1. EE 5340 Semiconductor Device Theory Lecture 9 - Fall 2003
Professor Ronald L. Carter
L 09 Sept 23

2. Band diagram for p+-n jctn* at Va = 0Ec
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 09 Sept 23

3. Band diagram for p+-n at Va=0 (cont.)
A total band bending of qVbi = q(fn-fp) = kT ln(NdNa/ni2) is necessary to set EFp = Efn
For -xp < x < 0, Efi - EFP < -qfp, = |qfp| so p < Na = po, (depleted of maj. carr.)
For 0 < x < xn, EFN - EFi < qfn, so n < Nd = no, (depleted of maj. carr.)
-xp < x < xn is the Depletion Region
L 09 Sept 23

4. Depletion approx. charge distribution
+Qn’=qNdxn
+qNd
[Coul/cm2]
-xp x
-xpc xn
xnc
-qNa
Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn
Qp’=-qNaxp
[Coul/cm2]
L 09 Sept 23

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
L 09 Sept 23

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
L 09 Sept 23

7. Soln to Poisson’s Eq in the D.R.Ex
-xp xn x
-xpc
xnc
-Emax
L 09 Sept 23

8. Soln to Poisson’s Eq in the D.R. (cont.)
L 09 Sept 23

9. Soln to Poisson’s Eq in the D.R. (cont.)
L 09 Sept 23

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
L 09 Sept 23

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
L 09 Sept 23

12. Sample calculations What are Neff, W ?
Neff, = 1.97E15/cm3 W = micron
What is xn ?
= micron
What is Emax ? 2.14E4 V/cm
L 09 Sept 23

13. 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
L 09 Sept 23

14. Effect of V  0
L 09 Sept 23

15. 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
L 09 Sept 23

16. 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-
L 09 Sept 23

17. 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 09 Sept 23

18. 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
L 09 Sept 23

19. 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
L 09 Sept 23

20. 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”
L 09 Sept 23

21. 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
L 09 Sept 23

22. 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 09 Sept 23

23. 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 09 Sept 23

24. 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 09 Sept 23

25. 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.
L 09 Sept 23

26. Junction C (cont.)
The C-V relationship simplifies to
L 09 Sept 23

27. Junction C (cont.)
If one plots [C’j]-2 vs. Va Slope = -[(C’j0)2Vbi]-1 vertical axis intercept = [C’j0]-2 horizontal axis intercept = Vbi
C’j-2
C’j0-2 Va
Vbi
L 09 Sept 23

28. Test 1 - 25Sept03 8 AM Room 206 Activities Building
Open book - 1 legal text or ref. Only
Calculator allowed
A cover sheet will be included with full instructions. See for examples from Fall 2002.
L 09 Sept 23