Ellipsometry for measuring the
thickness of thin films
Author:
Sourabh Singh Chauhan
School of Physical
Sciences
NISER,Bhubaneswar
Contents
1 Introduction
2
2 Principle of ellipsometry
2
3 Methods to determine ψ and ∆
3.1 Photometric ellipsometry . . . . . . . . . . . . . . . . . . . . .
3.2 Null ellipsometry . . . . . . . . . . . . . . . . . . . . . . . . .
6
6
8
4 Observations and results
4.1 Null ellipsometry . . . . . . . .
4.1.1 Thickness of the film . .
4.1.2 Error analysis . . . . . .
4.2 Photometric ellipsometry . . . .
4.2.1 Thickness of the film . .
4.2.2 Error analysis . . . . . .
4.3 Measurement of refractive index
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5 Outlook
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6 Conclusive remarks
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7 Acknowledgement
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1
1
Introduction
Ellipsometry is a an optical measurement technique to measure the transmission and reflection properties after light is incident on some material.
The name ellipsometry comes from the fact that most often light becomes
elliptically after passing through the medium. In ellipsometry change in polarization state is studied to infer properties of medium. There are already
several other methods to measure thickness of the thin films. But ellipsometric measurements have advantage of their own.
There are several advantages of using ellipsometry to measure properties of
thin films and other objects:• High precision in wavelength (for spectroscopic ellipsometry)
• Non destructive measurement (No need to disturb the system)
• Fast measurement
• Various properties can be measured like thickness and refractive index
of material
Some disadvantages:• Necessity of an optical model
• Data analysis becomes complicated
• Low spatial resolution
• Not good for materials with low absorption coefficient
Through out the calculations we are going to assume the medium to be homogeneous. We are a Aluminum thin film (assumed to be semi- infinite)with
Si as a substrate made by thermal method. We took refractive index for thin
film1 as 1.371 − i7.618[3].
2
Principle of ellipsometry
We can divide the electric field component of light coming from laser in two
components i.e. s-polarized and p- polarized component[1]. In ellipsometry
s-polarized and s-polarized light is made to incident at Brewster angle of the
1
We assumed that the refractive index does not change much when we make thin film
2
Figure 1: Principle used in Ellipsometry[2]
sample. Then the final polarization state of light is observed. In fig. 1 2
we have Eiσ and Eiπ are incident electric field vectors in s- polarized and
p- polarized plane respectively. The vectors on the incident and reflection
sides overlap completely when θ = 900 . The incident light is polarized +450
relative to the Eiπ axis.We have here Eiπ = Eiσ for this polarization since the
amplitudes of p- and s-polarizations are the same and the phase difference
between the polarizations is zero.
The only way that there can be a change in the reflection coefficients for
two different coefficients of the light can be due to contribution of electric
dipoles (dielectric) present in the sample. Thus, upon light reflection on a
sample, p- and s-polarizations show different changes in amplitude and phase.
Ellipsometry measures the two values (ψ, ∆) that express the amplitude ratio
and phase difference between p- and s-polarizations, respectively.If we take
sample structure to be simple (homogeneous), the amplitude ratio is characterized by the refractive index n and absorption is characterized by extinction
coefficient. Here the two values can be determined from a measurement by
applying the Fresnel equations. This is the principle behind ellipsometry.
We define the amplitude of reflection for s and p polarization as
P = tan(ψ)eI∆ =
2
s–σ, p–π
3
Erπ /Eiπ
Pπ
=
Pσ
Erσ /Eiσ
Figure 2: The two interface[2]
Let us consider an interface of two mediums with refractive indices as
n0 and n1 . The boundary conditions for normal and parallel components of
the electric fields in the material can be derived using Maxwell’s equations.
Boundary conditions give the reflection coefficients for the components of the
electric field that lie on the plane of incidence(π) and are normal to it(σ) as
ρπ =
n1 cos(θ0 ) − n0 cos(θ1 )
n1 cos(θ0 ) + n0 cos(θ1 )
ρσ =
n0 cos(θ0 ) − n1 cos(θ1 )
n0 cos(θ0 ) + n1 cos(θ1 )
Here n1 sin(θ1 ) = n0 sin(θ0 ). As can be noted refractive index can be
complex if there is absorption, hence the angles in general are complex.
Our system has three layers one layer of n0 and of n0 separated by a
medium of refractive index n1 thickness d. For multiple reflection, we get
the following for a given component of electric field.
Er = (ρ01 +τ10 ρ12 τ01 e−i2β +τ01 ρ12 ρ10 ρ12 τ01 e−i4β +τ01 ρ12 ρ10 ρ12 ρ10 ρ12 τ10 e−i6β ...)E
where Er is the reflected component,
E is the incident component of electric field,
ρij and τij are reflection and transmission coefficients respectively for a light
4
travelling from ith medium to j th medium. β = 2πdn1λcos(θ1 ) where λ is the
wavelength of the light considered. If the second medium is absorptive, then
β can be complex.
We can derive the following expression for P
P =
Er
ρ01 + ρ12 e−i2β
=
E
1 + ρ01 ρ12 e−i2β
In particular for σ and π components,
Pπ =
ρ01,π + ρ12,π e−i2β
Erπ
=
Eπ
1 + ρ01,π ρ12,π e−i2β
Pσ =
ρ01,σ + ρ12,σ e−i2β
Erσ
=
Eσ
1 + ρ01,σ ρ12,σ e−i2β
Pπ
Er
ρ01,π + ρ12,π e−i2β 1 + ρ01,σ ρ12,σ e−i2β
= π =
.
Pσ
Eπ
1 + ρ01,π ρ12,π e−i2β ρ01,σ + ρ12,σ e−i2β
ρ12,π ρ01,π ρ12,σ X 2 + (ρ01,π ρ01,σ ρ12,σ + ρ12,π )X + ρ01,π
=
ρ12,π ρ01,σ ρ12,σ X 2 + (ρ01,π ρ01,σ ρ12,π + ρ12,σ )X + ρ01,σ
AX 2 + BX + C
=
DX 2 + EX + F
P =
(1)
(2)
(3)
where X = e−i2β .
In general P is a complex quantity hence it can be written as P = tan(Ψ)ei∆
where Ψ and ∆ are real. The objective of ellipsometry is to deduce these two
angles experimentally.
Equation (3) gives a quadratic equation for X.
p
−(P E − B) ± (P E − B)2 − 4(P D − A)(P F − C)
X=
2(P D1 − A)
,
where A = ρ12,π ρ01σ ρ12,σ ,
B = ρ01,π ρ01σ ρ12,σ + ρ12,π ,
C = ρ01,π ,
D = ρ01,π ρ01π ρ01,σ ,
E = ρ01,π ρ12π ρ01,σ + ρ12,σ ,
F = ρ01,σ ,
Solving for this gives two values of X. We choose the value with lower
imaginary part.
5
3
Methods to determine ψ and ∆
There are many optical arrangements that can give us values of ψ and ∆.
This leads to several ellipsometric techniques. We are going to discuss two
techniques here:-
3.1
Photometric ellipsometry
Figure 3: Rotating analyzer Ellipsometry[2]
For photometric ellipsometry two polarizers, one before sample and one
after sample. We used a rotating analyzer set up where we kept the analyzer
at fixed and changed the polarizer angle to obtain intensity at different angle.
These intensities are then related to the parameters Ψ and ∆.[4] After a
linearly polarized light is made to incident on a sample because to different
refractive index in two different directions we get an elliptically polarized
light. Two methods of finding Ψ and ∆ will be discussed.
The electric field at the detector in terms of Jones matrix formalism can
be written as follows.
E0 =
E0 =
1 0
0 0
cos(α2 ) sin(α2 )
− sin(α2 ) cos(α2 )
ρπ 0
0 ρσ
cos(α2 )ρπ Ei cos(α1 ) + sin(α2 )ρσ Ei sin(α1 )
0
6
=
Ei cos(α1 )
Ei sin(α1 )
Eπ cos(α2 ) + Eσ sin(α2 )
0
I0 = E0† E0
(4)
2
2
∗)
= cos (α2 )Eπ Eπ∗ + sin (α2 )Eσ Eσ∗ + cos(α2 ) sin(α2 )(Eπ Eσ∗ + Eσ Eπ(5)
1
=
[s0 + s1 cos(2α2 ) + s2 sin(2α2 )]
(6)
2
where the three Stokes parameters, s0 , s1 and s2 are given by s0 =
Eπ Eπ∗ + Eσ Eσ∗ , s1 = Eπ Eπ∗ − Eσ Eσ∗ and s2 = Eπ Eσ∗ + Eσ Eπ∗ .
Using Eπ = ρπ cos(α1 )Ei , Eπ∗ = ρπ∗ cos(α1 )Ei , Eσ = ρσ sin(α1 )Ei and
Eσ∗ = ρσ∗ sin(α1 )Ei , we can get,
1
0
0
I0 = s0 [1 − cos(2Ψ ) cos(2α2 ) + sin(2Ψ ) cos(∆) sin(2α2 )]
2
0
(7)
0
tan Ψ
where Ψ is such that tan Ψ = tan
.
α1
0
0
Comparing we get, cos(2Ψ ) = − ss10 and sin(2Ψ ) cos(∆) =
We can see that,
1
I0 (0o ) = (s0 + s1 )
2
1
I0 (45o ) = (s0 + s2 )
2
1
I0 (90o ) = (s0 − s1 )
2
s2
.
s0
(8)
(9)
(10)
If we choose tan(α1 ) = 45o , we get
I0 (90) − I0 (0)
I0 (90) + I0 (0)
2I0 (45o ) − I0 (90) − I0 (0)
sin(2Ψ) cos(∆) =
I0 (90) + I0 (0)
cos(2Ψ) =
(11)
(12)
Hence with three experimental inputs, we can determine Ψ and ∆.
Another way of determining the Stokes parameters is,
Rπ
Rπ
2π(2 02 I0 (θ)dθ + π I0 (θ)dθ)
2
s0 =
(13)
3
Rπ
Rπ
R 3 3π
− π 4 I0 (θ)dθ + 02 I0 (θ)dθ + π I0 (θ)dθ
4
2
s1 =
(14)
3
Rπ
Rπ
(3 02 I0 (θ)dθ − π I0 (θ)dθ)
2
s2 =
(15)
3
7
Figure 4: Here values of a sample are assumed to be ψ = 450 and ∆ = 900 .
The detected light intensity will be zero.[2]
3.2
Null ellipsometry
For this technique we used two analyzers and a quarter wave plate. As shown
in fig. 4[2] the incident light goes through a polarizer which makes the light
linearly polarized. (We kept compensator’s slow axis at 450 w.r.t. π axis).
After passing through the compensate light becomes elliptically polarized
which then passes through the sample. For some polarizer angle the elliptical
polarization produced by sample can be compensated by the quarter wave
plate giving a linearly polarized light after sample. This light can be made
to obtain zero intensity at some angle of analyzer. The objective of the
experiment is to find this combination. Jones matrix formulation gives(fig.
5) :Lout = AR(A)SR(−C)CR(C)R(−P )P Lin
We obtain
EA = Pπ [cos(C) cos(P −C)−ρ1c sin(C) sin(P −C)]Pσ sin A[sin C sin(P −C)+ρ1c cos C sin(P −C)]
Here ρ1c is the phase shift by compensator e−iδ . The rotation angle for
quarter plate is 450 and phase shift is 900 . Here for we put zero intensity
EA = 0, C = 45, δ = 90 we get
P = tan(ψ)ei∆ =
Pπ
= tan(A)ei(−2P +90)
Pσ
From here we can get (ψ, ∆) values are determined by A and P. we get
ψ = −A(−A > 0), ∆ = −2P + 900
8
Figure 5: Setup for null ellipsometry[5]
and for
ψ = A(A > 0)∆ = −2P − 900
Similarly we will get two solutions for C = −450 . Therefore we get four sets
of P and A giving zero intensity. We take average for those to obtain (ψ, ∆)
and hence thickness of thin film.
4
Observations and results
Incident angle = 350
4.1
4.1.1
Null ellipsometry
Thickness of the film
In null ellipsometry as mentioned in above section we kept the compensator
at 450 and changed Analyzer and polarizer angle to obtain the minimum
intensity3 . We obtained null intensity at Pθ = 43o and A = 43o . Hence,
3
We did not observe completely dark spot. It may be due to alignment problem
9
Ψ = 0.750 and ∆ = 3.070
P =
Pπ
= −0.930 + 0.067i
Pσ
We used Mathematica for all the calculations. We expected the thickness to be real but we got thickness having imaginary part well. Therefore
we ended up getting two different values for thickness using argument and
absolute value of x.Two solutions for eiβ (X) give –
X = |X0 |eiΓ ,
Let d be the thickness. The thickness was calculated in two ways.
• d = log(|X|) 4π(cos(θ)r nλc +cos(θ)c nr ) =7.008 nm
• d = − 4π(nr (cos θ)Γλ
=40.78 nm
r −nc (cos θ)c )
The average of both the methods gave thickness as 23.89nm.
4.1.2
Error analysis
Error in calculated values of Ψ and ∆ can be calculated in the following way:Ψ=A
∆ = −2Pθ − 90
δ(Ψ) = δ(A)
This gives
δ(∆) = 2δ(Pθ )
Least count of the polarizer and analyzer were 0.2o (0.003rad).
Error in P = PPπσ due to the error in Ψ and ∆ is given for the real part
and imaginary part separately.
δ(Preal ) =
p
(δ(Ψ) sec2 Ψ cos ∆)2 + (δ(∆) tan Ψ sin ∆)2 = 0.00653rad
δ(Pimaginary ) =
p
(δ(Ψ) sec2 Ψ sin ∆)2 + (δ(∆) tan Ψ cos ∆)2 = 0.00651rad
.
Thickness as a function complex value part P and imaginary part of P
is plotted for a range provided by range of errors. It gives an estimate of
sensitivity of thickness to change of ψ and ∆
10
Figure 6: Sensitivity of thickness calculation(using argument of X- 2nd
method) to errors in calculation of P
11
Figure 7: Sensitivity of thickness calculation(using modulus of X- 1nst
method) to errors in calculation of P
12
From the graph, we get
d1 = 7 ± 2nm
and
d2 = 40 ± 10nm.
The average is 23 ± 6nm.
4.2
Photometric ellipsometry
Figure 8: Photometry-Polarizer angle= 45o
4.2.1
Thickness of the film
We plotted the intensity as a function of analyzer angle to get the values of ψ
and ∆. We chose intensity at 00 , 450 and 900 to get the value of ψ and ∆ and
finally getting the thickness of thin film. By using a code in Mathematica
we obtained the thickness of 5.23nm from absolute value of X. This value is
not expected from the way thin film has been made.
4.2.2
Error analysis
cos 2Ψ =
I0 (90) − I0 (0)
I0 (90) + I0 (0)
13
Figure 9: Photometry-Intensity distribution when incident light is π or σ
polarized.
δ(I)
δ(Ψ) =
sin 2Ψ
s
2
(I0 (90) − I0 (0))2
1
+
2
= 0.0004rad
(I0 (90) + I0 (0))4
(I0 (90) + I0 (0))2
sin 2Ψ cos ∆ =
1
δ(∆) =
sin(2ψ) sin(∆)
v"
u u
t
2I0 (45) − I0 (90) − I0 (0)
I0 (0) + I0 (90)
2
I0 (0) + I0 (90)
2
+2
2I0 (45)
(I0 (0) + I0 (90))2
2
− (2 cos(2ψ)δ(ψ) cos(∆))2
δ(∆) = 0.052rad ≈ 3o
The error in P = PPπσ due to the error in Ψ and ∆ is given for the real
part and imaginary part separately.
δ(Preal ) =
p
δ(Pimaginary ) =
(δ(Ψ) sec2 Ψ cos ∆)2 + (δ(∆) tan Ψ sin ∆)2 = 0.0106rad
p
(δ(Ψ) sec2 Ψ sin ∆)2 + (δ(∆) tan Ψ cos ∆)2 = 0.0539rad
14
Figure 10: The thickness values calculated using method 1 from the P value
obtained using photometry for P values lying within the error bars of real
and imaginary P around the obtained P value.
15
Figure 11: The thickness values calculated using method 2 from the P value
obtained using photometry for P values lying within the error bars of real
and imaginary P around the obtained P value.It may be noted that thickness
never becomes positive in this regime. Hence, this part of the experiment
should be considered to be a failure.
16
4.3
Measurement of refractive index of glass
We measured thickness of glass plate to be 0.12 cm and tried to obtain the
refractive index of the glass (we can approximate it to semi infinite medium).
Single interface formula gives:p
[ 1 − 4 sin2 (θ0 ) tan(ψ)ei∆ + 2 tan ψei∆ + tan2 (ψ)ei∆ ]n0 sin(θ0 )
n1 =
cos(θ0 )[1 + tan(ψ)ei∆ ]
4
Using null ellipsometry, the polarizer and analyzer angles which can
null(minimum) value at the detector were P = 44.6o and A = 57.6o .
The refractive index was obtained as,0.77. Using Photometric ellipsometry, the refractive index was obtained as, 0.91. (figure 12)
Figure 12: Photometry with glass-Polarizer angle= 45o
The expected value of refractive index is 1.5.
Error in the values obtained using photo ellipsometry and null ellipsometry
is 39% and 49% respectively. Simplistic model of homogeneity of glass or
misalignment could have given not good values for refractive index of glass.
4
we used cos(θ1 ) =
p
1 − (n0 sin(θ0 )/n1 )2 and equations for reflection coefficients.
17
5
Outlook
• We can combine both the ellipsometric technique to obtain both refractive index and thickness of thin film.
• Other than spectroscopic ellipsometry there are many other arrangements of similar set up can be used to determine the thickness of thin
film like:– Rotating compensator ellipsometry (compensator is rotated in
place of analyzer or polarizer)
– Phase modulated ellipsometry (input intensity is modulated using
another input supply.)
• One can use ellipsometric technique in infrared regime to study phonon
modes and free-Carrier absorption. Infrared ellipsometry is performed
using a Michelson interferometer combined with ellipsometry set up. It
is known as frequency transform infrared spectroscopy( FTIR) (page
108[2])
Figure 13: Frequency transform infrared spectroscopy set up
6
Conclusive remarks
• Major problems in the experiment are faced because of depolarization
effects due to different part of sample.
• Proper alignment can lead to better results.
18
Figure 14: Depolarization of light[2]
• Null ellipsometry gives thickness of thin film as 23±6nm and refractive
index of glass as 0.77 ± 0.37.
• Photometric ellipsometry did not give good results for thickness of thin
film. We get the refractive index of glass as 0.91 ± 0.36
• We have plotted graph over a range of real and imaginary part of P to
obtain the sensitivity of thickness because of change in ψ and ∆. Very
high sensitivity leads to absurd results in this case. (Figure 15,16)
19
Figure 15: Variation of thickness with real and imaginary part of P (when
calculated using Method 1)
7
Acknowledgment
This experiment was performed as a part of 8th semester lab course at NISER
along with Maneesha Sushama Pradeep. I would like to say thanks to Ritwick
sir for his guidance during the course of experiment. I also express my gratitude to Abhinav Agrawal, 4th yr SPS, NISER for providing the thin film to
us for performing the experiment.
References
[1] ’Optics’ by Hecht.
[2] ’Spectroscopic ellipsometry’ by Hiroyuki Fujiwara (Page 82,140...),
[3] www.filmetrics.com/refractive index database/Al/Alluminium.
20
Figure 16: Variation of thickness with real and imaginary part of P (when
calculated using Method 2)
[4] ’main.pdf’ Project report from Aalborg university by Jesper Jung Jakob
Bork, Tobias Holmgaard, Niels Anker Kortbek
[5] http://www.kw.igs.net/ jackord/ee/e7.html
21