Electromigration reliability: Exercises — Hinglish deep dive

6.4.8 · D4 · Hardware › Power, Thermal & Reliability › Electromigration reliability

Exercises se pehle, do pictures physics ko dimag mein fix karti hain. Pehli, electromigration actually hoti kya hai — electron wind aur mass kahaan jaata hai:

Figure dekho: conventional current (amber, upar) ek direction mein point karta hai, lekin electrons (cyan arrows) ulti direction mein stream karte hain. Woh electrons fixed metal atoms se collide karte hain aur unhe downstream drag karte hain — atoms cathode end se drain hote hain (ek void chodte hain, jo open circuit hai) aur anode end par pile up hote hain (ek hillock, jo neighbour ko short kar sakta hai). Neeche har problem ultimately is baare mein hai ki yeh drain kitni tezi se hoti hai.

Doosri, Black's law ki temperature ke saath shape — kyun ek hi wire hot vs cold mein wildly alag lifetimes jeeti hai:

Curve dekho: temperature badhne par MTTF girti hai. Yeh straight line nahi hai — yeh ek exponential cliff hai. Amber markers do operating points dikhate hain; unke beech ka vertical drop woh "acceleration factor" hai jo tum yahan almost har problem mein compute karte ho.

Poore note mein hum ek master law use karte hain, median-time-to-failure ke liye Black's Equation:

Kelvin kyun, Celsius kabhi nahi? Exponent $T$ se divide karta hai. $0^{\circ} C$ par tum zero se divide karte aur negative Celsius par sign flip ho jaata — physically bakwaas. Kelvin absolute zero se start hota hai ( $0 K = - 273.15^{\circ} C$ ), isliye yeh hamesha positive hota hai. Convert karo $T_{K} = T_{^{\circ} C} + 273.15$ se.

Level 1 — Recognition

Recall Solution

Electrons current carry karte hain, aur electrons conventional current ke ulte flow karte hain — toh yahan electron wind right se left blow karta hai. Wind atoms ko apne saath dhakelta hai, yaani atoms right se left migrate karte hain (anode ki taraf). Is page ki top figure ko reference picture ke roop mein use karo: cyan electron arrows amber current arrow ke against point karte hain.

Ise clearly anchor karte hain. Conventional current left (anode) se enter karta hai aur right (cathode) se bahar jaata hai. Electrons ulta karte hain: woh cathode (right) se enter karte hain aur anode (left) se bahar jaate hain. Wind, aur isliye atoms, cathode → anode = right → left move karte hain.

Atoms cathode (right) se drain away hote hain → void right/cathode side par banti hai.
Atoms anode (left) par pile up hote hain → left par hillock banta hai.

Jawab: atoms right→left move karte hain (anode ki taraf); void right (cathode/upstream) end par appear hoti hai.

Recall Solution

Temperature: exponent $E_{a} / (k_{B} T)$ hai. Jab $T$ badhta hai, $1/ T$ ghatta hai, toh exponent ghatta hai, toh $exp (E_{a} / (k_{B} T))$ ghatta hai → MTTF girti hai. Zyada garam = choti life. Yeh physics se match karta hai: heat atoms ko tezi se ucharhal deti hai (faster diffusion). Yeh exactly woh cliff hai jo doosri figure mein draw ki gayi hai.
Current density: MTTF $\propto J^{- n}$ hai jahan $n > 0$ hai. Bada $J$ → bada denominator $J^{n}$ → chota MTTF. Zyada current = choti life.

Jawab: zyada $T$ life choti karta hai; zyada $J$ life choti karta hai.

Level 2 — Application

Recall Solution

Hum kya karte hain: do temperatures par Black's equation ka ratio lete hain. Ratio kyun: constants $A$ aur $J^{- n}$ identical hain (same wire, same current), toh woh cancel ho jaate hain, sirf temperature part bachta hai.

Convert karo: $T_{1} = 110 + 273.15 = 383.15$ K, $T_{2} = 90 + 273.15 = 363.15$ K. $\frac{MTTF _{2}}{MTTF _{1}} = \frac{e x p ( E _{a} / ( k _{B} T _{2} ) )}{e x p ( E _{a} / ( k _{B} T _{1} ) )} = exp [\frac{E _{a}}{k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}})]$ Pieces compute karo:

$\frac{E _{a}}{k _{B}} = \frac{0.8}{8.617 \times 1 0 ^{- 5}} = 9284.0$ K.
$\frac{1}{363.15} - \frac{1}{383.15} = 2.7537 \times 1 0^{- 3} - 2.6099 \times 1 0^{- 3} = 1.437 \times 1 0^{- 4} K^{- 1}$ .
Exponent $= 9284.0 \times 1.437 \times 1 0^{- 4} = 1.334$ .
Factor $= e^{1.334} = 3.797$ .

$MTTF_{2} = 8 \times 3.797 = 30.4 years$ 20 °C thanda → lagbhag $3.8 \times$ lambi life. ✔

Recall Solution

MTTF $\propto J^{- n}$ hai. $J$ teen guna karne par lifetime $(3)^{- n} = (3)^{- 2} = \frac{1}{9}$ se multiply hoti hai. Jawab: lifetime original ki $1/9$ ( $\approx 11%$ ) ho jaati hai. Itna brutal kyun hai: square ka matlab hai current mein $3 \times$ push wear rate mein $9 \times$ punch hai — exactly isliye current-density design rules strict hoti hain.

Level 3 — Analysis

Recall Solution

Hum kya karte hain: phir ratio — same $T$ , same $A$ , toh Arrhenius term aur $A$ cancel ho jaate hain, sirf current part bachta hai. $\frac{MTTF _{P}}{MTTF _{Q}} = (\frac{J _{Q}}{J _{P}})^{n}$ Plug in karo: $\frac{100}{44.4} = (\frac{1.5 \times 1 0 ^{6}}{1.0 \times 1 0 ^{6}})^{n} \Rightarrow 2.2523 = (1.5)^{n}$ Dono sides ka logarithm lo — logs kyun? kyunki $n$ exponent mein fansa hua hai, aur log woh tool hai jo exponent ko neeche ek multiplier mein kheench laata hai: $ln (2.2523) = n ln (1.5) \Rightarrow n = \frac{0.8121}{0.4055} = 2.00$ Jawab: $n \approx 2$ → yeh wire void-growth limited hai.

Recall Solution

Idea (parent note se): atoms jo anode par push hote hain ek back-pressure build karte hain jaise compressed spring. Agar wire itni choti ho, toh woh back-stress electron wind ko balance kar leta hai void grow hone se pehle. Balance condition hai $J \cdot L < (J L)_{crit}$ . $L < \frac{( J L ) _{crit}}{J} = \frac{3000 A/cm}{8 \times 1 0 ^{5} A/cm ^{2}} = 3.75 \times 1 0^{- 3} cm$ Seedha micrometres mein convert karo. Kyunki $1 μ m = 1 0^{- 4} cm$ , $1 0^{- 4}$ se divide karo: $L < \frac{3.75 \times 1 0 ^{- 3} cm}{1 0 ^{- 4} cm / μ m} = 37.5 μ m$ Jawab: $37.5 μ m$ se chota koi bhi segment is current par immortal hai.

Neeche ki figure balance visible karti hai. Top panel mein (short wire) amber back-stress arrow itna lamba ho jaata hai ki woh cyan electron-wind arrow ko poora cancel kar deta hai — koi net atom flow nahi, toh wire immortal hai. Bottom panel mein (long wire) back-stress arrow chota rehta hai, wind jeet jaata hai, aur ek void (woh chota amber circle) cathode side par khulta hai.

Level 4 — Synthesis

Recall Solution

Hum kya karte hain: is baar poora ratio lete hain, DONO current term aur temperature term rakhte hain (sirf $A$ cancel hota hai). $\frac{MTTF _{2}}{MTTF _{1}} = (\frac{J _{1}}{J _{2}})^{n} exp [\frac{E _{a}}{k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}})]$

Current factor: $J_{2} = 2 J_{1}$ , toh $(\frac{J _{1}}{J _{2}})^{2} = (\frac{1}{2})^{2} = 0.25$ .

Temperature factor: $T_{1} = 373.15$ K, $T_{2} = 393.15$ K.

$\frac{E _{a}}{k _{B}} = \frac{0.9}{8.617 \times 1 0 ^{- 5}} = 10444.5$ K.
$\frac{1}{393.15} - \frac{1}{373.15} = 2.5436 \times 1 0^{- 3} - 2.6799 \times 1 0^{- 3} = - 1.3633 \times 1 0^{- 4} K^{- 1}$ (negative — zyada garam life choti karta hai).
Exponent $= 10444.5 \times (- 1.3633 \times 1 0^{- 4}) = - 1.4239$ .
Temperature factor $= e^{- 1.4239} = 0.2408$ .

Combine karo: $\frac{MTTF _{2}}{MTTF _{1}} = 0.25 \times 0.2408 = 0.06019$ $MTTF_{2} = 6 \times 0.06019 = 0.361 years \approx 4.3 months$ Lesson: current double karna ( $\times \frac{1}{4}$ ) aur +20 °C ( $\times 0.24$ ) milke ek 6-saal ki life ko lagbhag 4 mahine tak kaat dete hain — do penalties multiply hoti hain. Isliye current-density design rules aur thermal budgets ko saath milake respect karna zaroori hai.

Level 5 — Mastery

Recall Solution

Strategy: lab wire ko hard stress karta hai (high $J$ , high $T$ ) taki woh jaldi fail ho, phir hum Black's equation ke ratio use karke gentle operating point tak extrapolate karte hain. Yeh exactly wahi kaam hai jo Arrhenius reliability model ke liye hai. $\frac{MTTF _{op}}{MTTF _{test}} = (\frac{J _{test}}{J _{op}})^{n} exp [\frac{E _{a}}{k _{B}} (\frac{1}{T _{op}} - \frac{1}{T _{test}})]$

Current factor (lower operating $J$ → lambi life, factor $> 1$ ): $(\frac{2 \times 1 0 ^{6}}{7 \times 1 0 ^{5}})^{2} = (2.8571)^{2} = 8.1633$

Temperature factor ( $T_{op} = 368.15$ K, $T_{test} = 448.15$ K):

$\frac{E _{a}}{k _{B}} = 10444.5$ K.
$\frac{1}{368.15} - \frac{1}{448.15} = 2.7163 \times 1 0^{- 3} - 2.2314 \times 1 0^{- 3} = 4.8489 \times 1 0^{- 4} K^{- 1}$ .
Exponent $= 10444.5 \times 4.8489 \times 1 0^{- 4} = 5.0644$ .
Temperature factor $= e^{5.0644} = 158.16$ .

Acceleration factor: $8.1633 \times 158.16 = 1291.1$ .

$MTTF_{op} = 180 h \times 1291.1 = 232, 405 h$ Years mein convert karo ( $1 yr = 8766 h$ ): $232405/8766 = 26.5 years$ . Jawab: haan — $26.5$ years 10-year target se aaram se zyada hai. ✔

Recall Solution

$MTTF_{op} = 10 yr = 87660 h$ set karo aur same ratio ko $J_{op}$ ke liye solve karo. $\frac{87660}{180} = (\frac{J _{test}}{J _{op}})^{2} e^{5.0644}$ Left side $= 487.0$ . Temperature factor $e^{5.0644} = 158.16$ divide karo: $(\frac{J _{test}}{J _{op}})^{2} = \frac{487.0}{158.16} = 3.0791$ Square root lo ( $n = 2$ power undo karo): $\frac{J _{test}}{J _{op}} = 3.0791 = 1.7547$ . $J_{op} = \frac{J _{test}}{1.7547} = \frac{2 \times 1 0 ^{6}}{1.7547} = 1.14 \times 1 0^{6} A/cm^{2}$ Jawab: tum $\approx 1.14 \times 1 0^{6} A/cm^{2}$ tak push kar sakte ho aur phir bhi 10 years hit kar sakte ho. Note karo yeh L5.1 ke $7 \times 1 0^{5}$ se zyada hai — L5.1 ke comfortable 26.5-year margin se consistent hai.

Recall

Recall Quick self-check

Ratio method — do conditions ke beech kya cancel hota hai? ::: Jo bhi identical ho: $A$ hamesha; $J^{- n}$ agar current unchanged ho; Arrhenius term agar temperature unchanged ho. $n$ ko exponent se free kaise karte ho? ::: Dono sides ka logarithm lo — yeh exponent ko neeche ek product mein kheench laata hai. Current penalty aur temperature penalty multiply kyun hoti hain, add nahi? ::: Black's equation ek $J$ -term aur ek $T$ -term ka product hai, toh unke ratios multiply hote hain. Immortal-length condition kya hai? ::: $L < (J L)_{crit} / J$ ; is se neeche, back-stress electron wind ko balance karta hai. Wire ke kis end par void banti hai, aur kyun? ::: Cathode (upstream) end par, kyunki atoms wahan se anode ki taraf electron wind ke saath drain hote hain.