I know the conductivity of a dielectric material is:
Conductivity = 2(Polarization at Zero Field)/(Electric Field*Time Period)
OR
Conductivity = 2(Polarization at Zero Field)(Frequency)/(Electric Field)
Now I want to convert the obtained conductivity into "conductivity loss". Some of the references which have used such conversion but didn't mention the conductivity loss formulas are stated below:
1. P. Khanchaitit, K. Han, M.R. Gadinski, Q. Li, Q. Wang, Nat. Commun. 4 (2013) 2845.
2. Y. Wang, J. Cui, L. Wang, Q. Yuan, Y. Niu, J. Chen, Q. Wang, H. Wang, J. Mater. Chem. A 5 (2017) 4710–4718.
3. Q. Chen, Y. Wang, X. Zhou, Q.M. Zhang, S. Zhang, Appl. Phys. Lett. 92 (2008) 142909
4. Zhongbin Pan, Baihui Liu, Jiwei Zhai, Lingmin Yao Ke Yang, Bo Shen, Nano Energy 40 (2017) 587–595
If someone knows a formula please share your answer. Thanks.
Some definitions:
ec = e' - je'' , complex permittivity tand = e''/e' , loss tangent s = omega*e'' , conductivity (alternating)
From the above equations:
s = tand*omega*e' where, omega: frequency of use e': real part of complex permittivity
Dear Mohsin,
I found a paper, that can help you.
With Regard
R. Mitdank
The electrical conductivity (σ) data from the experimental dielectric data using the relation as, σ = ωe0εrtanδ, where, w is the angular f, eo is free space dielectric constant and εr is the dielectric constant of the materials. From this above equation you can find out dielectric loss (tanδ). Mohsin Ali Marwat
Respected Rüdiger Mitdank Debajyoti Nath Rajbanul Akhond, I think the mentioned reference or relations do not relate conductivity with conduction loss.
One of the other renowned equations is mentioned below for your kind observation. Thanks
Dielectric loss = tanδ = conduction loss + Ferroelectric Loss.
Ferroelectric loss is usually derived after finding the conduction loss and dielectric loss.
Conduction loss is derived from the Ohm's Law by considering a capacitor as a resistor and using effective conductivity (from the equation mentioned in question). This is answered by one of my collogues, however, still need to find the relation.