Dear colleagues,
I meet a problem of calculate the total dielectric constant. Now I have a Ge nanowire lying down on the Si wafer. Its top is coat by GeOx and bottom is by SiO2. The dielectric constant of GeOx and SiO2 are 5.6 and 3.9, respectively. What is the tatol dielectric constant of the surrounding? Just average these two values?
Thanks in advace.
Dear Song,
Welcome!
The dielectric constant averaging can be done according to some assumptions.
I assume that the outer side will be an equipotential surface. The capacitance of such surrounding is C= epsilon A/d. = epsilon1 A1/d + epsilon2 A2/d = epsilon average A/d, where d is the thickness of the oxide layer assumed equal for the two oxides.
From this relation we can get the the average dielectric consatnt.
epsilon average=( epsilon1 A1 + epsilon2 A2)/A
If A1=A2= A/2,
Then
epsilon average= (epsilon1 + epsilon2)/2
Which is the simple average of the two dielectric constants.
Best wishes
It depends on the frequency range (or/and the EM-wavelength) of your application and the spatial configuration of the (1D ?) Ge-nanowires//SiO2.
Yes i agree with Ioannis Samaras
It depends on the frequency range (or/and the EM-wavelength) of your application and the spatial configuration of the (1D ?) Ge-nanowires//SiO2.
Only for homogeneous dielectric media kept in series, and defect-free charge-less interface in between, the net capacitance and hence dielectric constants can be approximated by their average.
Dear Song,
Welcome!
The dielectric constant averaging can be done according to some assumptions.
I assume that the outer side will be an equipotential surface. The capacitance of such surrounding is C= epsilon A/d. = epsilon1 A1/d + epsilon2 A2/d = epsilon average A/d, where d is the thickness of the oxide layer assumed equal for the two oxides.
From this relation we can get the the average dielectric consatnt.
epsilon average=( epsilon1 A1 + epsilon2 A2)/A
If A1=A2= A/2,
Then
epsilon average= (epsilon1 + epsilon2)/2
Which is the simple average of the two dielectric constants.
Best wishes
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