EE 5340
Semiconductor Device Theory
Lecture 8 - Fall 2009
Professor Ronald L. Carter
ronc@uta.edu
http://www.uta.edu/ronc
Test 1 – Sept. 24, 2009
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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.
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2
Si and Al and model
(approx. to scale)
metal
n-type s/c
Eo
Eo
qfm,Al ~
4.1 eV
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EFm Ec
EFi
Ev
p-type s/c
Eo
qcsi~
4.05 eV
qcsi~
4.05 eV
qfs,n
qfs,p
EFn
EFp
Ec
EFi
Ev
3
Making contact between metal & s/c
• Equate the EF in
Eo
the metal and s/c
qc (electron
materials far from
affinity)
the junction
qf
• Eo(the free level),
(work function)
must be continuous
across the jctn.
Ec
E
N.B.: qc = 4.05 eV (Si), F
E
Fi
qf F
and qf = qc + Ec - EF
Ev
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4
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 = qcsemiconductor in all of the s/c.
• Eo - EF,metal = qfmetal throughout metal.
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5
Ideal metal to n-type
barrier diode (fm>fs,Va=0)
metal
n-type s/c
qcs
qfm
qfBn
qfi
qfs,n
EFm
Depl reg
L 08 Sept 17
qf’n
Eo
Ec
EFn
EFi
Ev
No disc in Eo
Ex=0 in metal
==> Eoflat
fBn=fm- cs =
elec mtl to
s/c barr
fi=fBn-fn= fm-
fs elect s/c
to mtl barr
6
Metal to n-type
non-rect cont (fm<fs)
n-type s/c
metal
qcs
qfm
qfB,nqfi
qfs,n
EFm
Acc reg
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qfn
Eo
No disc in Eo
Ex=0 in metal
==> Eo flat
fB,n=fm - cs
= elec mtl to
s/c barr
Ec
EFn f = f -f < 0
i
Bn n
EFi
Ev Accumulation
region
7
Ideal metal to p-type
barrier diode (fm<fs)
p-type s/c
metal
qfm
qfBn
EFm
qfBp
Depl reg
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qf i
qcs
qfs,p
qfi qf <0
p
Eo
Ec
EFi
EFp
Ev
No disc in Eo
Ex=0 in metal
==> Eoflat
fBn= fm- cs =
elec mtl to
s/c barr
fBp= fm- cs +
Eg = hole m
to s
fi = fBp-fs,p =
hole s/c to
mtl barr
8
Metal to p-type
non-rect cont (fm>fs)
metal
n-type s/c
qcs
qfm
qfBn
q(fi)
qfs,n
EFm
qfBp qfi
Accum reg
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qf p
Eo
Ec
EFi
EfP
Ev
No disc in Eo
Ex=0 in metal
==> Eo flat
fB,n=fm- fs,n =
elec mtl to
s/c barr
fBp= fm- cs +
Eg = hole m
to s
Accumulation
region
9
Metal/semiconductor
system types
n-type semiconductor
• Schottky diode - blocking for fm > fs
• contact - conducting for fm < fs
p-type semiconductor
• contact - conducting for fm > fs
• Schottky diode - blocking for fm < fs
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Real Schottky
band structure1
• Barrier transistion
region, d
• Interface states
above fo acc, p neutrl
below fo dnr, n neutrl
Ditd -> oo, qfBn = Eg- fo
Fermi level “pinned”
Ditd -> 0, qfBn = fm - c
Goes to “ideal” case
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11
Fig 8.41 (a) Image charge and electric field
at a metal-dielectric interface (b) Distortion
of potential barrier at E=0 and (c) E0
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12
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:


E


  E  , where,    E  


 x x

 is the permitivity   o r , with
o  8.85E  14, Fd/cm, and
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r  11.7 for silicon
13
Poisson’s Equation
• n = no + dn, and p = po + dp, in non-equil
• For n-type material, N = (Nd - Na) > 0,
no = N, and (Nd-Na+p-n)=-dn +dp +ni2/N
• For p-type material, N = (Nd - Na) < 0,
po = -N, and (Nd-Na+p-n) = dp-dn-ni2/N
2/N
• So neglecting
n
i

q
  E  dp  dn , for n - type and

p - type material with dp or dn  0
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Ideal metal to n-type
barrier diode (fm>fs,Va=0)
metal
0
n-type s/c
xn
xnc
qcs
qfm
qfBn
qfbi
qfs,n
EFm
Depl reg
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qf’n
Eo
Ec
EFn
EFi
Ev
No disc in Eo
Ex=0 in metal
==> Eoflat
fBn=fm- cs =
elec mtl to
s/c barr
fbi=fBn-fn=
fm-fs elect
s/c to mtl
barr
15
Depletion
Approximation
• 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
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Ideal n-type Schottky
depletion width (Va=0)
Ex

qNd
Q’d =
qNdxn
xn
d
x
-Em
xn
x
dEx Em qNd


dx
xn

(Sheet of negative charge on metal)= -Q’d
xn 
2fi qNd  ,
xn
-  Exdx  fi
0
fi  fBn  fn  fm  c s  Vt ln Nc / Nd 
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Debye
length
Nd
0
n
xn
x
• 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 = -df/dx, we have
-d2f/dx2 = q(Nd - n)/ to be solved
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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), 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/kT)f’, f’ << fo
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19
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”
Vt
kT
LD 
, Vt 
, a transition length.
qn + p
q
Note : n + p  Nd for n - type, Na for
p - type and 2ni for intrinsic material.
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20
Debye length
(cont)
• LD estimates the transition length of
a step-junction DR. Thus,
LD Nd 
Vt
d

WVa 0 
2fi
• For Va = 0, fi ~ 1V, Vt ~ 25 mV
d < 11%  DA assumption OK
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21
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
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22
Effect of V  0
The only change now is that
xn
  Exdx  fi  Va , since the field due
0
to Va tends to reduce Ex . Solutions are
2fi  Va 
xn  
, and

 qNd

Emax
2qfi  Va Nd 
 



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23
Ideal metal to n-type
Schottky (Va > 0)
metal
n-type s/c
qcs
qfm
q(fi-Va)
qfBn
qfs,n
EFm
Depl reg
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qf’n
qVa = Efn - Efm
Eo Barrier for
electrons
from sc to
m reduced
Ec
to q(fbi-Va)
EFn
EFi qfBn the same
Ev
DR decr
24
Schottky diode
capacitance

qNd
Q  Q' A, where
Q’d =
qNdxn
-Q-dQ
Ex
-Em
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dQ’
xn
xn
A  jctn. area
x
Q'n  Q'n + dQ'
fi  Va  fi  Va  dV 
x
dEx Em qNd


dx
xn

dQ
dQ
Cj 

 dV dV
25
Schottky Capacitance
(continued)
• The junction has +Q’n=qNdxn (exposed
donors), and Q’n = - Q’metal (Coul/cm2),
forming a parallel sheet charge
capacitor.
Q'n  qNdxn  qNd
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2i  Va 
,
qNd
 Coul 
 2qNd i  Va ,  2 
 cm 
26
Schottky Capacitance
(continued)
• This Q ~ (fi-Va)1/2 is clearly nonlinear, and Q is not zero at Va = 0.
• Redefining the capacitance,
qNd
dQ'n
C'j 

,
dVa
2fi  Va 

A
2
so C'j  , [Fd/cm ], and Cj 
, [Fd]
xn
xn
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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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28
Schottky Capacitance
(continued)
• The C-V relationship simplifies to
1

 2
 Va
Cj  Cj0 1   , a model equation
 fi 
qNd
2
where Cj0  A
, [Fd/cm ]
2fi
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29
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
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fi
Va
30
References
1Device
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 m model.
2Physics
of Semiconductor Devices, by S. M.
Sze, Wiley, New York, 1981.
3Semiconductor Physics & Devices, 2nd ed.,
by Neamen, Irwin, Chicago, 1997.
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