
The current
I_{A} flowing through the ohmic
resistor of the equivalent circuit diagram is in phase with the
voltage U; it corresponds to the imaginary part e'' of the dielectric function times w. 


The 90^{o} outofphase current
I_{R} flowing through the "perfect" capacitor is given by the real
part e' of the dielectric function times
w. 


The numerical values of both elements must depend on the
frequency, of course  for w = 0,
R would be infinite for an ideal (nonconducting)
dielectric. 


The smaller the angle d or
tg d, the better with respect to power
losses. 

Using such an equivalent circuit diagram (with
always "ideal" elements), we see that a real dielectric may be modeled by a fictitious
"ideal" dielectric having no losses (something that does not exist!)
with an ohmic resistor in parallel that represents the losses. The value of the
ohmic resistor (and of the capacitor) must depend on the frequency; but we can
easily derive the necessary relations. 


How large is R, the more interesting quantity,
or better, the conductivity s of the material that corresponds to R?
Easy, we just have to look at the equation for the current
from above. 


For the inphase component we simply
have 





Since we always can
express an inphase current by the conductivity s
via 





we have 




In other words: The dielectric losses occuring in
a perfect dielectric are completely contained in the imaginary part of the
dielectric function and express themselves as if the material would have a
frequency dependent conductivity s_{DK} as given by the formula above. 


This applies to the case where our dielectric is still a
perfect insulator at DC (w = 0 Hz), or, a bit more general, at low frequencies;
i.e. for s_{DK}(w ® 0) = 0. 

However, nobody is perfect! There is no perfect insulator, at best we have good insulators. But now it is easy to see what we
have to do if a real dielectric is not a perfect insulator at low frequencies, but has
some finite conductivity s_{0} even
for w = 0. Take water with some dissolved
salt for a simple and relevant example. 


In this case we simple add s_{0}
to s_{DK} to obtain the total
conductivity responsible for power loss 



s_{total} 
= 
s_{perfect} +
s_{real} 




= 
s_{DK} +
s_{0} 




Remember: For resistors in parallel, you add the conductivities (or1/R's) ; it is with
resistivities that you do the
1/R_{total} = 1/R_{1} + 1/R_{2} procedure. 

Since it is often difficult to
separate s_{DK} and s_{0}, it is convenient (if somewhat confusing
the issue), to use s_{total} in the
imaginary part of the dielectric function. We have 





We also have a completely general way now, to
describe the response of any material to an
electrical field, because we now can combine dielectric behavior and
conductivity in the complete dielectric function of the material. 


Powerful, but only important at high frequencies;
as soon as the imaginary part of the "perfect" dielectric becomes
noticeable. But high frequencies is where the action is. As soon as we hit the
high THz region and beyond, we start to call what we do "Optics", or "Photonics", but the material roots of those
disciplines we have right here. 

In classical electrical engineering at not too
large frequencies, we are particularily interested in the relative magnitude of
both current contributions, i.e in tgd. From
the pointer diagram we see directly that we have 




We may get an expression for tg
d by using for example the
Debye equations for
e' and e''
derived for the dipole relaxation mechanism: 


tg d =

e''
e' 
= 
(e_{s}
– e_{¥}) · w / w_{0}
e_{s} + e_{¥} ·
w^{2}/
w_{0}^{2} 




or, for the normal case of e_{¥} = 1 (or
, more correctly e_{0})



tg d =

(e_{s} – 1) ·
w/ w_{0}
e_{s} + w^{2}/
w_{0}^{2} 




This is, of course, only applicable to real perfect dielectrics, i.e. for real dielectrics with
s_{0} = 0. 

The total power loss, the really interesting quantity, then becomes (using
e'' = e' ·
tgd, because tgd is now seen as a material
parameter) . 


L_{A} = w ·
e' · E^{2} · tg
d 



This is a useful relation for a dielectric with a
given tg d (which, for the range of
frequencies encountered in "normal" electrical engineering is
approximately constant). It not only gives an idea of the electrical losses,
but also a very rough estimate of the breakdown strength of the material. If
the losses are large, it will heat up and this always helps to induce immediate
or (much worse) eventual breakdown.


We also can see now what happens if the dielectric
is not ideal (i.e. totally insulating), but
slightly conducting: 


We simply include s_{0} in the definition of tgd (and then automatically in the value of e''). 


tg d is then nonzero even
for low frequencies  there is a constant loss of power into the dielectric.
This may be of some consequence even for small tg d values, as the example will show: 

The tg d value
for regular (cheap) insulation material as it was obtainable some 20
years ago at very low frequencies (50 Hz; essentially DC) was
about tg d = 0,01. 


Using it for a highvoltage line (U = 300 kV)
at moderate field strength in the dielectric (E = 15MV/m;
corresponding to a thickness of 20 mm), we have a loss of 14
kW/m^{3} of dielectric, which translates into about 800 m
high voltage line. So there is little wonder that highvoltage lines were not
insulated by a dielectric, but by air until rather recently! 

Finally, some examples for the tg
d values for commonly used materials (and low
frequencies): 


Material 
e_{r} 
tg d
x 10^{4} 
Al_{2}O_{3}
(very good ceramic) 
10 
5....20 
SiO_{2} 
3,8 
2 
BaTiO_{3} 
500 (!!) 
150 
Nylon 
3,1 
10...0,7 
Poly...carbonate,
...ethylene
...styrol 
about 3 
PVC 
3 
160 


And now you understand how the
microwave oven works and
why it is essentially heating only the water contained in the food. 


