1EE130 Lecture 7, Slide 1Spring 2003Lecture #7Quiz #1 Results(undergrad. scores only)N = 73; mean = 21.6; σ = 2.1; high = 25; low = 14OUTLINE–Continuity equations– Minority carrier diffusion equations– Quasi-Fermi levelsReading: Chapter 3.4, 3.5EE130 Lecture 7, Slide 2Spring 2003Clarification: Direct vs. Indirect Band GapSmall change in momentum required for recombinationÆ momentum is conserved by photon emissionLarge change in momentum required for recombinationÆ momentum is conserved by phonon + photon emission2EE130 Lecture 7, Slide 3Spring 2003nndtdnτ∆−=ppdtdpτ∆−=Example: Relaxation to Equilibrium Statefor electrons in p-type materialfor holes in n-type materialConsider a semiconductor with no current flow in which thermal equilibriumis disturbed by the sudden creation of excess holes and electrons. Thesystem will relax back to the equilibrium state via R-G mechanism:EE130 Lecture 7, Slide 4Spring 2003Net Recombination Rate (General Case)• For arbitrary injection levels and both carrier types in a non-degenerate semiconductor, the net rate of recombination is:kTEEikTEEinpiTiiTenpennppnnnpndtpddtnd/)(1/)(1112 and where)()(−−≡≡+++−−=∆=∆ττ3EE130 Lecture 7, Slide 5Spring 2003Derivation of Continuity Equation• Accounting of carrier-flux into/out-of an infinitesimal volume:Jn(x) Jn(x+dx)dxArea A, volume Adx[]AdxnAdxxJAxJqtnAdxnnnτ∆−+−−=∂∂)()(1EE130 Lecture 7, Slide 6Spring 2003nnnnnnxxJqtndxxxJxJdxxJτ∆−∂∂=∂∂=>∂∂+=+)(1 )()()(LppLnnGpxxJqtpGnxxJqtn )(1 )(1+∆−∂∂−=∂∂+∆−∂∂=∂∂ττContinuityEquations:4EE130 Lecture 7, Slide 7Spring 2003Derivation of Minority-Carrier Diffusion Equations• Simplifying assumptions:–1-D– negligible electric field– n0, p0are independent of x– low-level injection conditionsEE130 Lecture 7, Slide 8Spring 20032PP22LpDpdxpdp∆=∆=∆τLPis the hole diffusion lengthpppDLτ≡Minority Carrier Diffusion Length• Consider the special case:– Constant minority-carrier (hole) injection at x=0– Steady state, no light5EE130 Lecture 7, Slide 9Spring 2003• Physically, LPand LNrepresent the average distance that minority carriers can diffuse into a sea of majority carriers before being annihilated.•Example: ND=1016cm-3; τp= 10-6sEE130 Lecture 7, Slide 10Spring 2003• Whenever ∆n= ∆p≠0,np≠ni2. However, we would like to preserve and use the relations:• These equations imply np = ni2, however. The solution is to introduce two quasi-Fermi levels FNand FPsuch thatkTEEcFceNn/)( −−=kTEEvvFeNp/)( −−=kTFEcNceNn/)( −−=kTEFvvPeNp/)( −−=Quasi-Fermi Levels6EE130 Lecture 7, Slide 11Spring 2003Consider a Si sample with ND = 1017 cm-3and ∆n = ∆p = 1014 cm-3.(a) Find n:n = n0+ ∆n = ND + ∆n ≈ 1017 cm-3(b) Find p:p = p0+ ∆p = ( ni2 / ND) + ∆p ≈ 1014 cm-3(c) Find the np product:np ≈ 1017 × 1014= 1031 cm-6>> ni2Example: Quasi-Fermi LevelsEE130 Lecture 7, Slide 12Spring 2003(d) Find FN:n = 1017 cm-3=Ec- FN= kT × ln(Nc/1017)= 0.026 eV × ln(2.8×1019/1017)= 0.15 eV(e) Find FP:p = 1014cm-3=FP–Ev= kT × ln(Nv/1017)= 0.026 eV × ln(1019/1014)= 0.30 eVkTEFvvPeN/)( −−kTFEcNceN/)(
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