Lecture #14Qualitative Electrostatics“Game Plan” for Obtaining r(x), E(x), V(x)Built-In Potential VbiVbi for “One-Sided” pn JunctionsThe Depletion ApproximationElectric Field in the Depletion LayerElectrostatic Potential in the Depletion LayerSlide 9Depletion Layer WidthOne-Sided JunctionsExampleSummaryEE130 Lecture 14, Slide 1Spring 2007Lecture #14OUTLINE• pn junction electrostaticsReading: Chapter 5EE130 Lecture 14, Slide 2Spring 2007Qualitative ElectrostaticsBand diagramElectrostatic potentialElectric fieldCharge densityEE130 Lecture 14, Slide 3Spring 2007“Game Plan” for Obtaining (x),E(x),V(x)•Find the built-in potential Vbi•Use the depletion approximation (x)(depletion-layer widths xp, xn unknown)•Integrate (x) to find E(x)–boundary conditions E(-xp)=0, E(xn)=0•Integrate E(x) to obtain V(x)–boundary conditions V(-xp)=0, V(xn)=Vbi•For E(x) to be continuous at x=0, NAxp = NDxn solve for xp, xnEE130 Lecture 14, Slide 4Spring 2007Built-In Potential VbiiAisidepFinNkTnpkTEEln ln)(sideniFsidepFisiden Ssidep Sbi)()( EEEEqViDisideniFnNkTnnkTEEln ln)(For non-degenerately doped material:EE130 Lecture 14, Slide 5Spring 2007Vbi for “One-Sided” pn JunctionssideniFsidepFibiEEEEqV )()(p+n junction n+p junctionEE130 Lecture 14, Slide 6Spring 2007The Depletion ApproximationOn the p-side, = –qNAOn the n-side, = qNDsAqNdxdE)()(1x-qNACxqNxssAE)()(sDxxn-qNxExpEE130 Lecture 14, Slide 7Spring 2007The electric field is continuous at x = 0 NAxp = NDxnElectric Field in the Depletion LayerEE130 Lecture 14, Slide 8Spring 2007On the p-side:(arbitrarily choose the voltage at x = xp to be 0)On the n-side:12)(2)( DxxqNxVpsA222)(2)(2)( xxqNVDxxqNxVnsDbinsDElectrostatic Potential in the Depletion LayerEE130 Lecture 14, Slide 9Spring 2007• At x = 0, expressions for p-side and n-side must be equal:• We also know that NAxp = NDxnEE130 Lecture 14, Slide 10Spring 2007Depletion Layer Width•Eliminating xp, we have:•Eliminating xn, we have:•Summing, we have:)(2DADAbisnNNNNqVx)(2DAADbispNNNNqVxDAbispnNNqVWxx112EE130 Lecture 14, Slide 11Spring 2007If NA >> ND as in a p+n junction:What about a n+p junction?where density dopantlighter NNNAD1111nDbisxqNVW 20ADnpNNxxqNVWbis2One-Sided JunctionsEE130 Lecture 14, Slide 12Spring 2007A p+n junction has NA=1020 cm-3 and ND =1017cm-3. What is a) its built in potential, b)W , c)xn , and d) xp ?Solution:a) b)c)d)V 1ln2iDGbinNqkTqEVμm 12.010106.111085.812222/1171914DbisqNVWμm 12.0Wxn0Å 2.1μm102.14ADnpNNxxExampleEE130 Lecture 14, Slide 13Spring 2007Summary•For a non-degenerately-doped pn junction at equilibrium:–Built-in potential –Depletion-layer width•For a one-sided (p+n or pn+) junction at equilibrium:–Built-in potential –Depletion-layer width2lniADbinNNqkTV DAbispnNNqVxxW112nDpAxNxN iGbinNqkTEV
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