Electromigration reliability: Worked examples — Hinglish deep dive

6.4.8 · D3 · Hardware › Power, Thermal & Reliability › Electromigration reliability

Parent: Electromigration reliability

Yeh page ek drill sheet hai. Kuch bhi compute karne se pehle, hum har tarah ke sawaal list karte hain jo electromigration (EM) pooch sakta hai, phir har box ke liye ek example solve karte hain taaki koi bhi case tum se anjaan na rahe.

Do laws yahan sab kaam karti hain. Pehle unhe plain words mein restate karte hain, taaki koi bhi symbol bina announcement ke na aaye.

Formula Woh do tools jo hum har jagah use karenge

Black's Equation (lifetime ka law): $MTTF = A J^{- n} e^{E_{a} / (k_{B} T)}$

$MTTF$ = median time to failure (kitne time baad half wires mar jayenge).
$A$ = ek constant jo wire ki geometry aur material se set hota hai.
$J$ = current density = current $\div$ cross-section area (amps per cm $^{2}$ ). Yeh batata hai ki electron traffic kitni dense hai.
$n$ = current exponent, ya toh $\approx 1$ ya $\approx 2$ (hum neeche explain karte hain kaunsa kab).
$E_{a}$ = activation energy electron-volts (eV) mein: ek atom ko hop karne ke liye jo "energy toll" dena padta hai. Bada toll = slower migration = zyada lamba life.
$k_{B} = 8.617 \times 1 0^{- 5}$ eV/K, Boltzmann's constant, temperature ko energy mein convert karta hai.
$T$ = absolute temperature kelvin (K) mein. Exponent ke andar hamesha kelvin, kabhi °C nahi.

The Blech condition (short-line immortality): $J \cdot L < (J L)_{cr i t} \Rightarrow wire never fails$ $L$ = wire ki length (cm). Agar product $J \cdot L$ ek critical threshold se neeche rahe, toh pile-up ho rahe atoms ka back-pressure electron wind ko cancel kar deta hai.

Definition Main kaunsa

$n$ choose karun? ( $n \approx 1$ vs $n \approx 2$ ) Exponent $n$ koi free choice nahi hai — yeh batata hai ki void formation ke kis stage mein bottleneck hai:

$n \approx 1$ — nucleation-limited. Wire tab marta hai jab void paida hota hai. Yeh tab choose karo jab ek void, nucleate hone ke baad, turant wire ko span kar le — jaise bahut narrow / near-bamboo lines, single-grain-width interconnects, ya short segments jahan growth trivial ho. Yahan lifetime nucleation rate ke saath track karti hai, jo $J$ mein linear hai.
$n \approx 2$ — growth-limited. Ek void jaldi nucleate hota hai lekin use killing size tak dhheere dhheere grow karna padta hai. Yeh wider lines, vias, aur current-carrying power rails ke liye choose karo jahan growth time dominate karta hai. Lifetime accumulated mass transport $\propto J^{2}$ ke saath track karti hai.
Practical rule: agar failure mode bata diya gaya hai, use use karo. Agar nahi, toh wide power/clock nets → $n = 2$ ; thin signal lines / conservative qualification → $n = 1$ . Jab doubt ho, $n = 1$ design margin ke liye pessimistic (shorter-life) choice hai, isliye vendors aksar isi se qualify karte hain.

The scenario matrix

Har EM homework ya exam question inhi cells mein se ek mein aata hai. Hum har ek ko cover karenge.

#	Cell class	Kya badalta hai	Kaunsa tool	Example
C1	Temperature down (cooling)	Sirf $T$	Arrhenius ratio	Ex 1
C2	Temperature up (self-heating)	Sirf $T$	Arrhenius ratio	Ex 2
C3	Current density scaling, $n = 2$	Sirf $J$	Power law	Ex 3
C4	Current density scaling, $n = 1$ (wrong- $n$ trap)	Sirf $J$	Power law	Ex 4
C5	Dono $J$ aur $T$ ek saath badlein	$J$ aur $T$	Full Black ratio	Ex 5
C6	Degenerate: $J \to 0$ ya $L \to 0$ (limits)	limiting input	formula interpret karo	Ex 6
C7	Blech immortality — max length dhundho	geometry	Blech product	Ex 7
C8	Real-world word problem (data-center)	mixed	Black + Arrhenius	Ex 8
C9	Exam twist: do lifetimes se $E_{a}$ nikalo	ulta solve karo	log of ratio	Ex 9

Example 1 — Wire ko cool karna (Cell C1)

Worked example Statement

Ek wire $T_{1} = 105^{\circ} C$ par $5$ saal survive karti hai. Agar hum ise $T_{2} = 85^{\circ} C$ tak cool karein, toh yeh kitne der chalegi? $E_{a} = 0.7$ eV lo.

Forecast: cooler = slower atoms = lamba life. Guess: 5 saal se zyada — lekin 2× ya 10×?

Step 1 — Kelvin mein convert karo. $T_{1} = 105 + 273.15 = 378.15$ K, $T_{2} = 85 + 273.15 = 358.15$ K. (Neeche hum $378$ K aur $358$ K round karte hain.) Yeh step kyun? Exponent $e^{E_{a} / (k_{B} T)}$ sirf absolute (kelvin) temperature ke saath physically sensible hai — $0$ K par diffusion sach mein ruk jaati hai. Celsius se nonsense aata (tumhare paas negative °C ho sakta hai lekin kabhi negative K nahi).

Step 2 — Ratio lo taaki $A$ aur $J$ gayab ho jayein. $\frac{MTTF _{2}}{MTTF _{1}} = exp [\frac{E _{a}}{k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}})]$ Yeh step kyun? Same wire, same current, isliye $A$ aur $J^{- n}$ dono mein identical hain aur division par cancel ho jaate hain. Sirf temperature term bachi hai.

Step 3 — Numbers plug in karo ( $T_{1} = 378$ K, $T_{2} = 358$ K use karke). $\frac{E _{a}}{k _{B}} = \frac{0.7}{8.617 \times 1 0 ^{- 5}} = 8123.5$ K. $\frac{1}{358} - \frac{1}{378} = 1.4776 \times 1 0^{- 4}$ K $^{- 1}$ . Exponent $= 8123.5 \times 1.4776 \times 1 0^{- 4} = 1.200$ . Factor $= e^{1.200} = 3.32$ .

Step 4 — Multiply karo. $MTTF_{2} = 5 \times 3.32 = 16.6$ saal.

Verify: $T_{2} < T_{1}$ , isliye $1/ T_{2} > 1/ T_{1}$ , isliye bracket positive hai, isliye factor $> 1$ hai, isliye life badhi. Forecast se match karta hai. ✔

Example 2 — Self-heating life ko shortens karta hai (Cell C2)

Worked example Statement

Ek power rail $85^{\circ} C$ par $10$ saal ke liye rated hai, lekin woh Joule heating & self-heating ki wajah se $125^{\circ} C$ par chalta rehta hai. Naya lifetime? $E_{a} = 0.9$ eV (copper) use karo.

Forecast: hotter = shorter. Aur zyada $E_{a}$ temperature ko zyada hard bite deta hai, isliye steep drop expect karo.

Step 1 — Kelvin. $T_{1} = 358$ K (cool baseline), $T_{2} = 398$ K (hot). Kyun: same reason jaise Ex 1.

Step 2 — Arrhenius ratio, hot over cool. $\frac{MTTF _{h o t}}{MTTF _{coo l}} = exp [\frac{E _{a}}{k _{B}} (\frac{1}{398} - \frac{1}{358})]$ Yeh step kyun? Hume hot lifetime chahiye, isliye numerator ke $1/ T$ mein hot temperature daalo. Kyunki $398 > 358$ hai, bracket negative jaata hai → factor 1 se neeche → life girti hai. Woh sign check tumhara safety rail hai.

Step 3 — Numbers. $\frac{E _{a}}{k _{B}} = \frac{0.9}{8.617 \times 1 0 ^{- 5}} = 10444$ K. $\frac{1}{398} - \frac{1}{358} = - 2.806 \times 1 0^{- 4}$ . Exponent $= 10444 \times (- 2.806 \times 1 0^{- 4}) = - 2.931$ . Factor $= e^{- 2.931} = 0.0533$ .

Step 4. $MTTF_{h o t} = 10 \times 0.0533 = 0.533$ saal $\approx 6.4$ mahine.

Verify: $40^{\circ} C$ ki rise ne life ko ~19× cut kar diya. Arrhenius reliability model ka rule of thumb dekho: $E_{a} \approx 0.9$ ke liye, life roughly har $\sim 8$ – $9^{\circ} C$ par half hoti hai; $40^{\circ} C \approx 4.5$ halvings $\approx 2^{4.5} \approx 23 \times$ — same order. ✔

Example 3 — Current double karna, $n = 2$ (Cell C3)

Worked example Statement

Ek damascene wire ki width half kar di jaati hai (ek denser routing track fit karne ke liye), lekin woh same current carry karti hai. Void-growth-limited $n = 2$ ke saath, MTTF ka kya hoga?

Forecast: narrow wire, same current → traffic do guni dense → life giregi. Kitni?

Step 1 — Naya $J$ nikalo. $J = \frac{I}{area}$ . Width half karne se area half ho jaata hai, isliye $J_{2} = 2 J_{1}$ . Yeh step kyun? Current density Black's law mein actual driver hai, raw current nahi. Designers width control karte hain; physics $J$ feel karta hai. Dekho Design rules & current density limits.

Step 2 — Ratio sirf $J$ change hone ke saath. $\frac{MTTF _{2}}{MTTF _{1}} = (\frac{J _{1}}{J _{2}})^{n} = (\frac{1}{2})^{2}$ Yeh step kyun? Same temperature hai, isliye exponential cancel ho jaata hai; sirf $J^{- n}$ bachta hai. Note karo ki $J^{- n}$ ka matlab hai divide karne par smaller- $J$ -on-top.

Step 3 — Evaluate karo. $(1/2)^{2} = 1/4$ .

Verify: MTTF $\propto J^{- 2}$ , isliye $J$ double karne par life $2^{- 2} = 0.25$ se multiply hoti hai. Life ka quarter — textbook reason ki EM rules current per unit width cap karte hain. ✔

Example 4 — Same doubling, lekin $n = 1$ (Cell C4, wrong- $n$ trap)

Worked example Statement

Example 3 repeat karo, lekin failure nucleation-limited hai, isliye $n = 1$ . Answer kaise badlega, aur galat $n$ pick karna kyun matter karta hai?

Forecast: chhota exponent ke saath drop milder hona chahiye.

Step 1 — Same $J$ change. $J_{2} = 2 J_{1}$ . Kyun: geometry Ex 3 se identical hai; sirf physics regime ( $n$ ) differ karta hai.

Step 2 — Power law ko $n = 1$ ke saath apply karo. $\frac{MTTF _{2}}{MTTF _{1}} = (\frac{1}{2})^{1} = 1/2$ Yeh step kyun? $n$ literally current penalty ki steepness hai. $n = 1$ (void nucleation life limit karta hai) $n = 2$ (void growth life limit karta hai) se gentler hai. Upar "Which $n$ do I pick?" box use karo: ek near-bamboo narrow line nucleation-limited hai, isliye $n = 1$ yahan sahi regime hai.

Step 3 — Ex 3 se compare karo. $n = 2$ ne $1/4$ diya; $n = 1$ $1/2$ deta hai. $n = 2$ use karna jab reality $n = 1$ ho toh half true lifetime predict karta — yahan 2× error, aur decades par extrapolate karte waqt orders of magnitude. Dekho MTTF and FIT rates.

Verify: general $n$ ke liye ratio $(1/2)^{n}$ hai. $n = 1 \Rightarrow 0.5$ , $n = 2 \Rightarrow 0.25$ . Consistent. ✔

Example 5 — Dono $J$ aur $T$ badlein (Cell C5)

Worked example Statement

Ek wire redesign hoti hai: current density $1.5 \times$ badhti hai aur operating temperature $90^{\circ} C$ se $110^{\circ} C$ tak chadhti hai. $n = 2$ , $E_{a} = 0.8$ eV ke saath, combined lifetime multiplier kya hai?

Forecast: do harmful effects multiplicatively stack hote hain. Ek bada fall expect karo.

Step 1 — Kelvin. $T_{1} = 363$ K, $T_{2} = 383$ K.

Step 2 — FULL Black ratio likho. $\frac{MTTF _{2}}{MTTF _{1}} = (\frac{J _{1}}{J _{2}})^{n} exp [\frac{E _{a}}{k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}})]$ Yeh step kyun? Ab $A$ ke alawa kuch bhi cancel nahi hota — $J$ aur exponential dono differ karte hain. Do factors multiply hote hain kyunki Black's law ek $J$ -term aur ek $T$ -term ka product hai.

Step 3 — Current factor. $(1/1.5)^{2} = 0.4444$ .

Step 4 — Temperature factor. $\frac{E _{a}}{k _{B}} = \frac{0.8}{8.617 \times 1 0 ^{- 5}} = 9284$ K. $\frac{1}{383} - \frac{1}{363} = - 1.4386 \times 1 0^{- 4}$ . Exponent $= 9284 \times (- 1.4386 \times 1 0^{- 4}) = - 1.336$ . Temperature factor $= e^{- 1.336} = 0.2630$ .

Step 5 — Multiply karo. $0.4444 \times 0.2630 = 0.1169$ .

Verify: life ~11.7% tak gir jaati hai — roughly $8.6 \times$ reduction. Akela har effect (0.44 aur 0.26) 0.11 se upar hai, aur unka product dono se chhota hai, jaise do penalties compound hone chahiye. ✔

Example 6 — Degenerate limits: $J \to 0$ aur $L \to 0$ (Cell C6)

Worked example Statement

Black's law kya predict karta hai jab (a) $J \to 0$ , aur (b) Blech condition kya kehti hai jab $L \to 0$ ? Kya yeh physically sensible hain?

Forecast: koi current nahi ya koi length nahi → koi wear-out nahi → "immortal."

Step 1 — Black's law ka limit jab $J \to 0$ . $MTTF = A J^{- n} e^{E_{a} / (k_{B} T)} J \to 0 \infty$ Yeh step kyun? $J^{- n} = 1/ J^{n}$ blow up karta hai jab $J \to 0$ . Zero current density ka matlab zero electron wind hai, isliye koi atom nahi dhakele jaate — wire EM se kabhi wear out nahi hoti. Math aur picture agree karte hain.

Step 2 — Us infinity ka reality check. Kyun care karein? Infinity sirf EM prediction hai. Bina current wali wire phir bhi Stress migration se age hoti hai (thermomechanical stress, koi current nahi chahiye). Isliye "EM-immortal" ≠ "immortal." Kabhi bhi $\infty$ ko real lifetime mat quote karo.

Step 3 — Blech limit jab $L \to 0$ . $J \cdot L \to 0 < (J L)_{cr i t}$ kisi bhi finite $J$ ke liye. Yeh step kyun? Ek infinitesimally short wire hamesha Blech inequality satisfy karti hai, isliye woh hamesha immortal hai. Yahi reason hai ki short via-to-via jumpers high current "for free" carry kar sakte hain.

Verify: dono degenerate inputs immortality ki taraf jaate hain, physics se consistent (koi driving force nahi, ya koi room nahi killing void accumulate karne ka). ✔

Example 7 — Blech immortal length (Cell C7)

Worked example Statement

Ek test structure $(J L)_{cr i t} = 4000$ A/cm deta hai. Ek signal wire $J = 8 \times 1 0^{5}$ A/cm $^{2}$ par chalti hai. Immortality guaranteed karne wali sabse lambi wire kya hai? Phir: agar design ko $70 μ m$ wire chahiye, toh kya woh safe hai?

Forecast: threshold ko $J$ se divide karo length cm mein nikalne ke liye; phir microns mein convert karo.

Step 1 — Blech inequality ko $L$ ke liye solve karo. $L < \frac{( J L ) _{cr i t}}{J} = \frac{4000}{8 \times 1 0 ^{5}} = 5 \times 1 0^{- 3} cm$ Yeh step kyun? Immortality ka knob product $J \cdot L$ hai. $J$ fix karke, sabse bada safe $L$ threshold ko $J$ se divide karna hai.

Step 2 — Microns mein convert karo. $5 \times 1 0^{- 3} cm \times \frac{1 0 ^{4} μ m}{1 cm} = 50 μ m$ . Yeh step kyun? $1$ cm $= 1 0^{4} μ$ m — not $1 0^{3}$ . Reason: $1$ cm $= 1 0^{- 2}$ m aur $1 μ$ m $= 1 0^{- 6}$ m, isliye exponent difference $(- 2) - (- 6) = 4$ hai, $1 0^{4}$ deta hai. (Trap yeh hai ki "milli-to-micro is $1 0^{3}$ " sochna; lekin yeh centi-to-micro hai, jo $1 0^{4}$ hai.) Layout tools microns mein bolte hain, isliye translate karna zaroori hai.

Step 3 — $70 μ m$ wire judge karo. $70 μ m > 50 μ m$ , isliye Blech product exceed ho gaya → immortal nahi; ise ek real Black's-law lifetime check chahiye.

Verify: exactly $L = 50 μ m = 5 \times 1 0^{- 3}$ cm par, $J \cdot L = 8 \times 1 0^{5} \times 5 \times 1 0^{- 3} = 4000$ A/cm — precisely threshold. Isse lamba matlab over. ✔

Neeche figure dekho: violet line is wire ka $J \cdot L$ length ke saath grow karta hai, magenta dashed line fixed threshold $(J L)_{cr i t}$ hai, aur orange marker $50 μ m$ crossover hai. Crossover se left ka peach-shaded region immortal hai; $70 μ m$ navy dot failing region mein baithta hai.

Example 8 — Real-world word problem (Cell C8)

Worked example Statement

Ek data-center chip ka power grid $85^{\circ} C$ aur current density $J_{0}$ par $10$ saal ke liye qualified hai. Performance boost karne ke liye vendor clock raise karta hai, jo (i) grid current density ko $1.3 J_{0}$ tak badhata hai aur (ii) grid temperature ko $100^{\circ} C$ tak le jaata hai. $n = 2$ , $E_{a} = 0.9$ eV ke saath, kya grid abhi bhi $7$ -saal service target meet karta hai?

Forecast: do stressors — likely 10 saal se gir jaayega, lekin kya 7 clear karta hai?

Step 1 — Kelvin. $T_{1} = 358$ K, $T_{2} = 373$ K.

Step 2 — Full Black ratio (dono factors). $\frac{MTTF _{2}}{MTTF _{1}} = (\frac{1}{1.3})^{2} exp [\frac{0.9}{k _{B}} (\frac{1}{373} - \frac{1}{358})]$ Yeh step kyun? Do cheezein badlin, isliye hum current power-law aur Arrhenius term dono rakhte hain. Grid rail → growth-limited → $n = 2$ (dekho "Which $n$ " box).

Step 3 — Current factor. $(1/1.3)^{2} = 0.5917$ .

Step 4 — Temperature factor. $\frac{0.9}{8.617 \times 1 0 ^{- 5}} = 10444$ K. $\frac{1}{373} - \frac{1}{358} = - 1.1229 \times 1 0^{- 4}$ . Exponent $= 10444 \times (- 1.1229 \times 1 0^{- 4}) = - 1.1728$ ; factor $= e^{- 1.1728} = 0.3094$ .

Step 5 — Combine karo aur baseline par apply karo. Multiplier $= 0.5917 \times 0.3094 = 0.1831$ . New MTTF $= 10 \times 0.1831 = 1.83$ saal.

Step 6 — Decision. $1.83 < 7$ saal ⇒ FAIL. Clock boost free nahi hai; grid ko wider ya cooler karna padega.

Verify: har factor $< 1$ (dono stressors harmful hain), product aur bhi chhota, aur result 7 se kaafi neeche. Sanity intact. ✔

Example 9 — Exam twist: do lifetimes se $E_{a}$ extract karo (Cell C9)

Worked example Statement

Identical wires par do accelerated tests (same $J$ ): $200^{\circ} C$ par $MTTF = 200$ h aur $175^{\circ} C$ par $MTTF = 800$ h. Activation energy $E_{a}$ nikalo.

Forecast: cooler test 4× zyada chala, isliye exponential 4× jump kiya; ulta solve karne par mid-range $E_{a}$ milni chahiye (~0.7–1.0 eV).

Step 1 — Dono temperatures ko kelvin mein convert karo. $T_{1} = 200 + 273 = 473$ K (hot test), $T_{2} = 175 + 273 = 448$ K (cool test). Yeh step kyun? Arrhenius exponent ko absolute temperature chahiye; aur hume track rakhna hai ki kaunsa MTTF kaunse $T$ ke saath pair karta hai — $200$ h with $473$ K, $800$ h with $448$ K.

Step 2 — Ratio likho aur exponential isolate karo. $\frac{MTTF _{2}}{MTTF _{1}} = exp [\frac{E _{a}}{k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}})] = \frac{800}{200} = 4$ Yeh step kyun? Same $J$ aur material → $A$ aur $J^{- n}$ cancel ho jaate hain, pure Arrhenius bachta hai. Left side numerically pata hai (yeh 4 hai) aur $E_{a}$ chahiye, jo exponent mein chhupa hai.

Step 3 — $E_{a}$ ko exponent se neeche laane ke liye natural log lo. $ln 4 = \frac{E _{a}}{k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ Yeh step kyun? Unknown $E_{a}$ $e^{(\dots)}$ ke andar baith gaya hai. Natural log $ln$ $e^{x}$ ka exact inverse hai (yeh exponential ko "undo" karta hai), isliye ise apply karne par equation $E_{a}$ mein plain linear ban jaati hai. Isliye hum logarithm reach karte hain, say, square root nahi.

Step 4 — $E_{a}$ ke liye solve karo. $\frac{1}{448} - \frac{1}{473} = 1.1799 \times 1 0^{- 4}$ K $^{- 1}$ (ek positive bracket, kyunki $T_{2} < T_{1}$ ). $ln 4 = 1.3863$ . $E_{a} = k_{B} \cdot \frac{l n 4}{( 1/ T _{2} - 1/ T _{1} )} = 8.617 \times 1 0^{- 5} \cdot \frac{1.3863}{1.1799 \times 1 0 ^{- 4}} = 1.012 eV$ Yeh step kyun? Hum sirf dono sides ko bracket se divide karte hain aur $k_{B}$ se multiply karte hain — pure algebra, ab jab log ne $E_{a}$ free kar diya.

Verify: $\sim 1.0$ eV parent note mein quoted copper surface/interface range mein hai — Cu test ke liye plausible. Back plug karo: $exp [(1.012/8.617 \times 1 0^{- 5}) (1.1799 \times 1 0^{- 4})] = e^{1.386} = 4.00$ , jo measured $800/200$ ratio recover karta hai. ✔

Recall Quick self-test

Fixed T ke saath n=2 par J double karna MTTF ko kitne se multiply karta hai ::: $1/4$ Wire ko cool karna (lower T) MTTF ko karta hai ::: lamba (positive bracket, factor > 1) Jab $J \to 0$ , Black's MTTF jaata hai ::: $\infty$ tak (koi EM nahi), lekin stress migration phir bhi ise age karta hai $E_{a}$ ko exponential se bahar nikalne ke liye tum apply karte ho ::: natural log $ln$ Wire Blech-immortal hoti hai jab ::: $J \cdot L < (J L)_{cr i t}$ Wide power/clock rail ke liye, kaunsa current exponent choose karein? ::: $n \approx 2$ (growth-limited) $1$ cm mein kitne $μ$ m hote hain? ::: $1 0^{4} μ$ m (centi-to-micro $1 0^{4}$ hai, $1 0^{3}$ nahi)