1
\$\begingroup\$

Could someone knowledgeable explain why fiber optics could or could not be used for power transmission large or small?

The formula for power in optical fiber is shown below. X is photons per second, lambda is wavelength, light speed is c (speed of light is reduced significantly in fiber ~30% reduction from vacuum speed), h term is Planck constant.

power equation

Therefore we are transmitting power, but is there a converter out there to take this power and make it useful to electrical systems? How would one convert the light power to power useful to electronics? This would probably be just supplying a voltage to a circuit of resistance R.

\$\endgroup\$
4
  • \$\begingroup\$ You most definitely can convert light back to power (as in photovoltaics). Now you have to take into account conversion loss and loss over the medium. I don't know the values of either, but my hunch is that this is a lot less efficient that over copper. What's your use case? \$\endgroup\$
    – jcaron
    Jun 13, 2021 at 12:46
  • 1
    \$\begingroup\$ like you'd do with any other kind of light-to-electricity conversion - you need a photocell. These exist. It's what you know as "solar cells", but they also exist in other shapes and sizes and for more selective spectra. Was this really your question? \$\endgroup\$
    – Marcus Müller
    Jun 13, 2021 at 13:04
  • \$\begingroup\$ You could let the photons be absorbed in a chamber which contains water and let the resulted effects do some work. The idea unfortunately has been already known some time. Guys like Thomas Newcomen and James Watt constructed practical machines that way. Optical fibers nor powerful enough photon sources for them were not generally available then, but those guys had the necessary photon source just below their steam boilers. A machine could run a generator. \$\endgroup\$
    – user136077
    Jun 13, 2021 at 13:18
  • \$\begingroup\$ Laser cutters often use fibers to deliver power to whatever they're cutting. That's thermal power though not electric. \$\endgroup\$
    – user1850479
    Jun 13, 2021 at 14:48

3 Answers 3

Reset to default
5
\$\begingroup\$

Efficiency of light-to-electric converters (aka "solar cells") is in the 15-25% range, and efficiency of the emitter is below 50%. Nothing stops you from trying, but with this low end-to-end efficiency it isn't going to compete with wires.

If you had a requirement to supply a low power circuit with extra high voltage isolation, or you need a floating power supply with very low capacitance to everything around, or something like that, then perhaps it would be worth considering.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ Another scenario where it might be used: explosive environments. \$\endgroup\$
    – The Photon
    Jun 13, 2021 at 15:08
2
\$\begingroup\$

Loss in a fiber cable can be as low as 0,2dB per kilometer if the wavelength is optimal. That means 15km cable would dissipate half of the power. I guess 50...60Hz AC in copper or aluminium cables get attenuated substantially less if the voltage is high enough. This article http://insideenergy.org/2015/11/06/lost-in-transmission-how-much-electricity-disappears-between-a-power-plant-and-your-plug/ claims that typically 6% of the AC power is dissipated in wire losses between the power plant and users. So, the transmission efficiency in fibers at least is poor.

Then there's some difficulties to generate power plant strength narrow band radiation. Also the other end still needs some engineering. Consumers cannot today buy light operated vacuum cleaners etc. appliances.

Finally: To be reliable the fiber cannot be thick. I'm afraid (not calculated it, only guessed) fiber materials cannot stand the needed electric field strengths which can carry the output of a power plant.

\$\endgroup\$
1
\$\begingroup\$

I do not have experience with fibre optic cables to draw upon. I can, however, link a similar thread: here ...that deals with attenuation of power in long versus short fibre optic cables. While the question posed in the other thread is poorly supported (and the measurements given flimsy at best, as many of the answers agree), there is quite a bit of good material in the answers presented. They not only cover theory, but also how to go about testing your hypothesis. Hope this helps.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.