FreeLists / si-list / [SI-LIST] Re: Ground bounce (was VOH datasheet question)

["Muranyi, Arpad" <arpad.muranyi@xxxxxxxxx>]

Edi,

Not necessarily.  Consider this situation:

The capacitor acts as an AC short.  This means that
as a (fast) buffer switches, it will initially have=20
to drive a full short circuit current.  Contrast=20
that with a resistive load, where the output voltage
will always follow a divider ratio as determined
by the buffer and the load resistor.  In this case
the current will never reach the levels that you
would see with a capacitive load.

Arpad Muranyi
Intel Corporation
------------------------------------------------------

-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] =
On Behalf Of Edi Fraiman
Sent: Tuesday, March 14, 2006 3:43 AM
To: hmurray@xxxxxxxxxxxxxxx; si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: Ground bounce (was VOH datasheet question)

Hi,

I concur with you.
Increased capacitive load reduce the slew rate on the driver outputs.
The slower the output slew rate, the lower the ground bounce will be.

Edi
=3D20


> -----Original Message-----
> From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]
> On Behalf Of Hal Murray
> Sent: Tuesday, March 14, 2006 12:26 PM
> To: si-list@xxxxxxxxxxxxx
> Cc: hmurray@xxxxxxxxxxxxxxx
> Subject: [SI-LIST] Ground bounce (was VOH datasheet question)
>=3D20
>=3D20
> > You could clarify this issue from Dr. Johnson article named "Big
> > Buffer" (http://www.sigcon.com/Pubs/news/8_07.htm).
>=3D20
> As usual, his stuff is very good reading.
>=3D20
> This paragraph caught my attention.
>=3D20
> > The main factors that affect SSO noise are (1) the aggregate amount
of
> > current switched by your IC, (2) the rise time of that current, and
> > (3) the number and quality of the power and ground connections
> > provided in the IC package. Most relevant to this discussion is the
> > total amount of current--more capacitive loading enlarges that
> > current, increasing the amplitude of SSO noise.
>=3D20
> Why does more capacitive loading make SSO worse?
>=3D20
> The previous discussion showed that you can compute the output
resistance
> of
> the driver from the short circuit current.  That doesn't depend upon
the
> capacitive loading.
>=3D20
> I think that capacitive loading will make the duration of the output
> current
> pulse longer.  It will slow down the rise time of the signal, but
that's
> the
> voltage rise time.  The current rise time is what causes ground bounce
and
> that won't change if you add more load capacitance.
>=3D20
> What am I missing?
>=3D20
> Is the turn-on time for a driver slow rather than instantaneous?
"Slow"
> meaning roughly the same as the (normal) rise time so the normal
> capacitance
> gets charged while the driver is turning on and the driver never
delivers
> the
> full short circuit current?
>=3D20
>=3D20
>=3D20
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>=3D20
>=3D20
>=3D20
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