UNIVERSITY OF CALIFORNIA
College of Engineering
Department of Electrical Engineering and Computer Sciences
EE 130/230M
Spring 2013
Prof. King & Dr. Xu
Solution to Homework Assignment #4
Problem 1: Metal-semiconductor energy-band diagrams
From Lecture 7 Slide 12, one can find the Schottky barrier heights for NiSi contacting Si (ΦBn=0.65eV
and ΦBp=0.47eV). EcEF in the charge-neutral (“flat-band”) region of the semiconductor (away from the
depleted region) is determined by using the equation EcEF=kT ln(Nc/ND) for n-type Si and EFEv =kT
ln(Nv/NA) for p-type Si. The built-in potential is qVbi= ΦBn  (EcEF) for n-type Si and qVbi = ΦBp  (EFEv)
for p-type material. The width of the depleted region is
2es (Vbi -VA )
W=
qN
a) Current is conducted by majority carriers in a Schottky contact. The Schottky barrier height for electrons
(ΦBn=0.65eV) limits electron flow from the metal into n-type Si.
EcEF =0.026∙ln(3.2x1019/1017)=0.15eV⇒ qVbi = 0.65-0.15=0.5eV
W=
2 ´12 ´ 8.85´10-14 (0.5)
= 8.14 10-6 cm= 0.0814μm.
-19
17
1.6 ´10 ´10
b) ΦBn, EcEF and Vbi are unchanged from (a) since the contacting materials are the same. For an applied
voltage equal to -1V the Schottky junction is reverse biased, so the barrier for electron flow from the n-type
Si into the metal is increased. (The barrier for electron flow from the metal into the Si is unchanged.) The
depletion width is increased (in order to drop a larger voltage across the depleted region):
W=
2 ´12 ´ 8.85´10-14 (0.5 +1)
= 1.4  10-5 cm= 0.14μm.
-19
17
1.6 ´10 ´10
c) ΦBn, EcEF and Vbi are unchanged from (a) since the contacting materials are the same. For an applied
voltage equal to +0.2V the Schottky junction is forward biased, so the barrier height for electron flow from
the n-type Si into the metal is decreased. (The barrier for electron flow from the metal into the Si is
unchanged.) The depletion width is decreased (in order to drop a smaller voltage across the depleted
region):
W=
2 ´12 ´ 8.85´10-14 (0.5 - 0.2)
= 6.3  10-6 cm= 0.063μm.
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17
1.6 ´10 ´10
d) . The Schottky barrier height for holes (ΦBp=0.47eV) limits hole flow from the metal into p-type Si.
EFEv =0.026·ln(1.8x1019/1017)=0.135eV⇒ qVbi = 0.47-0.135=0.335eV
W=
2 ´12 ´ 8.85´10-14 (0.33)
= 6.6 10-6 cm= 0.066μm.
-19
17
1.6 ´10 ´10
The energy-band diagrams are drawn below:
Problem 2: Metal-semiconductor electrostatics
a) If ND(x)=ax for x>0, the depletion approximation leads to ρ(x)=qax for 0<x<W
Using Poisson’s equation, one can find the electric field distribution:
d

qa


x for 0<x<W
dx K s 0 K s 0
At the edge of the depleted region (x=W), the electric field is 0.
0
W
qa
qa
d


xd x   ( x )  
(W 2  x 2 ) 0<x<W


K s 0 x
2 K s 0
 ( x)
The electrostatic potential is obtained by integrating the electric field:
0
qa W 2 2
qa
ò dV = 2K e ò (w - x )dx Þ V (x) = 6K e (2W 3 - 3W 2 x + x3 ) 0<x<W
s 0 x
s 0
V ( x)
Since V=-(Vbi  VA) at x=0,
-(Vbi -VA) = -
qa
W3
3K se 0
Solving for the width of the depleted region:
é 3K e
ù
W = ê s 0 (Vbi -VA )ú
ë qa
û
Vbi is equal to the difference in work function, between the metal and the quasi-neutral semiconductor:
1
Vbi =  Bn  ( Ec  E F ) xw 
q
1/3
(EcEF)x=W = (EG/2) kT·ln(ND/ni)  ( EG / 2)  kT ln( aW / ni )
Substituting Equation (2) into Equation (1), we can obtain an equation for Vbi:
(1)
(2)
1é
E
aW ù
Vbi = êF B - ( G - kT ln(
)ú
që
2
ni û
Note that Vbi must be found by iteration, since W is a function of Vbi
b) The small-signal capacitance of this M-S diode is
CJ =
dQdep K se 0 A
K se0 A
=
=
1/3
dVA
W
é 3K se 0
ù
(Vbi -VA )ú
ê
ë qa
û
Problem 3: Practical ohmic contact
For significant tunneling to occur, the barrier to tunneling conduction must be very narrow, i.e. the width of
the depleted region in the Si must be very small. This means that the doping concentration in the Si must be
very high. We can approximate the Fermi level to be located at the conduction band edge, for such heavily
doped (degenerate) Si. At equilibrium, the built-in potential would be equal to the Schottky barrier height
(see the band diagram below). We can thus determine the doping concentration required, by using the
equation for depletion width:
1/ 2
 2K 

W   s 0  Bn 
 qN D


 2K 
  2  12  8.85  1014
 N D   s 20  Bn   
 (0.65)  8.6  1018 cm - 3
19
7 2
 qW
 1.6  10 (10  10 )

Problem 4: Small-signal capacitance of a Schottky contact
The small-signal capacitance of a Schottky contact to uniformly doped n-type Si is
dQdep K s 0 A
K s 0 A
CJ 


1/ 2
dVA
W
 2 K s 0

 qN Vbi  VA 
D

 .
 2V  VA  
1
  2 bi
2
C J  A qN D K s 0 
a) The x-intercept is the built-in potential Vbi= ΦBn(EcEF). Since ΦBn=ΦM-χ, a metal with a larger work
function will result in larger Vbi. From the plot, Vbi for metal 2 is ~1V while Vbi for metal 1 is ~0.4V. Since
Vbi for Metal 2 is larger, Metal 2 has the larger work function.

b) The slope of the plot is inversely proportional to the dopant concentration, i.e. lower ND corresponds to a
steeper slope. Thus, the contact with Metal 1 corresponds to more lightly doped silicon.