[SI-LIST] Re: Displacement vs conduction current Q & A from D r Archambeault
- From: Sitar Moniker <si_monkey2@xxxxxxxxx>
- To: gstokes@xxxxxxxxx, "'si-list@xxxxxxxxxxxxx'" <si-list@xxxxxxxxxxxxx>
- Date: Wed, 13 Oct 2004 11:35:36 -0700 (PDT)
Thank you for setting up the stage for what it takes to discuss this thread. In
particular, I liked your statement:
"For full understanding it is necessary to talk about electric field and
displacement, and the relation to charge flow and magnetic field."
Sorry, this is not what I asked for. Dr. Archambeault addressed what I asked
for. But, can we say that unconditionally? If Yes, end of discussion. If No,
under what conditions:
1. Displacement return current = conduction signal current ?
2. Above two currents flow at same speed ?
John Philips
Geoff Stokes <gstokes@xxxxxxxxx> wrote:
Just a fundamental point, if you will excuse me, addressing Sitar's
questions, which need to be answered properly for a deeper understanding of
electromagnetic theory or SI practice which is an application. Dr.
Archambeault has answered the practical side of this question, but which can
be answered more fundamentally. Dr. Archambeault's entertaining
introduction on the RMCEMC site is really *excellent* and encapsulates some
important guiding principles:
http://www.ewh.ieee.org/r5/denver/rockymountainemc/archive/2004/Sep/Maxwell.
pdf
In this he makes a point about "messy math". I believe the maths becomes
messy if you let it, but is essential to the understanding. =20
This answer is partly in the conservation of charge, divJ equals
-d(rho)/d(t) (I mean partial derivative everywhere in this discussion) where
J is the conduction current intensity (sometimes less helpfully called
"current density") and rho is the charge density. This means that where
static charge is not accumulated (the normal case in highly conductive
media), divJ must be zero. (The maths is crucial and incomplete in my
description here.) If divJ is zero everywhere, then the volume integral of
divJ inside a closed surface is zero and a current intensity integrated over
that surface is zero. The integration represents total current across the
surface. That closed surface could contain just a single connection point
node. Hence with accuracy for DC and low frequency applications, we then
have Kirchoff's Law of current, stating that the sum of currents into a node
must be algebraically zero. These currents are real currents involving flow
of charge on a conductor. For less conductive media, such as a dielectric
printed circuit board, and varying currents, divJ is non-zero. In fact we
say divJ' is zero where J' is the total current intensity, including both
conduction current J and "displacement current" dD/dt (partial derivative).
The debate on displacement current has raged for a century. According to
Bleaney and Bleaney, Maxwell introduced it to make the wave equations fit
the truth, perhaps an honourable reason, and a truth which seems to apply
for a wide range of conditions including non-ionizing radiation (DC to 300
GHz). If div(total current intensity) is zero everywhere, then again total
current into a finite volume is zero - hence the current in a via is the
same as the current on the track which feeds it and is the same as the
current passing through the capacitor connected to the track (provided
nothing else is connected to those nodes). *However*, there is a
displacement current in the surrounding dielectric, so again, the conduction
current gives an incomplete picture: there is a conduction current gathering
effect, so there appears to be a conduction current loss on the conductors.
This loss could be relatively large or small, depending on the material
properties, dimensions and frequency. An extreme example is a radiating
wire antenna, where the input conduction current decays along the wire
finally coming to zero at the wire end. A PCB also radiates to some extent,
producing coupling between tracks, vias and components and other PCBs and
systems.
For full understanding it is necessary to talk about electric field and
displacement, and the relation to charge flow and magnetic field. The idea
of Displacement field was introduced to account for polarization of a
dielectric - a sort of polarization intensity. Otherwise Displacement field
is directly proportional to electric field except in "funny" materials like
lithium niobate or titanium dielectrics (chip capacitors). In MKS or
Standard International (SI) units, the electric field E does not equal D in
a vacuum. That's just an accident. Displacement current is defined as
dD/dt. Taken together, J and dD/dt are equal to curlH according to Ampere's
Law, which is needed together with the laws of Faraday and Lenz (curlE
equals -dB/dt) to consider the complete electromagnetic wave and its
interaction with conduction current.
Sitar, I wonder if this is what you asked for.
Please excuse me if I have wasted anyone's time.
All best wishes
______________________________________________=20
Geoff Stokes
Systems Engineer, Signal Management Group
Zetex plc=20
Lansdowne Road, Chadderton, Oldham, OL9 9TY,=A0 UK=20
Tel direct:=A0 +44-161-622-4857=A0=A0 Switchboard: +44-161-622-4444
Fax:=A0 +44-161-622-4469=20
http://www.zetex.com=20
e-mail:=A0 gstokes@xxxxxxxxx=20
> -----Original Message-----
> From: Grasso, Charles [mailto:Charles.Grasso@xxxxxxxxxxxx]
> Sent: 12 October 2004 19:32
> To: 'si-list@xxxxxxxxxxxxx'
> Subject: [SI-LIST] Re: Displacement vs conduction current Q &=20
> A from Dr
> Archambeault
>=20
>=20
> Question:
> =20
> Hello Dr. Archambeault,
> =20
> Mr. Grasso provided your email and suggested to contact you directly.
> After going through your presentation slides, I have couple=20
> questions on
> displacement currents vs conduction currents.=20
>=20
> 1. Do they have comparable magnitudes?
> 2. Do they flow at comparable speeds?
> =20
> I appreciate any ideas you could share with me. I will not=20
> divulge any info
> without your permission.
>=20
> Thanks,
> Sitar
> =20
> =20
> Answer:
> Sitar,=20
>=20
> I am not sure exactly what you are asking, but your first=20
> question about
> comparable amplitudes........if you are talking about the=20
> effects where
> signal current flows through a via (for example), and there=20
> is no nearby
> 'ground' via, then all the signal current must flow through=20
> the dielectric
> as displacement current. So, in this case, the amplitudes=20
> are the same. If
> there is a 'ground' via near enough to have an effect (within=20
> ~ 200 mils)
> then the return current will split itself between the=20
> 'ground' via and the
> displacement current. the total will be the same as the=20
> original signal
> current. Th ebottom line is that current MUST return to its
> source.....somehow....and using the path of least inductance.=20
>=20
> the speed of these currents is the same, since the dielectric=20
> (around the
> conductor for the conduction current, or thru the dielectric for
> displacement current) is the same in both cases.=20
>=20
> I hope this helps=20
>=20
> bruce=20
>=20
> --------------------------------------------------------------
> --------------
> --------------------------------------------------
>=20
>=20
> Bruce Archambeault, Ph. D.=20
> Senior Technical Staff Member
> EMC CoC
> 919-486-0120
> t/l 526-0120
>=20
> =20
> Best Regards
> Charles Grasso
> Senior Compliance Engineer
> Echostar Communications Corp.
> Tel: 303-706-5467
> Fax: 303-799-6222
> Cell: 303-204-2974
> Email: charles.grasso@xxxxxxxxxxxx;=20
Email Alternate: chasgrasso@xxxxxxxx =20
=20
=20
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