Re: [SI-LIST] : Effect of low Zo for unterminated lines

Donald Telian (donaldt@cadence.com)
Tue, 19 Oct 1999 10:29:49 -0700

Chris,

During a signal transition, you set up a voltage divider between your
driver's impedance and the 29 Ohm Zo. As such, as Zo gets lower you cause
the driver to source/sink more dynamic current which sets up larger voltage
drops across the output transistors. The driver is working harder and
dissipating more power. Whether that "hurts" the driver is a function of
its datasheet ratings. Some devices are made to handle quite a bit of even
DC current, and at 10MHz I'd expect that the percentage of time it is
switching the larger currents is rather low.

Donald Telian
CADENCE

At 05:23 PM 10/18/1999 -0700, you wrote:
>Hi,
>
>Because of several constraints on our stackup and line width, a 6mil
>stripline trace yields a Zo = 29ohms. When I simulate a 29ohm,
>unterminated line on our board (@ 10MHz), I see a pretty nice waveform,
>albeit with slower rise/fall times (which is ok for the nets in
>question). When I increase the Zo to 60 ohms, for example, I see a lot
>of ringing on the same net.
>
>>From a theoretical point of view, my guess is that this happens because
>as you widen the trace, there is more capacitance, which slows down the
>rise/fall times. Is this a correct assumption? Are there other factors
>as well that are going on?
>
>What I am most concerned about is the effect of a low Zo net on its
>driver. For example, if I have an LVCH245 driver (3.3V) driving a 29ohm
>line, will the low Zo "hurt" the driver? Or, is the only negative
>effect the slower rise time? This question is a general question for
>most ICs.
>
>Thanks for your help,
>
>Chris
>
>
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>
>
Donald Telian
Cadence Design Systems
phone: 408-944-7791
email: donaldt@cadence.com

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