Electromigration reliability: Worked examples — Hardware deep dive

Parent: Electromigration reliability

This page is a drill sheet. Before we compute anything, we lay out every kind of question electromigration (EM) can ask, then we solve one example per box so you never meet a case you haven't already practised.

Two laws do all the work here. Let us re-state them in plain words first, so no symbol arrives unannounced.

Formula The two tools we will use everywhere

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

$MTTF$ = median time to failure (how long till half the wires die).
$A$ = a constant set by the wire's geometry and material.
$J$ = current density = current $\div$ cross-section area (amps per cm $^{2}$ ). It is how crowded the electron traffic is.
$n$ = current exponent, either $\approx 1$ or $\approx 2$ (we explain which, below).
$E_{a}$ = activation energy in electron-volts (eV): the "energy toll" an atom must pay to hop. Bigger toll = slower migration = longer life.
$k_{B} = 8.617 \times 1 0^{- 5}$ eV/K, Boltzmann's constant, converts temperature into energy.
$T$ = absolute temperature in kelvin (K). Always kelvin, never °C, inside the exponent.

The Blech condition (short-line immortality): $J \cdot L < (J L)_{cr i t} \Rightarrow wire never fails$ $L$ = length of the wire (cm). If the product $J \cdot L$ stays under a critical threshold, back-pressure from piled-up atoms cancels the electron wind.

Definition Which

$n$ do I pick? ( $n \approx 1$ vs $n \approx 2$ ) The exponent $n$ is not a free choice — it names what stage of void formation is the bottleneck:

$n \approx 1$ — nucleation-limited. The wire dies as soon as a void is born. Pick this when a void, once nucleated, immediately spans the wire — e.g. very narrow / near-bamboo lines, single-grain-width interconnects, or short segments where growth is trivial. Here lifetime tracks the nucleation rate, which is linear in $J$ .
$n \approx 2$ — growth-limited. A void nucleates early but must slowly grow to a killing size. Pick this for wider lines, vias, and current-carrying power rails where growth time dominates. Lifetime tracks the accumulated mass transport $\propto J^{2}$ .
Practical rule: if you're told the failure mode, use it. If not, wide power/clock nets → $n = 2$ ; thin signal lines / conservative qualification → $n = 1$ . When uncertain, $n = 1$ is the pessimistic (shorter-life) choice for design margin, so vendors often qualify with it.

The scenario matrix

Every EM homework or exam question falls into one of these cells. We will hit each one.

#	Cell class	What changes	Which tool	Example
C1	Temperature down (cooling)	$T$ only	Arrhenius ratio	Ex 1
C2	Temperature up (self-heating)	$T$ only	Arrhenius ratio	Ex 2
C3	Current density scaling, $n = 2$	$J$ only	Power law	Ex 3
C4	Current density scaling, $n = 1$ (wrong- $n$ trap)	$J$ only	Power law	Ex 4
C5	Both $J$ and $T$ change at once	$J$ and $T$	Full Black ratio	Ex 5
C6	Degenerate: $J \to 0$ or $L \to 0$ (limits)	limiting input	interpret formula	Ex 6
C7	Blech immortality — find max length	geometry	Blech product	Ex 7
C8	Real-world word problem (data-center)	mixed	Black + Arrhenius	Ex 8
C9	Exam twist: extract $E_{a}$ from two lifetimes	solve backwards	log of ratio	Ex 9

Example 1 — Cooling a wire (Cell C1)

Worked example Statement

A wire survives $5$ years at $T_{1} = 105^{\circ} C$ . If we cool it to $T_{2} = 85^{\circ} C$ , how long does it last? Take $E_{a} = 0.7$ eV.

Forecast: cooler = slower atoms = longer life. Guess: more than 5 years — but 2× or 10×?

Step 1 — Convert to kelvin. $T_{1} = 105 + 273.15 = 378.15$ K, $T_{2} = 85 + 273.15 = 358.15$ K. (We round to $378$ K and $358$ K below.) Why this step? The exponent $e^{E_{a} / (k_{B} T)}$ only makes physical sense with an absolute (kelvin) temperature — at $0$ K diffusion truly stops. Celsius would give nonsense (you can have negative °C but never negative K).

Step 2 — Take the ratio so $A$ and $J$ vanish. $\frac{MTTF _{2}}{MTTF _{1}} = exp [\frac{E _{a}}{k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}})]$ Why this step? Same wire, same current, so $A$ and $J^{- n}$ are identical in both and cancel on division. Only the temperature term is left.

Step 3 — Plug in numbers (using $T_{1} = 378$ K, $T_{2} = 358$ K). $\frac{E _{a}}{k _{B}} = \frac{0.7}{8.617 \times 1 0 ^{- 5}} = 8123.5$ 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. $MTTF_{2} = 5 \times 3.32 = 16.6$ years.

Verify: $T_{2} < T_{1}$ , so $1/ T_{2} > 1/ T_{1}$ , so the bracket is positive, so the factor $> 1$ , so life increased. Matches the forecast. ✔

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

Worked example Statement

A power rail rated for $10$ years at $85^{\circ} C$ is left running hot at $125^{\circ} C$ due to Joule heating & self-heating. New lifetime? Use $E_{a} = 0.9$ eV (copper).

Forecast: hotter = shorter. And higher $E_{a}$ makes temperature bite harder, so expect a steep drop.

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

Step 2 — Arrhenius ratio, hot over cool. $\frac{MTTF _{h o t}}{MTTF _{coo l}} = exp [\frac{E _{a}}{k _{B}} (\frac{1}{398} - \frac{1}{358})]$ Why this step? We want the hot lifetime, so put the hot temperature in the numerator's $1/ T$ . Because $398 > 358$ , the bracket goes negative → factor below 1 → life drops. That sign check is your safety rail.

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. $MTTF_{h o t} = 10 \times 0.0533 = 0.533$ years $\approx 6.4$ months.

Verify: a $40^{\circ} C$ rise cut life by ~19×. See the Arrhenius reliability model rule of thumb: for $E_{a} \approx 0.9$ , life roughly halves every $\sim 8$ – $9^{\circ} C$ ; $40^{\circ} C \approx 4.5$ halvings $\approx 2^{4.5} \approx 23 \times$ — same order. ✔

Example 3 — Doubling current, $n = 2$ (Cell C3)

Worked example Statement

A damascene wire's width is halved (to fit a denser routing track), but it carries the same current. With void-growth-limited $n = 2$ , what happens to MTTF?

Forecast: narrower wire, same current → traffic twice as dense → life falls. By how much?

Step 1 — Find the new $J$ . $J = \frac{I}{area}$ . Halving width halves the area, so $J_{2} = 2 J_{1}$ . Why this step? Current density is the actual driver in Black's law, not the raw current. Designers control width; the physics feels $J$ . See Design rules & current density limits.

Step 2 — Ratio with only $J$ changing. $\frac{MTTF _{2}}{MTTF _{1}} = (\frac{J _{1}}{J _{2}})^{n} = (\frac{1}{2})^{2}$ Why this step? Same temperature, so the exponential cancels; only $J^{- n}$ remains. Note $J^{- n}$ means smaller- $J$ -on-top when we divide.

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

Verify: MTTF $\propto J^{- 2}$ , so doubling $J$ multiplies life by $2^{- 2} = 0.25$ . A quarter of the life — the textbook reason EM rules cap current per unit width. ✔

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

Worked example Statement

Repeat Example 3, but the failure is nucleation-limited, so $n = 1$ . How does the answer change, and why does picking the wrong $n$ matter?

Forecast: the drop should be milder with a smaller exponent.

Step 1 — Same $J$ change. $J_{2} = 2 J_{1}$ . Why: geometry is identical to Ex 3; only the physics regime ( $n$ ) differs.

Step 2 — Apply the power law with $n = 1$ . $\frac{MTTF _{2}}{MTTF _{1}} = (\frac{1}{2})^{1} = 1/2$ Why this step? $n$ is literally the steepness of the current penalty. $n = 1$ (void nucleation limits life) is gentler than $n = 2$ (void growth limits life). Use the "Which $n$ do I pick?" box above: a near-bamboo narrow line is nucleation-limited, so $n = 1$ is the right regime here.

Step 3 — Compare to Ex 3. $n = 2$ gave $1/4$ ; $n = 1$ gives $1/2$ . Using $n = 2$ when reality is $n = 1$ would predict half the true lifetime — a 2× error here, and orders of magnitude when extrapolating over decades. See MTTF and FIT rates.

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

Example 5 — Both $J$ and $T$ change (Cell C5)

Worked example Statement

A wire is redesigned: current density rises by $1.5 \times$ and operating temperature climbs from $90^{\circ} C$ to $110^{\circ} C$ . With $n = 2$ , $E_{a} = 0.8$ eV, what is the combined lifetime multiplier?

Forecast: two harmful effects stack multiplicatively. Expect a big fall.

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

Step 2 — Write the FULL Black ratio. $\frac{MTTF _{2}}{MTTF _{1}} = (\frac{J _{1}}{J _{2}})^{n} exp [\frac{E _{a}}{k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}})]$ Why this step? Now nothing except $A$ cancels — both $J$ and the exponential differ. The two factors multiply because Black's law is a product of a $J$ -term and a $T$ -term.

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. $0.4444 \times 0.2630 = 0.1169$ .

Verify: life drops to ~11.7% — about a $8.6 \times$ reduction. Each effect alone (0.44 and 0.26) is above 0.11, and their product is smaller than either, as two penalties should compound. ✔

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

Worked example Statement

What does Black's law predict as (a) $J \to 0$ , and (b) what does the Blech condition say as $L \to 0$ ? Are these physically sensible?

Forecast: no current or no length → no wear-out → "immortal."

Step 1 — Limit of Black's law as $J \to 0$ . $MTTF = A J^{- n} e^{E_{a} / (k_{B} T)} J \to 0 \infty$ Why this step? $J^{- n} = 1/ J^{n}$ blows up as $J \to 0$ . Zero current density means zero electron wind, so no atoms are shoved — the wire never wears out by EM. The math and the picture agree.

Step 2 — Reality check on that infinity. Why care? Infinity is the EM prediction only. A wire with no current still ages by Stress migration (thermomechanical stress, no current needed). So "EM-immortal" ≠ "immortal." Never quote $\infty$ as a real lifetime.

Step 3 — Blech limit as $L \to 0$ . $J \cdot L \to 0 < (J L)_{cr i t}$ for any finite $J$ . Why this step? An infinitesimally short wire always satisfies the Blech inequality, so it is always immortal. This is exactly why short via-to-via jumpers can carry high current "for free."

Verify: both degenerate inputs push toward immortality, consistent with the physics (no driving force, or no room to accumulate a killing void). ✔

Example 7 — Blech immortal length (Cell C7)

Worked example Statement

A test structure gives $(J L)_{cr i t} = 4000$ A/cm. A signal wire runs at $J = 8 \times 1 0^{5}$ A/cm $^{2}$ . What is the longest wire guaranteed immortal? Then: if the design needs a $70 μ m$ wire, is it safe?

Forecast: divide the threshold by $J$ to get a length in cm; then convert to microns.

Step 1 — Solve the Blech inequality for $L$ . $L < \frac{( J L ) _{cr i t}}{J} = \frac{4000}{8 \times 1 0 ^{5}} = 5 \times 1 0^{- 3} cm$ Why this step? The immortality knob is the product $J \cdot L$ . Fixing $J$ , the largest safe $L$ is the threshold divided by $J$ .

Step 2 — Convert to microns. $5 \times 1 0^{- 3} cm \times \frac{1 0 ^{4} μ m}{1 cm} = 50 μ m$ . Why this step? $1$ cm $= 1 0^{4} μ$ m — not $1 0^{3}$ . Reason: $1$ cm $= 1 0^{- 2}$ m and $1 μ$ m $= 1 0^{- 6}$ m, so the exponent difference is $(- 2) - (- 6) = 4$ , giving $1 0^{4}$ . (The trap is thinking "milli-to-micro is $1 0^{3}$ "; but this is centi-to-micro, which is $1 0^{4}$ .) Layout tools speak microns, so we must translate.

Step 3 — Judge the $70 μ m$ wire. $70 μ m > 50 μ m$ , so the Blech product is exceeded → not immortal; it needs a real Black's-law lifetime check.

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

Look at the figure below: the violet line is this wire's $J \cdot L$ growing with length, the magenta dashed line is the fixed threshold $(J L)_{cr i t}$ , and the orange marker is the $50 μ m$ crossover. The peach-shaded region left of the crossover is immortal; the $70 μ m$ navy dot sits in the failing region.

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

Worked example Statement

A data-center chip's power grid is qualified for $10$ years at $85^{\circ} C$ and current density $J_{0}$ . To boost performance the vendor raises the clock, which (i) increases the grid current density to $1.3 J_{0}$ and (ii) raises grid temperature to $100^{\circ} C$ . With $n = 2$ , $E_{a} = 0.9$ eV, does the grid still meet a $7$ -year service target?

Forecast: two stressors — likely falls below 10 years, but does it clear 7?

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

Step 2 — Full Black ratio (both factors). $\frac{MTTF _{2}}{MTTF _{1}} = (\frac{1}{1.3})^{2} exp [\frac{0.9}{k _{B}} (\frac{1}{373} - \frac{1}{358})]$ Why this step? Two things changed, so we keep both the current power-law and the Arrhenius term. Grid rail → growth-limited → $n = 2$ (see the "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 and apply to baseline. Multiplier $= 0.5917 \times 0.3094 = 0.1831$ . New MTTF $= 10 \times 0.1831 = 1.83$ years.

Step 6 — Decision. $1.83 < 7$ years ⇒ FAILS the target. The clock boost is not free; the grid must be widened or cooled.

Verify: each factor $< 1$ (both stressors harmful), product smaller still, and the result well under 7. Sanity intact. ✔

Example 9 — Exam twist: extract $E_{a}$ from two lifetimes (Cell C9)

Worked example Statement

Two accelerated tests on identical wires (same $J$ ): $MTTF = 200$ h at $200^{\circ} C$ and $MTTF = 800$ h at $175^{\circ} C$ . Find the activation energy $E_{a}$ .

Forecast: the cooler test lasted 4× longer, so the exponential jumped by 4×; solving backwards should give a mid-range $E_{a}$ (~0.7–1.0 eV).

Step 1 — Convert both temperatures to kelvin. $T_{1} = 200 + 273 = 473$ K (the hot test), $T_{2} = 175 + 273 = 448$ K (the cool test). Why this step? The Arrhenius exponent needs absolute temperature; and we must keep track of which MTTF pairs with which $T$ — $200$ h with $473$ K, $800$ h with $448$ K.

Step 2 — Write the ratio and isolate the exponential. $\frac{MTTF _{2}}{MTTF _{1}} = exp [\frac{E _{a}}{k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}})] = \frac{800}{200} = 4$ Why this step? Same $J$ and material → $A$ and $J^{- n}$ cancel, leaving pure Arrhenius. We know the left side numerically (it's 4) and want $E_{a}$ , which is buried in the exponent.

Step 3 — Take the natural log to bring $E_{a}$ down from the exponent. $ln 4 = \frac{E _{a}}{k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ Why this step? The unknown $E_{a}$ sits inside $e^{(\dots)}$ . The natural log $ln$ is the exact inverse of $e^{x}$ (it "undoes" the exponential), so applying it turns the equation into a plain linear one in $E_{a}$ . This is why we reach for the logarithm and not, say, a square root.

Step 4 — Solve for $E_{a}$ . $\frac{1}{448} - \frac{1}{473} = 1.1799 \times 1 0^{- 4}$ K $^{- 1}$ (a positive bracket, since $T_{2} < T_{1}$ ). $ln 4 = 1.3863$ . $E_{a} = k_{B} \cdot \frac{l n 4}{( 1/ T _{2} - 1/ T _{1} )} = 8.617 \times 1 0^{- 5} \cdot \frac{1.3863}{1.1799 \times 1 0 ^{- 4}} = 1.012 eV$ Why this step? We simply divide both sides by the bracket and multiply by $k_{B}$ — pure algebra, now that the log has freed $E_{a}$ .

Verify: $\sim 1.0$ eV lies in the copper surface/interface range quoted in the parent note — plausible for a Cu test. Plug back: $exp [(1.012/8.617 \times 1 0^{- 5}) (1.1799 \times 1 0^{- 4})] = e^{1.386} = 4.00$ , which recovers the measured $800/200$ ratio. ✔

Recall Quick self-test

Doubling J at fixed T with n=2 multiplies MTTF by ::: $1/4$ Cooling a wire (lower T) makes MTTF ::: longer (positive bracket, factor > 1) As $J \to 0$ , Black's MTTF tends to ::: $\infty$ (no EM), but stress migration still ages it To pull $E_{a}$ out of the exponential you apply ::: the natural log $ln$ A wire is Blech-immortal when ::: $J \cdot L < (J L)_{cr i t}$ For a wide power/clock rail, pick which current exponent? ::: $n \approx 2$ (growth-limited) $1$ cm equals how many $μ$ m? ::: $1 0^{4} μ$ m (centi-to-micro is $1 0^{4}$ , not $1 0^{3}$ )