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 >> > > >> > > >> > > >> > > >> > >------------------------------------------------------------------ >> > >To unsubscribe from si-list: >> > >si-list-request@xxxxxxxxxxxxx with 'unsubscribe' in the Subject field >> > > >> > >or to administer your membership from a web page, go to: >> > >//www.freelists.org/webpage/si-list >> > > >> > >For help: >> > >si-list-request@xxxxxxxxxxxxx with 'help' in the Subject field >> > > >> > >List archives are viewable at: >> > > //www.freelists.org/archives/si-list >> > >or at our remote archives: >> > > http://groups.yahoo.com/group/si-list/messages >> > >Old (prior to June 6, 2001) list archives are viewable at: >> > > http://www.qsl.net/wb6tpu >> > > >> > >> > Rob Hinz >> > Senior Electromagnetics Specialist >> > SiQual Corporation >> > rob@xxxxxxxxxx >> > phone (503)885-1231 >> > fax (503)885-0550 >> > http://www.siqual.com >> > >> > ------------------------------------------------------------------ >> > To unsubscribe from si-list: >> > si-list-request@xxxxxxxxxxxxx with 'unsubscribe' in the Subject field >> > >> > or to administer your membership from a web page, go to: >> > //www.freelists.org/webpage/si-list >> > >> > For help: >> > si-list-request@xxxxxxxxxxxxx with 'help' in the Subject field >> > >> > List archives are viewable at: >> > //www.freelists.org/archives/si-list >> > or at our remote archives: >> > http://groups.yahoo.com/group/si-list/messages >> > Old (prior to June 6, 2001) list archives are viewable at: >> > http://www.qsl.net/wb6tpu >> > >> > >>------------------------------------------------------------------ >>To unsubscribe from si-list: >>si-list-request@xxxxxxxxxxxxx with 'unsubscribe' in the Subject field >> >>or to administer your membership from a web page, go to: >>//www.freelists.org/webpage/si-list >> >>For help: >>si-list-request@xxxxxxxxxxxxx with 'help' in the Subject field >> >>List archives are viewable at: >> //www.freelists.org/archives/si-list >>or at our remote archives: >> http://groups.yahoo.com/group/si-list/messages >>Old (prior to June 6, 2001) list archives are viewable at: >> http://www.qsl.net/wb6tpu >> > >Rob Hinz >Principal Engineer >SiQual Corporation >rob@xxxxxxxxxx >phone (503)885-1231 >fax (503)885-0550 >http://www.siqual.com > >------------------------------------------------------------------ >To unsubscribe from si-list: >si-list-request@xxxxxxxxxxxxx with 'unsubscribe' in the Subject field > >or to administer your membership from a web page, go to: >//www.freelists.org/webpage/si-list > >For help: >si-list-request@xxxxxxxxxxxxx with 'help' in the Subject field > >List archives are viewable at: > //www.freelists.org/archives/si-list >or at our remote archives: > http://groups.yahoo.com/group/si-list/messages >Old (prior to June 6, 2001) list archives are viewable at: > http://www.qsl.net/wb6tpu > > > ---------End of Included Message---------- _____________________________________________________________ ------------------------------------------------------------------ To unsubscribe from si-list: si-list-request@xxxxxxxxxxxxx with 'unsubscribe' in the Subject field or to administer your membership from a web page, go to: //www.freelists.org/webpage/si-list For help: si-list-request@xxxxxxxxxxxxx with 'help' in the Subject field List archives are viewable at: //www.freelists.org/archives/si-list or at our remote archives: http://groups.yahoo.com/group/si-list/messages Old (prior to June 6, 2001) list archives are viewable at: http://www.qsl.net/wb6tpu