Re: [SI-LIST] : Return path for stripline, two ground planes or o ne power and one ground?

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From: Larry Smith (ldsmith@lisboa.eng.sun.com)
Date: Fri Apr 07 2000 - 15:54:10 PDT


Thats right, the termination resistor is the only place where you need
to resolve return currents. Drops along the way or an unterminated
(open circuit) transmission line do not have a return current issue.
Charge is conserved on the transmission line,top and bottom power
planes in the case of an open circuit.

regards,
Larry Smith
Sun Microsystems

>Larry Smith wrote:
>
>>
>
>I am imagining a multi-drop bus with a termination at the end. Your description
>seems to assume a single receiver co-located with the termination. I don't think
>the receivers on a multi-drop bus need a
>decoupling capacitor, only the termination resistor will need one.
>
>> It turns out that decoupling capacitance is also needed at the
>> receiver. We have return current flowing on both the PCB power and
>> ground planes. That return current must be resolved at the receiver.
>> The original problem stated that the driver made a full rail to rail
>> transition and the far end was properly terminated to Vdd/2. A good
>> way to do that is with 2 resistors of value 2*Z0, one up to Vdd, the
>> other down to ground. If the receiver switches fully from high to low,
>> the receiver has 0 volts, and all of the termination current comes from
>> Vdd. (A zero impedance driver is required to do this, but the return
>> current concepts are well illustrated by this example.)
>>
>> There must be decoupling capacitance at the receiver in order to
>> resolve the transmission line return current that is flowing on both
>> the Vdd and ground planes. If there is no decoupling capacitance, the
>> power and ground planes will bounce in the vicinity of the receiver in
>> order to satisfy the return currents, just like the driver.
>>
>> regards,
>> Larry Smith
>> Sun Microsystems
>>
>> > Date: Tue, 04 Apr 2000 12:45:51 -0700
>> > From: Vinu Arumugham <vinu@cisco.com>
>> >
>> > Larry Smith wrote:
>> > >
>> > > At the driver IC, the transient current tries to go through one power
>> > > rail or the other. For a high to low transition, current comes into
>> > > the chip from the signal line and exits through the chip ground pin.
>> > > For a low to high transition, current comes in through the chip power
>> > > pin and exits through the signal pin. This is the case unless there is
>> > > significant decoupling capacitance on the chip. On-chip capacitance
>> > > tends to put the power and ground inductors in parallel (AC ground) so
>> > > that current can go in and out of both paths for either transition.
>> > >
>> >
>> > How can current flow through the power pin for a high to low transition
>even
>> with on-chip capacitance? Since the power pin is always at a higher potential
>> than the signal pin (with the exception of
>> > overshoots), there will be no current flowing out of the power pin at any
>> time.
>> >
>> > >
>> > > You are correct in saying that current travels on both the power and
>> > > ground planes in a stripline configureation (power/signal/ground
>> > > stackup). Decoupling capacitance is required at both the driver and
>> > > receiver ends to make this happen.
>> >
>> > It should be clarified that decoupling capacitance is needed at the
>> termination and not the receiver.
>> >
>> > > If nothing else, displacement
>> > > current goes through the parallel plate capacitance between the power
>> > > and ground plane, causing plane bounce.
>> > >
>>
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