Department of EECS University of California, BerkeleyEECS 105 Spring 2005, Lecture 27Lecture 27: PN JunctionsProf. NiknejadDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 27 Prof. A. NiknejadDiffusionz Diffusion occurs when there exists a concentration gradientz In the figure below, imagine that we fill the left chamber with a gas at temperate Tz If we suddenly remove the divider, what happens?z The gas will fill the entire volume of the new chamber. How does this occur?Department of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 27 Prof. A. NiknejadDiffusion (cont)z The net motion of gas molecules to the right chamber was due to the concentration gradientz If each particle moves on average left or right then eventually half will be in the right chamberz If the molecules were charged (or electrons), then there would be a net current flowz The diffusion current flows from high concentration to low concentration:Department of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 27 Prof. A. NiknejadDiffusion Equationsz Assume that the mean free path is λz Find flux of carriers crossing x=0 plane)(λn)0(n)(λ−n0λ−λthvn )(21λthvn )(21λ−())()(21λλnnvFth−−=+−−=dxdnndxdnnvFthλλ)0()0(21dxdnvFthλ−=dxdnqvqFJthλ=−=Department of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 27 Prof. A. NiknejadEinstein Relationz The thermal velocity is given by kTkTvmthn212*21=cthvτλ=Mean Free TimedxdnqkTqdxdnqvJnth==µλnnqkTDµ=**2ncnccththmqqkTmkTvvτττλ===Department of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 27 Prof. A. NiknejadTotal Current and Boundary Conditionsz When both drift and diffusion are present, the total current is given by the sum:z In resistors, the carrier is approximately uniform and the second term is nearly zeroz For currents flowing uniformly through an interface (no charge accumulation), the field is discontinousdxdnqDnEqJJJnndiffdrift+=+=µ21JJ=2211EEσσ=1221σσ=EE)(11σJ)(22σJDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 27 Prof. A. NiknejadCarrier Concentration and Potentialz In thermal equilibrium, there are no external fields and we thus expect the electron and hole current densities to be zero:dxdnqDEqnJonnn+==000µdxdnkTqEnDdxdnoonno 00φµ=−=0000ndnVndnqkTdtho==φDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 27 Prof. A. NiknejadCarrier Concentration and Potential (2)z We have an equation relating the potential to the carrier concentrationz If we integrate the above equation we havez We define the potential reference to be intrinsic Si:)()(ln)()(000000xnxnVxxth=−φφinxnx==)(0)(0000φ0000ndnVndnqkTdtho==φDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 27 Prof. A. NiknejadCarrier Concentration Versus Potentialz The carrier concentration is thus a function of potentialz Check that for zero potential, we have intrinsic carrier concentration (reference). z If we do a similar calculation for holes, we arrive at a similar equationz Note that the law of mass action is upheldthVxienxn/)(00)(φ=thVxienxp/)(00)(φ−=2/)(/)(20000)()(iVxVxineenxpxnthth==−φφDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 27 Prof. A. NiknejadThe Doping Changes Potentialz Due to the log nature of the potential, the potential changes linearly for exponential increase in doping:z Quick calculation aid: For a p-type concentration of 1016cm-3, the potential is -360 mVz N-type materials have a positive potential with respect to intrinsic Si1000000010)(log10lnmV26)()(lnmV26)()(ln)(xnxnxnxnxnVxiith≈==φ100010)(logmV60)(xnx ≈φ100010)(logmV60)(xpx −≈φDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 27 Prof. A. Niknejadn-typep-typeNDNAPN Junctions: Overviewz The most important device is a junction between a p-type region and an n-type regionz When the junction is first formed, due to the concentration gradient, mobile charges transfer near junction z Electrons leave n-type region and holes leave p-type regionz These mobile carriers become minority carriers in new region (can’t penetrate far due to recombination)z Due to charge transfer, a voltage difference occurs between regionsz This creates a field at the junction that causes drift currents to oppose the diffusion currentz In thermal equilibrium, drift current and diffusion must balance−−−−−−+ + + + ++ + + + ++ + + + +−−−−−−−−−−−−−V+Department of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 27 Prof. A. NiknejadPN Junction Currentsz Consider the PN junction in thermal equilibriumz Again, the currents have to be zero, so we havedxdnqDEqnJonnn+==000µdxdnqDEqnonn−=00µdxdnnqkTndxdnDEnon00001−=−=µdxdppqkTndxdpDEpop00001−==µDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 27 Prof. A. NiknejadPN Junction Fieldsn-typep-typeNDNA)(0xpaNp=0diNnp20=diffJ0EaiNnn20=Transition RegiondiffJdNn=0––+ +0E0px−0nxDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 27 Prof. A. NiknejadTotal Charge in Transition Regionz To solve for the electric fields, we need to write down the charge density in the transition region:z In the p-side of the junction, there are very few electrons and only acceptors:z Since the hole concentration is decreasing on the p-side, the net charge is negative:)()(000 adNNnpqx −+−=ρ)()(00 aNpqx−≈ρ0)(0<xρ0pNa>00<<−xxpDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 27 Prof. A. NiknejadCharge on N-Sidez Analogous to the p-side, the charge on the n-side is given by:z The net charge here is positive since:)()(00 dNnqx+−≈ρ00nxx<<0)(0>xρ0nNd>aiNnn20=Transition RegiondiffJdNn=0––+ +0EDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad“Exact” Solution for Fieldsz Given the above approximations, we now have an expression for the charge densityz We also have the following result from electrostaticsz Notice that the potential appears on both sides of the equation… difficult problem to solvez A
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