From: Vinu Arumugham (email@example.com)
Date: Tue Apr 04 2000 - 12:45:51 PDT
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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