Via Current Capacity Calculator
Size a plated through-hole via for the current it has to carry. Enter the finished hole diameter, plating thickness and the temperature rise you allow, and get the current capacity per via plus the barrel copper cross-section, voltage drop and power loss. Based on the IPC-2221 conductor-sizing charts.
The board is cut through the via's axis. d is the finished hole diameter, t the copper plated on the barrel wall — the metal that carries the current — and L the via length, the finished board thickness.
Standard plated through-hole vias only. The barrel is treated as an IPC-2221 conductor of the same copper cross-section, using k = 0.048 — the usual assumption for a via, whose copper reaches both outer surfaces. Resistance uses copper resistivity at the barrel's operating temperature (ambient + rise) and ignores the annular rings and the connecting copper.
Stop counting vias by hand
A calculator tells you one via is not enough — you still have to place the stitching, keep it on the right net, and redo it when the current or the stackup changes. In Flux, you just ask. Tell Flux how much current a net carries and it sizes the vias, stitches enough of them and keeps the copper consistent — live on the canvas.
- Ask for current per net, not a via count per pad
- Flux picks the via size and stitches as many as the current needs
- Vias stay IPC-compliant when the board thickness or plating changes
In Flux
Your 0.3 mm vias with 25 µm plating carry about 1.9 A each at a 10 °C rise, so for 2 A:
- Vias needed: 2 (capacity 3.8 A)
- Drop through the board: 1.3 mV
I've placed 2 stitching vias at the layer change on VBAT_5V. If you go to 0.2 mm vias I'll bump that to 3.
How the via is sized
The IPC-2221 conductor-sizing formula applied to the via barrel, evaluated client-side as you type.
Barrel cross-section
Unrolled, the plating is a tube of copper: the annulus between the finished hole of diameter d and the drilled hole around it, t thick. Area in mils².
Current per via
The same IPC-2221 curve used for traces, solved for current instead of width, with k = 0.048. Divide your current by it and round up to get the number of vias.
Resistance & drop
The barrel is a conductor as long as the board is thick. Vias in parallel divide that resistance, and the voltage drop and power loss follow from your current.
Via current capacity calculator FAQ
A via carries current in the copper plated on its barrel wall, so its capacity follows the same IPC-2221 curve as a trace: I = k · ΔT0.44 · A0.725, where A is the barrel's copper cross-section in mils² and ΔT the temperature rise you allow. A typical 0.3 mm via with 25 µm (1 mil) of plating has about 39.6 mils² of copper and carries roughly 1.9 A at a 10 °C rise. Enter your own hole size and plating above for an exact number.
Unrolled, the plating is a thin tube: the area is the annulus between the finished hole and the drilled hole, A = π · t · (d + t), with d the finished hole diameter and t the plating thickness. It is the plating thickness — not the hole size — that dominates: doubling the plating roughly doubles the copper, while doubling the hole only adds proportionally to the circumference.
Divide your current by the capacity of one via and round up. With 0.3 mm vias, 25 µm plating and a 10 °C rise (≈1.9 A each), 1 A needs 1 via, 3 A needs 2 and 5 A needs 3. Stitching several vias also cuts the resistance and the voltage drop through the board, so power nets are usually given more vias than the current strictly requires.
IPC-6012 Class 2 asks for at least 20 µm (0.8 mil) of copper in the hole and Class 3 for 25 µm (1 mil); most fabs quote 25 µm as standard and hit a bit more. Use the class your board is built to rather than the best case, and remember the plating is drawn from the same bath as the outer-layer copper, so heavier outer copper usually means slightly heavier barrels.
No — it covers standard plated through-hole vias, whose barrel runs the full finished board thickness. Blind, buried and laser-drilled microvias are shorter and sit in a different thermal environment, and their capacity depends on the stackup and the fab's process, so they need their own numbers. Impedance (Z₀) is also out of scope here.
No. Assign a target current to a net in Flux and it sizes and stitches the vias for you on the canvas — no counting barrels, and no redoing the math when the board thickness, plating or current changes.