Via Current Capacity Calculator

Current a plated via can carry, and how many you need in parallel.

Example: A 0.3 mm via with 25 mm plating carries about 1.795 A at a 10 °C rise — equivalent to a 0.67 mm trace.

Inputs

Drill diameter(mm)

Barrel plating(mm)

25 µm (0.025 mm) is the IPC Class 2 minimum and a common default.

Temperature rise(°C)

Board thickness(mm)

Target current (optional)(A)

Result

Current capacity

1.795 A

Barrel cross-section: 0.0236 mm²
Via resistance: 1.141 mΩ
Equivalent 1 oz trace width: 0.673 mm
Vias needed: 2
Resistance of that many: 570.4 µΩ
Drop at target current: 1.711 mV

Carrying 3 A needs 2 vias in parallel at this size. Stitching several is standard practice for power connections between layers — one via is rarely enough.
A via is a plated tube, not a solid cylinder. Its conducting area is π × 0.3 × 0.025 = 0.0236 mm² — equivalent to only a 0.67 mm trace in 1 oz copper. Larger drills help less than you would expect, because area scales with diameter rather than its square.
IPC-2221 was derived for traces in free air, and a via barrel is buried in laminate with copper planes at each end. In practice those planes conduct heat away, so this figure tends to be conservative — but it is an extrapolation either way, not a measurement.

Capacity against drill size (25 mm plating)

Drill	Area	Current	Equivalent trace
0.2 mm	0.0157 mm²	1.338 A	0.45 mm
0.3 mm	0.0236 mm²	1.795 A	0.67 mm
0.4 mm	0.0314 mm²	2.211 A	0.9 mm
0.5 mm	0.0393 mm²	2.6 A	1.12 mm
0.8 mm	0.0628 mm²	3.655 A	1.8 mm
1 mm	0.0785 mm²	4.297 A	2.24 mm

Formula

area = π × d × t    then IPC-2221 for current capacity

Calculations follow IPC-2221B.

Worked example

A 0.3 mm via with 25 mm plating carries about 1.795 A at a 10 °C rise — equivalent to a 0.67 mm trace.

area = π × d × t
π × 0.3 × 0.025
0.0236 mm²
The annulus of the plated barrel, not the area of the hole.
I = k × ΔT^0.44 × A^0.725
0.048 × 10^0.44 × 36.52^0.725
1.795 A
R = ρ × length / area
1.68e-8 × 1.6 mm / 0.0236 mm²
1.141 mΩ

Frequently asked questions

How much current can a via carry?

Less than the drill size suggests. A 0.3 mm via with 25 µm plating has only about 0.024 mm² of copper — equivalent to a 0.67 mm trace in 1 oz copper — which works out around 1 to 2 A for a 10 °C rise.

Why does a bigger drill help so little?

Because the conducting area is an annulus: π × diameter × plating thickness. It scales linearly with diameter, not as its square. Doubling the drill only doubles the copper, whereas doubling a trace's width and thickness would quadruple it.

How many vias do I need for a power connection?

Divide your current by one via's capacity and round up — then add margin. Stitching a small array is standard practice for power planes, and it helps thermally too, since each via conducts heat between layers as well as current.

Is IPC-2221 accurate for vias?

It is an extrapolation. The standard was derived for traces in free air, while a via barrel sits in laminate with copper planes at each end that conduct heat away. That makes the figure conservative in most real boards, but it is not a measurement.

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