As discussed earlier in the guide, the annular ring connects to the traces and the various layers of the PCB. Thus, it should be sizeable enough to support a reliant connection. The thicker the annular ring, the stronger the connection around the via will be. If the annular ring is not thick enough, it may cause connection issues or breaks.
You may have heard the term “teardrop annular ring” before. Teardrop pads feature additional copper near the junction of the annular ring that connects to the PCB traces. They assist in creating a more secure connection that may be more resistant to stress.
The idea here is simple. The more copper around the via, the thicker the annular ring is, and the stronger the signal it can carry to and from the traces.
How to position via in annular ring?
Not only do your annular rings have to be ample in size, but you need to position the hole (via) correctly. If you have positioned the hole too close to the pad’s boundaries, it could cause a tangency. Additionally, a tangency may also occur if the annular rings are too thin.
Nevertheless, when you drill your hole, you want to drill through the center of the copper pad. That will give you the most structurally efficient annular rings. Drilling too close to the pad’s edges will give you a compromised or bad via, which will effectively end in a tangency.
Drill too far from the middle of the pad could end up in a breakout. This is when the drill offsets the hole and breaks the pad’s boundaries’ continuity, resulting in a structurally insufficient annular ring.
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.
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
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