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Sorry, but I couldn't find a better title. Consider the following: I have an address/data bus interface. Intended frequency of the D/A bus would be around 10-50MHz.

According to the datasheet of the device in question (OMAP-L138) the maximum clock frequency appears to be around 100MHz (tCLK(min) is specified as 10ns). I would guess that the rise time of the signals would be around 200ps. But this is really just a guess since I wasn't able to get the information from said manual.

Now I was wondering if there was a rule of thumb of how much (length) the traces may differ, e.g. how much A1 may be offset from A2.

Armandas
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Tom L.
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2 Answers2

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The more precise terms to use are setup time and hold time, since your clocking scheme may vary. That is, your signals may be sampled on both edges (DDR), and so on.

From what I understand, setup time is for how long your data needs to be stable before a sampling clock edge and hold time is after your clock edge.

With a falling-setup and rising-sample scheme you practically make both these times \$T/2\$. Given your clock frequency of \$50MHz\$, the period \$T = \frac{1}{50\times10^6} = 20ns\$, thus \$T/2 = 10ns\$.

According to this answer the worst case propagation speed through copper is \$5.3 ns/m\$. Plugging this into the thing gives you \$\tilde{}1.9m\$. This if of course totally wrong, thus I must have wronged something. Someone please correct me.

Community
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Dzarda
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    50MHz is a low enough frequency that 1.9m is a perfectly reasonable result. But the actual impedance of the trace will make the real number quite a bit lower. – Ignacio Vazquez-Abrams Aug 25 '14 at 18:42
  • @Dzarda: That seems like a quite large number to me as well. My signal is only a single-sampled signal (only on one edge). The current trace length is around 60mm, with traces going +/- 5mm. I would guess there need to be other restrictions. – Tom L. Aug 26 '14 at 07:37
  • @Ignacio: Could you elaborate on that a little? – Tom L. Aug 26 '14 at 07:37
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The key factor depends on the skew and sample and hold time. If it is about 200ps (and Er = 4.5):

\$v = c/{\sqrt(E_r)} = 141.323.520m/s ≈ 141\mu m/ps\$

\$TraceLength_{Max} = v(\mu m/ps)·skew_{Max}(ps) ≈ 28264\mu m = 28.264mm\$

APica
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