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EE 5340 Semiconductor Device Theory Lecture 8 - Fall 2009. Professor Ronald L. Carter [email protected] http://www.uta.edu/ronc. Test 1 – Sept. 24, 2009. 8 AM Room 108 Nedderman Hall Open book - 1 legal text or ref., only. You may write notes in your book. Calculator allowed - PowerPoint PPT Presentation
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EE 5340Semiconductor Device TheoryLecture 8 - Fall 2009
Professor Ronald L. [email protected]
http://www.uta.edu/ronc
L 08 Sept 18 2
Test 1 – Sept. 24, 2009• 8 AM Room 108 Nedderman Hall• 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 http://www.uta.edu/ronc/5340/tests/ for examples from previous semesters.
L 08 Sept 17 3
Si and Al and model (approx. to scale)
qm,Al ~ 4.1 eV
Eo
EF
mEFp
EFn
Eo
Ec
Ev
EFi
qs,n
qsi~ 4.05 eV
Eo
Ec
Ev
EFi
qs,p
metal n-type s/c p-type s/c
qsi~ 4.05 eV
L 08 Sept 17 4
Making contact be-tween metal & s/c• Equate the EF in the
metal and s/c materials far from the junction
• Eo(the free level), must be continuous across the jctn.
N.B.: q = 4.05 eV (Si),and q = qEc - EF
Eo
EcEF EFiEv
q (electron affinity)
qF
q(work function)
L 08 Sept 17 5
Equilibrium Boundary Conditions w/ contact• No discontinuity in the free level, Eo at
the metal/semiconductor interface.• EF,metal = EF,semiconductor to bring the
electron populations in the metal and semiconductor to thermal equilibrium.
• Eo - EC = qsemiconductor in all of the s/c.• Eo - EF,metal = qmetal throughout metal.
L 08 Sept 17 6
Ideal metal to n-typebarrier diode (m>s,Va=0)
EFn
Eo
Ec
Ev
EFi
qs,n
qs
n-type s/c
qm
EF
m
metal
qBnqi
q’n
No disc in Eo
Ex=0 in metal ==> Eoflat
Bn=m- s = elec mtl to s/c barr
i=Bn-n= m-s elect s/c to mtl barr Depl reg
L 08 Sept 17 7
Metal to n-typenon-rect cont (m<s)
EFn
Eo
Ec
Ev
EFi
qs,n
qs
n-type s/c
qm
EF
m
metal
qB,n
qn
No disc in Eo
Ex=0 in metal ==> Eo flat
B,n=m - s = elec mtl to s/c barr
i= Bn-n< 0
Accumulation region
Acc reg
qi
L 08 Sept 17 8
Ideal metal to p-typebarrier diode (m<s)
EFp
Eo
Ec
Ev
EFi
qs,p
qs
p-type s/c
qm
EF
m
metal
qBn
qi
qp<0
No disc in Eo
Ex=0 in metal ==> Eoflat
Bn= m- s = elec mtl to s/c barr
Bp= m- s + Eg = hole m to si = Bp-s,p =
hole s/c to mtl barr
Depl regqBp
qi
L 08 Sept 17 9
Metal to p-typenon-rect cont (m>s)
No disc in Eo
Ex=0 in metal ==> Eo flat
B,n=m- s,n = elec mtl to s/c barr
Bp= m- s + Eg = hole m to s
Accumulation region
EFi
Eo
Ec
Ev
EfP
qs,n
qs
n-type s/c
qm
EF
m
metal
qBnq(i)
qpAccum reg
qBpqi
L 08 Sept 17 10
Metal/semiconductorsystem typesn-type semiconductor• Schottky diode - blocking for m > s
• contact - conducting for m < s
p-type semiconductor• contact - conducting for m > s
• Schottky diode - blocking for m < s
L 08 Sept 17 11
Real Schottkyband structure1
• Barrier transistion region,
• Interface statesabove o acc, p neutrlbelow o dnr, n neutrl
Dit -> oo, qBn= Eg- oFermi level “pinned”
Dit -> 0, qBn= m - Goes to “ideal” case
L 08 Sept 17 12
Fig 8.41 (a) Image charge and electric field at a metal-dielectric interface (b) Distortion of potential barrier at E=0 and (c) E0
L 08 Sept 17 13silicon for 711
andFd/cm, ,14E858with , ypermitivit the is
xEE where, ,E
r
o
ro
x
..
Poisson’s Equation• The electric field at (x,y,z) is
related to the charge density =q(Nd-Na-p-n) by the Poisson Equation:
L 08 Sept 17 14
Poisson’s Equation• n = no + n, and p = po + p, in non-
equil• For n-type material, N = (Nd - Na) > 0,
no = N, and (Nd-Na+p-n)=-n +p +ni2/N
• For p-type material, N = (Nd - Na) < 0, po = -N, and (Nd-Na+p-n) = p-n-ni
2/N• So neglecting ni
2/N
0n or p with material type-pand type-n for ,npqE
L 08 Sept 17 15
Ideal metal to n-typebarrier diode (m>s,Va=0)
EFn
Eo
Ec
Ev
EFi
qs,n
qs
n-type s/c
qm
EF
m
metal
qBnqbi
q’n
No disc in Eo
Ex=0 in metal ==> Eoflat
Bn=m- s = elec mtl to s/c barr
bi=Bn-n= m-s elect s/c to mtl barr
Depl reg
0 xn xnc
L 08 Sept 17 16
DepletionApproximation• For 0 < x < xn, assume n << no = Nd,
so = q(Nd-Na+p-n) = qNd
• For xn < x < xnc, assume n = no = Nd, so = q(Nd-Na+p-n) = 0
• For x = 0-, there is a pulse of charge balancing the qNdxn in 0 < x < xn
L 08 Sept 17 17
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
L 08 Sept 17 18
Debye length
• The DA assumes n changes from Nd to 0 discontinuously at xn.
• In the region of xn, Poisson’s eq is E = / --> dEx/dx = q(Nd -
n), and since Ex = -d/dx, we have-d2/dx2 = q(Nd - n)/ to be solved
n
xxn
Nd
0
L 08 Sept 17 19
Debye length (cont)• Since the level EFi is a reference for equil, we
set = Vt ln(n/ni)• In the region of xn, n = ni exp(/Vt), so
d2/dx2 = -q(Nd - ni e/Vt), let = o + ’, where o = Vt ln(Nd/ni) so Nd - ni e/Vt = Nd[1 - e/Vt-o/Vt], for - o = ’ << o, the DE becomes d2’/dx2 = (q2Nd/kT)’, ’ << o
L 08 Sept 17 20
Debye length (cont)• So ’ = ’(xn) exp[+(x-xn)/LD]+con. and n
= Nd e’/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
L 08 Sept 17 21
Debye length (cont)• LD estimates the transition length
of a step-junction DR. Thus, i
t
0V
dD2V
WNLa
• For Va = 0, i ~ 1V, Vt ~ 25 mV < 11% DA
assumption OK
L 08 Sept 17 22
Effect of V 0• Define an external voltage source, Va,
with the +term at the metal contact and the -term at the n-type contact
• For Va > 0, the Va induced field tends to oppose Ex caused by the DR
• For Va < 0, the Va induced field tends to aid Ex due to DR
• Will consider Va < 0 now
L 08 Sept 17 23
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
L 08 Sept 17 24
Ideal metal to n-typeSchottky (Va > 0)
qVa = Efn - Efm
Barrier for electrons from sc to m reduced to q(bi-Va)
qBn the sameDR decr
EFn
Eo
Ec
Ev
EFi
qs,n
qs
n-type s/c
qm
EF
m
metal
qBnq(i-Va)
q’nDepl reg
L 08 Sept 18 25
Schottky diodecapacitance
xn
x
qNd
-Q-Q
Q’d = qNdxn
x
Ex
-Em
dnmx qN
xE
dxdE
xn
Q’
VQ
VQC
VVVQQQ
area jctn.A where AQQ
j
aiai
nn
'''
,'
L 08 Sept 18 26
Schottky Capacitance(continued)• The junction has +Q’n=qNdxn (exposed
donors), and Q’n = - Q’metal (Coul/cm2), forming a parallel sheet charge capacitor.
2aid
d
aidndn
cmCoul VqN2
qNV2qNxqNQ
,
,'
L 08 Sept 18 27
Schottky Capacitance(continued)• This Q ~ (i-Va)1/2 is clearly non-
linear, and Q is not zero at Va = 0.• Redefining the capacitance,
[Fd] xAC and ][Fd/cm xC so
V2qN
dVdQC
nj
2
nj
aid
an
j
,,,'
,''
L 08 Sept 18 28
Schottky Capacitance(continued)• So this definition of the capacitance
gives a parallel plate capacitor with charges Q’n and Q’p(=-Q’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.
L 08 Sept 18 29
Schottky Capacitance(continued)• The C-V relationship simplifies to
][Fd/cm 2qNAC herew
equation model a V1CC
2
id
0j
21
ia
0jj
,
,
L 08 Sept 18 30
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
L 08 Sept 17 31
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.