[SI-LIST] Re: Should the signals always return back through GND
- From: steve weir <weirsi@xxxxxxxxxx>
- To: mike_bihan@xxxxxxxxxxxx, lgreen22@xxxxxxxxxxxxxx
- Date: Wed, 24 Aug 2005 05:30:43 -0700
Bi, I see one of my sentences sounds circular: "That increases coupling
because ... is itself a coupler to the victim." To make this very clear,
when we insert that "guard" or any other trace, the field redistributes
increasing the flux density in the region of the "guard", which is to say
the region of the victim, and thus forming the coupling antenna.
Steve.
At 05:20 AM 8/24/2005 -0700, steve weir wrote:
>Bi, no that is backwards. It is at high frequencies where we can't ignore
>wave effects that parallel shield traces can offer little benefit or worse
>create resonant couplers.
>
>There is an excellent treatment on this subject in Eric Bogatin's book,
>"Signal Integrity Simplified"
>
>Let's start with a coaxial shield.all around the victim trace. As long as
>the shield is thick enough skin effect in the shield prevents any
>measurable aggressor energy from penetrating to the victim. The important
>point here is that the aggressor fields have to penetrate the thickness of
>the shield to reach the victim. An equivalent shield in a PCB would have
>to be in the Z axis, such as a via fence.
>
>Now, look at what happens with a shield or guard trace on a PCB. It
>doesn't surround the victim. What it does is to bring another chunk of
>wave guide closer to the victim. That increases coupling from the
>aggressor because unlike the coax shield where the aggressor fields have to
>penetrate the shield to reach the victim, the parallel shield is itself a
>coupler to the victim. Now if we don't attach the shield to anything, (
>the evil of floating metal ), it becomes an increasingly efficient antenna
>at rising frequencies until the point that it becomes a quarter wave
>resonator. So, to make it an inefficient antenna and to prevent it from
>resonating, we need to drive a bunch of vias through it spaced
>substantially closer than one quarter wavelength of any strong harmonic in
>the aggressor.
>
>Eric points-out, as has Dr. Johnson in his first book, that by the time you
>allow enough space for the via clearances, and assuming you add enough
>vias, the improvement in isolation is slight compared to what you get by
>leaving the guard trace out. But if you are going to pay for those vias,
>do yourself a favor and build a via fence inside of a cavity where both
>planes are the return for the aggressor. This effectively builds a coax
>shield around the aggressor. Leave the parallel trace out.
>
>Steve.
>At 04:54 AM 8/24/2005 -0700, Bi Han wrote:
> >Green:
> >
> >I agree that shielding could cause problem in "capacitive coupling case".
> >However, if the inductive coupling dominate, shieling should help a lot.
> >
> >Since the current will mainly return in adjacent shieling trace instead of
> >"relative" faraway victim trace, most of inductive coupling should be
> >shielded. It will not be reflected in mutual inductance matrix directly,
> >but will show its impact after matrix reduction.
> >
> >thanks,
> >Han
> >"Lynne D. Green" <lgreen22@xxxxxxxxxxxxxx> wrote:
> >Hello, Hermann,
> >
> >Shielding can cause new problems. A shield trace can pick up crosstalk, and
> >then couple that crosstalk onto a third trace. I know of at least one
> >design that failed in this manner. The secondary coupling was in an
> >unexpected area of the board, making it hard to debug.
> >
> >There are two common shielding approaches. First, one could remove the
> >shield trace and leave the larger trace spacing in place (decreasing the
> >capacitance between traces). Second, one could add enough vias (to the
> >desired DC voltage) to make sure the shield does not conduct crosstalk noise
> >very far.
> >
> >Good SI tools can be used to check the crosstalk conduction and secondary
> >coupling.
> >
> >Best regards,
> >Lynne
> >
> >PS: more history on GND: Breadboard circuits had much higher impedances in
> >the circuit that in the connecting wires, so students (and junior engineers)
> >treated all points along the wire as GND. Given this earlier training,
> >learning distributed transmission line theory was very difficult for most of
> >them.
> >
> >
> >
> >"IBIS training when you need it, where you need it."
> >
> >Dr. Lynne Green
> >Green Streak Programs
> >http://www.greenstreakprograms.com
> >425-788-0412
> >lgreen22@xxxxxxxxxxxxxx
> >
> >
> >-----Original Message-----
> >From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On
> >Behalf Of hermann.ruckerbauer@xxxxxxxxxxxx
> >Sent: Tuesday, August 23, 2005 2:14 AM
> >To: a.ingraham@xxxxxxxx; nikitanivan@xxxxxxxxxxx
> >Cc: si-list@xxxxxxxxxxxxx
> >Subject: [SI-LIST] Re: Should the signals always return back through GND
> >
> >=20
> >
> >
> >Hi Andy,
> >
> >Good answer!
> >
> >That points me to one question that I have had quite a while ago ...
> >Just from a high level point of view I would try to get the same current
> >returen for DC and AC. So If I do have a high level terminated signal (e. g.
> >to VDD) I would try to do the referencing/shielding also with VDD to avoid
> >any referencing/return crossing even if it is just between DC
> >and AC (btw. What is DC what is AC in this discussion ;-) )!
> >
> >Of course I might get some disadvantages by doing so. Usually the most
> >stable signal is called GND (whatever it really is ...). So any referencing
> >to VDD might disturb my signal by talking to it due to any noise on the
> >reference!
> >
> >What is your/the groups opinion on this ?
> >
> >Thanks
> >
> >Hermann
> >
> >=20
> >
> >
> > >
> >
> >-----Original Message-----
> >From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx]
> >On Behalf Of Andrew Ingraham
> >Sent: Monday, August 22, 2005 6:24 PM
> >To: nikitanivan@xxxxxxxxxxx
> >Cc: si-list@xxxxxxxxxxxxx
> >Subject: [SI-LIST] Re: Should the signals always return back through GND
> >
> > > Its always said that GND acts as a return path for a signal.
> >
> >Really? I wasn't aware of this. If everyone says that where you come from,
> >then maybe you should educate your teachers and/or co-workers that they are
> >wrong.
> >
> >What we call GND is often used as a reference point for VOLTAGE, but this
> >doesn't mean it is where CURRENT flows.
> >
> > > Do all the signals know that they
> > > have to return back through GND and not through anyother track in
> > > the=20 signal layer having lower impedance? What are the factors taken
> > > into=20 consideration to make sure that signals return back through GND.
> >
> >At low frequencies and DC, signal currents return to wherever they came
> >from, to complete the loop.
> >
> >If you take a signal source and connect one end of the source to GND and the
> >other end of it to a wire that goes off somewhere to some load, then the DC
> >return path will have to get back to GND to complete that loop.
> >
> >If you take that same source and connect the first end of it to VDD rather
> >than to GND, then the DC return path will have to end up at VDD and not GND.
> >It might go by way of GND in order to get back to VDD, but that depends on
> >the topology, the load, etc. The exact path (GND plane vs. VDD plane, etc.)
> >that it takes, depends on the relative DC resistances and low frequency
> >impedances of those paths.
> >
> >Thus, when the pull-up transistor in a pull-up/pull-down (or
> >"totem-pole") pair is on, the return path (at DC and low frequencies) will
> >have to get back to the VDD supply net that connects to that output driver.
> >GND might not be involved at all.
> >
> >At high frequencies, signal switching current return paths for a wire or a
> >trace, are by way of any and every conductor that is nearby, to which field
> >lines can be drawn. Most of the high frequency current chooses the path(s)
> >with the lowest impedance. The signal propagates via an electro-magnetic
> >field in the dielectric, which just happens to cause currents to flow in all
> >conductors that touch the E-M field.
> >
> >If a trace runs over a solid plane that connects to, say, some VTT voltage,
> >then return current will be in that VTT plane. If a trace runs between two
> >planes, the high frequency return current will be shared between both of
> >them, regardless of what voltage each one is connected to. If a trace runs
> >over one plane for several inches, then a different plane for several
> >inches, and so on, then the return current will have to (or try to) find its
> >way on each of these planes, in series.
> >
> >The DC path may take a totally different route. That is, the initial
> >switching wavefront has a return path associated with it, determined by the
> >E-M field around and between the conductors; but after a few nanoseconds the
> >return current may take a different route, eventually determined by DC
> >resistances and not the E-M field anymore.
> >
> >Regards,
> >Andy
> >
> >
> >
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