Exercises — Photonic and optical interconnects

Reading the chain left to right:

1. Laser source — makes the blank optical carrier. (Supplies photons; electrical power in, steady light out.)
2. Modulator — electrons → photons (imprints the bit pattern onto the light).
3. Waveguide / fiber — photons → photons (just carries the light).
4. Photodetector — photons → electrons (recovers a current).

This is the E-O-E (Electrical–Optical–Electrical) link. The two conversions (modulator, detector) are where the energy is spent; the middle (the waveguide) is nearly free.

False. Light in silicon travels at $v=c/n$ with $n\approx 3.5$, so it is actually slower per metre than a fast electrical signal on a good copper trace. The real advantages are bandwidth density (many colours in one waveguide via WDM) and energy per bit (roughly distance-independent). Latency only improves indirectly, by removing serial repeaters/retimers.

$\tau \propto L^2$. Replace $L\to 3L$: ${\tau }_{\text{new}}=rc(3L{)}^2=9rc{L}^2=9{\tau }_{\text{old}}.$ Factor = 9. This is the "$L^2$ wall": tripling length gives nine times the delay. A photonic transit ($t=nL/c$, linear in $L$) would only triple.

Independent colours add linearly (they don't steal each other's capacity): $B_{\text{total}}=N\times B=32\times 50\text{Gb/s}=1600\text{Gb/s}=1.6\text{Tb/s}.$

$E_{\text{bit}}=\frac{P}{B_{\text{total}}}=\frac{3.2\text{J/s}}{1.6\times 10^{12}\text{bit/s}}=2\times 10^{-12}\text{J/bit}=2\text{pJ/bit}.$ The seconds cancel (Watts = J/s, bit rate = bit/s), leaving J/bit. $2\text{pJ/bit}$ is in the realistic near-sub-pJ range for co-packaged optics.

Fiber: $B_{\text{total}}=40\times 25=1000\text{Gb/s}$. Ratio: $\frac{B_{\text{total}}}{B_{cu}}=\frac{1000}{50}=20.$ The fiber wins by $20\times$ on a single line — and the extra 39 channels add no crosstalk, because different colours don't interfere. That bandwidth-density advantage, not latency, is the whole selling point.

Plug each phase into ${\cos }^2(\Delta \varphi /2)$:

• $\Delta \varphi =0$: ${\cos }^2(0)=1$ → full power → logic 1.
• $\Delta \varphi =\frac{\pi }{2}$: ${\cos }^2(\frac{\pi }{4})=(\frac{1}{\sqrt{2}}{)}^2=0.5$ → half power (the mid-point of the curve).
• $\Delta \varphi =\pi$: ${\cos }^2(\frac{\pi }{2})=0$ → zero power → logic 0.

So the electrical drive must swing the phase from $0$ to $\pi$ to flip a bit. On the figure, the red dot at $\Delta \varphi =\pi$ sits on the curve's floor — that is destructive interference: the two arms arrive exactly out of step and cancel.

$I=R_{\lambda }P=0.8\frac{\text{A}}{\text{W}}\times 0.5\times 10^{-3}\text{W}=4\times 10^{-4}\text{A}=0.4\text{mA}.$ A photodiode outputs a current, not a voltage, and $0.4\text{mA}$ into a chip's tiny input impedance is a minuscule signal. You need a Transimpedance Amplifier (TIA) to convert that current into a usable voltage swing. That TIA is a real slice of the link's energy budget — the receiver is not "free."

(a) Need $N\times B\ge 1\times 10^{12}$: $N\ge \frac{1\times 10^{12}}{25\times 10^9}=40.$ So $N=40$ colours (exactly hits $1\text{Tb/s}$).

(b) Each colour costs $40\text{mW}$: $P=N\times P_{ch}=40\times 40\text{mW}=1600\text{mW}=1.6\text{W}.$

(c) $E_{\text{bit}}=\frac{P}{B_{\text{total}}}=\frac{1.6\text{W}}{1\times 10^{12}\text{bit/s}}=1.6\times 10^{-12}\text{J/bit}=1.6\text{pJ/bit}.$ Notice: because per-channel power and per-channel bandwidth both scale with $N$, the energy per bit is independent of $N$ here — it's set entirely by the per-channel efficiency $P_{ch}/B=40\text{mW}/25\text{Gb/s}=1.6\text{pJ/bit}$. That is why photonics engineers obsess over per-channel numbers, not aggregate ones.

$\frac{8}{1.6}=5\times \text{ more efficient.}$ Photonics is distance-flat because once light is launched into a low-loss waveguide, carrying it another metre costs almost nothing (see Energy per bit as an efficiency metric). The cost lives in the conversions at each end (modulator + TIA), which don't grow with length. Copper, by contrast, pays $RC$ losses and needs active repeaters that scale with distance.

Copper delay ${\tau }_{cu}=rc{L}^2$; photonic transit $t_{ph}=nL/c$ plus fixed conversion overhead $t_{conv}$ (modulator + detector + TIA latency).

• At tiny $L$, $L^2$ is negligible, so ${\tau }_{cu}$ is minuscule — and it beats photonics because $t_{conv}$ (a fixed cost) dominates the light path. Converting to light and back is simply not worth it for a $200\mu \text{m}$ hop.
• Copper only loses once $L$ is long enough that $rc{L}^2$ (and the repeaters it forces) exceeds the photonic overhead. That crossover is why photonics targets chip-to-chip, co-packaged, and rack-scale links first, and on-chip NoC links only later.

Conclusion: photonics is not universally better — it wins in the bandwidth × distance regime where copper's $L^2$ delay and per-channel serial limit bite hardest.

$B_{\text{total}}=1\times B=B.$ With a single colour, the WDM advantage vanishes — the fiber carries exactly one stream at rate $B$, no parallelism. In this degenerate case a single copper lane at comparable $B$ may match it, and copper avoids the laser/modulator/TIA overhead. The photonic win is earned only when $N\gg 1$: parallelism in wavelength is the lever. A fiber with one colour is just a slow, expensive wire.

Silicon has an indirect bandgap: an electron dropping across the gap must also swap momentum with a lattice vibration, which makes radiative (light-emitting) recombination extremely inefficient. Silicon is a fine light guide and detector, but a terrible emitter. So lasers are built from III–V materials (InP, GaAs) that have direct bandgaps, then bonded or co-packaged onto the silicon photonic die. The waveguides, modulators (rings / MZMs), and detectors can be silicon; the light source generally cannot.

Copper delay $:::\tau =rc{L}^2$ (quadratic); photonic transit $:::t=nL/c$ (linear) WDM aggregate $:::B_{\text{total}}=NB$ Energy per bit $:::E_{\text{bit}}=P/B_{\text{total}}$, set by per-channel efficiency, $N$-independent MZM output $:::P_{out}=P_{in}{\cos }^2(\Delta \varphi /2)$, with $\Delta \varphi =\pi \to$ logic 0 Photodiode current $:::I=R_{\lambda }P$, then a TIA makes it a voltage