[SI-LIST] Re: what is the conductivity of a dielectric?

Hi Rob,

After reading your nice explanation on displacement current, I wanted to 
ask for examples. Considering the vast chip/package/board development 
arena, which features(for example, vias) come to your mind whose 
function can be explained, wholly or partly, using displacement current? 
That would really help understand it.

Thanks,
Sainath

---------Included Message----------
>Date: Fri, 12 Apr 2002 00:15:45 -0700
>From: "Rob Hinz" <rob@xxxxxxxxxx>
>Reply-To: <rob@xxxxxxxxxx>
>To: <RayCaliendo@xxxxxxxxxx>
>Cc: <si-list@xxxxxxxxxxxxx>
>Subject: [SI-LIST] Re: what is the conductivity of a dielectric?
>
>
>Hi Roy,
>
>Actually that is incorrect. Conductivity, as related to conduction 
current, 
>is a useful and appropriate concept for any real material, even 
>dielectrics. Now conductivity may be practically zero, for good PCB 
>substrate material. Because of this, the conduction current in a good 
>dielectric is also very small, but the equations remain valid.
>
>Also, displacement current has nothing to do with moving charge in a 
>dielectric. It has to do with time varying electric fields producing 
>magnetic fields and conservation of charge. It is a little hard to 
explain 
>so I will state what it is and then walk you through Maxwell's 
postulation 
>of the effect.
>
> From Maxwell's equations:
>
>curl(H) = J + d(D)/dt
>
>The displacement current is the d(D)/dt part. Thus:
>
>curl(H) = J + Jd
>
>Here is how we/Maxwell arrive at that. For static fields, the field 
>equations are:
>
>1) div(D) = rho (charge density)
>2) div(B) = 0 (no magnetic charges)
>3) curl(E) = 0
>4) curl(H) = J (electric current density)
>
>Now conservation of charge demands that:
>
>5) div(J) = -d(rho)/dt
>
>What this equation is saying is that the current flowing out of a small 

>volume is equal to the negative of the time rate of change of the 
charge 
>within the volume. If we apply the divergence operation to equation 4 
about 
>we get:
>
>6) div(J) = div(curl(H)) = 0            (The divergence of the curl of 
any 
>vector is zero)
>
>This result is clearly at odds with eqn. 5, and violates the 
conservation 
>of charge. Maxwell recognized that eqn. 4 was not complete and proposed 
an 
>extension to the static field equations as follows:
>
>7) curl(H) = J + d(D)/dt
>
>Now when the divergence operation is applied to eqn. 7, we get:
>
>8a) div(J) + div(d(D)/dt) = 0
>8b) div(J) = -div(d(D)/dt) = -d(div(D))/dt = -d(rho)/dt
>
>And charge is conserved, as 8b is clearly the same as eqn. 5.
>
>It is the additional component that Maxwell added to the statics 
version of 
>eqn. 7 that is called displacement current. It is so named because is 
>arises from the displacement vector "D." The added term contributes to 
the 
>curl of the magnetic field in the same way as an actual conduction 
current 
>density "J" does. But, displacement current can be non-zero even in a 
>vacuum, where there is no charge at all.
>
>Regarding the units, as another writer indicated, the loss tangent is 
>unit-less. There are several different forms for the loss tangent 
depending 
>on what approximations you want to make and how you want to express 
things. 
>The formulation I used is very general. If you look at the numerator of 
the 
>loss tangent as I stated it:
>
>tan(delta) = (we''+sigma)/(we')   [or rearranged:  tan(delta) = (e'' + 

>sigma/w)/e']
>
>we'' is really indistinguishable from sigma mathematically. They add 
>directly. Physically, however, the we'' term arises from the work done 
to 
>move bound charges. This is how water heats in a microwave oven. These 

>charges move a very small distance within the material creating a 
dipole 
>moment in the material (Pe) that reduces the field strength within the 

>material (think about Q=CV, if capacitance goes up with increasing er, 

>voltage comes down, for fixed charge). If in that process, any work is 

>done, heat is generated and the we'' term becomes non-zero. Some 
references 
>do not make this distinction and lump the we'' term with sigma. There 
is no 
>real problem with this. It gives the same answer, but it can lead to 
>confusion because the loss tangent is defined a little differently in 
this 
>case, as follows:
>
>e = e' -j(sigma)/(w) = e' - je''
>
>and
>
>tan(delta) = e''/e'
>
>With regard to what Howard Johnson proposes for loss tangent, I suspect 
he 
>is trying to make things a little easier than I have. Capacitance, 
>capacitive reactance, and resistance are pretty ill-defined (or 
undefined) 
>terms relating to a general dielectric material. I would not attempt to 

>make a circuit based analogy in this way. Circuit theory is merely a 
>simplification of Maxwell's equations for low frequencies. I would have 
to 
>see Howard's treatment of loss tangent using R's and Xc's to be certain 
but 
>I have never seen it defined other than I have described, and I have 
>checked it across multiple EM references. That doesn't mean it hasn't 
been 
>done, for good or ill.
>
>Hopefully this is at least sort of clear...!
>
>Regards,
>
>Rob Hinz
>Principal Engineer
>SiQual Corporation
>rob@xxxxxxxxxx
>phone (503)885-1231
>fax   (503)885-0550
>http://www.siqual.com
>At 04:24 PM 4/10/2002 -0700, RayCaliendo@xxxxxxxxxx wrote:
>
>
>>         Rob et. al.,
>>
>>         I believe the word 'conductivity' (sigma) should be used  for 
a
>>conductor, while the movement of charge in a dielectric is the 
'Displacement
>>current' (D = eE), which, if I understand it correctly, behaves "like" 
a
>>conduction current.  Also, It looks to me that the units of some of 
the
>>equations' here don't seem to balance. What have I missed?  I found 
some
>>other explanations for loss tangent :
>>                 - Tan(delta) = er'' / er'
>>                 - Howard Johnson article "Dielectric Loss Tangents"
>>                         Theta = Im(Capacitance) / Re (Capacitance)
>>                 - Tan(delta) = Resistance / Reactance (parallel 
equivalent
>>circuit)
>>
>>         Regards,
>>
>>         Ray Caliendo
>>         Solectron Corp
>>         (408)956-6294
>>
>> > ----------
>> > From:         Rob Hinz[SMTP:rob@xxxxxxxxxx]
>> > Reply To:     rob@xxxxxxxxxx
>> > Sent:         Tuesday, April 09, 2002 2:29 PM
>> > To:   Patrick_Carrier@xxxxxxxx
>> > Cc:   si-list@xxxxxxxxxxxxx
>> > Subject:      [SI-LIST] Re: what is the conductivity of a 
dielectric?
>> >
>> >
>> >
>> > Patrick,
>> >
>> > The definition of loss tangent, tan(delta) is:
>> >
>> > tan(delta) = (we'' + cond)/(we')
>> >
>> > Where:
>> >
>> > w = 2*pi*freq
>> > e' = eo*er (dielectric constant real part) This is the one we are 
used to
>> > seeing...
>> > e'' = imaginary (and therefore loss generating) part of the 
dielectric
>> > constant
>> > cond = electrical conductivity of the material.
>> >
>> > Thus, in general, the dielectric constant is expressed as a complex 
number
>> > as:
>> >
>> > e = e'-je''
>> >
>> > Now to your question, if you assume that the dielectric is 
otherwise
>> > lossless, that is, e''=0, then conductivity is:
>> >
>> > cond = tan(delta)*2*pi*freq*eo*er.
>> >
>> > So I would agree with the equation you propose except that it is 
missing a
>> >
>> > key term eo=8.854e-12. The should correct the scale problem you 
are
>> > noting...
>> >
>> > cond = .02*2*pi*100e6*8.854e-12*4 = 4.5e-4 S/m
>> >
>> > On background, the loss tangent equation is easily understood from 
first
>> > principles. If you recall the relationship between Electric flux 
(D) and
>> > Electric field (E) in free space:
>> >
>> > D = eo*E;
>> >
>> > the addition of a material to the space causes a polarization of 
the
>> > molecules of that material resulting in additional electric flux 
that can
>> > be represented as a polarization vector as:
>> >
>> > D = eo*E + Pe    (the same can be said of the magnetic field, for 
that, Pm
>> >
>> > is used)
>> >
>> > Pe is consequence of the applied E field and for linear materials,
>> > (generally true for the material we use in SI work), Pe = eo*Xe*E. 
Xe is
>> > the relative electric susceptibility of the material. In general, 
it may
>> > be
>> > complex resulting in the following:
>> >
>> > D = eo*E + Pe = eo*(1+Xe)*E = eo*er*E = e*E
>> >
>> > e = eo*(1+Xe) = e'-je''
>> >
>> > The complex part accounts for damping effects on the polarizing 
dipole
>> > vibrations. Like a finite Q tank circuit or a spring and dash pot, 
this
>> > loss is generally in the form of heat. You might ask why -je'' and 
not
>> > +je''? This is because choosing +je'' would violate the 
conservation of
>> > energy by allowing the dielectric to add energy to the system.
>> >
>> > Finally the equation for loss tangent can be arrived at using 
Maxwell's
>> > equations for time harmonic fields. I should point out that this is 
a
>> > sticky issue for those of us doing SI analysis in the time domain 
and wish
>> >
>> > to use the concept of loss tangent for that analysis. The 
assumption of
>> > constant loss tangent, brings with it all sorts of complex and 
probably
>> > non-causal time domain behavior. So BE CAREFUL!
>> >
>> > curl(H) = jwD + J   (J is electric current density, J = cond *E)
>> > curl(H) = jweE + cond*E
>> > curl(H) = jwe'E + (we'' + cond)*E
>> > curl(H) = jw(e'-je''-j(cond/w))*E
>> >
>> > As you can see here the e' term is the lossless part and 
j(e''+cond/w) is
>> > the "lossy" part and if we think of the lossless part, e', as being 
on the
>> >
>> > real axis and the "lossy" part (e'' + cond/w) as being on the 
imaginary
>> > axis and we take the ratio of imaginary and real parts to get a 
"tangent"
>> > that gives us a loss perfomance metric:
>> >
>> > tan(delta) = (we''+cond)/(we')
>> >
>> > for a SINGLE frequency!
>> >
>> > I hope this helps your understanding.
>> >
>> > Rob Hinz
>> > Principal Engineer
>> > SiQual Corporation
>> > rob@xxxxxxxxxx
>> > phone (503)885-1231
>> > fax   (503)885-0550
>> > http://www.siqual.com
>> >
>> >
>> >
>> >
>> > At 01:33 PM 4/9/2002 -0500, Patrick_Carrier@xxxxxxxx wrote:
>> >
>> > >Transmission line gurus and people who love dielectrics--
>> > >
>> > >I am trying to figure out the conductivity of a dielectric.
>> > >I have an equation that gives me:
>> > >tanD = 1/(2*pi*Freq*Er*rd) where rd is the resistivity of the 
dielectric
>> > >I assume that 1/rd is the conductivity of the dielectric.  Is that 
an
>> > >erroneous assumption?
>> > >That gives me the equation:
>> > >conductivity of dielectric = 2*pi*Freq*Er*tanD
>> > >
>> > >This second equation makes sense to me in that increasing your 
frequency
>> > >increases the dielectric conductivity, causing more "leakage" of 
your
>> > >transmitted energy.  However, using this equation, that would 
indicate
>> > that
>> > >the conductivity of a dielectric with Er=4 and tanD=0.02 would 
have a
>> > >conductivity approaching that of copper at 100MHz.  Now that does 
not
>> > make
>> > >sense.
>> > >
>> > >Is there a such thing as non-frequency-dependent conductivity of 
a
>> > >dielectric?  How would I obtain such a number?
>> > >Is there something else I am missing?
>> > >
>> > >Any guidance would be greatly appreciated.  Thanks.
>> > >--Pat
>> > >
>> > >
>> > >
>> > >
>> > 
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>> >
>> > Rob Hinz
>> > Senior Electromagnetics Specialist
>> > SiQual Corporation
>> > rob@xxxxxxxxxx
>> > phone (503)885-1231
>> > fax   (503)885-0550
>> > http://www.siqual.com
>> >
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>
>Rob Hinz
>Principal Engineer
>SiQual Corporation
>rob@xxxxxxxxxx
>phone (503)885-1231
>fax   (503)885-0550
>http://www.siqual.com
>
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