Semiconductor physics
Winter term 2012/2013
Prof. Dr. M. S. Brandt
Exercise 13, to be discussed in the week 04.02.-08.02.2013
Exercise 1: p-Si MOS
One can look at a metal-oxide-semiconductor junction as a plate capacitor, where one of the
plates is a semiconductor. The gate voltage, which is applied to the metal, drops not only
across the oxide, but also across the depletion region of the semiconductor VG = VOx + ΨS
(here it is assumed that the work functions for the metal and the semiconductor are equal).
The capacitance of the whole system
Ctot = COx /(1 + COx /CS )
is in first approximation defined by the width of the depletion region
r
2s 0 Ψs
,
Wd =
eNa
(1)
(2)
where S is the dielectric constant of the semiconductor, ΨS is the band bending, e is the
elementary charge and Na is the dopant concentration. Furthermore the potential drop across
the oxide can be estimated by VOx = S 0 ES /COx , where the electric field
ES = ±
1/2
−eΨ
S
eΨ
2kB T
S
−1
pp0 e kB T +
S 0
kB T
(3)
for the case of a p-type semiconductor. The + sign applies to ΨS > 0 and the - sign to ΨS < 0.
We furthermore assume that there are no trapped charges in the oxide or at the interface.
Assume a MOS capacitor on p-Si at 300 K with the following parameters:
Na = 1016 cm−3 , WOx = 30 nm, Ox = 2.4.
For equally spaced values of ΨS in the range - 0.2 V to 5.0 V, calculate the gate voltage VG ,
the oxide capacitance COx and the total capacitance Ctot . Make a plot of C/COx versus VG .
Figure 1: A MOS junction with the definition of the important quantities
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We assume that at 300 K pp0 = Na
S
S 0 ES
= ΨS +
WOx ES
VG = VOx + ΨS = ΨS +
COx
Ox
1/2
S
2kB T
−βeΨS
= ΨS +
WOx ±
Na (e
+ βeΨS − 1)
Ox
0 S
1/2
−38.7
38.7
−1
= ΨS ± 1.3186 · 10 V exp(
ΨS ) +
ΨS − 1
V
V
(4)
(5)
(6)
The oxide capacitance is constant over the whole gate voltage range:
COx = Ox 0 /WOx
But as the width of the depletion region changes with ΨS
r
2S 0 ΨS
Wd =
eNa
(7)
(8)
also the capacitance of the depletion layer changes:
Ctot =
COx
Ox 0 /WOx
=
Ox Wd
1 + COx /CS
1+ W
Ox S
Ox 0 /WOx
q
(10)
7.0832 · 10−4 mF2
p
1 + 1.0949 ΨS /V
(11)
=
1+
=
(9)
Ox
WOx S
2
2S 0 ΨS
eNa
Figure 2: Total capacitance of a MOS diode against applied gate voltage only considering the
variable depletion width
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Exercise 2: Schottky Diode
A Schottky diode is composed of a semiconductor in contact with a metal. The exact barrier
height can vary depending on the surface or interface condition and additional (deep) defect
states. A capacitance voltage (CV) measurement can give important information about a
semiconductor-metal junction.
The depletion-layer capacitance C per unit area [F / cm2 ]−2 can be calculated by
2(Vbi − Va )
1
=
2
C
qs 0 ND
(12)
where Vbi is the built-in potential of the junction, Va is the applied bias, S is the semiconductor
dielectric constant and ND is the dopant concentration.
In a measurement of a device with an area of 10−1 cm2 , the capacitance against Va is found to
be:
1
5
5 Va
(13)
=
1.57
· 10 − 2.12 · 10 ·
(C/µF)2
V
Calculate the built-in potential, the barrier height and the dopant concentration.
Take care that often, as here, the capacitance is considered in F/cm2 . But in a measurement
of a certain device the capacitance is measured in F. So first the unit of C needs to be changed
from µF to F/cm2 , so
1/C 2 = 1.57 · 1015 − 2.12 · 1015 Va (cm2 /F)2 .
(14)
We can obtain the built-in potential at 1/C 2 = 0
1.57 · 1015
= 0.74V
Vbi =
2.12 · 1015
(15)
Differentiating Eq. 12 and solving for ND :
2
−1
ND =
q0 S d(1/C 2 )/dVa
(16)
With
d (1/C 2 )
= −2.12 · 1015 (cm2 /F)2 /V
dVa
(17)
we find
ND =
2
1.6 · 10−19 · 11.9 · 8.85 · 10−14
·
1
2.12 · 1015
= 5.6 · 1015 cm−3
(18)
(19)
The potential distance between the Fermi level in the semiconductor and the conduction band
is eVn
Vn =
kB T NC
ln
e
ND
where NC = 2.82 · 1019 /cm−3 is the effective density of states.
2.82 · 1019
Vn = 0.0259 · ln
= 0.221V
5.6 · 1015
(20)
(21)
With that also the Schottky barrier height can be calculated
ΦBn = Vbi + Vn = 0.74 + 0.221 = 0.961V
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(22)
As an answer to a question concerning Eq. 20:
The formula EF = 0.5(ED + EC ) + 0.5kT ln(0.5 ∗ ND /NC ) is no use for us here, because ED is
unknown. Eq. 20 is false for T going to zero, as we replaced nn0 by ND , which is only valid at
sufficiently high temperatures. For the derivation of the formula, we refer to the first few lines
of the solution of sheet number 12.
Name
e-mail
Tel.Nr.
Office
Prof. Dr. M. S. Brandt MartinS.Brandt@wsi.tum.de 289-12758 S301
Patrick Altmann
patrick.altmann@wsi.tum.de 289-11527 S310
Benedikt Stoib
benedikt.stoib@wsi.tum.de 289-11525 S304
Konrad Klein
konrad.klein@wsi.tum.de
289-11565 S310
Florian Hrubesch
florian.hrubesch@wsi.tum.de 289-11380 C201
The exercises take place on Tue 12:20 - 13:30 (WSI S101), Wed 15:00-17:00 (ZNN 0th floor
seminar room), Thu 08:30-10:00 (WSI S101) and Thu 15:45-17:15 (WSI S101).
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