PCB annular ring and pad size: the IPC-2221 formula, breakout, drill size, aspect ratio, and a worked example
The annular ring is the copper left around a plated hole, and it is what connects the barrel to the trace and holds the solder joint of a through-hole part. On the drawing it is simply half the difference between the pad and the hole; on the board the hole is a little larger than nominal, a little off centre, and the pad a little smaller than drawn, so the ring that matters is the one left after all of that. This calculator takes the finished hole and its tolerance, the pad, the ring you need, the fab's allowance, the plating and the board, and gives the drill size, the nominal and worst-case rings on the outer and inner layers with a breakout check, the IPC-2221 minimum pad, the hole's aspect ratio and the solder-mask opening, with the pad drawn as designed and at its worst and the hole in cross-section.
How to use the annular ring calculator
- Type the finished hole size, its tolerance (±0.05 mm is usual for small holes, ±0.08 to ±0.1 for larger) and the pad diameter on the outer layers.
- Pick the rings you need on the outer and inner layers, the fabrication level (IPC-2221's A/B/C, or a modern fab's allowance) and the plating in the menu row. The nominal and worst-case rings and the minimum pad appear.
- Add the inner-layer pad if it differs and the board thickness for the drill's aspect ratio. The Drill table tab lists the minimum pads for every standard hole; "Use the minimum outer pad" writes it into the values.
What the ring must survive
Three things eat into the ring you drew. The hole has a size tolerance: the fab drills, plates and finishes within about ±0.05 mm, and the largest hole leaves the least copper. The drill is registered to the copper image with some error, so the hole sits off the pad's centre, and on the near side the ring loses the whole offset. Etching trims the pad's edge a little. IPC-2221 lumps the registration and the etch into one standard fabrication allowance on the diameter, 0.6 mm for Level A (general), 0.5 for Level B and 0.4 for Level C (high density); a modern fab with optical registration holds 0.3 mm, laser-registered drilling about 0.15. The ring at the worst case is the pad, less the largest hole, less the allowance, halved: the near side loses half the allowance, split between the hole's shift and the pad's etch. When it goes negative the hole breaks out of the pad: tolerated by IPC-6012 Class 2 on inner layers and on outer pads away from the trace, never by Class 3.
The IPC-2221 pad formula
Dmin = a + 2b + c
The same arithmetic solved for the pad: a is the maximum finished hole (nominal plus tolerance), b the minimum annular ring you need, c the fabrication allowance. The ring b is the fab's minimum for a sound connection: 0.05 mm on outer layers for Class 3, 0.025 mm on inner layers, and more where the joint is stressed, 0.15–0.25 mm for through-hole parts that are plugged and unplugged. The formula tells you why a 0.3 mm via needs a 0.6 mm pad at a tight fab and nearly 1 mm at Level A, and why the fab's quoted capability, not the IPC table, is the number to design to.
Finished hole and drill
Specify holes as finished sizes, after plating. The fab drills larger by the copper it plates on each wall, usually 20–30 µm, plus a little for variation, and picks a stocked drill: a 0.3 mm finished hole is drilled at about 0.4 mm, a 0.8 mm lead hole at 0.9. For a part lead the finished hole should be the lead's diameter plus 0.2–0.3 mm so that it inserts and the solder wicks up the barrel; for a via, as small as the fab makes cheaply, 0.2–0.3 mm. The solder-mask opening is the pad plus about 0.05 mm each side; vias of 0.3 mm and under are usually tented, mask over the pad with no opening at all.
Aspect ratio
The board's thickness over the drill. The plating solution has to flow through the hole and the current has to reach its middle, and both get harder as the hole gets deeper and narrower: the barrel is thinnest at mid-depth, and that is where it cracks when the board expands at reflow. Up to 8:1 is routine, 10:1 is most fabs' limit for a standard process, beyond it is a special. A 0.3 mm finished hole (0.4 mm drill) through 1.6 mm is 4:1; through 3.2 mm it is 8:1, and a 0.2 mm hole through the same board would be 10.7:1.
Your hole, step by step
- Drill: 0.3 mm + 2 × 25 µm + 0.05 mm = 0.4 mm, 4.0:1 through 1.6 mm.
- Ring as drawn: (0.6 mm − 0.3 mm) / 2 = 0.15 mm; worst case (0.6 mm − 0.35 mm − 0.5 mm) / 2 = −0.125 mm, breakout against 0.05 mm.
- Minimum pad: 0.35 mm + 2 × 0.05 mm + 0.5 mm = 0.95 mm outer (draw 0.95 mm), 0.9 mm inner; inner worst case −0.125 mm.
Worked example: a 0.3 mm via with a 0.6 mm pad
A 0.3 mm finished via, ±0.05 mm, with a 0.6 mm pad on a 1.6 mm board, 25 µm plating, at IPC-2221 Level B (0.5 mm allowance), needing a 0.05 mm outer ring and 0.025 mm inner. The drill is 0.3 + 0.05 + 0.05 = 0.4 mm, 4.0:1 through the board. As drawn the ring is (0.6 − 0.3) / 2 = 0.15 mm. At the worst case the hole is 0.35 mm, off centre by 0.125 mm with the pad etched by the same, so the ring is (0.6 − 0.35 − 0.5) / 2 = −0.125 mm: breakout, and IPC-2221 wants a pad of 0.35 + 0.1 + 0.5 = 0.95 mm. That is the old film-and-pin allowance. At a modern fab holding 0.3 mm the same pad leaves −0.025 mm at the worst case, still breakout, and the minimum pad is 0.75 mm; at a fab with laser-registered drilling holding 0.15 mm the worst-case ring is 0.05 mm, it meets the ring exactly, and a 0.6 mm pad is enough. The lesson of the example: ask the fab for its annular-ring capability and design to that, because the difference between 0.5 and 0.15 mm of allowance is the difference between a 0.95 mm pad and a 0.6 mm one.
Questions
What pad do I draw for a 0.3 mm via?
With a ±0.05 mm hole and a 0.05 mm ring: 0.6 mm at a fab with laser-registered drilling (0.15 mm allowance), 0.75 mm at an ordinary modern fab (0.3 mm), 0.95 mm by the IPC-2221 Level B table. Check the fab's capability sheet for its minimum annular ring and registration and design to those.
Does the ring matter on a via nobody solders?
Yes: it is the connection between the barrel and the trace, and a breakout on the side the trace enters from is an open circuit. On the other sides a breakout is cosmetic, which is what Class 2 tolerates.
Why are inner pads smaller?
They carry no joint and only need 0.025 mm of ring, and every tenth of a millimetre saved lets another trace pass between pins on the inner layers. Their registration is poorer, though, so do not shrink them below the formula.
What is a teardrop for?
A fillet where the trace meets the pad, so that a hole that breaks out toward the trace still has copper under it. It is the cheap fix for marginal rings on dense boards, and most layout tools add them automatically.