Annular Ring: Why the Pad Is Bigger Than You Drew It

The hole you get is not the hole you asked for

You specify a 0.3 mm drill. The fabricator drills 0.3 mm, then plates the barrel with 25 µm of copper. That copper goes on every wall, so it eats 25 µm from each side of the diameter.

Finished hole diameter: $0.3 - 2 \times 0.025 = 0.25$ mm.

This catches people constantly. If a component lead needs 0.3 mm of clearance, you must specify a finished hole of 0.3 mm and let the fab pick the drill, or specify a 0.35 mm drill yourself. Datasheets for connectors and headers quote finished hole size; CAD tools usually store drill size. They are not the same number.

What the annular ring is actually for

The annular ring is the copper between the hole edge and the pad edge. Its job is to survive misregistration.

Three things move the hole relative to the pad:

• Drill wander. The bit deflects entering the stack, more on thick boards and small drills.
• Layer-to-layer registration. Inner layers are imaged separately and laminated. They do not land perfectly.
• Material movement. The laminate shrinks and stretches during lamination, and not uniformly across a large panel.

IPC-6012 does not ask you to model these. It asks for a minimum ring that survives them, and it sets different numbers for external and internal layers because inner layers accumulate more registration error.

For Class 2, that is 0.05 mm external and 0.025 mm internal. Class 3 tightens it and adds a no-breakout requirement. Class 1 permits breakout entirely.

Working it through

Start from the finished hole and add rings on both sides:

For the 0.3 mm drill above, Class 2, with a 0.05 mm fabrication allowance:

• Finished hole: 0.25 mm
• Minimum external pad: 0.45 mm
• Minimum internal pad: 0.40 mm

So a 0.3 mm drill wants a 0.45 mm pad on the outer layers. Draw a 0.4 mm pad — which looks generous next to a 0.3 mm drill — and you are already under the external minimum.

The asymmetry between external and internal is worth internalising: internal pads need a smaller ring by spec but face larger registration error in practice. The spec accounts for this by also requiring less of them — internal layers are allowed to break out under Class 2, external layers are not.

Aspect ratio is the other constraint

The same padstack has a plating constraint that has nothing to do with the pad:

A 0.3 mm drill through a 1.6 mm board is 5.33:1. Standard fabrication handles 10:1, and most shops quote it without comment. Above that you are into pulse plating and a capability conversation.

Aspect ratio is why thick boards force larger drills. On a 3.2 mm backplane, that same 0.3 mm drill is 10.7:1 — past the standard limit — and you would move to 0.4 mm to get back under it. Which in turn grows the pad, which in turn eats routing channel.

Where the cost actually is

Class 3 is not uniformly more expensive. What drives cost is the combination:

• Small drills raise both aspect ratio and wander. Below 0.2 mm you leave mechanical drilling for laser, and the price step is real.
• Tight rings on large panels raise scrap, because registration error scales with panel size. The same design on a small panel yields better.
• No-breakout requirements on inner layers force tighter registration control across the whole panel, not just at that padstack.

A design that stays at 0.3 mm drills with 0.45 mm pads and Class 2 will be quoted by anyone. Push any one of those three and you narrow your supplier list.

Practical guidance

Specify finished hole size, not drill size, for anything a lead or pin passes through. Let the fab back-calculate the drill. Do not shrink pads to gain routing channel without checking the annular ring. The pad is not decoration — it is the misregistration budget. If you need the channel, use a smaller drill and accept the aspect ratio, or move to a different layer. Check inner and outer separately. A padstack that passes on layer 1 can fail on layer 4, and CAD tools often apply one pad size to all layers. Ask before assuming Class 3. Most commercial hardware is Class 2. Class 3 exists for equipment where field failure is unacceptable, and it costs both money and routing density.

The padstack and annular ring calculator works finished hole, minimum pad for both external and internal layers, land area, and aspect ratio against your IPC class in one pass.