1. EE 5340 Semiconductor Device Theory Lecture 8 - Fall 2003
Professor Ronald L. Carter
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2. Ideal n-type Schottky depletion width (Va=0)r Ex xn
qNd x
Q’d = qNdxd x xn
-Em d
(Sheet of negative charge on metal)= -Q’d
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3. Ideal metal to n-type Schottky (Va > 0)
qVa = Efn - Efm
Barrier for electrons from sc to m reduced to q(fbi-Va)
qfBn the same (data - p.166)
DR smaller Eo
qcs
qfm
q(fi-Va)
qfs,n
qfBn Ec
EFm
EFn
EFi Ev
Depl reg
qf’n
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4. Effect of V  0
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5. Schottky diode capacitancer
qNd
Q’d = qNdxn
dQ’ x
-Q-dQ xn Ex xn x
-Em
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6. Schottky Capacitance (continued)
The junction has +Q’n=qNdxn (exposed donors), and Q’n = - Q’metal (Coul/cm2), forming a parallel sheet charge capacitor.
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7. Schottky Capacitance (continued)
This Q ~ (i-Va)1/2 is clearly non-linear, and Q is not zero at Va = 0.
Redefining the capacitance,
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8. Schottky Capacitance (continued)
So this definition of the capacitance gives a parallel plate capacitor with charges dQ’n and dQ’p(=-dQ’n), separated by, L (=xn), 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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9. Schottky Capacitance (continued)
The C-V relationship simplifies to
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10. Schottky Capacitance (continued)
If one plots [Cj]-2 vs. Va Slope = -[(Cj0)2Vbi]-1 vertical axis intercept = [Cj0]-2 horizontal axis intercept = fi
Cj-2
Cj0-2 Va fi
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11. Profiling dopants in a Schottky diode
qNd
Q’d = qNdxn
dQ’ x
-Q-dQ xn Ex xn x
-Em
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12. Arbitrary doping profile
If the net donor conc, N = N(x), then at xd, the extra charge put into the DR when Va->Va+dVa is dQ’=-qN(xn)dx
The increase in field, dEx =-(qN/e)dx, by Gauss’ Law (at xn, but also const).
So dVa=-xndEx= (W/e) dQ’
Further, since qN(xn)dx = -dQ’metal, we have the dC/dx as ...
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13. Arbitrary doping profile (cont.)
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14. Ideal metal to n-type Schottky (Va > 0)
qVa = Efn - Efm
Barrier for electrons from sc to m reduced to q(fbi-Va)
qfBn the same
DR decr Eo
qcs
qfm
q(fi-Va)
qfs,n
qfBn Ec
EFm
EFn
EFi Ev
Depl reg
qf’n
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15. Ideal m to n s/c Schottky diode curr
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16. Metal to n-type non-rect cont (fm<fs)
n-type s/c
No disc in Eo
Ex=0 in metal ==> Eo flat
fB,n=fm - cs = elec mtl to s/c barr
fi= fBn-fn< 0
Accumulation region Eo
qcs
qfm
qfs,n
qfi
qfB,n Ec
EFm
EFn
EFi Ev
qfn
Acc reg
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17. Metal to n-type accum region (fm<fs)
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18. Metal to s/c contact resistance
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19. Energy bands for p- and n-type s/c
p-type
n-type Ec Ev Ec
EFi
EFN
qfn= kT ln(Nd/ni)
EFi
qfP= kT ln(ni/Na)
EFP Ev
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20. Making contact in a p-n junction
Equate the EF in the p- and n-type materials far from the junction
Eo(the free level), Ec, Efi and Ev must be continuous
N.B.: qc = 4.05 eV (Si),
and qf = qc + Ec - EF Eo
qc (electron affinity) qf
(work function) Ec EF
EFi
qfF Ev
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21. 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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22. 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
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23. 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
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24. 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]
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25. 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
h 0
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26. 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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27. 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
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28. Soln to Poisson’s Eq in the D.R.Ex
-xp xn x
-xpc
xnc
-Emax
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29. Test 1 - 25Sept03 8 AM Room 206 Activities Building
Open book - 1 legal text or ref., only.
You may write notes in your book.
Calculator allowed
A cover sheet will be included with full instructions. See for examples from Fall 2002.
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