Slide 1Metal/semiconductor system typesReal Schottky band structure1Slide 4Poisson’s EquationPoisson’s EquationIdeal metal to n-type barrier diode (fm>fs,Va=0)Depletion ApproximationIdeal n-type Schottky depletion width (Va=0)Debye lengthDebye length (cont)Debye length (cont)Debye length (cont)Effect of V 0Effect of V 0Ideal metal to n-type Schottky (Va > 0)Schottky diode capacitanceSchottky Capacitance (continued)Schottky Capacitance (continued)Schottky Capacitance (continued)Schottky Capacitance (continued)Schottky Capacitance (continued)Slide 23Energy bands for p- and n-type s/cMaking contact in a p-n junctionBand diagram for p+-n jctn* at Va = 0Band diagram for p+-n at Va=0 (cont.)Depletion ApproximationDepletion approx. charge distributionInduced E-field in the D.R.Induced E-field in the D.R.Induced E-field in the D.R. (cont.)Soln to Poisson’s Eq in the D.R.Soln to Poisson’s Eq in the D.R. (cont.)Soln to Poisson’s Eq in the D.R. (cont.)Comments on the Ex and VbiSample calculationsSample calculationsReferencesReferencesEE 5340Semiconductor Device TheoryLecture – Spring 2011Professor Ronald L. [email protected]://www.uta.edu/ronc©rlc L11-24Feb20112Metal/semiconductorsystem typesn-type semiconductor•Schottky diode - blocking for fm > fs•contact - conducting for fm < fsp-type semiconductor•contact - conducting for fm > fs•Schottky diode - blocking for fm < fs©rlc L11-24Feb20113Real Schottkyband structure1•Barrier transistion region, d•Interface statesabove fo acc, p neutrlbelow fo dnr, n neutrlDitd -> oo, qfBn = Eg- foFermi level “pinned”Ditd -> 0, qfBn = fm - cGoes to “ideal” case©rlc L11-24Feb20114Fig 8.41 (a) Image charge and electric field at a metal-dielectric interface (b) Distortion of potential barrier at E=0 and (c) E0©rlc L11-24Feb20115silicon for 711andFd/cm, ,14E858with , ypermitivit the is xEE where, ,Erorox..--Poisson’s Equation•The electric field at (x,y,z) is related to the charge density r=q(Nd-Na-p-n) by the Poisson Equation:©rlc L11-24Feb20116Poisson’s Equation•n = no + dn, and p = po + dp, in non-equil•For n-type material, N = (Nd - Na) > 0, no = N, and (Nd-Na+p-n)=-dn +dp +ni2/N•For p-type material, N = (Nd - Na) < 0, po = -N, and (Nd-Na+p-n) = dp-dn-ni2/N•So neglecting ni2/N 0n or p with material type-pand type-n for ,npqE-©rlc L11-24Feb20117Ideal metal to n-typebarrier diode (fm>fs,Va=0)EFnEoEcEvEFiqfs,nqcsn-type s/cqfmEFmmetalqfBnqfbiqf’nNo disc in EoEx=0 in metal ==> EoflatfBn=fm- cs = elec mtl to s/c barrfbi=fBn-fn= fm-fs elect s/c to mtl barr Depl reg0 xn xnc©rlc L11-24Feb20118DepletionApproximation•For 0 < x < xn, assume n << no = Nd, so r = q(Nd-Na+p-n) = qNd •For xn < x < xnc, assume n = no = Nd, so r = q(Nd-Na+p-n) = 0•For x = 0-, there is a pulse of charge balancing the qNdxn in 0 < x < xn©rlc L11-24Feb20119Ideal n-type Schottky depletion width (Va=0)xnxqNddQ’d = qNdxnxrEx-EmdnmxqNxEdxdExn(Sheet of negative charge on metal)= -Q’d dctsmnBniix0xdinNNVdxE- , qN2xn/ln©rlc L11-24Feb201110Debye length•The DA assumes n changes from Nd to 0 discontinuously at xn.•In the region of xn, Poisson’s 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 solvednxxnNd0©rlc L11-24Feb201111Debye 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) so Nd - ni ef/Vt = Nd[1 - ef/Vt-fo/Vt], for f - fo = f’ << fo, the DE becomes d2f’/dx2 = (q2Nd/ekT)f’, f’ << fo©rlc L11-24Feb201112Debye length (cont)•So f’ = f’(xn) exp[+(x-xn)/LD]+con. and n = Nd ef’/Vt, x ~ xn, where LD is the “Debye length” material. intrinsic for 2n and type-p for N type,-n for N pn :Notelength. transition a ,qkTV ,pnqVLiadttD©rlc L11-24Feb201113Debye length (cont)•LD estimates the transition length of a step-junction DR. Thus, it0VdD2VWNLa•For Va = 0, i ~ 1V, Vt ~ 25 mV d < 11% DA assumption OK©rlc L11-24Feb201114Effect of V 0•Define an external voltage source, Va, with the +term at the metal contact and the -term at the n-type contact•For Va > 0, the Va induced field tends to oppose Ex caused by the DR•For Va < 0, the Va induced field tends to aid Ex due to DR•Will consider Va < 0 now©rlc L11-24Feb201115 dimaxdinxaaix0xNVa2qEand ,qNVa2xare Solutions .E reduce to tends V todue field the since ,VdxEthat is now change only ThenEffect of V 0©rlc L11-24Feb201116Ideal metal to n-typeSchottky (Va > 0)qVa = Efn - EfmBarrier for electrons from sc to m reduced to q(fbi-Va) qfBn the sameDR decrEFnEoEcEvEFiqfs,nqcsn-type s/cqfmEFmmetalqfBnq(fi-Va)qf’nDepl reg©rlc L11-24Feb201117Schottky diodecapacitancexnxqNd-Q-dQQ’d = qNdxnxrEx-EmdnmxqNxEdxdExndQ’ VQVQCVVVQQQarea jctn.A where AQQjaiainn''','©rlc L11-24Feb201118Schottky Capacitance(continued)•The junction has +Q’n=qNdxn (exposed donors), and Q’n = - Q’metal (Coul/cm2), forming a parallel sheet charge capacitor. 2aiddaidndncmCoul VqN2qNV2qNxqNQ,,'©rlc L11-24Feb201119Schottky Capacitance(continued)•This Q ~ (i-Va)1/2 is clearly non-linear, and Q is not zero at Va = 0.•Redefining the capacitance, [Fd] xAC and ][Fd/cm xC soV2qNdVdQCnj2njaidanj,,,',''©rlc L11-24Feb201120Schottky Capacitance(continued)•So this definition of the capacitance gives a parallel plate capacitor with charges dQ’n and dQ’p(=-dQ’n), separated by, L (=xn), 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.©rlc L11-24Feb201121Schottky Capacitance(continued)•The C-V relationship simplifies to][Fd/cm 2qNAC herewequation model a V1CC2id0j21ia0jj,,©rlc L11-24Feb201122Schottky Capacitance(continued)•If one plots [Cj]-2 vs. VaSlope = -[(Cj0)2Vbi]-1vertical axis intercept = [Cj0]-2 horizontal
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