EE 5340 Semiconductor Device Theory Lecture 11 - Fall 2010

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EE 5340 Semiconductor Device Theory Lecture 11 - Fall 2010 Professor Ronald L. Carter [email protected] http://www.uta.edu/ronc
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EE 5340 Semiconductor Device Theory Lecture 11 - Fall 2010. Professor Ronald L. Carter [email protected] http://www.uta.edu/ronc. E FN. Band diagram for p + -n jctn* at V a = 0. E c. qV bi = q( f n - f p ). q f p < 0. E c. E Fi. E FP. E v. E Fi. q f n > 0. - PowerPoint PPT Presentation

Transcript of EE 5340 Semiconductor Device Theory Lecture 11 - Fall 2010

  • EE 5340Semiconductor Device TheoryLecture 11 - Fall 2010Professor Ronald L. [email protected]://www.uta.edu/ronc

  • L11 27Sep10Band diagram forp+-n jctn* at Va = 0EcEFiEvEcEFPEFiEv0xnx-xp-xpcxncqfp < 0qfn > 0qVbi = q(fn - fp)*Na > Nd -> |fp| > fnp-type for x0*

  • L11 27Sep10Depletion approx.charge distributionxnx-xp-xpcxncr+qNd-qNa+Qn=qNdxnQp=-qNaxpDue to Charge neutrality Qp + Qn = 0, => Naxp = Ndxn [Coul/cm2][Coul/cm2]*

  • L11 27Sep10Induced E-fieldin the D.R.xnx-xp-xpcxncDepletion region (DR)p-type CNRExExposed Donor ionsExposed Acceptor Ionsn-type chg neutral regp-contactN-contactW0*

  • L11 27Sep10Soln to PoissonsEq in the D.R.xnx-xp-xpcxncEx-Emax*

  • L11 27Sep10Soln to PoissonsEq in the D.R. (cont.)*

  • L11 27Sep10Effect of V 0Define an external voltage source, Va, with the +term at the p-type contact and the -term at the n-type contactFor Va > 0, the Va induced field tends to oppose Ex due to DRFor Va < 0, the Va induced field tends to add to Ex due to DRWill consider Va < 0 now*

  • L11 27Sep10*Band diagram forp+-n jctn* at Va 0EcEFNEFiEvEcEFPEFiEv0xnx-xp-xpcxncqfp < 0qfn > 0q(Vbi-Va)*Na > Nd -> |fp| > fnp-type for x0q(Va)

  • L11 27Sep10*Soln to PoissonsEq in the D.R.xnx-xp-xpcxncEx-Emax(V)-Emax(V-dV)W(Va)W(Va-dV)

  • L11 27Sep10Effect of V 0*

  • L11 27Sep10Effect of V 0Lever rule, Naxp = Ndxn, still appliesVbi = Vt ln(NaNd/ni2), still appliesW = xp + xn, still appliesNeff = NaNd/(Na + Nd), still appliesQn = qNdxn = -Qp = qNaxp, still applies

    For Va < 0, W increases and Emax increases*

  • L11 27Sep10One-sided p+n or n+p jctnsIf p+n, then Na >> Nd, and NaNd/(Na + Nd) = Neff --> Nd, andW --> xn, DR is all on lightly d. sideIf n+p, then Nd >> Na, and NaNd/(Na + Nd) = Neff --> Na, andW --> xp, DR is all on lightly d. sideThe net effect is that Neff --> N-, (- = lightly doped side) and W --> x-*

  • L11 27Sep10Depletion Approxi-mation (Summary)For the step junction defined by doping Na (p-type) for x < 0 and Nd, (n-type) for x > 0, the depletion width W = {2e(Vbi-Va)/qNeff}1/2, where Vbi = Vt ln{NaNd/ni2}, and Neff=NaNd/(Na+Nd). Since Naxp=Ndxn, xn = W/(1 + Nd/Na), and xp = W/(1 + Na/Nd).*

  • L11 27Sep10Debye lengthThe DA assumes n changes from Nd to 0 discontinuously at xn, likewise, p changes from Na to 0 discontinuously at -xp.In the region of xn, Poissons eq is E = r/e --> dEx/dx = q(Nd - n), and since Ex = -df/dx, we have-d2f/dx2 = q(Nd - n)/e to be solved*

  • L11 27Sep10Debye length (cont)Since the level EFi is a reference for equil, we set f = Vt ln(n/ni)In the region of xn, n = ni exp(f/Vt), so d2f/dx2 = -q(Nd - ni ef/Vt), letf = fo + f, where fo = Vt ln(Nd/ni) soNd - ni ef/Vt = Nd[1 - ef/Vt-fo/Vt], for f - fo = f
  • L11 27Sep10Debye length (cont)So f = f(xn) exp[+(x-xn)/LD]+con. and n = Nd ef/Vt, x ~ xn, whereLD is the Debye length

    *

  • L11 27Sep10Debye length (cont)LD estimates the transition length of a step-junction DR (concentrations Na and Nd with Neff = NaNd/(Na +Nd)). Thus,For Va=0, & 1E13 < Na,Nd < 1E19 cm-313% < d < 28% => DA is OK*

  • L11 27Sep10Junction C (cont.)xnx-xp-xpcxncr+qNd-qNa+Qn=qNdxnQp=-qNaxpCharge neutrality => Qp + Qn = 0, => Naxp = Ndxn dQn=qNddxndQp=-qNadxp*

  • L11 27Sep10JunctionCapacitanceThe junction has +Qn=qNdxn (exposed donors), and (exposed acceptors) Qp=-qNaxp = -Qn, forming a parallel sheet charge capacitor.*

  • L11 27Sep10JunctionC (cont.)So this definition of the capacitance gives a parallel plate capacitor with charges dQn and dQp(=-dQn), separated by, L (=W), with an area A and the capacitance is then the ideal parallel plate capacitance.Still non-linear and Q is not zero at Va=0.*

  • L11 27Sep10JunctionC (cont.)This Q ~ (Vbi-Va)1/2 is clearly non-linear, and Q is not zero at Va = 0.Redefining the capacitance,*

  • L11 27Sep10JunctionC (cont.)The C-V relationship simplifies to*

  • L11 27Sep10JunctionC (cont.)If one plots [Cj]-2 vs. VaSlope = -[(Cj0)2Vbi]-1vertical axis intercept = [Cj0]-2 horizontal axis intercept = Vbi*