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Conductor-to-Pad Maximum Width Calculator

Find the widest conductor that can reach a pad through the channel between its neighbours, from pad diameter, pitch and your fabricator’s minimum clearance.

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Formula

Wmax=PDpad(N+1)CNW_{max} = \frac{P - D_{pad} - (N+1)\,C}{N}

Reference: IPC-2221B clearance practice; the clearance value itself is a fabricator rule and is an input here.

W_maxWidest conductor that fits (mm)
PCentre-to-centre pitch to the neighbouring pad (mm)
D_padPad diameter (mm)
CMinimum copper-to-copper clearance (mm)
NConductors routed through the channel

How It Works

The widest conductor that can reach a pad is almost never set by the pad. It is set by the gap either side of the entry path, and that gap belongs to the neighbouring pads and the fabricator's clearance rule.

Two adjacent pads on a given pitch leave a channel of the pitch less one full pad diameter, because each pad contributes only its radius to the gap between them. Routing conductors through that channel costs one clearance to the pad on each side, plus one clearance between each adjacent pair of conductors — so N conductors need N+1 clearances in total. Whatever remains divides equally between them.

The clearance is a shop rule, not a physical constant, which is why it is an input here rather than a constant buried in the arithmetic. Fabricators differ, and a house rule hidden inside a formula produces a number your own shop will reject with nothing on the page explaining why. Put your fabricator's minimum in and the answer is theirs.

When the clearances alone exceed the channel the geometry has no solution. That case is reported as not fitting, naming which constraint binds, rather than as a negative width — a negative dimension rendered as a number is how an unbuildable board reaches fabrication.

The same arithmetic answers the reverse question. If the pad diameter is the free variable rather than the trace width, the maximum pad diameter calculator solves it from the same geometry.

Worked Example

Take 0.6 mm pads on a 1.27 mm pitch — a common through-hole connector footprint — with a fabricator minimum clearance of 0.15 mm, routing one conductor between them.

The channel between the two pads is the pitch less one pad diameter:

1.27 − 0.6 = 0.67 mm

One conductor needs a clearance to the pad on each side, so two clearances:

(1 + 1) × 0.15 = 0.30 mm

Leaving:

0.67 − 0.30 = 0.37 mm

So a 0.37 mm trace is the widest that fits, and the copper occupies 55.2% of the channel.

That is generous — comfortably above any standard process. Now route two conductors through the same channel instead. The clearance demand rises to three clearances, 0.45 mm, leaving 0.22 mm to share between two traces: 0.11 mm each. Still buildable at a fine-line process, but no longer routine, and the second conductor cost far more than its own width.

Tighten the pitch to 0.8 mm with the same 0.6 mm pads and the channel drops to 0.2 mm. Two clearances of 0.15 mm need 0.3 mm, which is more than the channel holds, so nothing fits — and the constraint that binds is the clearance, not the trace.

Practical Tips

  • Check this before committing a connector footprint, not after routing. Pad diameter is easy to change early and hard to change late.
  • If the result is under about 0.1 mm, expect a fine-line surcharge; under 0.075 mm, expect the fabricator to decline or quote as advanced technology.
  • Where two conductors will not fit between pads, consider routing one on an inner layer rather than shrinking both to the process limit.
  • Shrink the pad before tightening the clearance. The clearance is what the shop can hold; the pad is what you chose.
  • Take the pitch from the nearest neighbour in any direction, including diagonals on a staggered grid — that is the one that binds.
  • Use the same fabricator clearance figure across this calculator, the BGA breakout calculator and the maximum pad diameter calculator so the three agree.

Common Mistakes

  • Subtracting the pad diameter twice. Each pad contributes only its radius to the gap between them, so the pitch loses one full diameter, not two.
  • Forgetting that each extra conductor costs an additional clearance as well as its own width. On a fine pitch that extra clearance is often larger than the trace itself.
  • Using a house clearance rule from a different fabricator. The number is a shop capability and it varies; use the one that will build the board.
  • Reading a negative result as a very thin trace. If the arithmetic goes negative the geometry does not exist, which is why this calculator refuses rather than reporting it.
  • Applying the result to a soldermask-defined pad without accounting for the mask opening, which may be smaller than the copper.

Frequently Asked Questions

Because the channel runs between two pads and each contributes only its radius to the gap. Pitch minus one full diameter is the edge-to-edge distance between them.
Yes. Fabricators differ, and a house rule buried in a formula produces an answer your own shop will reject with nothing on the page explaining why. Put your fabricator's number in and the result is theirs.
It tells you which constraint binds. Usually the fix is a smaller pad rather than a tighter clearance, since the clearance is what your shop can actually hold and the pad is what you chose.
Yes — the geometry is the same. For a BGA array specifically, use the breakout calculator, which adds the diagonal escape case and reports how many rows a layer can escape.
Because it needs clearance on both sides of itself. Going from one conductor to two adds one more clearance as well as one more trace width, which on a fine pitch is the larger of the two costs.

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