Solution for FYS4310 Test May 11, 2005
Duration: Two hours; Do as much you can.
(Open books test.)
All help items are allowed except direct communication between the students.
Deliver the answers on numbered sheets of paper with your name.
1)
We will make a simple ohmic contact for p-type Si for laboratory use, that is, we want to
perform electrical measurements, which require an ohmic contact and where the
influence of this upon the measurement result is minimal. We assume the influence is at
a minimum when the contact resistance is the smallest. We have the choice of
evaporating Al or Pt. Which metal would give the least contact resistance? Give a short
explanation.
Solution:
We look up the barrier heights for Al and for Pt. Figure 15.20 in the textbook gives the
barrier height to n-type Si. We read : Bn(Pt) = 0.85 eV and Bn(Al) = 0.6 eV
We have Bn +Bp =Eg, so that Bp(Pt) = 0.3 eV, Bp (Al) = 0.55 eV.
The contact resistance increases with the barrier height, so that the contact resistance for
the Pt contact will have the least value.
2)
We want to make a thin film of gold. It should have a certain thickness for our purpose
(For example a semitransparent electrode for the measurement of electroluminescence).
The thickness monitor is broken, so we have to do without. We have a gold wire with a
diameter of 0.5 mm. We want to cut a certain length of the gold wire and place it in a Mo
(molybdenum) boat and apply current through it until all the gold has evaporated.
How long do we need the Au wire to be for a film of 15 nm thickness? -The distance
from the Mo boat to the substrate for the film is 45 cm.
Solution:
Assume for simplicity equal evaporation rate in all directions (- alternatively equal evap.
rate in the upper half sphere.) So to evaporate a film of thickness t at a distance h require
the following volume of the metal evaporated
V = 4  h2 t
( eq. 2 a)
The volume of the wire before it was completely evaporated was
V =  (D/2)2 L ( eq. 2 b)
where D and L is the diameter and length of the wire respectively.
We set the equations 2a and 2b equal to each and solve for L, the length of wire;
L = 16*h^2*t / D^2
Answer is 19.4 cm
3)
We consider a Schottky barrier to n-Si. Let the donor concentration be 1015 cm-3 and the
temperature 22 °C. Without image force reduction, the barrier height is 0.75 eV.
a) Calculate how large the reduction in the barrier height is due to image force lowering
for an applied bias of 0V.
b) Calculate the distance from the metallurgical junction to the maximum of the electron
potential energy.
c) Repeat the calculations for a) and b) for an applied bias of 0.2 V in the forward
direction.
Solution
What is image force lowering again? An electron in the depletion region of the
semiconductor will induce charges on the metal. How to find: the charge distribution, the
electric field, and, what we are most interested in, the electrostatic potential resulting
from the induced charge, -is a well known problem from elementary electrostatics.
Assume a point charge outside a metal
Is
equiv
to
The metal is an equipotential surface, so the electric field lines will be perpendicular to
the surface. The field distribution outside the metal will be the same as if we replace the
surface charges by an imaginary mirror charge in the same distance from the surface as
the charge outside the surface. We can then calculate the potential from the Coulomb
attraction between two point charges. The force for a separation 2x in vacuum is
1
q2
F 
(2x) 2 4 0
where the other symbols have their usual meaning.
The Coulomb potential energy for this situation is now found for example by calculating
the work done in bringing the electron from infinity to a separation 2x.

x
q2
 c x    F(t)dt 
160 x

( This is the work done by the electron )
The total potential energy of an electron in the depletion region is given by the sum of
this c, ( where you put in the appropriate dielectric constant , let’s assume 0Ks, where
 is the static dielectric constant found in most tables) and the usual depletion zone
K
s
potential obtained from Poissons equation and depletion zone approximation.
Alternatively one could do the calculation with a constant electrical field, the electrical
field is largest at the metal/Si junction, We here choose this approximation and write the
total potential as
-tot(x)
2
-tot(x) = -qE x -q /(160Ksx)
qE x
approx
-c(x)
where E is the electrical field, and the zero is arbitrary chosen as the
potential at x=0 in the idealized case ignoring c (i.e. setting it to zero, peak of black
curve)
To find when -ot is maximum,
we set dtot/dx = 0; giving

q
qE
 x m 
and tot (x m )  q
  
160K s E
40K s

This is now the image force barrier height lowering To set in values we need to calculate
a decent value for E, the maximum electrical field. By use of Poissons equation I get

E  q

N 
2qN D 
B 0  qVF  kTln C 
0K s 
N D 
Where ND is the donor concentration, NC effective density of states in conduction band,
Bo the barrier height (without image force lowering, of course). We put this into the
expression for xm. VF is the forward bias voltage.
a) VF = 0: E = 12290 V/cm, ∆ = 0.0123002 eV, b) xm = 5.00413 e-7 cm = 5 nm,
b) VF=0.2: E = 9442.82, xm = 5.7089 e-7 cm = 5.71 nm, ∆ = 0.01078 eV,
4)
When we evaporate a thin metal film on a substrate, the film will be poly-crystalline. The
film will consist of many crystallites. It is possible to find out how large are these grains.
One popular method for doing so is X-ray diffraction. In the X-ray diffraction method,
the film is exposed to a parallel beam of monochromatic X-rays. Each crystallite – every
grain- will contribute to the intensity in the diffraction pattern, which we in principle
takes a picture of. The diffraction pattern can be viewed as an addition of Fourier
transforms of all the grains. For a polycrystalline film the diffraction pattern will consist
of rings of a certain width. The width tells about the grain size and is the most widespread
method for measuring the grain size.
a) Explain why the width of the intensity in a ring tells about the grain size.
One could say that for a small grain there are few planes that contribute to the pattern.
Bearing in mind that the pattern is the Fourier transform, the sharpness will then decrease
according to the Fourier theorem. This way of explaining is familiar for people knowing
DSP and Fourier analysis.
b) What does the ratio of the intensity of different rings tell?
It can tell about preferential orientation of grains. The ratios for random oriented grains
are known for a known crystal structure, and deviation from this tell about preferential
orientations.
5)
We will estimate the value of the specific contact resistance for a particular contact to Si:
We have a donor concentration of 2E18cm-3 and have a metal contact of NiSi with a
barrier height of 0.68 eV. The specific contact resistance is defined as Rc=(diff(J, V) )-1
where J is the current density (It is usual, albeit confusing, to say resistance pr area)
a) Estimate the value (and units) of the specific contact resistance.
i)For thermiionic emission
J = RT2exp(B/kT)exp(qV/kT – 1); Rc= 1/Diff(J,V) = k/(qRT)*exp(B/kT)
Rc=600 Ω cm2
ii) for tunneling
Rc= A0*exp(C2/PhiB/sqrt(ND)) where A0 is unknown. It can be estimated from fits to
experimental data, but you will see that for 2e18 for n-Si you are at most an order of
magnitude away from the value calculated for thermionic emission, so we take i) as a
reasonable estimate here
b) Calculate the resistance if the area of the contact is 1x1µm
i) R=Rc/A = 600/1e-4/1e-4 = 6 e10 Ω
c) Assume that you measure the total resistance from the contact, including the
contact resistance and the resistance of a wafer that is 500 µm thick and has a
similar back contact with an area of 75 cm2. What is the total resistance?
We can ignore the backside contact since the area is much larger than that of the front.
We must estimate the resistance of the wafer. Without calculating the spreading
resistance exactly we can say it will between that of a cylinder with a diam of 1µm and 9
cm. Lets calculate resistance of the small one: Rwafer<< rho*Pi*(0.5*1e-4)^2*500e-4;
The resistivity of Si doped 2e18 rho=2e-2 Ωcm and yields as an upper limit 1.2 e6 Ω
which anyhow is much smaller than the contact resistance calculated, so the value
calculated for the contact resistance will dominate.
6)
Diffusion will be involved in a reaction between a metal and a semiconductor to form
silicide. Assume the metal Ni deposited onto Si. Heat treatment will induce a planar
layer of Ni2Si between Ni and Si. The thickness of the Ni2Si layer increases with
annealing time. This is due to diffusion.
In an experiment we use radioactive metal atom labeling in order to study the diffusion
mechanisms. Assume that we deposit 50 nm inactive Ni, then deposit 5 nm radioactive
Ni*. We can measure how Ni* is distributed after diffusion by measuring the activity
after successive short ion etches of the material. Assume for simplicity that the diffusivity
of Ni in Ni is small.
a) Sketch how the radioactive Ni* will distribute if the diffusion proceeds by grain
boundary diffusion of Ni.
Ni*/Ni/Si -> Ni*/Ni2Si/Si -> Ni2Si/Ni2*Si/Si
b) Sketch how the radioactive Ni* will distribute if the diffusion proceeds by grain
boundary diffusion of Si.
Ni*/Ni/Si -> Ni*/Ni2Si/Si -> Ni2*Si/Ni2Si/Si
c) Sketch how the radioactive Ni* will distribute if the diffusion proceeds by lattice
diffusion of Ni.
d)
Ni*/Ni/Si -> Ni*/Ni2Si/Si -> [Ni*Ni]2Si/Si with activity gradient in silicide
e) Sketch how the radioactive Ni* will distribute if the diffusion proceeds by lattice
diffusion of Si.
Ni*/Ni/Si -> Ni*/Ni2Si/Si -> Ni2*Si/Ni2Si/Si
7)
Grain boundary diffusion is important in thin films. The diffusivity is given by a constant
Do and an activation energy Ea. Which will have the largest value, you think, the
diffusivity for grain boundary diffusion or that for lattice diffusion? Tell briefly why.
The diffusivity is defined from Ficks 1st law: J=-D diff(C,x), so it is a matter of flux.
The diffusivity usually follows something like D = Do exp (-Ea/kT), Ea will be smaller for
grain boundary diffusion because the grain boundary is more open structure than bulk (
does not apply for H in Pd ). If we consider Do from an atomic/microscopic way then the
Do’s should not be much different, though grain boundary diff is likely to be a 2D process
while bulk is 3D. So defining diffusivity in a microscopic atomistic way will make that of
grain boundary diffusion largest. Whether the macroscopic transport is larger depends
also on the grain size. So we cannot tell which process will dominate transport without
knowing.
Will lattice diffusion or grain boundary diffusion dominate atomic transport in the film at
high temperatures ( say close to the melting point )
Lattice diffusion will dominate because there will be a larger volume accessible to jump
in and out of for bulk diffusion. So it is likely we have something like
Bulk
ln(flux)
Grain boundary
1/T
The total flux is the sum of the fluxes, if the flux in bulk is larger than that carried in h
grainboundary then the bulk lattice diffision dominates.
8)
The grain size in a thin film will depend upon various parameters(surprise!) Assume that
we deposit a film of Ta (tantalum) and one film of In (indium) using the same deposition
rate ( 0.3 nm/sec) onto a substrate at room temperature. Which film will yield the largest
grains? Explain your reasoning. ( Hint: Check the melting temperatures of the metals,
try to explain what it has to do with the matter, without rigorous mathematical
deductions)
Tantalum is hard, has a high melting point, indium is soft and has a low melting point.
Diffusivity generally scales with melting point. So Ta is expected to have lower
diffusivity than In. Thus the saturation of nucleation sites will be much lower for Ta.
Also the coalescence will be less likely for Ta, Both these factors contribute to a smaller
grain size for Ta in agreement with experiments.
9)
Assume you have some experimental laboratory Schottky diodes to n-type Si. For one of
them, diode A, you have made an ohmic backside contact by heavy doping with
Phosporous followed by deposition of Ni. For the other diode, the backside is pressed
towards a chuck. The area of the backside contact is much larger than the Schottky
contact. Make a sketch of the diode characteristics in a ln(I) vs. V diagram. Assume that
diode A is suitable to measure the barrier height, while for diode B this is difficult
because of a high backside contact.
For A is ln(I) vs V a straight line for a large range of V. For B it bends over, increasing
much slower than exponential. So there may not be any region where you can
extrapolate the curve to 0 bias to get the barrier height,
10)
A p-type Si Schottky diode can be used for an IR detector. The response will vanish at a
certain wavelength. Make a rough estimate of how far into the infrared (maximum
wavelength) a p-Si/Pt Schottky diode can be used.
The process is internal photo exitation; IR radiation is hitting the diode from the
backside. It is not absorbed in Si. It is absorbed in the metal and excites a hole over the
Schottky barrier. The barrier for holes is 0.3 eV if we read off the barrier height for
electrons 0.85 eV, 0.3 eV corresponds to ( 1.24/0.3=) 4 µm. The radiation should have a
wavelengh shorter than this in order to get a response.
11)
There are different methods one can use to measure the specific contact resistance. The
most popular is called the Kelvin probe method and the transmission line method. The
geometry of the last one is sketched below. We there have pads, which can be ‘probed’.
(Put a micro probe onto, for measuring voltage or applying bias, or current) The distances
between the pads are arranged with different distances. Measurement of the resistance
between different pads then gives the same contribution from the contact resistance, but
different contributions from the semiconductor resistance. Thereby one can separate the
two contributions to the measured resistance.
We measure the resistance between different pairs of electrodes such as
R12,R13,R14,R15,R23,R24,R25,R34,R45.
Make a formulae for the specific contact resistance expressed by the measured
resistances.
1
3a
2a
a
2
3
4a
4
n-type
p-type
5
3a
n-type
0.001a
a
Call Ra the resistance of the semiconductor in a rectangle ax3a
Then
R12= 2*Rc+Ra=2Rc+Ra
R13=3*Rc+Ra+2*Ra=3Rc+3Ra
R14=4*Rc+Ra+2*Ra+3*Ra=4Rc+6Ra
R15=5Rc+10Ra etc for R23,
We have an overdetermined set, we can find the best fit to the data for Ra and Rc
The resistivity of Si and the contact resistance is thus found.
From a plot of (R12,a) (R23,2a) (R34,3a) (R45,4a) one can determine Rc and Ra, A
graph might look like..
R
12)
In a certain IC process we havea a thick oxide ( A so called ‘FOX’-field oxide) say 1.5 µm
thick on Si. We want to define areas where we have stripes of silicides in contact with
silicon and with the SiO2 oxide at the edges. We make the silicide by evaporating a metal
on a photo-lithographically patterned substrate and the heating. We want to have silicide
lines that are 1 µm wide. Our process steps, starting with the wafer with FOX, will
consist of spinning on negative photo-resist, baking, development, and baking. The oxide
will be etch with buffered HF, with the photo-resist as an etching mask. Sketch the mask
suitable for a stripe of silicide that is 1 µm wide and 10 µm long. Give dimensions and
explain.
Take into consideration underetching and compensation. The etch is isotropic. So
etching 1.5 µm down also etches 1.5 µm to the sides at the top. You are the designer and
make some of the rules. You could say you make the mask itself 1x10 µm, and the
etching has to stop exactly when the bottom etched out feature matches the edge of the
mask. If it is etched too long it will be wider, if too short it will be to narrow, so it will be
a rather critical process.
13)
What are the main general advantages and weaknesses of positive and negative resist
respectively. Also explain why ( qualitatively and in particular which properties of the
resist is playing a part in this)
Read the book. Comment on dimensional presicion and swelling of negative resist.
Comment on speed. Comment on robustness.