6.102 Spring 2007 Lecture 5 1Lecture 5PN Junction and MOS Electrostatics(II)PN JUNCTION IN THERMAL EQUILIBRIUMOutline1. Introduction2. Electrostatics of pn junction in thermal equilibrium3. The depletion approximation4. Contact potentialsReading Assignment:Howe and Sodini, Chapter 3, Sections 3.3-3.66.102 Spring 2007 Lecture 5 21. Introduction• pn junction– p-region and n-region in intimate contactWhy is the p-n junction worth studying?It is present in virtually every semiconductor device!Example: CMOS cross-sectionUnderstanding the pn junction is essential to understanding transistor operation6.102 Spring 2007 Lecture 5 32. Electrostatics of p-n junction in equilibriumFocus on intrinsic region:Doping distribution of an abruptp-n junction6.102 Spring 2007 Lecture 5 4What is the carrier concentration distribution in thermal equilibrium?First think of the two sides separately:Now bring the two sides together. What happens?6.102 Spring 2007 Lecture 5 5Resulting carrier concentration profile in thermal equilibrium:• Far away from the metallurgical junction: nothing happens–Two quasi-neutral regions• Around the metallurgical junction: diffusion of carriers must counter-balance drift– Space-charge region6.102 Spring 2007 Lecture 5 6On a linear scale:We can divide semiconductor into three regions• Two quasi-neutral n- and p-regions (QNR’s)• One space-charge region (SCR)Now, we want to know no(x), po(x), ρ(x), E(x) and φ(x).We need to solve Poisson’s equation using a simple but powerful approximationThermal equilibrium: balance between drift and diffusionJn(x) = Jndrift(x)+Jndiff(x)=0Jp(x) = Jpdrift(x)+Jpdiff(x)=06.102 Spring 2007 Lecture 5 73. The Depletion Approximation• Assume the QNR’s are perfectly charge neutral• Assume the SCR is depleted of carriers – depletion region• Transition between SCR and QNR’s sharp at–-xpoand xno (must calculate where to place these)x <−xpo; po(x) = Na, no(x) =ni2Na−xpo< x < 0; po(x), no(x) << Na0 < x < xno; no(x), po(x) << Ndx > xno; no(x) = Nd, po(x) =ni2Nd6.102 Spring 2007 Lecture 5 8Space Charge Densityρ(x) = 0; x<−xpo=−qNa; −xpo< x < 0= qNd;0< x < xno= 0; x > xno6.102 Spring 2007 Lecture 5 9Electric Fieldx<−xpo;E(x) = 0−xpo< x < 0; E(x) − E(−xpo) =1εs−qNad ′ x − xpox∫=−qNaεsx⎡ ⎣ ⎢ ⎤ ⎦ ⎥ −xpox=−qNaεsx + xpo()0 < x < xno; E(x) =qNdεsx − xno()x > xno; E(x) = 0Integrate Poisson’s equationE(x2) − E(x1) =1εsρ(x) dxx1x2∫6.102 Spring 2007 Lecture 5 10Electrostatic Potential(with φ=0 @ no=po=ni)φ=kTq• lnnoniφ=−kTq• lnponiIn QNRs, noand poare known ⇒ can determine φin p-QNR: po=Na⇒in n-QNR: no=Nd⇒φp=−kTq• lnNaniφn=kTq• lnNdniBuilt-in potential:φB=φn−φp=kTq• lnNdNani2This expression is always correct in TE! We did not use depletion approximation.6.102 Spring 2007 Lecture 5 11To obtain φ(x) in between, integrate E(x)φ(x2) −φ(x1) =− E( ′ x )d′ x x1x2∫x<−xpo;φ(x) =φp−xpo< x < 0;φ(x) −φ(−xpo) =− −qNaεs′ x + xpo()d′ x −xpox∫=qNa2εsx + xpo()2φ(x)=φp+qNa2εsx + xpo()20 < x < xno;φ(x) =φn−qNd2εsx − xno()2x > xno;φ(x) =φnAlmost done ….6.102 Spring 2007 Lecture 5 12Still do not know xnoand xpo⇒ need two more equations1. Require overall charge neutrality:2. Require φ(x) to be continuous at x=0;Two equations with two unknowns — obtain solution:nodpoaxqNxqN=2222nosdnposapxqNxqNεφεφ −=+() ()adadBspoddaaBsnoNNNqNxNNNqNx+=+=φεφε 22Now problem is completely solved!6.102 Spring 2007 Lecture 5 13Solution SummarySpace Charge DensityElectrostatic FieldElectrostatic Potential6.102 Spring 2007 Lecture 5 14Other results:Width of the space charge region:()dadaBsnopodoNqNNNxxx+=+=φε2Field at the metallurgical junction:()dasdaBoNNNNqE+=εφ26.102 Spring 2007 Lecture 5 15Three Special Cases• Symmetric junction: Na= Nd• Asymmetric junction: Na> Nd• Strongly asymmetric junction–p+n junction: Na>> Ndnopoxx=nopoxx<xpo<< xno≈ xdo≈2εsφBqNdEo≈2qφBNdεsThe lightly-doped side controls the electrostatics of the pn junction6.102 Spring 2007 Lecture 5 164. Contact PotentialPotential distribution in thermal equilibrium so far:Question 1: If I apply a voltmeter across the pn junction diode, do I measure φB?yesno it dependsQuestion 2: If I short terminals of pn junction diode, does current flow on the outside circuit?yesno sometimes6.102 Spring 2007 Lecture 5 17We are missing contact potential at the metal-semiconductor contacts:Metal-semiconductor contacts: junction of dissimilar materials⇒ built-in potentials at contacts φmnand φmp.Potential difference across structure must be zero⇒ Cannot measure φB.φB=φmn+φmp6.102 Spring 2007 Lecture 5 185. PN Junction-Reverse Biasxdo= xpo+ xno=2εs(φB− VD) Na+ Nd()qNaNdAssume: No Current Flows ;;;;;;;;;;;;;;;;pnxnWn−Wp−xpohmic contact top sideohmic contactto n side−xpxnφj φpφpmφmnφ(x)φpmφmnφjx(a)(b)Wn−Wpφn++−−+−− +− +− ++− xVD < 0 VVD (< 0 V)+−ID ≈ 0 ASame Analysis applies:Substitute φj=φB−VD6.102 Spring 2007 Lecture 5 19What did we learn today?• Electrostatics of pn junction in equilibrium–A space-charge region surrounded by two quasi-neutral regions formed.• To first order, carrier concentrations in space-charge region are much smaller than the doping level– ⇒ can use Depletion Approximation• From contact to contact, there is no potential build-up across the pn junction diode– Contact potential(s).Summary of Key
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