From: Larry Smith (email@example.com)
Date: Tue Apr 04 2000 - 15:09:03 PDT
Vinu - The on-chip decoupling capacitance causes current to flow out the
power pin of a driving chip when the signal transition is from high to
The most obvious current path for the high to low transition is from
the signal pin, through the pull down device (NFET) and out the ground
pin. But, the ground pin has inductance and develops a voltage across
it. This is usually called "ground bounce". The chip internal ground
bounces up compared to the local PCB ground.
Charge is stored in the on-chip decoupling capacitor, Q=CV. The
capacitor attempts to keep the voltage between chip-gnd and chip-power
constant. As chip-gnd bounces up, so does chip-power with respect to
PCB nodes. Chip-power is now at a higher voltage than PCB-power and
current begins to flow _out_ of the chip power pin. The system behaves
as if the inductance of the power pin and ground pin are in parallel.
Displacement current is flowing through the chip decoupling capacitor
and diminishes the voltage across it by I=C*dV/dt. Charge stored in
the capacitor diminishes.
Note that if the signal is attached to a PCB stripline, sandwiched
between a power and ground plane, we have exactly what we want.
Assuming symmetry in the stackup, half of the PCB return current is on
the ground plane and half of the return current is on the power plane.
Half of the current comes out the driver ground pin and half of the
driver current comes out the power pin, assuming symetrical inductance.
If there had been no decoupling capacitance on the chip, all of the
current would have come out the ground pin. Displacement current would
flow between the PCB power and ground planes to enable the return
current to flow on the power plane, thus causing power plane bounce.
Discrete capacitors mounted on the power planes are generally not fast
enough to help much during the fall (rise) time of the driver.
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.
> Date: Tue, 04 Apr 2000 12:45:51 -0700
> From: Vinu Arumugham <firstname.lastname@example.org>
> 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
> > 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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