[SI-LIST] Re: Lumped capacitance estimation
- From: steve weir <weirsi@xxxxxxxxxx>
- To: sushil.gupta@xxxxxx, a.ingraham@xxxxxxxx
- Date: Thu, 07 Apr 2005 06:05:07 -0700
Sushil, that capacitance you are trying to model does not exist in a vacuum
w/o incremental inductance that oppose the changing currents needed to
charge the line.
For purposes of argument, assume that the propagation velocity is
6"/ns. Every fs of that 600ps rising edge, the wave propagates 0.006 mils
away from the driver, and does so through an incremental
inductance.equivalent to the partial inductance for that bit of the
transmission line length. So, while you increase the capacitance of the
wire by lengthening it, each incremental bit of capacitance is decoupled
more and more from the driver by the inductance along the way. Let us
assume the inductance is about 6.8nH / ". Would it make anymore sense for
you to ask what the equivalent inductance of the line should be to get the
same loaded wave form as your present question concerning the
capacitance? Both effects are definitely present and interact with each
other. Surely if you break the line into fs long LC stages you can get an
accurate representation of the behavior. You might be unhappy with the
solution time for a 500,000 stage LC network. Perhaps you can approximate
with fewer stages and still get a decent answer.
Steve
At 04:55 PM 4/7/2005 +0530, Sushil Kumar GUPTA wrote:
>Hi Andrew,
> Perhaps I couldn't explain the problem in right way.
> Suppose you have a transmission line (cable) which is
>very long let's say 30ns delay. If we assume cable has no loss then
>whatever waveform you apply, you will get FAR-END wave after 30ns delay.
>I am assuming both end are perfectly matched.
> Now if you cansider the NEAR-END rise/fall time (let's assume
>avg value 600ps) , it will not depend on the length of the cable. So my
>question is that waht will be the approximate value of capacitive load
>at the NEAR-END (this time no cable is used) which will provide the same
>rise/fall time achieved with cable.
> I can state the problem in a different way also. Let's assume
>that IO driver has 600ps rise/fall time with 10pf capacitive load. If I
>increase the capacitive load obviously rise/fall time will increase.
> The PCB trace has 2.72pf/inch capacitive load. If now I connect cable(
>PCB trace) with IO driver, it will see 2.72pf/inch capacitve load. The
>driver has 45Ohm dc impedance w.r.t ground. So when driver starts
>charging PCB trace, every inch of trace will be charged, but only that
>part of PCB trace will impact the rise/fall time which is travelled
>during transition time. So how will I know how many inches to be
>considered.
>Andrew Ingraham wrote:
>
> >> I need to calculate equivalent lumped capacitance seen by an IO
> >>driver which is connected to PCB trace (approx 3 Inches) and then cable.
> >>The average rise/fall time is 600ps. I have information about the PCB
> >>trace capacitance/inches. Will my assumption be correct if I take
> >>PCB-trace length (for lumped-capacitance calculation) which provides 300ps
> >>delay.
> >>
> >>
> >
> >For lumped capacitance, just multiply capacitance/length times the length.
> >
> >The rise/fall time doesn't make any difference as far as equivalent lumped
> >capacitance is concerned. However, if the rise/fall time is fast enough,
> >you shouldn't be using a lumped capacitance in simulations.
> >
> >Are you sure your PCB trace delay is only 100ps/inch?
> >
> >
> >
> >>What would be the best way to simulate IO driver as far as rise/fall time
> >>accuracy is concerned assuming lumped model for Tx line.
> >>
> >>
> >
> >For accuracy, best way = abandon the lumped model for the line.
> >
> >What did you mean by "best way" anyway? If you have an I/O driver model and
> >a transmission line model, just simulate them and look at the results.
> >
> >Regards,
> >Andy
> >
> >
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