[SI-LIST] Re: SSN Vs Load cap
- From: Jai Shanker <jswarrier@xxxxxxxxx>
- To: starsilic@xxxxxxxxx
- Date: Thu, 8 Feb 2007 01:47:06 -0800 (PST)
Hi Canes,
I think you may have missed some components in the
overall model.
I = C dv/dt
only when there is no other element involved. You are
considering a ckt I think which is at least an LC ckt
with L being the pin inductance, possibly an RLC ckt
with some parasitic resistances thrown in as well.
Eqn 1 is then no longer complete or correct for the I
in that ckt in response to a step driver output Vo.
Or factor out the drop in L (or L+R) and restrict
yourself to the voltage across the cap alone for dv/dt
and eqn 1 is back to valid.
regards,
Jai Shanker
--- Canes Venatici <starsilic@xxxxxxxxx> wrote:
> sso noise = L.di/dt = LCd2v/dt2 implies, for a
> constant L and once we'd reached the
> peak driving current, the increase in cap, reduces
> the dv/dt. So noise should be constant.
> But, since noise is prop. to d2v/dt2, the increase
> in cap makes, d2v/dt2 term to reduce much more than
> dv/dt.
> So the SSO noise reduces.
> This is simple as i and v are dual to each other.
> Correct me if I'm wrong.
> ----- Original Message ----
> From: Ihsan Erdin <erdinih@xxxxxxxxx>
> To: starsilic@xxxxxxxxx
> Cc: sabbu1981@xxxxxxxxxxx; si-list@xxxxxxxxxxxxx
> Sent: Friday, February 2, 2007 6:33:17 PM
> Subject: Re: [SI-LIST] Re: SSN Vs Load cap
>
> 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
> >
>
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