, and negative charge is in the p-region:nodpoaxqNxqN =16Reverse Biased PN Junction What happens if we “reverse-bias” the PN junction? Since no current is flowing, the entire reverse biased potential is dropped across the transition region To accommodate the extra potential, the charge in these regions must increase If no current is flowing, the only way for the charge to increase is to grow (shrink) the depletion regions+−DbiV+−φDV0<DV917Voltage Dependence of Depletion Width Can 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)(018Charge Versus Bias As we increase the reverse bias, the depletion region grows to accommodate more charge Charge is not a linear function of voltage This is a non-linear capacitor We can define a small signal capacitance for small signals by breaking up the charge into two termsbiDaDpaDJVqNVxqNVQφ−−=−= 1)()()()()(DDJDDJvqVQvVQ +=+1019Derivation of Small Signal Capacitance From last lecture we found Notice
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