1. EE 5340 Semiconductor Device Theory Lecture 11 - Fall 2010
Professor Ronald L. Carter

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
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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]
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4. 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
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5. Soln to Poisson’s Eq in the D.R.Ex
-xp xn x
-xpc
xnc
-Emax
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6. Soln to Poisson’s Eq in the D.R. (cont.)
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7. 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
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8. Band diagram for p+-n jctn* at Va  0Ec
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
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9. 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)
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10. Effect of V  0
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11. 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
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12. 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-
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13. 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).
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14. 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
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15. 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
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16. 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”
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17. 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
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18. 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
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19. Junction Capacitance
The junction has +Q’n=qNdxn (exposed donors), and (exposed acceptors) Q’p=-qNaxp = -Q’n, forming a parallel sheet charge capacitor.
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20. 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.
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21. Junction C (cont.)
This Q ~ (Vbi-Va)1/2 is clearly non-linear, and Q is not zero at Va = 0.
Redefining the capacitance,
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22. Junction C (cont.)
The C-V relationship simplifies to
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23. 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
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