Cylindrical capacitor dissipation factor - QuickField simulation example
This is an example of the cylindrical capacitor simulation, performed with QuickField software. Capacitor consists of ceramic tube with silver electrodes mounted at the surface.
How to find permittivity effect in layered insulation?
Engineering answer Typical applications Geometry
Given
Task
Solution
Results
Field distribution in cylindrical capacitor:
Engineering question
Set up an axisymmetric QuickField AC Conduction problem for a layered cylindrical capacitor structure and evaluate the permittivity effect on electric field from computed field results.
layered dielectric structures, ceramic capacitor insulation systems, multilayer insulation components
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Simulation problem
Problem Type
Axisymmetric problem of AC conduction.
Relative permittivity of ceramic ε = 6.
Conductivity of ceramic g = 10 nS/m;
Voltage U = 10 V.
Frequency f = 1 kHz.
Determine capacitance and dissipation factor of the capacitor.
The value of dissipation factor can be calculated as tg(δ) = PA/PR. Capacitance can be calculated as C = q/U, where U is potential difference between electrodes and q - is a charge on electrodes.
Charge q = 27.7 pC.
Capacitance C = 2.77 pF.
Active power PA = 24.5 nW.
Reactive power PR = 817 nW.
Dissipation factor tg(δ) = 0.03.
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