Re: [SI-LIST] : Parallel Plate Capacitance for Bypass

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From: Bob Lewandowski ([email protected])
Date: Fri Mar 24 2000 - 15:32:19 PST


All "Hi-K" dielectrics (i.e., X7R, Y5V, Z5U) have decreasing capacitance with
increasing frequency. X7R is the best of this group. Larger valued caps are
usually characterized at 1 kHz, the reduction in value is most significant from
1k to a few MHz, above that the frequency reduction effect is less.

I'm not sure what specs you're referring to concerning performance above 150 -
200 MHz, but at higher frequencies the series inductance of the cap and the
traces that get to and from it are the dominant effect with bypass caps. A
typical 0805 cap will have an effective body inductance of about 1 nH. At 150
MHz the inductive reactance is about 1 ohm which will be approximately equal to
the capacitive reactance of a 1 nF cap. Consequently, the cap alone is series
resonant at that frequency. If you add two traces and vias to the cap that are
each about 0.1 in. long, the inductance of the structure increases to about 3 nH
and it has >2 ohms of inductive reactance at 150 MHz. The cap's reactance is
almost out of the picture and all you have is an effective 3 nH inductor in its
place.

"Hansen, Chris" wrote:

> According to specifications, X7R type chip capacitors used for power
> subsystem bypass seem to become ineffective above 150 to 200 Mhz. For
> frequencies higher up, you are reliant upon the internal parallel plate
> capacitance. I have two questions. 1. Is the calculation for the required
> capacitance that is needed for the parallel plate capacitance the same as that
> used for a discrete chip capacitor bypass network, or is there a conservation
> of charge situation where the real answer is Cpp*dVnoise = Cload*dVchange?
> (where Cpp = parallel plate capacitance, dVnoise = tolerable change in power
> supply voltage, Cload = sum of load capacitance being switched simultaneously,
> dVchange = voltage change through driver output switch). 2. If you have a
> power plane sandwiched between 2 ground planes, does the parallel plate
> capacitance double for a given area?
> From "Reference Data for Radio Engineers", published by ITT, Copyright 1956:
>
> "For parallel-plate capacitor
>
> C=0.225 * Er * (N - 1) * A" / t" micromicrofarads { (pF) :-) }
>
> Where: Er = the dielectric constant relative to air
> N = the number of plates
> A" = the area of one side of one plate in square inches
> t" = the thickness of the dielectric in inches.
> This formula neglects "fringing" at the edges of the plates."
> Thank you,Chris HansenSr. Design EngineerDPT & Adaptec Companies

---Bob Lewandowski
    Vixel Corp.

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