EE 5340 Semiconductor Device Theory Lecture 11 - Fall 2010 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc

Band diagram for p+-n jctn* at Va = 0 Ec qVbi = q(fn - fp) qfp < 0 EFi Ec EFP EFN Ev EFi qfn > 0 *Na > Nd -> |fp| > fn Ev p-type for x<0 n-type for x>0 x -xpc -xp xn xnc L11 27Sep10

Depletion approx. charge distribution +Qn’=qNdxn +qNd [Coul/cm2] -xp x -xpc xn xnc -qNa Due to Charge neutrality Qp’ + Qn’ = 0, => Naxp = Ndxn Qp’=-qNaxp [Coul/cm2] L11 27Sep10

Induced E-field in the D.R. Ex p-contact N-contact O - O + p-type CNR n-type chg neutral reg O - O + O - O + Exposed Acceptor Ions Depletion region (DR) Exposed Donor ions W x -xpc -xp xn xnc L11 27Sep10

Soln to Poisson’s Eq in the D.R. Ex -xp xn x -xpc xnc -Emax L11 27Sep10

Soln to Poisson’s Eq in the D.R. (cont.) L11 27Sep10

Effect of V  0 Define an external voltage source, Va, with the +term at the p-type contact and the -term at the n-type contact For Va > 0, the Va induced field tends to oppose Ex due to DR For Va < 0, the Va induced field tends to add to Ex due to DR Will consider Va < 0 now L11 27Sep10

Band diagram for p+-n jctn* at Va  0 Ec q(Va) q(Vbi-Va) qfp < 0 EFi Ec EFP EFN Ev EFi qfn > 0 *Na > Nd -> |fp| > fn Ev p-type for x<0 n-type for x>0 x -xpc -xp xn xnc L11 27Sep10

Soln to Poisson’s Eq in the D.R. Ex W(Va-dV) W(Va) -xp xn x -xpc xnc -Emax(V) -Emax(V-dV) L11 27Sep10

Effect of V  0 L11 27Sep10

Effect of V  0 Lever rule, Naxp = Ndxn, still applies Vbi = Vt ln(NaNd/ni2), still applies W = xp + xn, still applies Neff = NaNd/(Na + Nd), still applies Q’n = qNdxn = -Q’p = qNaxp, still applies For Va < 0, W increases and Emax increases L11 27Sep10

One-sided p+n or n+p jctns If p+n, then Na >> Nd, and NaNd/(Na + Nd) = Neff --> Nd, and W --> xn, DR is all on lightly d. side If n+p, then Nd >> Na, and NaNd/(Na + Nd) = Neff --> Na, and W --> xp, DR is all on lightly d. side The net effect is that Neff --> N-, (- = lightly doped side) and W --> x- L11 27Sep10

Depletion 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 27Sep10

n x xn Nd Debye length The 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, 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 solved L11 27Sep10

Debye 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), let f = 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 L11 27Sep10

Debye length (cont) So f’ = f’(xn) exp[+(x-xn)/LD]+con. and n = Nd ef’/Vt, x ~ xn, where LD is the “Debye length” L11 27Sep10

13% < d < 28% => DA is OK Debye 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-3 13% < d < 28% => DA is OK L11 27Sep10

Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn Junction C (cont.) r +Qn’=qNdxn +qNd dQn’=qNddxn -xp x -xpc xn xnc -qNa Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn dQp’=-qNadxp Qp’=-qNaxp L11 27Sep10

Junction Capacitance The junction has +Q’n=qNdxn (exposed donors), and (exposed acceptors) Q’p=-qNaxp = -Q’n, forming a parallel sheet charge capacitor. L11 27Sep10

Junction C (cont.) So this definition of the capacitance gives a parallel plate capacitor with charges dQ’n and dQ’p(=-dQ’n), 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 27Sep10

Junction C (cont.) This Q ~ (Vbi-Va)1/2 is clearly non-linear, and Q is not zero at Va = 0. Redefining the capacitance, L11 27Sep10

Junction C (cont.) The C-V relationship simplifies to L11 27Sep10

Junction C (cont.) If one plots [Cj]-2 vs. Va Slope = -[(Cj0)2Vbi]-1 vertical axis intercept = [Cj0]-2 horizontal axis intercept = Vbi Cj-2 Vbi Va Cj0-2 L11 27Sep10