[SI-LIST] Re: Copper Fill --- correction
- From: "Jason D Leung" <Jason.d.Leung@xxxxxxxxxxx>
- To: scott@xxxxxxxxxxxxx, RayCaliendo@xxxxxxxxxx
- Date: 14 Jun 2002 08:59:51 -0400
Hi guys,
I understand that if the fill is too close to the signal trace it is going
to increase the capacitance and decrease the impedance in turn.
But my question will be why do we need to put a fill on the pcb. ( this may
be a basic question , and I always heard my colleagues saying it, but it
helps me alot, if there's a kind soul to help me to understand it)
thanks in advance
Regards
Jason
Scott McMorrow wrote:
> Okay, now I really need some coffee ... and it's 5:30 in the evening.
> Here's yet another correction and more elaboration.
>
> A square patch of copper on top of a dielectric and plane forms a crude
> cavity resonator which will have a number of natural resonant
> eigenmodes. The first resonant point is a half-wave resonance. The
> resonance frequency is equal to the round trip of a wave propagating
> back and forth across the region, much like what would happen in a
> square pool of water between the two walls. For a one square inch patch
> of copper, the resonant frequency on FR-4 is somewhere between 2.7 and
> 3.3 GHz
>
> A very crude square patch
>
> w
> --------------
> | | resonant frequency = 1/( 2 * w * Vp)
> | | Vp = velocity of propagation of wave in
> dielectric medium = c/sqrt(Er)
> | | w = width and height of square
> | |
> | |
> ---------------
>
> Adding vias to the patch effectively pins the structure down and raises
> it's resonant frequency. The spacing of the vias control the resonance,
> with a half-wave resonance being the lowest generally supported by the
> structure. Stitching the patch at all four corners actually does nothing
> to change the resonant frequency of the cavity. However, placing a
> fifth via at the center will push the resonance up by a factor of 2.
>
> w
> @------------@
> | | resonant frequency = 1/( 4 * w * Vp)
> | | Vp = velocity of propagation of wave in
> dielectric medium
> | @ | w = width and height of square
> | | @ = Via stitch
> | |
> @------------@
>
> In essence, the via spacing has changed from w to w/2. These formulas
> are quite approximate, but will place us in the general playing field. A
> full-wave field solver such as PowerSI from Sigrity, or SIwave and HFSS
> from Ansoft, will provide much more exact resonance frequency analysis.
>
> So, the "rough" formula I gave for via spacing should be:
>
> via spacing =1/(2 * (f * Vp)) not 2/(f * Vp) as I had indicated.
>
> In my previous example, this would amount to a via spacing of 0.59
> inches, not 1.18.
>
> Sorry about my confusion.
>
> Scott
>
> --
> Scott McMorrow
> Teraspeed Consulting Group LLC
> 2926 SE Yamhill St.
> Portland, OR 97214
> (503) 239-5536
> http://www.teraspeed.com
>
> Scott McMorrow wrote:
>
> > My equation below has an error.
> >
> > The maximum via spacing should be:
> >
> > 2/(f * Vp)
> >
> > This accounts for the half-wave resonance which occurs between vias.
> >
> >
> > regards,
> >
> > Scott
> >
> >
> > Scott McMorrow wrote:
> >
> >> Ray,
> >>
> >> When the separation of the copper fill from any signal is > 20 times
> >> the signal-to-plane spacing there is no effect on the signals. So,
> >> for 5 mil dielectric thickness, the fill area to signal trace
> >> clearance should be 100 mils or greater. I would also suggest
> >> grounding each fill area at multiple points to eliminate possible EMI
> >> coupling problems.. At a minimum, use ground stitch vias at four
> >> corners to contain resonances of the fill area. For extremely high
> >> speed multi-gigahertz systems, the vias should be spaced a maximum
> >> distance apart of 1/(f * Vp) apart. Where f = the upper frequency of
> >> operation and Vp is the velocity of propagation of signals within the
> >> dielectric.
> >>
> >> For FR4 Vp is around 180ps/in. For 3.125 Gbps signallling the upper
> >> frequency you are concerned with is the 3rd harmonic of the switching
> >> frequency, 4.68 GHz. In this case, the maximum stitch via spacing
> >> would be
> >> 1/(4.68e9 * 180e-12) = 1.18 inches.
> >>
> >> Best regards,
> >>
> >> Scott
> >>
> >>
> >>
> >>
> >
>
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