[SI-LIST] Re: Return currents

  • From: olaney@xxxxxxxx
  • To: venki.pras@xxxxxxxxx
  • Date: Mon, 5 Nov 2007 08:36:47 -0800

A1: Yes.  For that short pulse example, the return current is a local
phenomenon that travels with the forward pulse, and it will exist on both
nearby planes, split according to proximity but in sum equal to the
forward current.  If the trace is not in isolation, some of the current
will also be induced in nearby traces = crosstalk.
A2: Yes, image current will exist simultaneously on every adjacent plane
(split as per above), regardless of where each fits into the scheme of
the PDN.  The PDN design must account for this by keeping a low HF
impedance between adjacent planes, not just at the power supply, but
continuously along the path of each signal trace placed between them. 
This is done by keeping the separation small, and adding properly
implemented bypass caps.  In the case of diff pairs, return currents
still exist to the extent that the finite distance between conductors
results in image currents that are also slightly separated, and therefore
sum to zero only in the aggregate.  For side by side, the aggregate
cancellation happens on each image plane.  For over/under, the aggregate
cancellation requires local current flow between planes, provided by
local interplane capacitance.

On Mon, 5 Nov 2007 10:03:48 -0500 "Pras venki" <venki.pras@xxxxxxxxx>
writes:
Thanx a lot for replying..i guess it is a bit more clear 2 me than b4...

do u mean 2 say that at an instant of time the current pulse can exist on
the signal trace (flowing away frm the driver) and right at that instant
the 'reverse' current pulse can exist on the Vdd plane (flowing into the
driver)? 

also that an image current can exist on the Vdd plane at the same time
when the forward current is flowing (in case the forward current pulse is
wide enough). Just that they have 2 sum 2 give the current actually
required to charge the load (KCL)? 

Thanx a lot,
Pras


 
On 11/5/07, olaney@xxxxxxxx <olaney@xxxxxxxx> wrote: 
Others are doing a fine job of explanation, but to help clarify, keep in
mind the basic principle that current can and does "flow" through 
insulators -- in the form of capacitance charging and discharging
currents.  This applies to interplane capacitance as well as discrete
bypass caps.  It is not necessary to think of current as arriving from
and returning to the power supply unless you are at a frequency lower 
than the bandwidth of the regulator loop.  Past a few tens of kilohertz,
the PS only supplies average current, and capacitance does all the real
work.  The closer the capacitor is to the load, the harder it works.  At 
high enough frequencies, even the storage caps in the power supply cease
to matter.  HF performance is often determined almost entirely by the
capacitance in the package and within a few inches of the pin.  In fact, 
HF currents never reach the power supply for properly designed boards,
and it is not a meaningful participant in the current demands of fast
edges.

The term "return current" in context means reverse current flows induced 
by signals propagating in the forward direction.  By KCL, these must sum
to the same magnitude, though the finite speed of propagation means that
this is on an instantaneous, local basis.  For short pulses, it is 
possible to have current (and corresponding return current) flowing in
that part of the path where the pulse is at a given instant, but zero (or
some other value) on either side.  The fact that the planes on either 
side might be at different DC potentials is irrelevant.  That only enters
in by way of requiring dielectric separation between layers, i.e.
capacitance, however implemented.  It is the adequacy of that
implementation that is at issue, not whether any given plane is ground or

other reference in the DC sense.  The return current will split according
to proximity to nearby planes without regard for the DC use of those
planes.  The corresponding local voltages are a matter of the impedance 
presented by the local PDN.

Orin Laney

On Sun, 4 Nov 2007 21:44:27 -0500 "Pras venki" <venki.pras@xxxxxxxxx>
writes:
> Hello Guys,
> I have this confusion regarding "*Return currents*"- 
>
> 1) In the paper "SSN & power plane bounce in CMOS technology" by
> Larry Smith
> ( http://www.csee.umbc.edu/vlsi/reports/ssn_pwr_planes.pdf 
> <http://www.csee.umbc.edu/vlsi/reports/ssn_pwr_planes.pdf+>).  This
> is
> available online for free, so i m pasting it here. (I hope i can) 
>
> In the following excerpt-
>
> "Suppose the transition is from low to high and the cross-section of
> the
> package has the transmission line located above a Vdd plane as shown 
> in
> figure 1.The driver connects the Vdd plane to the transmission line
> through
> a low impedance.Current flows from the Vdd plane onto the
> transmission line
> which is low, say ground potential.As the wave front propagates down
> the
> transmission line, charge flows into the capacitance between the
> trace and
> the Vdd plane, raising the potential on the trace up to Vdd. The
> current
> path is complete because charge from the Vdd plane flows in a
> complete loop
> from the Vdd plane, through the driver and onto the transmission
> line that
> is referenced to the Vdd plane. If there is a ground plane 
> underneath the
> Vdd plane, it is not disturbed because it is not part of the current
> loop."
>
> Where does the return current flow? Does the return current flow
> through the 
> inter-plane capacitance? Doesn't it need to flow thru a reference
> plane?
>
> If it can't flow thru the Gnd plane, is it possible for it to flow
> thru the
> same power plane which is supplying the current (i hope not coz it 
> will
> totally screw my fundamentals on current flow).
>
>  2)What if the package does not have an explicit power or ground
> plane
> i.e. Power
> & Gnd are normal, thin traces (like other signal traces), 
> distributed
> sporadically along with other I/Os, signal, clock traces etc.
> (Although the
> power & ground traces are de-coupled inside the chip & if the I/O
> traces r
> driven using a CMOS buffer) 
>
> How is the return current going to flow now, when the I/O traces r
> driven
> using the buffer?
>
> Will the return current still look to flow through the nearest Power
> or Gnd 
> reference trace it finds, given they are randomly routed in the
> package like
> ordinary traces?
>
> I'd really appreciate if somebody can clear this nagging doubt.
>
> Thanx in advance. 
>
> Regards,
> pras
>
>
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