From: Ray Anderson (firstname.lastname@example.org)
Date: Wed Jul 26 2000 - 12:18:53 PDT
Just to restate what Ron said, a stub used as a capacitance
is a narrowband solution. From zero length to 1/4 lambda the
stub will look capacitive. From 1/4 lambda to 1/2 lambda it
will look inductive and so on where the sign of the reactance
changes every quarter wavelength. In microwave parlance a
quarter wave line is often termed an impedance inverter.
The input impedance of an open circuited lossless line
is Zin = -j * Zo cot(B*l) . From this you can see the
periodic change in the impedance's sign is caused by the
cotangent term as l varies.
So if your field solver tells you than your stub will exhibit
20pF/inch and you have a 1000 inch stub don't expect your
circuit to see 20,000 pF unless 1/4 lambda is > 1000 inches.
> From: Ron Miller <rmiller@Brocade.COM>
> To: "'Mendelsohn, Joseph P (Joseph)'" <email@example.com>,
"'Prasanna S'" <firstname.lastname@example.org>, "'email@example.com'"
> Subject: RE: [SI-LIST] : Stub Capacitance calculation
> Date: Wed, 26 Jul 2000 11:41:53 -0700
> There is a problem with this simplistic approach, and that is that
the quarter wave effect
> becomes more important than the capacitance as frequencies get into
> range. If you must use SPICE then capacitance probably is the best
you can do, but
> regardless of the capacitance, as the stub approaches 1/4 wavelength,
it approaches a
> short in the frequency domain. This is real, not just theory.
> Ron Miller
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