EE 5340
Semiconductor Device Theory
Lecture 11 – Spring 2011
Professor Ronald L. Carter
ronc@uta.edu
http://www.uta.edu/ronc
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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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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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
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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
7
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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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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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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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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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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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
16
Schottky diode
capacitance

qNd
-Q-dQ
Q  Q' A, where
Q’d =
qNdxn
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
17
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
2i  Va 
,
qNd
 Coul 
 2qNd i  Va ,  2 
 cm 
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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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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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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
22
Diagrams for ideal metal-semiconductor Schottky diodes. Fig. 3.21 in Ref 4.
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Energy bands for
p- and n-type s/c
p-type
n-type
Ec
EFi
EFP
Ev
qfn= kT ln(Nd/ni)
qfP= kT ln(ni/Na)
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Ec
EFN
EFi
Ev
24
Making contact
in a p-n junction
• Equate the EF in
Eo
the p- and n-type
qc (electron
materials far from
affinity)
the junction
qf
• Eo(the free level),
(work function)
Ec, Efi and Ev must
be continuous
Ec
E
N.B.: qc = 4.05 eV (Si), F
E
Fi
qf F
and qf = qc  Ec - EF
Ev
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Band diagram for
p+-n jctn* at Va = 0
Ec
EFi
EFP
Ev
qVbi = q(fn
qf p < 0
-xpc
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-xp 0
fp)
qf n > 0
*Na > Nd -> |fp| > fn
p-type for x<0
-
n-type for x>0
xn
Ec
EFN
EFi
Ev
xnc
x
26
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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Depletion
Approximation
• 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
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Depletion approx.
charge distribution

+Qn’=qNdxn
[Coul/cm2]
+qNd
-xp
-xpc
Qp’=-qNaxp
[Coul/cm2]
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xn
xnc
x
-qNa Due to Charge
neutrality
Qp’ + Qn’ = 0,
=> Naxp = Ndxn
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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
 x  xn




  E  dS   dV  A  Exdx   A Qn'  Qp'
 x   xp 
S
V


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 


Induced E-field
in the D.R.
Ex
p-contact
p-type
CNR
- O
+
O
n-type chg
+
O
O
neutral reg
+
- O
O
Depletion
Exposed
Acceptor Ions region (DR)
W
-xpc
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N-contact
-xp 0
xn
Exposed
Donor ions
xnc
x
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Induced E-field
in 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
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Soln to Poisson’s
Eq in the D.R.
Ex
-xp
xn
-xpc
qNa
dEx

dx

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xnc
-Emax
x
dEx qNd

dx

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Soln to Poisson’s
Eq in the D.R. (cont.)
Now, the relationship of Vbi to the
soln of the P.E. in the D.R. is that
x


n
kT  NaNd 
ln
 Vbi    Exdx,
q  n2 
 xp
 i 

dV 

 Ex  

dx 


q
kT
2
2

Ndxn  Na xp , (note
 Vt )
2
q
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Soln to Poisson’s
Eq in the D.R. (cont.)
Ndxn  Na xp , and let W  xn  xp , then
 2Vbi 
NaNd
W 
.
, where Neff 
Na  Nd
 qNeff 
1
Since we must also have Vbi  EmaxW,
2
2qVbiNeff  2Vbi
then Emax  



W


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Comments on the
Ex 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=0dx  Ex(x=0dx)  Emax
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Sample
calculations
• Vt  25.86 mV at 300K
•  = ro = 11.7*8.85E-14 Fd/cm
= 1.035E-12 Fd/cm
• If Na5E17/cm3, and Nd2E15 /cm3,
then for ni1.4E10/cm3, then what is
Vbi = 757 mV
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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
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References
1Device
Electronics for Integrated Circuits, 2 ed., by
Muller and Kamins, Wiley, New York, 1986. See
Semiconductor Device Fundamentals, by Pierret, AddisonWesley, 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.
4Device Electronics for Integrated Circuits, 3/E by
Richard S. Muller and Theodore I. Kamins. © 2003
John Wiley & Sons. Inc., New York.
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References
1 and M&KDevice
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.
3 and **Semiconductor Physics & Devices, 2nd ed.,
by Neamen, Irwin, Chicago, 1997.
Fundamentals of Semiconductor Theory and
Device Physics, by Shyh Wang, Prentice Hall,
1989.
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