[SI-LIST] Re: SSN Vs Load cap

Ihsan,
Your question confuses me. The physical meanings are
the laws behind the Maxwell's equations. I don't know
any new law more than that.

Yu
--- Ihsan Erdin <erdinih@xxxxxxxxx> wrote:

> Yu,
> 
> I'm guessing you're referring to the equation
> involving the 2nd time
> derivative of current when you say it has no
> problem. No one here is
> saying it has mathematically any problem but if you
> can explain the
> physical meaning of the 2nd derivative of current
> wrt time you're
> going to teach me something.
> 
> Regards
> 
> Ihsan
> 
> On 2/2/07, Yu Wang <wangy_km@xxxxxxxxx> wrote:
> > Ihsan & Canes,
> > To my knowledge of physics and math, Canes's
> equation
> > has no problem. However, the equation is a classic
> > boundary problem which means the solution of
> equation
> > is very dependent on the start status and boundary
> > conditions.  so di/dt can be more or less. In
> general,
> > because of the driving capability of the driver is
> > limited, so increasing C can make di/dt smaller.
> >
> > Regards,
> > Yu
> >
> >
> > --- Ihsan Erdin <erdinih@xxxxxxxxx> wrote:
> >
> > > Canes,
> > >
> > > Sabayachi's explanation is plain and accurate.
> > > i=Cdv/dt is a lumped
> > > approximation to Maxwell-ampere eq.; so is
> v=Ldi/dt
> > > to
> > > Maxwell-Faraday. As such, both have physical
> > > connotations. Your
> > > di/dt=Cdv2/dt2 is mathematically correct but
> doesn't
> > > have any useful
> > > physical connotations.
> > >
> > > Regards
> > >
> > > Ihsan
> > >
> > > On 2/2/07, Canes Venatici <starsilic@xxxxxxxxx>
> > > wrote:
> > > > Upto the maximum current driving capability of
> > > driver, the increase in load capacitance causes
> > > > the current rate to increase and noise
> increases,
> > > so SSO index should reduce.
> > > > Once its reached the maximum drive capability
> at a
> > > particular load cap, the increase of cap
> > > > causes the rate of voltage to fall, as per the
> > > max. current spec. But since the current remains
> > > same (as the driver
> > > > cant supply more than that), I expect SSO
> index to
> > > be unchanged.
> > > > Equations :
> > > > i  = c. dv/dt
> > > > di/dt = c.(d2v/dt2) ....(its d square v by dt
> > > square)
> > > > sso noise = L.di/dt = L.c.(d2v/dt2).
> > > >
> > > > Please clarify, if I'm missing anything.
> > > >
> > > > Regards
> > > > Canes
> > > >
> > > > ----- Original Message ----
> > > > From: Sabyasachi Mohapatra
> <sabbu1981@xxxxxxxxxxx>
> > > > To: si-list@xxxxxxxxxxxxx
> > > > Sent: Thursday, February 1, 2007 6:13:12 PM
> > > > Subject: [SI-LIST] Re: SSN Vs Load cap
> > > >
> > > > It is pretty simple.
> > > > as u increase the load cap with same drive
> > > strength of driver,
> > > > rise or fall slew will degrade.
> > > > As the buffers will be switching slower now,
> dI/dt
> > > will be less,
> > > > rate of change of current though package
> inductor
> > > will be less.
> > > > So less SSO noise.
> > > >
> > > > --- In si-list@xxxxxxxxxxxxxxx, Canes Venatici
> > > <starsilic@...> wrote:
> > > > >
> > > > > Hi all,
> > > > > I couldn't understand why if load
> capacitance is
> > > increased, SSN is
> > > > reduced (as the recommendation is increased
> SSO
> > > index, more signal
> > > > pads can be accommodated by power pads).
> > > > > As di/dt is directly proportional to load
> cap,
> > > increasing the load
> > > > causes more drop in parasitic inductance and
> SSN
> > > should increase.
> > > > > Could anyone clarify.
> > > > > Thanks,
> > > > > Canes
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > >
> > >
> >
>
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=== message truncated ===





 
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