From: Ingraham, Andrew (Andrew.Ingraham@compaq.com)
Date: Tue Jan 09 2001 - 05:29:33 PST
> We would like to find how close to the end do we need to connect the
> Since we cannot change the location of the resistor, we change the rise
> time of
> a pulse generator. We are using a HP TDR. As we increase the rise time, we
> get a better
> behaviour of termination, and get a smaller sink on the TDR screen.
> For each sink, TDR measures an excess capacitance created by stub.
It is not just capacitance. You have a short length of transmission line,
the "stub" from the termination resistor to the far end. A capacitance has
a purely imaginary impedance. The input impedance of the stub is complex
(real + imaginary).
In the time domain, it looks like a pure resistance (=Zo, its characteristic
impedance) until a round-trip delay time later, and then it looks like an
open circuit if you have nothing on the end of the stub.
> For a perfect termination, there is
> no excess capacitance. Is there any relation (equation) between this
> excess capacitance and
> the rise-time or the stub length ? We need some analytic formulas, if any.
The capacitance of a short length of ideal transmission line (i.e., your
stub) is given by the formula
C = Td / Zo = (Len / vel) / Zo
where Td is the one-way delay of the line, Zo is its characteristic
impedance, Len is its length, and vel is the line's propagation velocity (in
consistent units). vel = c * sqrt(1/Er) where c is 2.998e+8 m/s and Er is
the effective dielectric constant.
When the round-trip delay (2 * Td) is short compared to your signal's
risetime, then it may be OK to treat the stub as a lumped capacitance like
this. Don't forget to add the input capacitance of your receiving IC that's
on the end.
When the round-trip delay is not short compared to the risetime, then you
should treat it as a length of transmission line and not a lumped
capacitance. For that, SPICE simulations are useful for seeing what will
happen. I would avoid formulas in that case.
The first approximation (total capacitance = Td/Zo + Cin of your receiving
IC) is OK at low frequencies only, but under-estimates the capacitance at
higher frequencies, until the stub is quarter-wave resonant where the
approximation completely breaks down and the effective capacitance
approaches infinity. This happens at a frequency lower than that where the
stub length is 1/4 wave long, due to the capacitance (Cin) on the end of the
stub. "Frequency" here is not the repetition rate of your signal; it is the
frequency content of your signal due to its risetime.
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