Homework problems FYS4310 and FYS9310* May, 2010
Problems with * are obligatory for FYS9310 only
1500-1)
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. (You may use text book chapter 15 as reference)
Solution:
We look up the barrier heights for Al and for Pt. Figure 15.19 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.
1500-2)*
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
(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 Ks
 the static dielectric constant found in most tables) and the usual depletion zone
is
potential obtained from Poisson’s 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)
qE x
2
approx
-tot(x) = -qE x -q /(160Ksx)
F 
-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

NC 
2qN D 


qV

kTln

 
F
0K s  B 0
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,
c) VF=0.2: E = 9442.82, xm = 5.7089 e-7 cm = 5.71 nm, ∆ = 0.01078 eV,
1500-3)*
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.
1500-4)
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) Calculate an estimate for 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.
1500-5)
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*. Schematically we can notate the structure as ‘Si/Ni/Ni*’ where the surface is to the
right. 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
1500-6)
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. For small grains and low temperatures it will dominate.
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 the
grain-boundary then the bulk lattice diffusion dominates.
1500-7)
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.
1500-8)
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 excitation; 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
wave length shorter than this in order to get a response.
800-1)
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 precision and swelling of negative resist.
Comment on speed. Comment on robustness.
900-1)
See problem 9.2 in text book
Consider the analogy with classical binary scattering problem. Scattering through a large
angle is dependant upon the masses ratios of the particles colliding. Electrons are much
lighter than the atoms they collide against. In ion implantation the projectiles can even be
heavier than the atoms of the material they are implanted into i.e P, As, Sb into Si. The
latter would be like colliding a bowling ball with a football – The bowling ball will not
scatter back even for a head on collision, while if you scatter a pea from a football, it will
bounce back.
900-2)
See problem 9.5 in text book
The thermal expansion is given by
∆x/x = CTE*∆x
For Si3N4 CTE=2.7e-6/K
Here x is 2.5 cm so ∆x= 2.5 cm(2.7e-6/K)*10K=6.75e-5 cm = 0.025 µm
If the minimum feature size is 0.1 µm and the minimum distortion is
∆T = ∆x/x*1/CTE= 2,5e-6 cm/(2.5cm)/2.7e-6/K =0.37 °C