Lecture 5 PN Junction and MOS Electrostatics(II) PN JUNCTION IN THERMAL EQUILIBRIUM Outline 1. Introduction 2. Electrostatics of pn junction in thermal equilibrium 3. The depletion approximation 4. Contact potentials Reading Assignment: Howe and Sodini, Chapter 3, Sections 3.3-3.6 6.012 Spring 2009 Lecture 5 11. Introduction • pn junction – p-region and n-region in intimate contact Why is the p-n junction worth studying? It is present in virtually every semiconductor device! Example: CMOS cross-section Understanding the pn junction is essential to understanding transistor operation 6.012 Spring 2009 Lecture 5 2 pnn+n+n+p+p+p+p-MOSFET n-MOSFETFigure by MIT OpenCourseWare.Doping distribution of an abrupt p-n junction 2. Electrostatics of p-n junction in equilibrium Focus on intrinsic region: 6.012 Spring 2009 Lecture 5 3What is the carrier concentration distribution in thermal equilibrium? First think of the two sides separately: Now bring the two sides together. What happens? 6.012 Spring 2009 Lecture 5 4Resulting 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 region 6.012 Spring 2009 Lecture 5 5On a linear scale: 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 approximation We can divide semiconductor into three regions • Two quasi-neutral n- and p-regions (QNR’s) • One space-charge region (SCR) Thermal equilibrium: balance between drift and diffusion J n (x) = J n drift (x) + J n diff (x) = 0 J p (x) = J p drift (x) + J p diff (x) = 0 6.012 Spring 2009 Lecture 5 63. 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 – -x po and x no (must calculate where to place these) x < −x po; po(x) = Na, no (x) = ni 2 Na −x po < x < 0; po (x), no(x) << Na 0 < x < xno; no (x), po(x) << Nd x > xno; no (x) = Nd , po (x) = ni 2 Nd 6.012 Spring 2009 Lecture 5 7Space Charge Density ρρρρ(x) = 0; x <−xpo = − qNa; −xpo< x < 0 = qNd ; 0 < x < xno = 0; x > xno 6.012 Spring 2009 Lecture 5 8Electric Field Integrate Poisson’s equation 1 x2 E(x2) − E(x1) = ∫ρρρρ(x) dxεεεεs x1 x < − x po ; E(x) = 0 − x po < x < 0; E(x ) − E( −x po ) = 1 εεεεs −qN a d ′x − x po x ∫ = − qN a εεεεs x −x po x = − qNa εεεεs x + x po( ) 0 < x < xno ; E(x) = qN d εεεεs x − xno( ) x > xno ; E(x) = 0 6.012 Spring 2009 Lecture 5 9This expression is always correct in TE! We did not use depletion approximation. Electrostatic Potential (with φφφφ=0 @ no=po=ni) φφφφ= kT • ln no φφφφ= − kT • ln po q ni q ni In QNRs, no and po are known ⇒ can determine φ aφφφφ= − kT • ln N in p-QNR: po=N a ⇒ pq ni kT Nd= • lnin n-QNR: no=Nd ⇒φφφφnq ni Built-in potential: kT NdNaφφφφB =φφφφn −φφφφp = • ln 2q ni 6.012 Spring 2009 Lecture 5 10To obtain φφφφ(x) in between, integrate E(x) x2 φφφφ(x2) −φφφφ(x1) = − ∫ E(x′)dx′ x1 x < −x po; φφφφ(x) = φφφφp −x po < x < 0; φφφφ(x) − φφφφ(−x po ) = − − qNa εεεεs ′x + xpo( )d′x −x po x ∫ = qNa 2εεεεs x + x po( )2 φφφφ(x) = φφφφp + qNa 2εεεεs x + xpo( )2 0 < x < xno; φφφφ(x) = φφφφn − qNd 2εεεεs x − xno( )2 x > xno; φφφφ(x) = φφφφn Almost done …. 6.012 Spring 2009 Lecture 5 11Still do not know xno and x po ⇒⇒⇒⇒ need two more equations 1. Require overall charge neutrality: qNaxpo = qNdxno 2. Require φ(x) to be continuous at x=0; φφφφ + qNa x 2 = φφφφ − qNd x 2 p po n no2εεεε2εεεε s s Two equations with two unknowns — obtain solution: 2εεεε φφφφ N 2εεεε φφφφ N s B a s B d x = no po q(Na + Nd )Nd x = q(Na + Nd )Na Now problem is completely solved! 6.012 Spring 2009 Lecture 5 12Solution Summary Space Charge Density Electrostatic Field Electrostatic Potential 6.012 Spring 2009 Lecture 5 13Other results: Width of the space charge region: 2εεεεs φφφφB (Na + Nd ) x = x + x = do po no qNaNd Field at the metallurgical junction: 2qφφφφBNaNdE = o εεεεs (Na + Nd ) 6.012 Spring 2009 Lecture 5 14The lightly-doped side controls the electrostatics of the pn junction Three Special Cases • Symmetric junction: Na = Nd • Asymmetric junction: Na > Nd • Strongly asymmetric junction – p+n junction: Na >> Nd x = x po no x < x po no 2εεεεs φφφφB xpo << xno ≈ xdo ≈ qNd 2qφφφφBNdE ≈o εεεεs 6.012 Spring 2009 Lecture 5 154. Contact Potential Potential distribution in thermal equilibrium so far: Question 1: If I apply a voltmeter across the pn junction diode, do I measure φB? yes no it depends Question 2: If I short terminals of pn junction diode, does current flow on the outside circuit? yes no sometimes 6.012 Spring 2009 Lecture 5 16We are missing contact potential at the metal-semiconductor contacts: Metal-semiconductor contacts: junction of dissimilar materials ⇒ built-in potentials at contacts φ mn and φ mp . Potential difference across structure must be zero ⇒ Cannot measure φB. φφφφB = +φφφφmn φφφφmp 6.012 Spring 2009 Lecture 5 175. PN Junction-Reverse Bias Assume: No Current Flows --ohmic contact to ZDsOAt -w,~///////////A+ %wn p side + -v, (<OW -O -PI ++j X --ohmic contact %-qmnI + to n side (a>t+'"' Same Analysis applies: (b) Substitute 6.012Spring 2009 Lecture BWhat did we learn today? Summary of Key Concepts • 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). 6.012 Spring 2009 Lecture 5 19MIT OpenCourseWarehttp://ocw.mit.edu 6.012 Microelectronic Devices and Circuits Spring 2009 For information about citing these materials or our Terms of Use, visit:
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