Electromigration reliability: Exercises — Hardware deep dive

Before the exercises, two pictures fix the physics in your mind's eye. First, what electromigration actually is — the electron wind and where mass ends up:

Look at the figure: conventional current (amber, top) points one way, but the electrons (cyan arrows) stream the opposite way. Those electrons collide with the fixed metal atoms and drag them downstream — atoms drain from the cathode end (leaving a void, an open circuit) and pile up at the anode end (a hillock, which can short to a neighbour). Every problem below is ultimately about how fast this drain happens.

Second, the shape of Black's law in temperature — why the same wire lives wildly different lifetimes hot vs cold:

Look at the curve: MTTF plunges as temperature rises. It is not a straight line — it is an exponential cliff. The amber markers show two operating points; the vertical drop between them is the "acceleration factor" you compute in almost every problem here.

Throughout we use one master law, Black's Equation for the median-time-to-failure:

Why kelvin, never Celsius? The exponent divides by $T$ . At $0^{\circ} C$ you would divide by zero and at negative Celsius you would flip the sign — physical nonsense. Kelvin starts at absolute zero ( $0 K = - 273.15^{\circ} C$ ), so it is always positive. Convert with $T_{K} = T_{^{\circ} C} + 273.15$ .

Level 1 — Recognition

Recall Solution

Electrons carry the current, and electrons flow opposite to conventional current — so here the electron wind blows right to left. The wind shoves atoms along with it, i.e. atoms migrate right to left (toward the anode). Use the top figure of this page as the reference picture: cyan electron arrows point against the amber current arrow.

Let us anchor it cleanly. Conventional current enters at the left (anode) and leaves at the right (cathode). Electrons do the reverse: they enter at the cathode (right) and exit at the anode (left). The wind, and therefore the atoms, move cathode → anode = right → left.

Atoms drain away from the cathode (right) → void forms on the right/cathode side.
Atoms pile up at the anode (left) → hillock on the left.

Answer: atoms move right→left (toward anode); void appears on the right (cathode/upstream) end.

Recall Solution

Temperature: the exponent is $E_{a} / (k_{B} T)$ . As $T$ grows, $1/ T$ shrinks, so the exponent shrinks, so $exp (E_{a} / (k_{B} T))$ shrinks → MTTF falls. Hotter = shorter life. This matches physics: heat makes atoms jump around faster (faster diffusion). It is exactly the cliff drawn in the second figure above.
Current density: MTTF $\propto J^{- n}$ with $n > 0$ . Bigger $J$ → bigger denominator $J^{n}$ → smaller MTTF. More current = shorter life.

Answer: higher $T$ shortens life; higher $J$ shortens life.

Level 2 — Application

Recall Solution

What we do: take the ratio of Black's equation at the two temperatures. Why a ratio: the constants $A$ and $J^{- n}$ are identical (same wire, same current), so they cancel, leaving only the temperature part.

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

$\frac{E _{a}}{k _{B}} = \frac{0.8}{8.617 \times 1 0 ^{- 5}} = 9284.0$ 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 $= e^{1.334} = 3.797$ .

$MTTF_{2} = 8 \times 3.797 = 30.4 years$ Cooler by 20 °C → about $3.8 \times$ longer life. ✔

Recall Solution

MTTF $\propto J^{- n}$ . Tripling $J$ multiplies lifetime by $(3)^{- n} = (3)^{- 2} = \frac{1}{9}$ . Answer: lifetime drops to $1/9$ ( $\approx 11%$ ) of the original. Why so brutal: the square means a $3 \times$ push in current is a $9 \times$ punch in wear rate — this is exactly why current-density design rules are strict.

Level 3 — Analysis

Recall Solution

What we do: ratio again — same $T$ , same $A$ , so the Arrhenius term and $A$ cancel, leaving only the current part. $\frac{MTTF _{P}}{MTTF _{Q}} = (\frac{J _{Q}}{J _{P}})^{n}$ Plug in: $\frac{100}{44.4} = (\frac{1.5 \times 1 0 ^{6}}{1.0 \times 1 0 ^{6}})^{n} \Rightarrow 2.2523 = (1.5)^{n}$ Take the logarithm of both sides — why logs? because $n$ is trapped up in an exponent, and the log is the tool that pulls an exponent down into a multiplier: $ln (2.2523) = n ln (1.5) \Rightarrow n = \frac{0.8121}{0.4055} = 2.00$ Answer: $n \approx 2$ → this wire is void-growth limited.

Recall Solution

The idea (from the parent note): atoms pushed to the anode build a back-pressure like a compressed spring. If the wire is short enough, that back-stress balances the electron wind before a void can grow. The balance condition is $J \cdot L < (J L)_{crit}$ . $L < \frac{( J L ) _{crit}}{J} = \frac{3000 A/cm}{8 \times 1 0 ^{5} A/cm ^{2}} = 3.75 \times 1 0^{- 3} cm$ Convert to micrometres directly. Since $1 μ m = 1 0^{- 4} cm$ , divide by $1 0^{- 4}$ : $L < \frac{3.75 \times 1 0 ^{- 3} cm}{1 0 ^{- 4} cm / μ m} = 37.5 μ m$ Answer: any segment shorter than $37.5 μ m$ is immortal at this current.

The figure below makes the balance visible. In the top panel (short wire) the amber back-stress arrow grows long enough to fully cancel the cyan electron-wind arrow — no net atom flow, so the wire is immortal. In the bottom panel (long wire) the back-stress arrow stays stubby, the wind wins, and a void (the small amber circle) opens on the cathode side.

Level 4 — Synthesis

Recall Solution

What we do: take the full ratio, keeping BOTH the current term and the temperature term this time (only $A$ cancels). $\frac{MTTF _{2}}{MTTF _{1}} = (\frac{J _{1}}{J _{2}})^{n} exp [\frac{E _{a}}{k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}})]$

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

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 — hotter shortens life).
Exponent $= 10444.5 \times (- 1.3633 \times 1 0^{- 4}) = - 1.4239$ .
Temperature factor $= e^{- 1.4239} = 0.2408$ .

Combine: $\frac{MTTF _{2}}{MTTF _{1}} = 0.25 \times 0.2408 = 0.06019$ $MTTF_{2} = 6 \times 0.06019 = 0.361 years \approx 4.3 months$ The lesson: doubling current ( $\times \frac{1}{4}$ ) and +20 °C ( $\times 0.24$ ) together cut a 6-year life to about 4 months — the two penalties multiply. This is why current-density design rules and thermal budgets must be respected together.

Level 5 — Mastery

Recall Solution

Strategy: the lab stresses the wire hard (high $J$ , high $T$ ) to fail it fast, then we extrapolate to the gentle operating point using the ratio of Black's equation. This is exactly what the Arrhenius reliability model is for. $\frac{MTTF _{op}}{MTTF _{test}} = (\frac{J _{test}}{J _{op}})^{n} exp [\frac{E _{a}}{k _{B}} (\frac{1}{T _{op}} - \frac{1}{T _{test}})]$

Current factor (lower operating $J$ → longer 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: $8.1633 \times 158.16 = 1291.1$ .

$MTTF_{op} = 180 h \times 1291.1 = 232, 405 h$ Convert to years ( $1 yr = 8766 h$ ): $232405/8766 = 26.5 years$ . Answer: yes — $26.5$ years comfortably exceeds the 10-year target. ✔

Recall Solution

Set $MTTF_{op} = 10 yr = 87660 h$ and solve the same ratio for $J_{op}$ . $\frac{87660}{180} = (\frac{J _{test}}{J _{op}})^{2} e^{5.0644}$ Left side $= 487.0$ . Divide out the temperature factor $e^{5.0644} = 158.16$ : $(\frac{J _{test}}{J _{op}})^{2} = \frac{487.0}{158.16} = 3.0791$ Take the square root (undo the $n = 2$ power): $\frac{J _{test}}{J _{op}} = 3.0791 = 1.7547$ . $J_{op} = \frac{J _{test}}{1.7547} = \frac{2 \times 1 0 ^{6}}{1.7547} = 1.14 \times 1 0^{6} A/cm^{2}$ Answer: you may push up to $\approx 1.14 \times 1 0^{6} A/cm^{2}$ and still hit 10 years. Note this is higher than the $7 \times 1 0^{5}$ of L5.1 — consistent with L5.1's comfortable 26.5-year margin.

Recall

Recall Quick self-check

Ratio method — what cancels between two conditions? ::: Whatever is identical: $A$ always; $J^{- n}$ if current is unchanged; the Arrhenius term if temperature is unchanged. How do you free $n$ from an exponent? ::: Take the logarithm of both sides — it pulls the exponent down into a product. Why do a current penalty and a temperature penalty multiply, not add? ::: Black's equation is a product of a $J$ -term and a $T$ -term, so their ratios multiply. What is the immortal-length condition? ::: $L < (J L)_{crit} / J$ ; below this, back-stress balances the electron wind. Which end of the wire grows a void, and why? ::: The cathode (upstream) end, because atoms drain away from it toward the anode with the electron wind.