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influence of voltage on dielectric loss
Posted by Tim on 3/18/2005, 6:21 am
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Re: influence of voltage on dielectric loss
Posted by Bert Hickman on 3/18/2005, 5:49 pm, in reply to "influence of voltage on dielectric loss " Hello Tim, Unfortunately, I really can't provide further insights into the phenomenon you are seeing since I am not an expert on the dielectric properties of water. My guess is that you may be seeing some nonlinear/breakdown effects at the electrode-water interface combined with conduction effects within the bulk water. For example, suppose that your experimental setup develops a thin insulating layer on one, or both, electrodes via anodization. In this case, most of the electrical stress will appear across the thin insulating layer(s). The layers may be able to support a 100 volt stress without breaking down, so you measure relatively low electrolytic conduction losses at this voltage. However, at 400 volts, you may be seeing some breakdown of the insulating layer(s) and thus significantly higher conduction losses through the bulk liquid. But this is only a guess... Good luck and best regards, -- Bert --
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calculation of dielectric loss
Posted by Raju Manna on 7/13/2006, 1:52 am, in reply to "Re: influence of voltage on dielectric loss "
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Re: calculation of dielectric loss
Posted by Bert Hickman on 7/13/2006, 3:46 pm, in reply to "calculation of dielectric loss" Hello Raju, I have comparatively little experience with microwave engineering, but I'll try to provide some starting points. The estimated power (in watts/cubic cm) that will be converted to heat within the dielectric due to losses can be estimated as: Pd = (55.61*10^-14*)*(Ed^2)*f*Er*tan(d) where: Pd = watts/cm^3 A simple plane, sinusoidal, TEM wave in free space (Zo = 377 ohms) will have an average E field (Eo) of: Eo = Sqrt(P*377) where P = watts/meter^2. The E field within the dielectric (Ed) will be about 1/10 of Eo because of the ceramic material's relatively high dielectric constant. However, it's quite unlikely that you have this simple of a case in reality. Finally, the total power in watts absorbed by your dielectric slab will be: W = Pd*V IN reality, the actual distribution of the electric field within the dielectric is most likely quite non-uniform and, as a result, you'll develop a number of "hot spots" wherever the E-field is more intense. A more exact solution will require using field solver/modelling software. Here are some sources of information you may find helpful: Hope this helped and best regards, Bert
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