analysis of nanostructural layers using low frequency impedance spectroscopy
DESCRIPTION
Analysis of nanostructural layers using low frequency impedance spectroscopy. Part 1: Background Physics. Hans G. L. Coster. Sinewave signal generator. voltage. Response of materials to AC currents. Current. Phase shift. Shift in relative magnitude. Electrical Circuit Equivalents. - PowerPoint PPT PresentationTRANSCRIPT
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Analysis of nanostructural layers using low frequency impedance spectroscopy
Hans G. L. Coster
Part 1: Background Physics
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Response of materials to AC currents
Sinewave signal generator
Phase shiftShift in relative magnitude
Current voltage
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Electrical Circuit Equivalents
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Impedance of capacitance elements
C = Q/V
Sinusoidal currents
1 jwheretjo eii
)sin( tii o
Same form as:
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Impedance of a capacitance element
TT
C
dti
C
dq
00
CV
Cj
iei
CjC
ei tjo
T tjo
1
0
CC ZV iBut we have that
HenceCj
1ZC
1 jwheretjo eii
(Ohms’s Law)
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Impedance of capacitance elements
C
j
Cj
1
ZC
C = Q/V
current
C
j
The voltage lags behind (phase lags) the current
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Parallel Circuits
TR
Vi
21
21
21
R
1
R
1V
R
V
R
V
iii
21T R
1
R
1
R
1 Hence
i V
i2
i1 R1
R2
i V
RT
21T GGGor
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Parallel impedances
ZC
ZR
ZT
RCT Z
1
Z
1
Z
1
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Circuit elements to represent Layers
1C
1G
ZC
ZR
ZT
1
RCT Z
1
Z
1Z
111
1
11T GC
R
1CZ
jj
21
21
211
11 GC
CG
GC
1
j
j
Real Part
Imaginary Part
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Phase relationships
222T
T
Yand
Z
1Y
GC
GCjT
We measure ZT and
ZT
Real part
Imaginary part
But it is simpler to work in terms of
Admittances (Y= 1/Z)
For a single “layer” containing parallel R and C elements :
cosYG
sinYC
T
T
YT
Real part
Imaginary part
jC
G
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Detecting substructure with Impedance Spectroscopy
Homogeneous film Film with two homogeneous sub-structural layers
1C
1G
1C
1G
2C
2G
C1= 0.006 F/m2
G1 = 0.003 S/m2
C2=0.059 F/m2
G2 =4.35 S/m2
Consider an example with the following dielectric parameters:
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0.1 1 10 100 1000
Frequency - Hz
100
10
1
0.1
Imp
edan
ce
Impedance Spectroscopy: Dispersion of Impedance with frequency
1C
1G
1G
C1
G2
C 2
Conclusion: The impedance as a function of frequency does not allow us to discriminate between the single and 2 layer structures.
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0.1 1 10 100 1000
Frequency - Hz
Cap
acit
ance
– m
F m
-2
0.0060
0.0059
0.0058
0.0057
0.0056
0.0055
Impedance Spectroscopy: Capacitance dispersion with frequency
1C
1G
1G
C1
G2
C 2
The capacitance as a function of frequency allows the single and 2 layered structure to be readily distinguished.
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0.1 1 10 100 1000
Frequency - Hz
Con
du
ctan
ce –
S m
-2
0.040
0.035
0.030
0.025
0.020
0.015
0.010
0.005
Impedance Spectroscopy: Conductance dispersion with frequency
1C
1G
1G
C1
G2
C 2
The conductance as a function of frequency also allows the single and 2 layered structure to be readily distinguished.
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A two layer sandwich
1C
1G
2C
2G
221
2221
21
22
22
21
212
221
221
2221
212
221
22
21
2221
2
CCGG
CGCGGGGGG
CCGG
GCGCCCCCC
Whilst C1, G1, C2, G2 are frequency independent, C and G for the combination is frequency dependent
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Alkane layers on silicon
Silicon Wafer
SiO2
Silicon Wafer
SiH40% NH4F
Silicon Wafer
SiH
UV
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Self Assembled Organic Films
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Sil
icon
[11
1]
monolayerC
solutionG
monolayerG
Decane layer attached to Si
The Born energy for ion partitioning into the alkane layer is very high. Therefore the electrical conductance of this layer will be very low.
Organic films on Silicon substartes
More info on Born Energy
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The Spectrometer
Impedance range: 0.1 -1010 Frequency: < 10-2 – 106 Hz
Impedance precision: 0.002% Phase resolution: 0.001 o
Inphaze.com.au
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inphaze.com.au
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+ +
Ion in water Image of ion in dielectric
m ~ 2 -3
for hydrophobic films
Organic filmExternal solution
w=80
Ion Partitioning into an organic film
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The Born Energy arising from image forces on the ions is:
wmo
e11
R8
22zBW
For a K+ ion in a alkane layer, WB ~ 3 eV
Ion partitioning into molecular films
Recall that kT~ 0.025 eV at room temperature
So that the partitioning goes as ~ e-120 !!
It will be a very poor electrical conductor!Return to Main