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I'm not remembering my education very well. I know when a signal encounters an impedance discontinuity there will be some reflection, I just want to know if in my case that is a practical concern.

I have two 50 ohm single ended signals that come onto a board via 50 ohm SMA connectors. They are to be fed into a 50 ohm differential load. My substrate is FR-4 with a thickness of 62 mils, and a maximum operating frequency of about 100 MHz.

schematic

simulate this circuit – Schematic created using CircuitLab

I only have control over what is in the dotted rectangle. I can't change the differential input impedance or the amplifier (which happens to have differential output)

The distance between the SMA connector and the differential amplifier is less than 1/20 of a wavelength. Does that render the discontinuity insignificant?

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  • \$\begingroup\$ I assume that you mean "fed from a 50 Ohm differential source" ? \$\endgroup\$ – Russell McMahon Feb 21 '14 at 19:35
  • \$\begingroup\$ Well this is where things seem to get murky. There is an unknown differential impedance driving 2 separate 50 ohm coax cables. What is interesting is that the datasheet says both "50 ohm input termination" and "50 ohm differential input impedance" \$\endgroup\$ – HL-SDK Feb 21 '14 at 20:32
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I only have control over what is in the dotted rectangle.

If that is true you'll need two 25 ohm resistors in series with each feed to the 50 ohm terminator to avoid reflections (placed local to the 50ohm). The sending end impedance can be zero ohms providing the receiving end is perfect (which at the moment it isn't without the two 25 ohm resistors previously mentioned). If the receiving end were not perfect, a reflection would occur but this would be swallowed by the sending end if it had exactly 100 ohms terminating it: -

enter image description here

So what would a 1st reflection cause at the receiving end - the waveform the receiver receives could be put "out of shape" (digital transmissions) or changed in amplitude (for CW type transmissions). Neither will be earth shattering and probably livable with but, the 2nd reflection could be a different story. The 2nd reflection arises from the first reflection bouncing off the incorrectly terminated sending end and by the time it reaches the receiving end it may be enough to turn a digital 1/0 signal into a digital 0/1 symbol i.e. it causes inter-symbol interference.

So, for digital transmissions, longer cables can make the 2nd reflection trespass the timeslot of "later" bit and shorter cables have less of a problem. On the other hand, longer cables are likely to attenuate a reflection more so it's a little difficult to be general about this problem.

If in doubt, as a first measure, ensure the receiving end is accurately terminated. I hardly ever feed out to a coax or twinax via a line termination impedance so this should be OK. If you can't fix the receiving end then hope that the reflections are not significant enough to cause inter-symbol corruptions. For a CW signal I don't think you'll see total blackouts with a 100 ohm line being terminated in 50 ohms.

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  • \$\begingroup\$ Thanks -- I'll be adding pads for resistors. I just learned there is another source connected to these differential lines -- it supplies a bias. Thanks for your explanation, the coax cables happen to be relatively longer (~1/8 lambda) \$\endgroup\$ – HL-SDK Feb 21 '14 at 23:32
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Yes, distance to a pickup that is not impedance matched does matter. At 1/20 wavelength, you should be OK though.

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    \$\begingroup\$ Is reflection less of a concern when the length is just a fraction of the wavelength (if so, why) or is it just about that the small distance has a small overall impact? What distance would be critical and why? \$\endgroup\$ – Rev1.0 Feb 21 '14 at 19:17
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    \$\begingroup\$ @Rev1.0 It's all about propagation time. For a fixed wave velocity, it takes longer for a signal to reach the far end. Likewise, a higher frequency signal has smaller lengths between signal variations. If you can't assume a conductor has a uniform value (ignoring resistive drops), then you have a transmission line and need to worry about impedance matching. You can play around with this online transmission line simulator I wrote to see what happens as time delay changes vs. signal rise time (a.k.a. frequency) \$\endgroup\$ – helloworld922 Feb 21 '14 at 20:33
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    \$\begingroup\$ @Rev1.0 didn't quite fit in one comment, but interesting values to try (from default settings): rise time = 1ns, tDelay = 100ps (~1/10th wavelength). Notice not much reflection even with mismatched impedances. rise time = 1ns, tDelay = 1ns (~1 wavelength). Notice that reflections are quite dramatic, and impedance mismatch will ruin your day. \$\endgroup\$ – helloworld922 Feb 21 '14 at 20:38
  • \$\begingroup\$ @helloworld922: Thanks for the feedback, the simulator is interesting. \$\endgroup\$ – Rev1.0 Feb 22 '14 at 13:11

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