**From:** Ingraham, Andrew (*[email protected]*)

**Date:** Tue Jan 09 2001 - 05:29:33 PST

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*> We would like to find how close to the end do we need to connect the
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*> end-terminator.
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*> Since we cannot change the location of the resistor, we change the rise
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*> time of
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*> a pulse generator. We are using a HP TDR. As we increase the rise time, we
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*> get a better
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*> behaviour of termination, and get a smaller sink on the TDR screen.
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*> For each sink, TDR measures an excess capacitance created by stub.
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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
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*> no excess capacitance. Is there any relation (equation) between this
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*> excess capacitance and
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*> the rise-time or the stub length ? We need some analytic formulas, if any.
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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.

Regards,

Andy

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**Next message:**Hans Mellberg: "[SI-LIST] : Reminder; Today, SCV EMC Society presents:"**Previous message:**Grossman, Brett: "RE: [SI-LIST] : Web Resource for Material Properties"**Maybe in reply to:**Goferman Stas: "[SI-LIST] : End Termination"

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