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When looking at IPC-7251 Table 3-3 (see the table in Sergei Gorbikov's answer at How to determine annular ring width for thru-hole pads?), for a through-hole pad in Level B the annular ring is given as 0.35mm. Is that the actual annular ring width or the total width to add to the hole size? What I mean is suppose you have a 0.4mm diameter hole, would you have a 0.75mm diameter pad or a 1.1mm diameter one?

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Annular ring is the width of the ring. So for a 0.35mm minimum annular ring the hole is no more than 0.7mm smaller diameter than the pad diameter.

There seem to be some differences whether "minimum annular ring" should account for misalignment of the holes, and whether the drilled hole size or the finished hole size should be used.

PCB manufacturers generally accept a minimum annular ring which assumes perfect hole alignment and using finished hole size, in my experience.

IPC has a definition of minimum annular ring that includes all the factors, so it does not necessarily correspond to what you would set as a DRC limit in an EDA program. In other words they are referring to the minimum ring you would actually measure under a microscope with misalignment and other manufacturing tolerances and not the nominal difference between hole and pad radius.

According to IPC, for Level B there is a 0.25mm fabrication allowance for 1oz or less copper thickness and 8 or fewer layers.

So according to IPC, the minimum pad diameter for external layers is the maximum diameter of the finished hole plus double the minimum annular ring (including etch back allowance) plus the fabrication allowance. For internal layers, the same numbers are used but the maximum drill diameter is used.

The fabrication allowance increases for more than 8 layers, and for heavier copper. See IPC-2221A.

Refer to the actual standards if the slight differences are important.

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  • \$\begingroup\$ Thanks for the very thorough answer. So then, to confirm, if you have a part with for example 0.2mm diameter leads on a 4-layer 1oz copper board, per IPC Level B you should have a 0.2+0.2=0.4mm diameter hole, and a 0.4+2*0.35+0.25=1.35mm diameter pad? \$\endgroup\$
    – John Arg
    Commented Jan 11 at 20:40
  • \$\begingroup\$ Reading through IPC-2221 section 9 I see that the minimum annular ring is 0.05mm, and the fabrication tolerance for level B is 0.25mm. Then that means I would add 0.25 + 0.05*2 = 0.35mm to the hole diameter? \$\endgroup\$
    – John Arg
    Commented Jan 11 at 21:34
  • \$\begingroup\$ Yes, by my reading, 0.35mm + finished hole diameter for a supported hole (eg. PTH), level B, external pad, 8 or fewer layers, 1 oz or thinner copper. Of course a bit more is probably better from a manufacturing pov. Slightly larger for internal pads because drill size is larger than finished hole size. \$\endgroup\$ Commented Jan 11 at 22:31

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