[SI-LIST] Re: Loss Tangent question

Doug,

They have to give the same value of the imaginary part
of the dielectric constant.
A simplified deduction form Maxwell:
curl H=J+j*omega*D = sigma*E +j*omega*Kr*E =
j*omega*(Kr+sigma/j*omega)*E = j*omega*(Kr-j*Ki)*E

Kr, Ki ? real, imaginary part of the dielectric
constant (equivalent or effective)
Thus:
Ki=sigma/omega
And also by definition Ki=Kr*tand  which gives:
tand=sigma/Kr*omega
Sigma=2*pi*f*Kr*tand
(include eps0 in Kr when calculating
this,Kr=eps0*epsrelative)


Note that from the physics point of view, the term
loss tangent includes more than actual conductive
losses, like for example losses due to reorientation
of the crystal domains in feroelectric materials and
various other loss mechanisms at atomic and molecular
level. 
But for Maxwell?s equations they are equivalent to
some frequency dependent conductivity, and that is the
convention I used above (ie equivalent or effective
conductivity).

Agree with Steve on the frequency dependence. 


 
Peter

--- steve weir <weirsi@xxxxxxxxxx> wrote:
> Doug, yes, you can solve this as a phase plane
> problem:
> 
> The loss tangent expresses the ratio of the
> conductance to the reciprocal 
> capacitive reactance at the test frequency.
> 
> Loss tangent is frequency dependent, but changes
> slowly.  So, DC to 
> daylight is out.  You will want the loss tangent for
> the frequency range of 
> interest.
> 
> Steve.
> 
> At 12:05 PM 3/26/2005 -0800, Doug Brooks wrote:
> >The model for a "lossy" (microstrip or stripline)
> transmission line has
> >terms for series resistance (skin effect) and
> parallel conductance
> >(dielectric losses.)  If I have a specification for
> loss tangent (at a
> >particular frequency), is there a straightforward
> way to get from loss
> >tangent to conductance? That is, how do I calculate
> conductance given a
> >loss tangent?
> >
> >Secondly, if I know the loss tangent at one
> frequency, can I therefore
> >derive it for all other frequencies? Or are there
> too many variables involved?
> >
> >Thanks for your input.
> >
> >Doug Brooks
> >
> >
>
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