7 ms·

This seems like it ought to be much bigger than it is. Two-way single-cable fiber-optics, with cheaper components than photoresistors? What's the catch?

std_throwaway 8y ago
LEDs usually are not very good photodiodes. You optimize them differently for each usage. In abstract theory the perfect photodiode and the perfect LED would be the same but in reality there are more tradeoffs and you end up with different devices.
tlb 8y ago
Detectors for fiber optic data transmission need a very fast response time. For gigabit data, you need sub-nanosecond response. LEDs are much slower. They're also less sensitive, which means you need a higher-power transmitting LED to get the same range.
- monochromatic 8y ago
  What drives the response time, internal capacitance or something?
  - tlb 8y ago
    Yes, capacitance of the junction.
    - DoctorOetker 8y ago
      That's a rather vague statement, it is the textbook answer directed at the users of discretes: given a fixed discrete photodiode part and a fixed load resistor one can change the capacitance by reverse biasing and thus improve the bandwidth of the resultant RC filter. But this is not a full truth from the perspective of the photodiode designer: he can trivially increase or decrease the capacitance by designing a photodiode with a larger or smaller area. Suppose ratio of the incoming light power over photodiode area is unaffected by doubling the original photodiode area. If he doubles the area and hence capacitance, then he also doubles the injected photocurrent! Hence he can halve the load resistor, hence the bandwidth is unaffected by doubliing capacitance! Consider the source illumination, what is the smallest area you can concentrate the light in? If it is smaller than your photodiode's active area, it pays off to select a smaller photodiode since the unused area is contributing dead weight capacitance. But if decreasing the area does not increase the optical power / photodiode area ratio, you need better optics or illumination. EDIT: fixed poorly phrased sentence
sigstoat 8y ago
pretty sure the expense in BiDi transceivers (which aren’t that expensive) comes from the WDM coupler, not the receiving element.
- upofadown 8y ago
  I think the point is that with a LED to LED system you would not need any sort of optical multiplexing.
  - sigstoat 8y ago
    i felt like returning to half-duplex communications was enough of a non-starter to discount it as a possible suggestion.
  - jsjohnst 8y ago
    There’s a limit to how many pulses you can have per second. They stack more waves to get more capacity.

std_throwaway 8y ago

LEDs usually are not very good photodiodes. You optimize them differently for each usage. In abstract theory the perfect photodiode and the perfect LED would be the same but in reality there are more tradeoffs and you end up with different devices.

tlb 8y ago

Detectors for fiber optic data transmission need a very fast response time. For gigabit data, you need sub-nanosecond response. LEDs are much slower. They're also less sensitive, which means you need a higher-power transmitting LED to get the same range.

monochromatic 8y ago
What drives the response time, internal capacitance or something?
- tlb 8y ago
  Yes, capacitance of the junction.
  - DoctorOetker 8y ago
    That's a rather vague statement, it is the textbook answer directed at the users of discretes: given a fixed discrete photodiode part and a fixed load resistor one can change the capacitance by reverse biasing and thus improve the bandwidth of the resultant RC filter. But this is not a full truth from the perspective of the photodiode designer: he can trivially increase or decrease the capacitance by designing a photodiode with a larger or smaller area. Suppose ratio of the incoming light power over photodiode area is unaffected by doubling the original photodiode area. If he doubles the area and hence capacitance, then he also doubles the injected photocurrent! Hence he can halve the load resistor, hence the bandwidth is unaffected by doubliing capacitance! Consider the source illumination, what is the smallest area you can concentrate the light in? If it is smaller than your photodiode's active area, it pays off to select a smaller photodiode since the unused area is contributing dead weight capacitance. But if decreasing the area does not increase the optical power / photodiode area ratio, you need better optics or illumination. EDIT: fixed poorly phrased sentence

sigstoat 8y ago

pretty sure the expense in BiDi transceivers (which aren’t that expensive) comes from the WDM coupler, not the receiving element.

upofadown 8y ago
I think the point is that with a LED to LED system you would not need any sort of optical multiplexing.
- sigstoat 8y ago
  i felt like returning to half-duplex communications was enough of a non-starter to discount it as a possible suggestion.
- jsjohnst 8y ago
  There’s a limit to how many pulses you can have per second. They stack more waves to get more capacity.