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Would anyone please explain why you don't need a controlled impedance board when you produce a PCB dealing with 50Mhz digital signal? I thought you might need a controlled impedance board, since 10Mhz is already fast enough to make problems 'in measurement'. If one measure such digital signal using an oscilloscope, the impedance matching is important. But why isn't it the case for PCB board production?

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    \$\begingroup\$ I advice you to read the book called High Speed Digital Design a Book of Black Magic by Howard W. Johnson. I will provide full apreciation for what is going on in high speed signals and also describe what really is "high speed". Remember, it is the rise time and fall time of the signal that determines whether it will manifest high speed effects or not. Thus, if the rise time is fast enough compared with the signal path propagation delay time, we will still get high speed effects but it depends on the design if that will actually cause the design to work fail. \$\endgroup\$ – quantum231 Feb 28 '16 at 15:25
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In general, you don't need to worry about impedance matching at any frequency - AS LONG AS YOUR TRACE IS SHORT ENOUGH. The rule of thumb is that the time of flight along the trace must be less than 10% of the rise (or fall) time of the signal being propagated. This is slightly different than Matt Young's answer, and it's important if your signal is digital. A 50 MHz square wave with sharp edges will have a rise time much less than the period of the fundamental, and particularly in logic this can cause real problems if the reflected component is too large.

So a good, clean 50 MHz (20 nsec) square wave might have rise and fall times in the range of 2-3 nsec, and (at a rule of thumb of 8 inches/nsec for PCBs), you'd want to start worrying for trace lengths of 2-3 inches or so. See here, for example. This is, in some respects, a very conservative number, since it assumes that the transmitter has zero output impedance and the receiver has infinite input impedance, and the transmission line (pc trace) has no losses. While none of these are exactly true, being conservative is very, very good if you want to avoid unpleasant surprises.

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    \$\begingroup\$ +1 for pointing out the importance of edge rates rather than fundamental frequency. \$\endgroup\$ – uint128_t Feb 28 '16 at 16:16
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    \$\begingroup\$ For reference for the OP in conjunction with the excellent answer; signals propagate at about 160 ps/inch surface and 175 ps/inch internal for most flavours of FR-4. \$\endgroup\$ – Peter Smith Feb 28 '16 at 16:23
  • \$\begingroup\$ Would it help to smooth out the signal transmitted by the transmitter deliberately and then rectify it at the receiver side? \$\endgroup\$ – John Dvorak Feb 28 '16 at 17:46
  • \$\begingroup\$ @JanDvorak - I'm not sure what you mean, but the answer is probably no. \$\endgroup\$ – WhatRoughBeast Feb 28 '16 at 18:04
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\$\lambda=\dfrac{c}{f}=\dfrac{3\times10^8}{50\text{MHz}}=6\text m\$

A general rule of thumb is to not worry about transmission line effects until the transmission line reaches \$\frac{\lambda}{10}\$. In the 50MHz case, that is 60cm. That would take a pretty darn big PCB. I'm betting your scope probe lead is longer than that.

None of that is to say not to use controlled impedance tracks if you're already going to be doing so for higher speed signals. Going through the effort to do so for a couple 50MHz nets is not really worth the effort and expense in a lot of cases.

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    \$\begingroup\$ Speed of EM waves in copper is about 2/3 c. So it would be 4m and 40cm, respectively--not that the lambda/10 rule of thumb is necessarily exact enough to make much of a difference in practice. \$\endgroup\$ – Oleksandr R. Feb 28 '16 at 16:23
  • \$\begingroup\$ @OleksandrR. speed of EM waves in copper is irrelevant at frequencies below 10⁹ GHz. I suppose you mean the combined effective propagation speed of a wave in the surrounding dielectric, guided by currents in the copper down to skin depth. \$\endgroup\$ – leftaroundabout Feb 28 '16 at 22:05
  • \$\begingroup\$ @leftaroundabout yes, you are right. That will teach me for using excessively colloquial language. Thanks for your correction. \$\endgroup\$ – Oleksandr R. Feb 29 '16 at 3:52

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