PCB

Padstack & Annular Ring Calculator

Calculate minimum pad diameter and annular ring per IPC-6012 for through-hole vias and component pads.

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Formula

D_{pad} = D_{drill} + 2 \cdot AR_{min} + FA

D_pad— Minimum pad diameter (mm)

D_drill— Drill diameter (mm)

AR_min— Minimum annular ring (IPC-6012) (mm)

FA— Fabrication allowance for drill wander (mm)

How It Works

The annular ring is the radial width of copper pad remaining around a drilled hole after plating. It is the critical parameter ensuring mechanical integrity of the via-to-pad connection and reliable solder joints. IPC-6012 Rev E defines minimum annular ring requirements by product class.

Class 2 (dedicated service electronics — consumer, industrial) allows tangential breakout on internal layers with a minimum 0.025 mm annular ring internally and 0.050 mm externally. Class 3 (high reliability — aerospace, medical, military) requires no breakout at all with 0.050 mm minimum on all layers.

The fabrication allowance accounts for drill positional accuracy (drill wander). Modern CNC drills achieve ±0.025–0.050 mm positional accuracy. Adding this allowance to the minimum pad diameter ensures the annular ring requirement is met even at worst-case drill registration. Aspect ratio (board thickness / drill diameter) is also constrained — Class 2 allows up to 10:1, Class 3 up to 12:1 — to ensure reliable copper plating through the barrel.

Worked Example

Given: Drill diameter = 0.3 mm, plating thickness = 25 µm, IPC Class 2, fabrication allowance = 0.05 mm, board thickness = 1.6 mm Step 1: Finished hole diameter

D_{finished} = D_{drill} - 2 \times t_{plating} = 0.3 - 2 \times 0.025 = 0.25

mm Step 2: IPC-6012 minimum annular rings

External layers: $AR_{ext} = 0.050$ mm (Class 2 and 3)
Internal layers: $AR_{int} = 0.025$ mm (Class 2 allows breakout)

Step 3: Minimum pad diameter (external)

D_{pad,ext} = D_{drill} + 2 \times AR_{ext} + FA = 0.3 + 2 \times 0.05 + 0.05 = 0.45

mm Step 4: Minimum pad diameter (internal)

D_{pad,int} = D_{drill} + 2 \times AR_{int} + FA = 0.3 + 2 \times 0.025 + 0.05 = 0.40

mm Step 5: Aspect ratio check

AR = T / D_{drill} = 1.6 / 0.3 = 5.33:1

✓ (below Class 2 maximum of 10:1) Result: External pad ≥ 0.45 mm, internal pad ≥ 0.40 mm.

Practical Tips

✓Add 0.05 mm fabrication allowance as a minimum; for HDI processes with laser drilling, this can be reduced to 0.025 mm
✓For via-in-pad BGAs, the annular ring requirement still applies — ensure the BGA land pad is large enough to contain the via plus annular ring
✓Check your fabricator's drill accuracy specification — some budget fabs have ±0.075 mm wander, requiring larger allowance
✓Use IPC Class 2 as default for commercial products; only specify Class 3 when contractually required (adds 15–30% to board cost)
✓For microvias (laser-drilled, < 0.15 mm), IPC-6012 Class 2 allows 0.0 mm annular ring (tangential) — but verify with your fabricator

Common Mistakes

✗Using finished hole diameter instead of drill diameter when calculating pad size — the pad must cover the drill hit, not the plated hole
✗Forgetting fabrication allowance — the annular ring minimum alone is not sufficient; drill wander can shift the hole off-centre
✗Applying Class 3 rules to consumer products — over-specifying wastes board area and reduces routing density
✗Ignoring aspect ratio limits — high aspect ratios prevent uniform plating in the via barrel, causing voids

Frequently Asked Questions

What is the minimum annular ring for standard PCB vias?

Per IPC-6012 Class 2: 0.050 mm on external layers, 0.025 mm on internal layers (with allowed tangential breakout). Class 3 requires 0.050 mm on all layers with no breakout.

How does plating thickness affect the annular ring?

Plating thickness determines the finished hole diameter (drill minus 2× plating) but does not directly affect the required annular ring. However, thicker plating reduces the finished hole size, which may require a larger drill if minimum hole diameter is specified.

What aspect ratio limits apply to vias?

IPC-6012 limits aspect ratio (board thickness / drill diameter) to 10:1 for Class 2 and 12:1 for Class 3. Higher ratios make it difficult to achieve uniform electroless copper plating in the barrel, risking voids and reliability failures.

Can I use smaller pads with controlled-depth drilling?

Blind and buried vias typically have tighter positional tolerance (laser drilling) and shorter barrel length, allowing smaller pads. However, the IPC annular ring requirement still applies based on the via class.

How do I calculate pad size for a press-fit connector?

Press-fit holes need interference fit — typically drill = pin diagonal + 0.05 mm. The pad must be large enough for the annular ring plus allowance for the press insertion force without pad delamination. Consult connector datasheet for recommended hole and pad sizes.

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