From: Charles R. Patton (firstname.lastname@example.org)
Date: Thu Oct 21 1999 - 16:51:36 PDT
Although the velocity is 1/sqrt(LC) and indeed does slow down with increasing C, it would seem to me that the important point is that the effective impedance of this "transmission line" is also going down in proportion to this
velocity since impedance,
Zo =sqrt(L/C). It is the impedance seen at this point that you need to lower in order to reduce power supply noise., which was the original reason we added capacitance in the first place. So if we could get the velocity to
zero, we would have a short – and a very effective bypass., other than that the power supply would be a bit unhappy looking into that hypothetical board!
Charles R. Patton
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Charles R. Patton, Sr. Princ. Eng
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Western Digital Corp.
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