Download presentation
Presentation is loading. Please wait.
Published byJayson Goodwin Modified over 5 years ago
1
EE 5340 Semiconductor Device Theory Lecture 11 - Fall 2003
Professor Ronald L. Carter L 11 Sept 30
2
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 L 11 Sept 30
3
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 L 11 Sept 30
4
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 L 11 Sept 30
5
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). L 11 Sept 30
6
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) L 11 Sept 30
7
Effect of V 0 L 11 Sept 30
8
Junction Capacitance The junction has +Q’n=qNdxn (exposed donors), and (exposed acceptors) Q’p=-qNaxp = -Q’n, forming a parallel sheet charge capacitor. L 11 Sept 30
9
Junction C (cont.) This Q ~ (Vbi-Va)1/2 is clearly non-linear, and Q is not zero at Va = 0. Redefining the capacitance, L 11 Sept 30
10
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 L 11 Sept 30
11
Junction Capacitance Estimate CJO Define y Cj/CJO
Calculate y/(dy/dV) = {d[ln(y)]/dV}-1 A plot of r y/(dy/dV) vs. V has slope = -1/M, and intercept = VJ/M L 11 Sept 30
12
dy/dx - Numerical Differentiation
L 11 Sept 30
13
Practical Junctions Junctions are formed by diffusion or implantation into a uniform concentration wafer. The profile can be approximated by a step or linear function in the region of the junction. If a step, then previous models OK. If linear, let the local charge density r=qax in the region of the junction. L 11 Sept 30
14
Practical Jctns (cont.)
Na(x) N N Shallow (steep) implant Na(x) Linear approx. Box or step junction approx. Nd Nd Uniform wafer con x (depth) x (depth) xj L 11 Sept 30
15
Linear graded junction
Let the net donor concentration, N(x) = Nd(x) - Na(x) = ax, so r =qax, -xp < x < xn = xp = xo, (chg neu) r = qa x r Q’n=qaxo2/2 -xo x xo Q’p=-qaxo2/2 L 11 Sept 30
16
Linear graded junction (cont.)
Let Ex(-xo) = 0, since this is the edge of the DR (also true at +xo) L 11 Sept 30
17
Linear graded junction (cont.)
Ex -xo xo x -Emax |area| = Vbi-Va L 11 Sept 30
18
Linear graded junction (cont.)
L 11 Sept 30
19
Linear graded junction, etc.
L 11 Sept 30
20
Doping Profile If the net donor conc, N = N(x), then at x, the extra charge put into the DR when Va->Va+dVa is dQ’=-qN(x)dx The increase in field, dEx =-(qN/e)dx, by Gauss’ Law (at x, but also all DR). So dVa=-xddEx= (W/e) dQ’ Further, since qN(x)dx, for both xn and xn, we have the dC/dx as ... L 11 Sept 30
21
Arbitrary doping profile (cont.)
L 11 Sept 30
22
Arbitrary doping profile (cont.)
L 11 Sept 30
23
Arbitrary doping profile (cont.)
L 11 Sept 30
24
Arbitrary doping profile (cont.)
L 11 Sept 30
25
Example An assymetrical p+ n junction has a lightly doped concentration of 1E16 and with p+ = 1E18. What is W(V=0)? Vbi=0.816 V, Neff=9.9E15, W=0.33mm What is C’j0? = 31.9 nFd/cm2 What is LD? = 0.04 mm L 11 Sept 30
26
Reverse bias junction breakdown
Avalanche breakdown Electric field accelerates electrons to sufficient energy to initiate multiplication of impact ionization of valence bonding electrons field dependence shown on next slide Heavily doped narrow junction will allow tunneling - see Neamen*, p. 274 Zener breakdown L 11 Sept 30
27
Effect of V 0 L 11 Sept 30
28
Ecrit for reverse breakdown (M&K**)
Taken from p. 198, M&K** L 11 Sept 30
29
Reverse bias junction breakdown
Assume -Va = VR >> Vbi, so Vbi-Va-->VR Since Emax~ 2VR/W = (2qN-VR/(e))1/2, and VR = BV when Emax = Ecrit (N- is doping of lightly doped side ~ Neff) BV = e (Ecrit )2/(2qN-) Remember, this is a 1-dim calculation L 11 Sept 30
30
Junction curvature effect on breakdown
The field due to a sphere, R, with charge, Q is Er = Q/(4per2) for (r > R) V(R) = Q/(4peR), (V at the surface) So, for constant potential, V, the field, Er(R) = V/R (E field at surface increases for smaller spheres) Note: corners of a jctn of depth xj are like 1/8 spheres of radius ~ xj L 11 Sept 30
31
BV for reverse breakdown (M&K**)
Taken from Figure 4.13, p. 198, M&K** Breakdown voltage of a one-sided, plan, silicon step junction showing the effect of junction curvature.4,5 L 11 Sept 30
32
References * Semiconductor Physics and Devices, 2nd ed., by Neamen, Irwin, Boston, 1997. **Device Electronics for Integrated Circuits, 2nd ed., by Muller and Kamins, Wiley, New York, 1986. L 11 Sept 30
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.