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L10 22Sep103 Ideal n-type Schottky depletion width (V a =0) xnxn x qN d Q’ d = qN d x n x ExEx -E m xnxn (Sheet of negative charge on metal)= -Q’ d
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EE 5340Semiconductor Device TheoryLecture 10 – Fall 2010
Professor Ronald L. [email protected]
http://www.uta.edu/ronc
L10 22Sep10 2
Test 1 – W 29Sep10• 11 AM Room 108 Nedderman Hall• Covering Lectures 1 through 10• 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. For examples see http://www.uta.edu/ronc/5340/tests/.
L10 22Sep10 3
Ideal n-type Schottky depletion width (Va=0)
xn
x
qNd
Q’d = qNdxn
x Ex
-Em
dnmx qN
xE
dxdE
xn
(Sheet of negative charge on metal)= -Q’d
dctsmnBni
ix
0xdin
NNV
dxE- , qN2x n
/ln
L10 22Sep10 4
Debye length
n
xxn
Nd
0
material. intrinsic for 2npn and type,-p extrinsic for Npn
type,-n extrinsic for N pn :Note
length. transition a , ,
i
a
d
q
kTVpnq
VL tt
D
L10 22Sep10 5
dimax
d
in
xa
aix
0x
NVa2qE
and ,qNVa2x
are Solutions .E reduce to tends V to
due field the since ,VdxE
that is now change only Then
Effect of V 0
L10 22Sep10 6
Schottky diodecapacitance
xn
x
qNd
-Q-Q
Q’d =
qNdxn
x
Ex
-Em
dn
mx qNxE
dxdE
xn
Q’
2aid
d
aidndn
cmCoul VqN2
qNV2qNxqNQ
,
,'
[Fd] xAC and ][Fd/cm xC so
V2qN
dVdQC
nj
2
nj
aid
an
j
,,,'
,''
L10 22Sep10 7
Schottky Capacitance(continued)• If one plots [Cj]-2 vs. Va Slope = -
[(Cj0)2Vbi]-1 vertical axis intercept = [Cj0]-2 horizontal axis intercept = i
Cj-2
iVa
Cj0-2
Diagrams for ideal metal-semiconductor Schottky diodes. Fig. 3.21 in Ref 4.L10 22Sep10 8
L10 22Sep10
Energy bands forp- and n-type s/c
p-typeEc
Ev
EFi
EFP
qP= kT ln(ni/Na)
Ev
Ec
EFi
EFNqn= kT ln(Nd/ni)
n-type
9
L10 22Sep10
Making contactin 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.: q = 4.05 eV (Si),and q = qEc - EF
Eo
EcEF EFiEv
q (electron affinity)
qF
q(work function)
10
L10 22Sep10
Band diagram forp+-n jctn* at Va = 0
EcEFNEFi
Ev
Ec
EFP
EFi
Ev
0 xnx
-xp-xpc xnc
qp < 0
qn > 0
qVbi = q(n - p)
*Na > Nd -> |p| > n
p-type for x<0 n-type for x>0
11
L10 22Sep10
• A total band bending of qVbi = q(n-p) = kT ln(NdNa/ni
2) is necessary to set EFp = Efn
• For -xp < x < 0, Efi - EFP < -qp, = |qp| so p < Na = po, (depleted of maj. carr.)
• For 0 < x < xn, EFN - EFi < qn, so n < Nd = no, (depleted of maj. carr.)
-xp < x < xn is the Depletion Region
Band diagram forp+-n at Va=0 (cont.)
12
L10 22Sep10
DepletionApproximation• Assume p << po = Na for -xp < x < 0, so
= q(Nd-Na+p-n) = -qNa, -xp < x < 0, and p = po = Na for -xpc < x < -xp, so = q(Nd-Na+p-n) = 0, -xpc < x < -xp
• Assume n << no = Nd for 0 < x < xn, so = q(Nd-Na+p-n) = qNd, 0 < x < xn, and n = no = Nd for xn < x < xnc, so = q(Nd-Na+p-n) = 0, xn < x < xnc
13
L10 22Sep10
Depletion approx.charge distribution
xnx
-xp
-xpc xnc
+qNd
-qNa
+Qn’=qNdxn
Qp’=-qNaxp
Due to Charge
neutrality Qp’ + Qn’ = 0, => Naxp =
Ndxn
[Coul/cm2]
[Coul/cm2]14
L10 22Sep10
Induced E-fieldin 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 QQAdxEAdVdSE 'p'n
xx
xxx
VS
n
p
≈0
15
L10 22Sep10
Induced E-fieldin the D.R.
xnx
-xp-xpc xnc
O-O-O-
O+O+
O+
Depletion region (DR)
p-type CNR
Ex
Exposed Donor ions
Exposed Acceptor Ions
n-type chg neutral reg
p-contact N-contact
W
016
L10 22Sep10
Induced E-fieldin the D.R. (cont.)• Poisson’s Equation E = /, has the
one-dimensional form, dEx/dx = /, which must be satisfied for = -qNa, -xp < x < 0, and = +qNd, 0 < x < xn, with Ex = 0 for the remaining range
17
L10 22Sep10
Soln to Poisson’sEq in the D.R.
xn x-xp
-xpc xnc
Ex
-Emax
dx qN
dxdE
ax qN
dxdE
18
L10 22Sep10
Soln to Poisson’sEq in the D.R. (cont.)
)VqkT (note ,xNxN2
qdxdVE ,dxEV
nNNlnq
kTthat is D.R. the in P.E. the of solnthe to V of iprelationsh the Now,
t2pa2nd
xx
xxbi2
ida
bi
n
p
19
L10 22Sep10
Soln to Poisson’sEq in the D.R. (cont.)
WV2N2qVE then
,WE21V have also must we Since
.NNNNN where ,qN
V2W
then ,xxW let and ,xNxN
bieffbimax
maxbi
dada
effeffbi
pnpand
20
L10 22Sep10
Comments on theEx 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=0xEx(x=0x) Emax
21
L10 22Sep10
Sample calculations• Vt 25.86 mV at 300K• = ro = 11.7*8.85E-14 Fd/cm
= 1.035E-12 Fd/cm• If Na5E17/cm3, and Nd2E15
/cm3, then for ni1.4E10/cm3, then what is Vbi = 757 mV
22
L10 22Sep10
Sample calculations• What are Neff, W ?
Neff, = 1.97E15/cm3 W = 0.707 micron
• What is xn ?
= 0.704 micron• What is Emax ? 2.14E4 V/cm
23
L10 22Sep10 24
References1Device Electronics for Integrated Circuits, 2 ed., by
Muller and Kamins, Wiley, New York, 1986. See Semiconductor Device Fundamentals, by Pierret, Addison-Wesley, 1996, for another treatment of the model.
2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981.
3Semiconductor Physics & Devices, 2nd ed., by Neamen, Irwin, Chicago, 1997.
4Device Electronics for Integrated Circuits, 3/E by Richard S. Muller and Theodore I. Kamins. © 2003 John Wiley & Sons. Inc., New York.