Photonic and optical interconnects: Exercises — Hinglish deep dive

6.5.15 · D4 · Hardware › Advanced & Emerging Architectures › Photonic and optical interconnects

Level 1 — Recognition

(Kya tum piece ko naam de sakte ho aur picture padh sakte ho?)

L1.1 — Chaar blocks ko naam do

Ek optical link neeche ek chain ke roop mein draw ki gayi hai. Chain mein chaar missing block names order mein bharo, aur har ek ke liye batao ki woh electrons → photons, photons → photons, ya photons → electrons karta hai.

Figure — Photonic and optical interconnects

Recall Solution

Chain ko left se right padhte hain:

Laser source — blank optical carrier banata hai. (Photons supply karta hai; electrical power andar, steady light bahar.)
Modulator — electrons → photons (bit pattern ko light par imprint karta hai).
Waveguide / fiber — photons → photons (bas light carry karta hai).
Photodetector — photons → electrons (current recover karta hai).

Yeh E-O-E (Electrical–Optical–Electrical) link hai. Do conversions (modulator, detector) woh jagah hain jahan energy kharch hoti hai; beech wala (waveguide) almost free hai.

L1.2 — Kaunsa metric jeetta hai?

Sach ya jhooth: "Photonic interconnects ka copper par main advantage bahut kam latency (raw travel time) hai."

Recall Solution

Jhooth. Silicon mein light $v = c / n$ par travel karta hai jahan $n \approx 3.5$ , toh woh actually ek achhe copper trace par ek fast electrical signal se prति metre slower hai. Asli advantages hain bandwidth density (WDM ke zariye ek waveguide mein kai colours) aur energy per bit (roughly distance-independent). Latency sirf indirectly improve hoti hai, serial repeaters/retimers hataane se.

Level 2 — Application

(Derived formulas mein plug karo: $τ = r c L^{2}$ , $B_{total} = N B$ , $E_{bit} = P / B_{total}$ .)

L2.1 — RC delay scaling

Length $L$ ki ek copper wire ka delay $τ = r c L^{2}$ hai. Agar wire ko 3 guna longer banao, toh delay kitne factor se badhta hai?

Recall Solution

$τ \propto L^{2}$ . $L \to 3 L$ replace karo: $τ_{new} = r c (3 L)^{2} = 9 r c L^{2} = 9 τ_{old} .$ Factor = 9. Yeh " $L^{2}$ wall" hai: length teena karne se delay nau guna ho jaata hai. Ek photonic transit ( $t = n L / c$ , $L$ mein linear) sirf teena hota.

L2.2 — WDM aggregate bandwidth

Ek single waveguide $N = 32$ wavelength channels carry karta hai, har ek $B = 50 Gb/s$ par run karta hai. Aggregate bandwidth $B_{total}$ kya hai?

Recall Solution

Independent colours linearly add hote hain (woh ek doosre ki capacity nahi churaate): $B_{total} = N \times B = 32 \times 50 Gb/s = 1600 Gb/s = 1.6 Tb/s .$

L2.3 — Energy per bit

Ek photonic link $B_{total} = 1.6 Tb/s$ move karne ke liye $P = 3.2 W$ total electrical power burn karta hai. Picojoules per bit mein $E_{bit}$ nikalo.

Recall Solution

$E_{bit} = \frac{P}{B _{total}} = \frac{3.2 J/s}{1.6 \times 1 0 ^{12} bit/s} = 2 \times 1 0^{- 12} J/bit = 2 pJ/bit .$ Seconds cancel ho jaate hain (Watts = J/s, bit rate = bit/s), J/bit bachta hai. $2 pJ/bit$ co-packaged optics ke liye realistic near-sub-pJ range mein hai.

Level 3 — Analysis

(Compare karo, aur numbers ke peeche geometry / physics explain karo.)

L3.1 — Copper lane vs WDM fiber

Ek copper serial lane $B_{c u} = 50 Gb/s$ (ek stream) karta hai. Ek fiber WDM run karta hai $N = 40$ colours ke saath $B = 25 Gb/s$ har ek par. Fiber ek physical line par throughput mein copper lane ko kitne factor se beat karta hai?

Recall Solution

Fiber: $B_{total} = 40 \times 25 = 1000 Gb/s$ . Ratio: $\frac{B _{total}}{B _{c u}} = \frac{1000}{50} = 20.$ Fiber $20 \times$ jeetta hai ek single line par — aur extra 39 channels koi crosstalk add nahi karte, kyunki alag colours interfere nahi karte. Woh bandwidth-density advantage, latency nahi, poora selling point hai.

L3.2 — Mach–Zehnder output level

Ek Mach–Zehnder modulator (dekho Mach-Zehnder Modulator) ka output power $P_{o u t} = P_{in} cos^{2} (Δ ϕ /2)$ hai. Neeche transfer curve dekho. $Δ ϕ = 0, \frac{π}{2}, π$ ke liye $P_{o u t} / P_{in}$ compute karo aur batao kaunsa logic 0 encode karta hai.

Recall Solution

Har phase ko $cos^{2} (Δ ϕ /2)$ mein plug karo:

$Δ ϕ = 0$ : $cos^{2} (0) = 1$ → full power → logic 1.
$Δ ϕ = \frac{π}{2}$ : $cos^{2} (\frac{π}{4}) = (\frac{1}{2})^{2} = 0.5$ → half power (curve ka mid-point).
$Δ ϕ = π$ : $cos^{2} (\frac{π}{2}) = 0$ → zero power → logic 0.

Toh electrical drive ko phase $0$ se $π$ tak swing karna padta hai ek bit flip karne ke liye. Figure par, $Δ ϕ = π$ par red dot curve ke floor par baitha hai — woh destructive interference hai: dono arms exactly out of step pahunchte hain aur cancel ho jaate hain.

L3.3 — Detector ko TIA kyun chahiye?

Responsivity $R_{λ} = 0.8 A/W$ ka ek photodiode $P = 0.5 mW$ optical power receive karta hai. Yeh kaunsa current output karta hai? Yeh digital logic ko directly kyun drive nahi kar sakta, aur kya fix karta hai?

Recall Solution

$I = R_{λ} P = 0.8 \frac{A}{W} \times 0.5 \times 1 0^{- 3} W = 4 \times 1 0^{- 4} A = 0.4 mA .$ Ek photodiode current output karta hai, voltage nahi, aur chip ki tiny input impedance mein $0.4 mA$ ek bahut chota signal hai. Tumhe us current ko ek usable voltage swing mein convert karne ke liye ek Transimpedance Amplifier (TIA) chahiye. Woh TIA link ke energy budget ka ek real slice hai — receiver "free" nahi hai.

Level 4 — Synthesis

(Multiple derived results ko ek design answer mein combine karo.)

L4.1 — Bandwidth aur energy target saath mein meet karo

Tumhe ek waveguide link design karna hai jo kam se kam $1 Tb/s$ deliver kare. Har colour $B = 25 Gb/s$ par run karta hai. Har colour ki transmit+receive chain (laser share + modulator + TIA) $P_{c h} = 40 mW$ burn karti hai. (a) Minimum number of colours $N$ ? (b) Total power $P$ ? (c) Energy per bit $E_{bit}$ ?

Recall Solution

(a) $N \times B \geq 1 \times 1 0^{12}$ chahiye: $N \geq \frac{1 \times 1 0 ^{12}}{25 \times 1 0 ^{9}} = 40.$ Toh $N = 40$ colours (exactly $1 Tb/s$ hit karta hai).

(b) Har colour $40 mW$ laagat hai: $P = N \times P_{c h} = 40 \times 40 mW = 1600 mW = 1.6 W .$

(c) $E_{bit} = \frac{P}{B _{total}} = \frac{1.6 W}{1 \times 1 0 ^{12} bit/s} = 1.6 \times 1 0^{- 12} J/bit = 1.6 pJ/bit .$ Gaur karo: kyunki per-channel power aur per-channel bandwidth dono $N$ ke saath scale karte hain, energy per bit yahan $N$ se independent hai — yeh poori tarah per-channel efficiency $P_{c h} / B = 40 mW /25 Gb/s = 1.6 pJ/bit$ se set hoti hai. Isliye photonics engineers per-channel numbers par obsess karte hain, aggregate walon par nahi.

L4.2 — Long distance par kaunsa link?

$2 m$ chip-to-chip run ke liye, ek copper lane ki energy per bit distance ke saath badhti hai (equalizers/repeaters chahiye), near $8 pJ/bit$ land karti hai; photonic link $2 m$ ke bawajood near $1.6 pJ/bit$ rehta hai. Photonics yahan kitni baar zyada energy-efficient hai, aur iska number distance-flat kyun rehta hai?

Recall Solution

$\frac{8}{1.6} = 5 \times more efficient.$ Photonics distance-flat hai kyunki ek baar light ek low-loss waveguide mein launch ho jaaye, use ek aur metre carry karna almost kuch nahi lagta (dekho Energy per bit as an efficiency metric). Cost har end par conversions mein rehti hai (modulator + TIA), jo length ke saath nahi badhte. Copper, is ke viparit, $R C$ losses pay karta hai aur active repeaters chahiye jo distance ke saath scale karte hain.

Level 5 — Mastery

(Poore system mein reason karo, awkward cases including.)

L5.1 — Copper actually kab jeetta hai?

Bahut choti wires ke liye (ek core ke andar kuch sau µm), kya $L^{2}$ copper wall matter bhi karta hai? Do scaling laws ke saath argue karo, aur batao copper abhi bhi kahan photonics ko beat karta hai.

Recall Solution

Copper delay $τ_{c u} = r c L^{2}$ ; photonic transit $t_{p h} = n L / c$ plus fixed conversion overhead $t_{co n v}$ (modulator + detector + TIA latency).

Tiny $L$ par, $L^{2}$ negligible hai, toh $τ_{c u}$ minuscule hai — aur woh photonics ko beat karta hai kyunki $t_{co n v}$ (ek fixed cost) light path dominate karta hai. $200 μ m$ hop ke liye light mein convert karna aur wapas aana sirf worth it nahi hai.
Copper tabhi harta hai jab $L$ itna lamba ho ki $r c L^{2}$ (aur woh repeaters jo isse force karta hai) photonic overhead exceed kare. Yahi crossover reason hai kyun photonics chip-to-chip, co-packaged, aur rack-scale links pehle target karta hai, aur on-chip NoC links baad mein.

Conclusion: photonics universally better nahi hai — woh bandwidth × distance regime mein jeetta hai jahan copper ka $L^{2}$ delay aur per-channel serial limit sabse zyada bite karte hain.

L5.2 — Degenerate WDM: $N = 1$

$B_{total} = N B$ kya deta hai jab $N = 1$ ? Is limiting case ko physically interpret karo, aur connect karo kyun ek bare fiber automatically copper se better nahi hai.

Recall Solution

$B_{total} = 1 \times B = B .$ Ek single colour ke saath, WDM advantage gayab ho jaata hai — fiber exactly ek stream rate $B$ par carry karta hai, koi parallelism nahi. Is degenerate case mein comparable $B$ par ek single copper lane isse match kar sakta hai, aur copper laser/modulator/TIA overhead avoid karta hai. Photonic win sirf tabhi earn hoti hai jab $N ≫ 1$ : wavelength mein parallelism hi lever hai. Ek colour wala fiber sirf ek slow, mehenga wire hai.

L5.3 — Silicon mein laser kyun nahi grow kar sakte?

Ek designer propose karta hai ki ek bonding step save karne ke liye laser directly silicon CMOS die mein fabricate karein. Woh physics explain karo jo ise rokti hai, aur industry kya karta hai instead (dekho Silicon Photonics).

Recall Solution

Silicon ka indirect bandgap hai: gap ke across drop karne wale ek electron ko lattice vibration ke saath momentum bhi swap karna padta hai, jo radiative (light-emitting) recombination ko extremely inefficient banata hai. Silicon ek accha light guide aur detector hai, lekin ek bura emitter hai. Toh lasers III–V materials (InP, GaAs) se banaye jaate hain jinke direct bandgaps hote hain, phir silicon photonic die par bonded ya co-packaged kiye jaate hain. Waveguides, modulators (rings / MZMs), aur detectors silicon ho sakte hain; light source generally nahi ho sakta.

Recall Yaad karne ke liye ek-line summary, chhupa lo

Copper delay $::: τ = r c L^{2}$ (quadratic); photonic transit $::: t = n L / c$ (linear) WDM aggregate $::: B_{total} = N B$ Energy per bit $::: E_{bit} = P / B_{total}$ , per-channel efficiency se set hoti hai, $N$ -independent MZM output $::: P_{o u t} = P_{in} cos^{2} (Δ ϕ /2)$ , $Δ ϕ = π \to$ logic 0 Photodiode current $::: I = R_{λ} P$ , phir ek TIA use voltage banata hai