From: Vinu Arumugham ([email protected])
Date: Fri Apr 07 2000 - 13:37:59 PDT
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 <[email protected]>
> >
> > 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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