[SI-LIST] Re: Should the signals always return back through GND

Hello, Bi,

It is interesting to see the chip perspective on this reflector.  There is a
lot in common.

Metal on adjacent layers provides less effective shielding these days, since
the traces are now higher than they are wide.  But you could consider a
"wide" power/gnd trace on the layer above or below the critical trace for
providing a return path, without having a complete plane.  You would still
want to leave space between your critical trace and adjacent traces on the
same layer.

You might want to check some of the EDA sites, since chip-level RLC
extraction tools are getting better.

Best regards,
Lynne


-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On
Behalf Of steve weir
Sent: Wednesday, August 24, 2005 8:48 AM
To: Bi Han
Cc: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: Should the signals always return back through GND

Bi, yes I was looking at it as a PCB with the assumption that the aggressor
had some return path.  In your example, it seems there was no defined return
path to start with.  That is a bad place to be.  You are right, I would
prefer to view the "shield" as the second half of a CPW.
Regards,

Steve.
At 08:41 AM 8/24/2005 -0700, Bi Han wrote:
>Steve:
>Thanks for pointing out the problem with so detailed illustration.
>There a key difference between our standing point. You assumed that the 
>trace is on the PCB board, so there is GND plane below the signal trace.
>The GND plane will introduce a relative good return path for the signal 
>trace, introducing another nearby return path will not help too much on
EMI.
>I agree with your point on standing wave phenomenon in shielding trace, 
>even both ends are grounded.
>I assumed that the signal trace is on chip, where there is no good GND 
>plane on the chip. Introducing any nearby shielding trace will greatly 
>help "collecting" return current. You could view shielding and sig 
>trace as a pair of diff-trace, or a loop of current flow, differential 
>trace will greatly reduce EMI, if the victim is not too close to the pair.
>thanks,
>Han
>
>steve weir <weirsi@xxxxxxxxxx> wrote:
>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" 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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