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The 'amount' of conductive material in the via should be equal (or more) to the amount of conductive material in the trace.

For example, lets say the trace is 12mil which is 0.304mm. Now, find a hole (via) with a perimeter of 0.304.

Some math: P=2pir, 0.304 = 2 * pi * r --> r = 0.043mm --> d(iameter) = 2*r --> d = 0.096

A via of 0.1mm should be (theoretically) enough to deal with a 12mil trace. This assumes the via plating thickness is the same as the trace thickness, which is probably not valid, particularly for thicker boards. However, even doubling the 0.1 mm (4 mil) via diameter gives a pretty small hole.

The 'amount' of conductive material in the via should be equal (or more) to the amount of conductive material in the trace.

For example, lets say the trace is 12mil which is 0.304mm. Now, find a hole (via) with a perimeter of 0.304.

Some math: P=2pir, 0.304 = 2 * pi * r --> r = 0.043mm --> d(iameter) = 2*r --> d = 0.096

A via of 0.1mm should be (theoretically) enough to deal with a 12mil trace. This assumes the via plating thickness is the same as the trace thickness, which is probably not valid, particularly for thicker boards. However, even doubling the 0.1 mm (4 mil) via diameter gives a pretty small hole. You must verify with the board house that they can drill such a tiny diameter and insert or plate a via so small. All board houses have a minimum hole diameter and slot width.

The 'amount' of conductive material in the via should be equal (or more) to the amount of conductive material in the trace.

For example, lets say the trace is 12mil which is 0.304mm. Now, find a hole (via) with a perimeter of 0.304.

Some math: P=2pir, 0.304 = 2 * pi * r --> r = 0.043mm --> d(iameter) = 2*r --> d = 0.096

A via of 0.1mm should be (theoretically) enough to deal with a 12mil trace.

The 'amount' of conductive material in the via should be equal (or more) to the amount of conductive material in the trace.

For example, lets say the trace is 12mil which is 0.304mm. Now, find a hole (via) with a perimeter of 0.304.

Some math: P=2pir, 0.304 = 2 * pi * r --> r = 0.043mm --> d(iameter) = 2*r --> d = 0.096

A via of 0.1mm should be (theoretically) enough to deal with a 12mil trace. This assumes the via plating thickness is the same as the trace thickness, which is probably not valid, particularly for thicker boards. However, even doubling the 0.1 mm (4 mil) via diameter gives a pretty small hole.

The 'amount' of conductive material in the via should be equal (or more) to the amount of conductive material in the trace.

For example, lets say the trace is 12mil which is 0.304mm. Now, find a hole (via) with a perimeter of 0.304.

Some math: P=2pir, 0.304 = 2 * pi * r --> r = 0.043mm --> d(iameter) = 2*r --> d = 0.096

A via of 0.1mm should be (theoretically) enough to deal with a 12mm trace.

The 'amount' of conductive material in the via should be equal (or more) to the amount of conductive material in the trace.

For example, lets say the trace is 12mil which is 0.304mm. Now, find a hole (via) with a perimeter of 0.304.

Some math: P=2pir, 0.304 = 2 * pi * r --> r = 0.043mm --> d(iameter) = 2*r --> d = 0.096

A via of 0.1mm should be (theoretically) enough to deal with a 12mil trace.