For a electrical switching device which opens and closes frequently and forcefully. Would a combination of a soft polymer matrix and a conductive metal fibre be a good choice for this? While polymer in itself is not known for having good wear resistance, since it is opened and closed forcefully, using an elastomer could take the energy from the forceful closing and just deform very shortly. The metal matrix inside is protected and could be the charge carrier. Are my assumptions correct? What other ways are there to achieve what I mentioned?
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One problem is that you may not have good contact between adjacent metal fibres. In general resin/fibre composites aim to completely coat the fibres so it is entirely possible that a metal/plastic composite could have very low conductivity, depending on how the fibres are orientated. Equally fine metal fibres will have a layer of oxidation on the surface which will further decrease conductivity.
I can't really see any reason why a metal composite would be better than solid metal for this sort of application.
If you want the spring/damping characteristics of an elastomer it would probably be better and simpler to mount a brass contactor on a polymer leaf spring.
Mechanical (as opposed to semiconductor) electrical contacts are most commonly made from brass as it has a good balance of conductivity, mechanical properties (reasonable hardness, toughness, ductility and strength) and corrosion resistance.
answered Apr 17, 2016 at 14:49
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$\begingroup$ Is it possible to construct a composite of any (non-conductive) matrix and arrange/design the fibers in a way that they will make the resultant composite conductive? $\endgroup$– zeroCommented Apr 17, 2016 at 14:58
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$\begingroup$ You could do it by aligning all of the fibres in the direction of the current flow but that largely defeats the purpose of a composite and it seems unlikely that any polymer composite is going to have better wear resistance than a solid metal part. Using a semiconductor based switch would eliminate mechanical wear entirely. $\endgroup$ Commented Apr 17, 2016 at 15:03