4 ms·

Depending how you do the actual modulation, optical modulation does not add any significant latency (there can be no processing involved and the extra transmiss

Depending how you do the actual modulation, optical modulation does not add any significant latency (there can be no processing involved and the extra transmission length you'd need for the modulation (i.e. Electroopotic conversion, rf amplifiers...) is negligible. The big issue is really 1. Photonic waveguides are much larger than electronic ones (due to the wavelength) 2. You loose dynamic range and in EO conversion (shot noise is significant at optical frequencies) 3. Co integration of optics and photonics components is nontrivial due to the different materials and processes. 4. Power efficiency of EO conversion is also not that great. Where photonics shines is for transmission of high frequencies (i.e. a lot of data) over long distances and being immune to EM interference. So there is certainly a tradeoff for at what transmission distances to go optical and as data rates keep going up the tradeoff length has become shorter and shorter. Intel, Nvidia, AMD et al. All do research into optical interconnect.

aborsy 2y ago
Photonic wavelengths are shorter than electronic wavelengths. Why are photonics waveguides bigger then?
- murkt 2y ago
  What’s an electronic wavelength in this context? What’s its size? Photonic ones I assume are in near-IR, on the order of a micrometer.
  - aborsy 2y ago
    Optical fiber communication operates at 1.55 micro meters or 193THz. Electronics operates in the electromagnetic spectrum or at GHz. There might be no fixed size or wavelength in either case, but the shortest wavelength for radio signals that can be transmitted with acceptable loss with today’s technology is mili meters. Which brings the question: why operating wavelengths are smaller but “waveguides” are bigger in optical fiber communication. In fact, fiber itself is a waveguide and its diameter is tens of micro meters.
    - silizium 2y ago
      The full optical wave is contained in the dielectric conductor. This conductor needs it's minimum cross section such that the wave can propagate. If it is too small then the wave can not propagate. Also there is a maximum cross section if you want single mode operation. You get to this result if you take the electromagntic wave equation - a partial differential equation - and solve that for your transmission line configuration. The proper analogy in the realm of electrical waveguides is the hollow waveguide. The hollow waveguide supports TE- and TM-modes but not TEM modes just like a dielectric conductor. The size is also a function of the dielectric constant ε. What we mostly use are TEM waveguides like microstrips or coaxial cables. The difference between electrical waveguides that supports TEM modes and waveguides that supports TE/TM modes is that the former has two independent potential planes and the latter only one. Also TEM waveguides do not have a lower cutoff frequency. A TEM wave with any frequency can propagate on any microstrip configuration. This is not true for TE/TM waves. What's important to understand is that for microstrips/coaxial cables the power isn't transferred in the metal but in the space (dielectric) around the metal - see Poynting vector. So what happens if you have a second conductor in that space? You get crosstalk! So TEM transmission lines do not contain the wave like hollow waveguides or optical fibers (edit: ok coaxial cables do, microstrips don't) Now the question, how big is the microstrip? Is it just the width of the signal conductor? No, it is not. Edit: The width of the metal lines in a chip is given by the current it must carry - current density requirement, electro-migration issues. Power lines are wide because they have to supply power to the circuit but logic traces in CMOS technology only carry negligible amount of current. In circuits like RF power amplifiers with bipolar transistors the trace width is much larger because it has to carry a much larger current. But again, microstrip lines do not have a lower cutoff frequency.
- numpad0 2y ago
  electrical connections don't require electrons fly all the way across dielectric materials
- deleted 2y ago
  [deleted]
- cycomanic 2y ago
  Sorry this is a somewhat handwavy explanation that glosses over many details, but the size of components in electronic circuits (i.e. transistors and connections) fundamentally depends on the de Broglie wavelength of the electrons (some tens/hundreds of pm) why the size of a photonic circuit depends on the wavelength of light (order of 1 mu m). So one can make much more compact electronic circuits than photonic. Note that is somewhat different to transmission lines which are larger for electronics (depending on the RF carrier frequency) than optics.
lloeki 2y ago
I seem to recall information travels slower in fiber (optical) vs wires (electric), resp. ~2c/3 vs ~c, or am I remembering it wrong? Or is it a significantly different optical medium? If so, does that matter at all here? Dunno if that holds up for such kind of devices and/or at these scales (much shorter distance, but also much higher speed).
- Terr_ 2y ago
  IIRC the lower speed in fiber-optic cables has to do with the the refractive index of the glass, and maybe some bouncing introduced by curvature. I'm not sure what kind of refractive indices are possible in much smaller photonic circuits, particularly if it's not practical to develop and run everything in a permanent vacuum.
  - deleted 2y ago
    [deleted]
- cycomanic 2y ago
  In general that's correct, however it's not quite as simple. For latency you need to consider group velocity (not phase velocity), which depends on the waveguide and frequency of the EM wave. I'm actually not quite sure what it is for electric waves, but for most photonic structures it's very similar to the phase velocity. The phase velocity is about c/1.5 for silica, but more like c/3.5-c/2 for most materials used in photonic integration. I'm not an expert on integrated electronic circuits, but I guess the difference could matter depending on application.
  - hughesjj 2y ago
    I'm thankful you posted this and adjacent chain of responses and I enjoyed learning what you shared =) Viva la HN

aborsy 2y ago

Photonic wavelengths are shorter than electronic wavelengths. Why are photonics waveguides bigger then?

murkt 2y ago
What’s an electronic wavelength in this context? What’s its size? Photonic ones I assume are in near-IR, on the order of a micrometer.
- aborsy 2y ago
  Optical fiber communication operates at 1.55 micro meters or 193THz. Electronics operates in the electromagnetic spectrum or at GHz. There might be no fixed size or wavelength in either case, but the shortest wavelength for radio signals that can be transmitted with acceptable loss with today’s technology is mili meters. Which brings the question: why operating wavelengths are smaller but “waveguides” are bigger in optical fiber communication. In fact, fiber itself is a waveguide and its diameter is tens of micro meters.
  - silizium 2y ago
    The full optical wave is contained in the dielectric conductor. This conductor needs it's minimum cross section such that the wave can propagate. If it is too small then the wave can not propagate. Also there is a maximum cross section if you want single mode operation. You get to this result if you take the electromagntic wave equation - a partial differential equation - and solve that for your transmission line configuration. The proper analogy in the realm of electrical waveguides is the hollow waveguide. The hollow waveguide supports TE- and TM-modes but not TEM modes just like a dielectric conductor. The size is also a function of the dielectric constant ε. What we mostly use are TEM waveguides like microstrips or coaxial cables. The difference between electrical waveguides that supports TEM modes and waveguides that supports TE/TM modes is that the former has two independent potential planes and the latter only one. Also TEM waveguides do not have a lower cutoff frequency. A TEM wave with any frequency can propagate on any microstrip configuration. This is not true for TE/TM waves. What's important to understand is that for microstrips/coaxial cables the power isn't transferred in the metal but in the space (dielectric) around the metal - see Poynting vector. So what happens if you have a second conductor in that space? You get crosstalk! So TEM transmission lines do not contain the wave like hollow waveguides or optical fibers (edit: ok coaxial cables do, microstrips don't) Now the question, how big is the microstrip? Is it just the width of the signal conductor? No, it is not. Edit: The width of the metal lines in a chip is given by the current it must carry - current density requirement, electro-migration issues. Power lines are wide because they have to supply power to the circuit but logic traces in CMOS technology only carry negligible amount of current. In circuits like RF power amplifiers with bipolar transistors the trace width is much larger because it has to carry a much larger current. But again, microstrip lines do not have a lower cutoff frequency.
numpad0 2y ago
electrical connections don't require electrons fly all the way across dielectric materials
deleted 2y ago
[deleted]
cycomanic 2y ago
Sorry this is a somewhat handwavy explanation that glosses over many details, but the size of components in electronic circuits (i.e. transistors and connections) fundamentally depends on the de Broglie wavelength of the electrons (some tens/hundreds of pm) why the size of a photonic circuit depends on the wavelength of light (order of 1 mu m). So one can make much more compact electronic circuits than photonic. Note that is somewhat different to transmission lines which are larger for electronics (depending on the RF carrier frequency) than optics.

lloeki 2y ago

I seem to recall information travels slower in fiber (optical) vs wires (electric), resp. ~2c/3 vs ~c, or am I remembering it wrong? Or is it a significantly different optical medium? If so, does that matter at all here? Dunno if that holds up for such kind of devices and/or at these scales (much shorter distance, but also much higher speed).

Terr_ 2y ago
IIRC the lower speed in fiber-optic cables has to do with the the refractive index of the glass, and maybe some bouncing introduced by curvature. I'm not sure what kind of refractive indices are possible in much smaller photonic circuits, particularly if it's not practical to develop and run everything in a permanent vacuum.
- deleted 2y ago
  [deleted]
cycomanic 2y ago
In general that's correct, however it's not quite as simple. For latency you need to consider group velocity (not phase velocity), which depends on the waveguide and frequency of the EM wave. I'm actually not quite sure what it is for electric waves, but for most photonic structures it's very similar to the phase velocity. The phase velocity is about c/1.5 for silica, but more like c/3.5-c/2 for most materials used in photonic integration. I'm not an expert on integrated electronic circuits, but I guess the difference could matter depending on application.
- hughesjj 2y ago
  I'm thankful you posted this and adjacent chain of responses and I enjoyed learning what you shared =) Viva la HN