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I have a 4 layer PCB with a typical stackup (signal-gnd-vcc-signal) and on the top layer there's a CMOS sensor connected with a flat flex cable...

I was wondering, since there's no reference plane underneath the cable, What would the return paths for those signals look like ? is there a chance for coupling between the signals here ?

And how to minimize that loop ? if I add a reference plane just under the cable and connector on the top layer, would that help provide a low impedance return path for those signals ?

Update: a photo of the cable enter image description here enter image description here

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  • \$\begingroup\$ if i understand correctly,return currents in Flex cable should flow in GND wire in flex cable. \$\endgroup\$
    – user19579
    Jan 3, 2014 at 8:32
  • \$\begingroup\$ I wouldn't call that a Flat Flex Cable. That is flexible printed circuit (and pretty torn up, too!) \$\endgroup\$
    – Scott Seidman
    Jan 3, 2014 at 13:54
  • \$\begingroup\$ @ScottSeidman yeah I accidentally broke one sensor and then I took it apart :) \$\endgroup\$
    – mux
    Jan 3, 2014 at 17:42

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The return path for a trace in a single layer flex cable in free space is going to be the other traces. If this is a digital interconnect with high speed, you most likely have a layout like:

gnd - sig - gnd - sig - ...

or

gnd - sig - sig - gnd - ...

This is typical coplanar transmission lines and you can simulate that easily with a 2.5D simulator like Hyperlynx og SigXplorer (both expensive). I think you can get a good idea from the free TNT field solver as well. That will help you determine both what the impedance of your signal traces are and what amount of coupling you get between traces.

Now if you put this (very) close to other metal like your top layer Cu this may change. If and by how much depends on the geometries. Put Cu closer than the trace-to-trace distance in the flex and it will certainly do something.

Specifically it will reduce coupling (x-talk) like you mention and lower the trace impedance.

It it worth doing? Good question.

I would recommend you do some quick simulations to find out before you waste too much time and energy solving a problem that may not even be there to begin with :-)

Let me know if you want to provide more details, so I can give a more detailed answer?

Update

As can be seen on the photos you added, the flex is a 2-layer and some of the traces actually do have a reference plane running on the opposite side of the cable.

Provided the thing is well engineered, you should not try to change impedance by putting other metal real close to the cable. Don't however assume that it is well engineered just because some (well known) company is selling it :-)

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  • \$\begingroup\$ well it costs nothing to add a small ground pour under the connector and cable, but I just wanted to understand first if that would have any effect, if I understand correctly, you're saying the cable has inner layers ? if so, then I don't need to worry about that right ? \$\endgroup\$
    – mux
    Jan 3, 2014 at 13:13
  • \$\begingroup\$ and also, I don't think it's physically possible to place the pour that close to the cable, I could glue it, but I'd rather not :D \$\endgroup\$
    – mux
    Jan 3, 2014 at 13:15
  • \$\begingroup\$ @mux: the cable does not have inner layers, and still you don't need to worry about these things, moreover, there's definitely no point in putting "ground pour" under the cable - you're trying to solve a problem which is not there. \$\endgroup\$
    – Laszlo Valko
    Jan 3, 2014 at 13:31
  • \$\begingroup\$ @LaszloValko I'm also interested in understanding how it works, this is a subject that always confuses me. \$\endgroup\$
    – mux
    Jan 3, 2014 at 17:44
  • \$\begingroup\$ I just saw this reference plane when I took the picture, it's interesting, because the plane is under the slow I2C signals and not the clock (24-48MHz) and data signals which would've made more sense. \$\endgroup\$
    – mux
    Jan 3, 2014 at 17:47

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