1. EE 5340 Semiconductor Device Theory Lecture 12 – Spring 2011
Professor Ronald L. Carter

2. Depletion Approximation
Assume p << po = Na for -xp < x < 0, so r = q(Nd-Na+p-n) = -qNa, -xp < x < 0, and p = po = Na for -xpc < x < -xp, so r = q(Nd-Na+p-n) = 0, -xpc < x < -xp
Assume n << no = Nd for 0 < x < xn, so r = q(Nd-Na+p-n) = qNd, 0 < x < xn, and n = no = Nd for xn < x < xnc, so r = q(Nd-Na+p-n) = 0, xn < x < xnc
©rlc L12-01Mar2011

3. 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]
©rlc L12-01Mar2011

4. Induced E-field in the D.R.
The sheet dipole of charge, due to Qp’ and Qn’ induces an electric field which must satisfy the conditions
Charge neutrality and Gauss’ Law* require that Ex = 0 for -xpc < x < -xp and Ex = 0 for -xn < x < xnc ≈0
©rlc L12-01Mar2011

5. 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
©rlc L12-01Mar2011

6. Induced E-field in the D.R. (cont.)
Poisson’s Equation E = r/e, has the one-dimensional form, dEx/dx = r/e, which must be satisfied for r = -qNa, -xp < x < 0, and r = +qNd, 0 < x < xn, with Ex = 0 for the remaining range
©rlc L12-01Mar2011

7. Soln to Poisson’s Eq in the D.R.Ex
-xp xn x
-xpc
xnc
-Emax
©rlc L12-01Mar2011

8. Soln to Poisson’s Eq in the D.R. (cont.)
©rlc L12-01Mar2011

9. Soln to Poisson’s Eq in the D.R. (cont.)
©rlc L12-01Mar2011

10. Comments on the Ex and Vbi
Vbi is not measurable externally since Ex is zero at both contacts
The effect of Ex does not extend beyond the depletion region
The lever rule [Naxp=Ndxn] was obtained assuming charge neutrality. It could also be obtained by requiring Ex(x=0-dx) = Ex(x=0+dx) = Emax
©rlc L12-01Mar2011

11. Sample calculations Vt = 25.86 mV at 300K
e = ereo = 11.7*8.85E-14 Fd/cm = 1.035E-12 Fd/cm
If Na=5E17/cm3, and Nd=2E15 /cm3, then for ni=1.4E10/cm3, then what is Vbi = 757 mV
©rlc L12-01Mar2011

12. Sample calculations What are Neff, W ?
Neff, = 1.97E15/cm3 W = micron
What is xn ?
= micron
What is Emax ? 2.14E4 V/cm
©rlc L12-01Mar2011

13. 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)
©rlc L12-01Mar2011

14. Effect of V  0
©rlc L12-01Mar2011

15. 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
©rlc L12-01Mar2011

16. Junction Capacitance
The junction has +Q’n=qNdxn (exposed donors), and (exposed acceptors) Q’p=-qNaxp = -Q’n, forming a parallel sheet charge capacitor.
©rlc L12-01Mar2011

17. 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.
©rlc L12-01Mar2011

18. Junction C (cont.) The C-V relationship simplifies to
©rlc L12-01Mar2011

19. 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
©rlc L12-01Mar2011

20. 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
©rlc L12-01Mar2011

21. dy/dx - Numerical Differentiation
©rlc L12-01Mar2011

22. 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.
©rlc L12-01Mar2011

23. 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
©rlc L12-01Mar2011

24. 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
©rlc L12-01Mar2011

25. Linear graded junction (cont.)
Let Ex(-xo) = 0, since this is the edge of the DR (also true at +xo)
©rlc L12-01Mar2011

26. Linear graded junction (cont.)Ex
-xo xo x
-Emax
|area| = Vbi-Va
©rlc L12-01Mar2011

27. Linear graded junction (cont.)
©rlc L12-01Mar2011

28. Linear graded junction, etc.
©rlc L13-03Mar2011

29. References
1 and M&KDevice Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, See Semiconductor Device Fundamentals, by Pierret, Addison-Wesley, 1996, for another treatment of the m model.
2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981.
3 and **Semiconductor Physics & Devices, 2nd ed., by Neamen, Irwin, Chicago, 1997.
Fundamentals of Semiconductor Theory and Device Physics, by Shyh Wang, Prentice Hall, 1989.
©rlc L12-01Mar2011