1Department of EECS University of California, BerkeleyEECS 105 Spring 2004, Lecture 9Lecture 9: Diffusion, Electrostatics review, and CapacitorsProf. J. S. SmithDepartment of EECS University of California, BerkeleyEECS 105 Spring 2004, Lecture 9 Prof. J. S. SmithContextz In the last lecture, we looked at the carriers in a neutral semiconductor, and drift currentsz In this lecture, we will continue to study transport--- the motion of carriers due to diffusion, and the influence of charge distributions on the electric fields– Review of Electrostatics– Diffusion– IC MIM Capacitorsz In the next lecture, we will study P-N diodesDepartment of EECS University of California, BerkeleyEECS 105 Spring 2004, Lecture 9 Prof. J. S. SmithElectrostatics Reviewz The force between any two charges is given by Coulomb’s lawz Where is unit vector in the direction away from the other charge.z Since Maxwell’s equations are linear, we can add up all the forces from other charges, and define the electric field at a point:2214ˆrqqeFπε=reˆ21211114ˆ1nnnnrqeFqqEπε∑∑==rrDepartment of EECS University of California, BerkeleyEECS 105 Spring 2004, Lecture 9 Prof. J. S. SmithElectrostatic fieldsz Since we are going to be dealing with large numbers of charges, we can use a continuum model:– The charges are given as a smooth density:Coulombs/cm3– The electric field is a smooth vector field which diverges from positive charge, and converges on negative charges – Which is one of Maxwell’s equations),( trvρερ),(),(trtrEvrr=⋅∇++++-2Department of EECS University of California, BerkeleyEECS 105 Spring 2004, Lecture 9 Prof. J. S. SmithGauss’s Lawz Gauss’s Law state that the total amount of E field (flux) leaving a volume is equal to the net charge enclosed +++++∫=⋅εQdSE∫∫==⋅∇VVQdVdVEεερ/εQdSEdVESV∫∫=⋅=⋅∇Recall:Department of EECS University of California, BerkeleyEECS 105 Spring 2004, Lecture 9 Prof. J. S. SmithElectrostatics in 1Dz In EE105, we are almost always going to use 1-D models, so it simplifies:z Consider a uniform charge distributionερ==⋅∇dxdEEdxdEερ=')'()()(00dxxxExExx∫+=ερ)(xρxxdxxxExερερ00')'()( ==∫Zero fieldboundarycondition1x0ρ1x)(xE10xερDepartment of EECS University of California, BerkeleyEECS 105 Spring 2004, Lecture 9 Prof. J. S. SmithElectrostatic Potentialz The electric field (force) is related to the potential (energy):z Negative sign says that field lines go from high potential points to lower potential points (negative slope)z Note: Electrons “float” to a high potential point:dxdEφ−=dxdeqEFeφ−==1φ2φdxdeFeφ−=eThe E field is the (-) slope of the potential!Department of EECS University of California, BerkeleyEECS 105 Spring 2004, Lecture 9 Prof. J. S. SmithMore Potentialz Integrating this basic relation, we have that the potential is the integral of the field:z In 1D, this is a simple integral:z Since the derivative of the E field is the charge, we can integrate again to get Poisson’s equation in 1D:∫⋅−=−CldExxr)()(0φφ)(0xφ)(xφEldr∫−=−xxdxxExx0')'()()(0φφερφ)()(22xdxxd−=3Department of EECS University of California, BerkeleyEECS 105 Spring 2004, Lecture 9 Prof. J. S. SmithBoundary Conditionsz Potential must be a continuous function. If not, the fields (forces) would be infinite z Electric fields need not be continuous. We have already seen that the electric fields diverge on charges. In fact, across an interface we have:z Field discontinuity implies charge density at surface! )(11εE)(22εE∫=+−=⋅insideQSESEdSE2211εεεx∆00⎯⎯→⎯→∆xinsideQ02211=+− SESEεε1221εε=EESDepartment of EECS University of California, BerkeleyEECS 105 Spring 2004, Lecture 9 Prof. J. S. SmithBoundary conditions→Metalsz The presence of metals greatly simplifies boundary conditions.z Inside a metal, the E fields are very small (otherwise the current would be very large)z The inside of a metal has no free charges (if it had free charges, they are free to move and would very rapidly scatter to the edges of the metalz So all of the net charge on a metal occurs on its surface, andz The surface of a metal is therefore all nearly at the same potential (exception: long wires conducting a current)Department of EECS University of California, BerkeleyEECS 105 Spring 2004, Lecture 9 Prof. J. S. SmithNote: Band edge diagramsz We will often draw a diagram of the valence and the conduction band edges as a function of position.z The energy at the band edge corresponds to the potential energy that an electron has (which is the negative of the electrostatic potential). Thus the slope of the band edge with distance is the electric field. (Silicon)P typeN typeEnergyDistance+++++++++---------→Force on electrons←E fieldDepartment of EECS University of California, BerkeleyEECS 105 Spring 2004, Lecture 9 Prof. J. S. SmithDiffusionz Diffusion occurs when there exists a concentration gradientz In the figure below, imagine that we fill the left chamber with a perfume at temperate Tz If we suddenly remove the divider, what happens?z The perfume will fill the entire volume of the new chamber. How does this occur?4Department of EECS University of California, BerkeleyEECS 105 Spring 2004, Lecture 9 Prof. J. S. SmithDiffusionz Even though there is no force acting on the perfume molecules, because there are more on the left than on the right, their random motions take more from the left to the right than are going from right to left.z Electrons and holes do the same thing, but since they are charged, they carry current with them. z Diffusion moves particles in addition to the motion from electric forcesDepartment of EECS University of California, BerkeleyEECS 105 Spring 2004, Lecture 9 Prof. J. S. SmithDiffusion (cont)z The net motion of gas molecules to the right chamber was due to the concentration gradientz Diffusion will lead to a net flow of particles as long as the distribution of particles is not uniformz Diffusion causes a flow of particles from places of high concentration to places of lower concentration.Department of EECS University of California, BerkeleyEECS 105 Spring 2004, Lecture 9 Prof. J. S. SmithDiffusionz If the electrons move long distances without collisions, they would quickly spread out, diffusion would be large. →diffusion is proportional to , the mean free pathz If the larger the particles thermal velocities are, then the faster
View Full Document
Unlocking...