From: Dmitri Kuznetsov (firstname.lastname@example.org)
Date: Fri Oct 22 1999 - 12:42:43 PDT
I believe this discussion may be missing an important point.
The input impedance of a finite-length tline is not constant. It looks
like a comb filter with a period of 1/(2*delay). The value changes
between very small and a very high impedance for low-loss unterminated
lines (zero and infinity for lossless lines).
The impedance is very high at dc, and gradually reduces to almost zero
at f=1/(4*delay). Then it repeats with a period of 1/(2*delay). I
attached to this email an input impedance plot for a foot-long line with
1.3 ns delay. The dips in impedance are located at 192+n*384 Mhz, n =
0, 1, 2...
As you can observe, the load presented by an unterminated line depends
not so much on the characteristic impedance as on how the power spectrum
of the signal correlates with n/(4*delay). The average impedance will
be high for signal frequencies below 1/(4*delay) (shorter line, slower
driver case). It will also be high for majority of broad-spectrum
digital signals. But you could damage the driver by a periodic signal
(clock) with a frequency close to 1/(4*delay) + n/(2*delay).
And of course, the best thing to do is to run a simulation and verify
that the average and pick current is within the driver specs. That's
why we, EDA vendors, are in business.
Dmitri Kuznetsov, Ph.D.
ViewLogic Systems, Inc. e-mail: email@example.com
1369 Del Norte Rd. Tel: (805)278-6824
Camarillo, CA 93010 Fax: (805)988-8259
Chris Bobek wrote:
> I think I should restate my question. The original question wasn't just pertaining to an LVCH245, that was just an example. I'm trying to find out what the effects are on a driver when the Zo is low.
> As a general example, assume you had a 30ohm line being driven by a driver that was designed to drive a 50ohm load. The line is unterminated and suppose it's a foot long. Suppose there are multiple drops on the line. I've
> already seen simulations of this, but I'm wondering if this will damage the driver or not.
> Thanks for your help,
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