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When you drill a hole into a copper PCB pad, the resulting copper surrounding the hole is what we know as annular rings. It essentially acts as the wall of the drilled via. Your annular ring size will depend on the size of the via and the size of the copper pad’s size.

But what is the purpose of the annular rings? As you may or may not know, PCBs are multi-layered. Your circuit board traces will need to connect to various layers of your PCB. They usually achieve this by reaching a pad which routes them through to another layer by a via.

The annular ring surrounding the via facilitates the connection between the trace and the via. What this means is the thicker the annular ring, the stronger the connection. But how thick should the annular rings be, and how do you calculate it?

How to calculate annular ring?

To find an annular ring’s width, you need to use this formula: (pad diameter – via diameter) ÷ 2.

If the diameter of the pad is 25 mils (0.50mm) and the hole/via is 15 mils (0.38mm) then your equation will look like this:

(25 – 15) ÷ 2 = 5 mils

(0.50 – 0.38) ÷ 2 = 0.06mm.

Let us look at further real-world examples. That will include the annular ring’s width for standard vias, micro via, and component holes.

Standard Via

Generally, a standard via will feature a hole with a diameter of 0.20mm (7.87 mils). On a 0.40mm (15.74 mils) pad, your final equations will look like this:

(0.40mm – 0.20mm) ÷ 2 = 0.10mm

(15.74 mils – 7.87 mils) ÷ 2 = 3.93 mils


Micro vias are typically smaller than your standard holes. They will have a hole diameter of 0.10mm (3.93 mils). On a pad with a diameter of 0.30mm(11.82mm), your equation will look like this:

(0.30mm – 0.10mm) ÷ 2 = 0.10mm

(11.82 mils – 3.93 mils) ÷ 2 = 3.93 mils

Component Hole

As we discussed earlier, component holes tend to be larger than average vias. On a pad with a 1.20mm (47.24 mils) diameter and a 0.80mm (31.50 mils) hole, your equation will look like this:

(1.20mm – 0.80mm) ÷ 2 = 0.20mm

(47.24 mils – 31.50 mils) ÷ 2 = 7,87 mils

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