**From:** Larry Smith (*ldsmith@lisboa.eng.sun.com*)

**Date:** Fri Mar 17 2000 - 13:18:37 PST

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Ron - see comments in text.

*> Date: Fri, 17 Mar 2000 11:13:44 -0800
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*> From: Ron Miller <rmiller@Brocade.COM>
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*>
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*> Larry
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*>
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*> I have some issues as follows
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*>
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*> >
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*> > So, if we compare the two power plane pairs below with 4 and 40 mil
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*> > separations:
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*> >
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*> > The velocities are constant.
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*> >
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*> > The capacitance of the 4 mil planes is 10X the 40 mil plane.
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*>
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*> Capacitance as you mentioned above is proportional to the inverse of the
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*> square of the distance making 4 mils 100 X the 40 mil plane.
*

Capacitance between two point sources is proportional to the inverse of

the square of the distance. But the capacitance between two parallel

plates of a capacitor is proportional to the thickness of the

dielectric (neglecting the fringe field). This happens as you

integrate over the surface of point charges.

*> > The inductance of the 40 mil plane is 10X the 4 mil plane.
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*>
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*> The inductance does not change with dielectric thickness. In fact if you
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*> look at a smith chart you will note that the low impedance outside circles
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*> tend to approximate the outer circle as you go out. The outside circle
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*> follows the electrical length and cannot be made larger then the length
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*> of the line. This is a limit.
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I'm not sure about the Smith Chart argument, but the inductance between

planes definitely goes down as the dielectric gets thinner. Remember

the good old equation:

velocity = 1/sqrt(LC) ?

The velocity is constant and the capacitance goes up as the dielectric

between planes gets thinner. Therefor, inductance must go down.

regards,

Larry Smith

Sun Microsystems

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