1EE105 - Fall 2006Microelectronic Devices and CircuitsProf. Jan M. Rabaey (jan@eecs)Lecture 4: CapacitorsPN Junctions2Overview Last lecture– Diffusion currents– Overview of IC fabrication process– Review of electrostatics This lecture– Capacitances– pn Junctions23Administrativia Another Make-up Lecture Monday at 4pm (streamed)NO LECTURE ON TUESDAY4IC MIM Capacitor By forming a thin oxide and metal (or polysilicon) plates, a capacitor is formed Contacts are made to top and bottom plate Parasitic capacitance exists between bottom plate and substrateTop PlateBottom PlateBottom PlateContactsCVQ =Thin Oxide35Review of Capacitors For an ideal metal, all charge must be at surface Gauss’ law: Surface integral of electric field over closed surface equals charge inside volume+++++++++++++++++++++−−−−−−−−−−−−−−−+−Vs∫=⋅εQdSE∫−=⋅εQdSEsoxVtEdlE ==⋅∫0oxstVE =0∫==⋅εQAEdSE0εQAtVoxs=sCVQ =oxtACε=6Capacitor Q-V Relation Total charge is linearly related to voltage Charge density is a delta function at surface (for perfect metals)sCVQ =sVQy)(yQy+++++++++++++++++++++−−−−−−−−−−−−−−−47A Non-Linear Capacitor We’ll soon meet capacitors that have a non-linear Q-V relationship If plates are not ideal metal, the charge density can penetrate into surface)(sVfQ =sVQy)(yQy+++++++++++++++++++++−−−−−−−−−−−−−−−8What’s the Capacitance? For a non-linear capacitor, we have We can’t identify a capacitance Imagine we apply a small signal on top of a bias voltage: The incremental charge is therefore:ssCVVfQ ≠= )(sVVsssvdVVdfVfvVfQs=+≈+=)()()(Constant chargesVVsvdVVdfVfqQQs=+≈+=)()(059Small Signal Capacitance Break the equation for total charge into two terms:sVVsvdVVdfVfqQQs=+≈+=)()(0ConstantChargeIncrementalChargessVVvCvdVVdfqs===)(sVVdVVdfC=≡)(10Example of Non-Linear Capacitor Next lecture we’ll see that for a PN junction, the charge is a function of the reverse bias: Small signal capacitance:bpajVxqNVQφ−−= 1)(ConstantsCharge At N Side of JunctionVoltage Across NPJunctionbjbbpajjVCVxqNdVdQVCφφφ−=−==1112)(0611PN Junctions (Diodes)12Carrier Concentration and Potential In thermal equilibrium, there are no external fields and we thus expect the electron and hole current densities to be zero:dxdnqDEqnJonnn+==000μdxdnkTqEnDdxdnoonno00φμ⎟⎠⎞⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛−=0000ndnVndnqkTdtho=⎟⎟⎠⎞⎜⎜⎝⎛=φ713Carrier Concentration and Potential (2) We have an equation relating the potential to the carrier concentration If we integrate the above equation we have We define the potential reference to be intrinsic Si:)()(ln)()(000000xnxnVxxth=−φφinxnx == )(0)(0000φ0000ndnVndnqkTdtho=⎟⎟⎠⎞⎜⎜⎝⎛=φ14Carrier Concentration Versus Potential The carrier concentration is thus a function of potential Check that for zero potential, we have intrinsic carrier concentration (reference). If we do a similar calculation for holes, we arrive at a similar equation Note that the law of mass action is upheldthVxienxn/)(00)(φ=thVxienxp/)(00)(φ−=2/)(/)(20000)()(iVxVxineenxpxnthth==−φφ815The Doping Changes Potential Due to the log nature of the potential, the potential changes linearly for exponential increase in doping: Quick calculation aid: For a p-type concentration of 1016cm-3, the potential is -360 mV N-type materials have a positive potential with respect to intrinsic Si()1000000010)(log10lnmV26)()(lnmV26)()(ln)(xnxnxnxnxnVxiith⋅⋅≈==φ100010)(logmV60)(xnx ≈φ100010)(logmV60)(xpx −≈φ16PN Junction: Overview917PN Junction: Overview18PN Junction: Overview Present in most IC structures1019n-typep-typeNDNAPN Junctions: Overview 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 balance−−−−−−+ + + + ++ + + + ++ + + + +−−−−−−−−−−−−−V+20PN Junction Currents Consider the PN junction in thermal equilibrium Again, the currents have to be zero, so we havedxdnqDEqnJonnn+==000μdxdnqDEqnonn−=00μdxdnnqkTndxdnDEnon00001−=−=μdxdppqkTndxdpDEpop00001−==μ1121PN Junction Fieldsn-typep-typeNDNA)(0xpaNp =0diNnp20=diffJ0EaiNnn20=Transition RegiondiffJdNn =0––+ +0E0px−0nx22Total Charge in Transition Region To solve for the electric fields, we need to write down the charge density in the transition region: In the p-side of the junction, there are very few electrons and only acceptors: Since the hole concentration is decreasing on the p-side, the net charge is negative:)()(000 adNNnpqx −+−=ρ)()(00 aNpqx −≈ρ0)(0<xρ0pNa>00<<− xxp1223Charge on N-Side Analogous to the p-side, the charge on the n-side is given by: The net charge here is positive since:)()(00 dNnqx +−≈ρ00nxx <<0)(0>xρ0nNd>aiNnn20=Transition RegiondiffJdNn =0––+ +0E24“Exact” Solution for Fields Given the above approximations, we now have an expression for the charge density We also have the following result from electrostatics Notice that the potential appears on both sides of the equation… difficult problem to solve A much simpler way to solve the problem…⎩⎨⎧<<−<<−−≅−0/)(/)(00)(0)()(00nVxidpoaVxixxenNqxxNenqxththφφρsxdxddxdEερφ)(0220=−=1325Depletion Approximation Let’s assume that the transition region is completely depleted of free carriers (only immobile dopants exist) Then the charge density is given by The solution for electric field is now easy⎩⎨⎧<<+<<−−≅0000)(ndpoaxxqNxxqNxρsxdxdEερ)(00=)(')'()(00000pxxsxEdxxxEp−+=∫−ερField zero outsidetransition region26Depletion Approximation (2) Since charge density is a constant If we start from the n-side we get the following result)(')'()(000
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