Worked solutionOne step at a time
What the thinnest wire is made of
At a ~3 nm-class node the lowest metal is about 12 nm wide and 24 nm tall. A barrier liner of about 1.5 nm takes part of that, and copper this thin conducts about 2× worse than bulk copper, because electrons scatter off the surfaces and grain boundaries. Taking 3.6 µΩ·cm over the 9 × 22.5 nm copper core:
A distributed RC line: delay grows as L²
Every micron adds resistance and every micron adds capacitance that has to be charged through all the resistance before it. The Elmore estimate of the 50% delay of the wire alone is
Add the driver and the receiving gate (a repeater-sized inverter: 1.2 kΩ drive, 1.25 fF input, 1 fF own output) and you get 12.46 ns. Simulating the wire as a 100-section RC ladder and solving the step response exactly gives 12.45 ns. A FO4 logic gate here takes about 5 ps: the wire is about 2,500 gates long. Doubling the length to 2 mm multiplies the delay by 3.95.
One repeated stage
Cut the wire into 10 pieces of 100 µm. Each repeater drives one piece and the next repeater's input:
= 16.8 ps + 121.8 ps + 15.4 ps ≈ 154 ps (ladder simulation: 156.4 ps)
The wire part fell 100× (from 12.2 ns to 0.12 ns), but the repeater and load parts did not shrink much.
Answer: B, about 8×
Ten stages: 10 × 156.4 ps = 1.56 ns, against 12.45 ns for the unbroken wire. The speed-up is 7.96×. The L² term only drops by the number of pieces (10×), and each repeater adds about 30 ps on top, so the real gain is a little under 10×.
What the model showedDelay against the number of pieces
The reel's model (c3-wire-delay/model.py) simulates every stage as an RC ladder driven through the repeater's resistance and measures the 50% time. Total delay for the 1 mm wire:
- 1 piece (no repeaters): 12.45 ns
- 2 pieces: 6.40 ns
- 5 pieces: 2.77 ns
- 10 pieces: 1.56 ns
- 20 pieces: 0.98 ns
- 50 pieces: 0.67 ns
The gains shrink as the repeaters' own delay starts to dominate. These are order-of-magnitude numbers from stated assumptions, not from a specific foundry's design kit.
Wrong answersWhy each one is tempting
| Pick | What went wrong |
|---|---|
| A · about 2× | Assumed the repeaters eat most of the gain. At 10 pieces they add only about a fifth of the total. |
| C · about 30× | A guess between 10× and 100×. Ten pieces can never beat 10× on the wire term alone. |
| D · about 100× | Used the per-piece saving (1/k²) and forgot that the signal crosses all k pieces. |
Follow-ups an interviewer asks nextTry these before opening them
What is the best number of repeaters?
Balance the wire term (falls as 1/k) against the repeater cost (grows as k). The optimum is k ≈ √(0.38·r·c·L² / (0.69·R_D·(C_par + C_in))), which is in the tens for this wire and repeater. Past that the delay rises again; in practice designers also count area and power and use fewer.
With optimal repeaters, how does delay grow with length?
Linearly. Each segment has a fixed best length, so a longer wire just has more of the same segments. Repeaters turn L² into L.
Why not just route it on a thicker metal layer?
That is the usual first fix. Upper layers are many times wider and taller, so r falls by 10× to 100× while c stays near 0.2 fF/µm. Long routes go up; the thinnest layers are kept for short local connections.
Why does copper get worse as it gets thinner?
Electrons in copper travel about 40 nm between collisions. In a 12 nm wire they hit the surfaces and grain boundaries much more often, and the barrier liner, which barely conducts, takes a bigger share of the cross-section. This is why alternatives such as ruthenium and molybdenum are studied for the thinnest layers.
