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Semiconductor Device Modeling and Characterization – EE5342 Lecture 09– Spring 2011. Professor Ronald L. Carter [email protected] http://www.uta.edu/ronc/. First Assignment. e-mail to [email protected] In the body of the message include subscribe EE5342 - PowerPoint PPT Presentation
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Semiconductor Device Modeling and
Characterization – EE5342 Lecture 09– Spring 2011
Professor Ronald L. [email protected]
http://www.uta.edu/ronc/
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First Assignment
• e-mail to [email protected]– In the body of the message include
subscribe EE5342 • This will subscribe you to the
EE5342 list. Will receive all EE5342 messages
• If you have any questions, send to [email protected], with EE5342 in subject line.
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Second Assignment
• Submit a signed copy of the document that is posted at
www.uta.edu/ee/COE%20Ethics%20Statement%20Fall%2007.pdf
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Additional University Closure Means More Schedule
Changes• Plan to meet until noon some days in the next few weeks. This way we will make up for the lost time. The first extended class will be Monday, 2/14.
• The MT changed to Friday 2/18• The P1 test changed to Friday 3/11.• The P2 test is still Wednesday 4/13• The Final is still Wednesday 5/11.
MT and P1 Assignment on Friday, 2/18/11
• Quizzes and tests are open book – must have a legally obtained copy-no
Xerox copies.– OR one handwritten page of notes.– Calculator allowed.
• A cover sheet will be published by Wednesday, 2/16/11.
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Energy bands forp- and n-type s/c
p-typeEc
Ev
EFi
EFpqfp= kT ln(ni/Na)
Ev
Ec
EFi
EFnqfn= kT ln(Nd/ni)
n-type
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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.: qc = 4.05 eV (Si),
and qf = qc + Ec - EF
Eo
EcEf EfiEv
qc (electron affinity)
qfF
qf(work function)
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Band diagram forp+-n jctn* at Va = 0
EcEfNEfi
Ev
Ec
EfP
Efi
Ev
0 xnx
-xp-xpc xnc
qfp < 0
qfn > 0
qVbi = q(fn - fp)
*Na > Nd -> |fp| > fn
p-type for x<0 n-type for x>0
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• A total band bending of qVbi = q(fn-fp) = kT ln(NdNa/ni
2)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
Band diagram forp+-n at Va=0 (cont.)
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DepletionApproximation• 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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Poisson’sEquation• The electric field at (x,y,z) is
related to the charge density r=q(Nd-Na-p-n) by the Poisson Equation:
silicon for 7.11andFd/cm, ,14E85.8
with , ypermitivit the is xEE where, ,E
r
o
ro
x
r
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Poisson’sEquation• For n-type material, N = (Nd - Na) >
0, no = N, and (Nd-Na+p-n)=-dn +dp +ni
2/N• For p-type material, N = (Nd - Na) <
0, po = -N, and (Nd-Na+p-n) = dp-dn-ni
2/N• So neglecting ni
2/N, [r=(Nd-Na+p-n)]
carriers. excess with material type-pand type-n for ,npqE dd
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Quasi-FermiEnergy
used. be must levelFermi-quasi the then ,nnn i.e.,
m,equilibriu not in ionconcentrat the IfkT
EEexpnn and , n
nlnkTEE
:by given are level Energy Fermi the andconc carrier mequilibriu the m,equilibriu In
o
fifio
io
fif
d+
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Quasi-FermiEnergy (cont.)
+
+
kTEE
nnn
nnnkTEE
fifn
i
o
i
ofifn
exp
:is density carrier the and
, ln
:defined is (Imref) level Fermi-Quasi The
d
d
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Quasi-FermiEnergy (cont.)
d+
d+
kTEE
npp
nppkTEE
fpfi
i
o
i
ofpfi
exp
:is density carrier the and
, ln
:as defined is (Imref) level Fermi-Quasi the holes, For
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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+
r
h 0
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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
W0
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Depletion approx.charge distribution
xnx
-xp
-xpc xnc
r+qNd
-qNa
+Qn’=qNdxn
Qp’=-qNaxp
Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn
[Coul/cm2]
[Coul/cm2]
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1-dim soln. ofGauss’ law
nx
nnax
ppax
px
ndpada
daeff
npeff
bi
xx ,0E ,xx0 ,xxNq E
,0xx ,xxNq- E
xx ,0E
,xNxN ,NNNNN
,xxW ,qNVaV2W
+
+
+
xxn xn
c
-xpc-xp
Ex
-Emax
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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 = {2(Vbi-Va)/qNeff}1/2, where Vbi = Vt ln{NaNd/ni
2}, and Neff=NaNd/(Na+Nd). Since Naxp=Ndxn, xn = W/(1 + Nd/Na), and xp = W/(1 + Na/Nd).
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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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JunctionC (cont.)
xnx-xp
-xpc xnc
r+qNd
-qNa
+Qn’=qNdxn
Qp’=-qNaxp
Charge neutrality => Qp’ + Qn’ = 0,
=> Naxp = Ndxn
dQn’=qNddxn
dQp’=-qNadxp
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JunctionC (cont.)• The C-V relationship simplifies to
][Fd/cm ,NNV2NqN'C herew
equation model a ,VV1'C'C
2dabi
da0j
21
bia0jj
+
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JunctionC (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
VbiVa
C’j0-2
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Arbitrary dopingprofile• If the net donor conc, N = N(x),
then at xn, the extra charge put into the DR when Va->Va+dVa is dQ’=-qN(xn)dxn
• The increase in field, dEx =-(qN/)dxn, by Gauss’ Law (at xn, but also const).
• So dVa=-(xn+xp)dEx= (W/) dQ’• Further, since N(xn)dxn = N(xp)dxp
gives, the dC/dxn as ...
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Arbitrary dopingprofile (cont.)
+
+
+
pn
j
3j
j
j
n
j
nd
ndj
pn
2j
np
2n
j
xNxN1
dV'dC
q
'C'Cd
Vdq'C
xd'Cd
N with
, dV'Cd
dC'xdqNdV
xdqNdVdQ''C further
,xNxN1'C
dxdx1
Wdx'dC
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Arbitrary dopingprofile (cont.)
,VV2qN'C where , junctionstep
sided-one to apply Now .dV
'dCq
'C xN
profile doping the ,xN xN orF
abij
3j
n
pn
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Arbitrary dopingprofile (cont.)
bi0j
bi23
bia0j
23
bia30j
V2qN'C when ,N
V1
VV12
1'qC
VV1'C
N so
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Arbitrary dopingprofile (cont.)
)( and ,1
2
and
when area),(A and V, , ' ,quantities measured of in terms So,
22
0
VCxN
dVC
dqA
NxNxNN
CAC
jnd
j
rapnd
jj
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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, the 1-dim Poisson equation is dEx/dx = q(Nd - n), and since Ex = -df/dx, the potential is the solution to -d2f/dx2
= q(Nd - n)/
n
xxn
Nd
0
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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), letf = 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/kT)f’, f’ << fo
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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”
material. intrinsic for 2n and type-p for N type,-n for N pn :Note
length. transition a ,qkTV ,pnq
VL
iad
ttD
+
+
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Debye length (cont)• LD estimates the transition length
of a step-junction DR (concentrations Na and Nd with Neff = NaNd/(Na +Nd)). Thus,
biefft
da0VdDaD
V2NV
N1
N1
WNLNL
a
+
+d
• For Va=0, & 1E13 < Na,Nd < 1E19 cm-3
13% < d < 28% => DA is OK
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Example
• An assymetrical p+ n junction has a lightly doped concentration of 1E16 and with p+ = 1E18. What is W(V=0)?
Vbi=0.816 V, Neff=9.9E15, W=0.33mm
• What is C’j? = 31.9 nFd/cm2
• What is LD? = 0.04 mm
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References *Fundamentals of Semiconductor Theory and
Device Physics, by Shyh Wang, Prentice Hall, 1989.
**Semiconductor Physics & Devices, by Donald A. Neamen, 2nd ed., Irwin, Chicago.
M&K = Device Electronics for Integrated Circuits, 3rd ed., by Richard S. Muller, Theodore I. Kamins, and Mansun Chan, John Wiley and Sons, New York, 2003.
• 1Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986.
• 2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981.
• 3 Physics of Semiconductor Devices, Shur, Prentice-Hall, 1990.
