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PCB Via Estimator

Estimate plated via resistance and power loss

mm

Diameter of the empty hole after plating, not the drill diameter.

mm

Copper thickness in the barrel; 0.025 mm = 25 µm.

mm

Length of the plated barrel carrying the current.

A

Try an example: 0.4 mm hole, 35 µm plating, 1 A · 0.3 mm hole, 25 µm plating, 2 A

More about this tool

How it works

Estimate a plated copper via barrel’s resistance, voltage drop, and power loss at 20 °C.

  • For finished hole diameter D and plating thickness t, the cylindrical barrel cross-section is the exact annulus A = π(D + t)t.
  • Converts millimeters to meters and estimates R = ρL/A with nominal copper resistivity 1.72 × 10⁻⁸ Ω·m at 20 °C. Computes DC voltage drop I × R and loss I² × R.

Ideal DC copper barrel at 20 °C. This is not a current-capacity, thermal, signal-integrity or fabrication check. It excludes pads, traces, contact resistance, plating variation and heating. Numerical bounds are safeguards, not manufacturing limits; confirm the finished plating and design with your fabricator.

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Related tools

Estimate a plated PCB via

Estimate the resistance, voltage drop and power loss of a copper via barrel.

What it is for

Use the PCB Via Estimator to estimate the DC resistance and electrical loss of a single plated copper via barrel. Enter the finished hole diameter, finished copper plating thickness, the conducting length and the current. The tool reports the copper cross-section, resistance, voltage drop and power loss at 20 °C.

This is a simplified cylinder with uniform copper plating. Its calculation uses the same annular cross-section as the KiCad PCB Calculator. It does not estimate safe current capacity, temperature rise, inductance, signal integrity or fabrication compliance. Confirm actual dimensions and tolerances with the board fabricator and validate the assembled design.

Worked example

Enter a finished hole diameter of 0.4 mm, plating thickness of 0.035 mm, length of 1.6 mm, and current of 1 A. The copper annulus has an area of about 0.04783 mm². At the assumed resistivity of 1.72 × 10⁻⁸ Ω·m, the resistance is approximately 0.000575 Ω. The voltage drop is about 0.000575 V, and the loss is about 0.000575 W at 1 A.

If the current is doubled while the dimensions remain the same, resistance stays constant, voltage drop doubles, and power loss quadruples. These figures describe the ideal barrel at 20 °C; they do not predict the barrel’s operating temperature.

How to read the result

The finished hole diameter is the empty hole after plating. It is not the drill diameter. The plating thickness is the copper deposited on the wall of that hole, not the thickness of the surface copper layer. For example, 25 µm is 0.025 mm. Ask your fabricator which finished barrel thickness is appropriate for your design and tolerances.

For finished diameter D and plating thickness t, the annular area is π(D + t)t. The tool converts dimensions to SI units and calculates resistance as resistivity times length divided by area. Voltage drop is current times resistance; power loss is current squared times resistance.

Tips and common mistakes

Use the actual conducting barrel length, which may differ from the nominal board thickness for blind or buried vias. Avoid entering a finished drill size as the empty hole diameter unless the fabrication information says they are the same.

The model excludes resistance in traces, pads and connections, variations in plating, heating, and frequency-dependent effects. Copper resistivity changes with temperature. Numerical input bounds prevent extreme calculations; they do not imply that a via can be manufactured at those dimensions or safely carry the entered current.

Parallel vias may not share current evenly. Their thermal and electrical behavior depends on the layout, copper planes, nearby components and fabrication. Examine the complete board and consult its fabricator before making a design decision.

FAQ

Does the result show how much current a via can safely carry?

No. It evaluates electrical loss at the current you enter and does not calculate temperature rise or current capacity.

Is a larger hole always better?

A larger hole increases the modeled copper cross-section for the same plating thickness, reducing this simple barrel resistance. However, pad size, spacing and manufacturing constraints also matter and are not checked.

Why may the measured resistance differ?

Actual copper thickness, geometry, temperature and contact resistance vary. The estimate includes only an ideal uniform barrel and excludes pads, traces and connections.

Guide last updated:

For scripts and AI agents

  • JSON API: POST https://ohmforgepcbviaestimator.usefulweb.app/api/v1/calculate (same calculation as this page)
  • Interface description: OpenAPI 3.1 · AI guide: llms.txt
  • MCP server (Streamable HTTP): https://ohmforgepcbviaestimator.usefulweb.app/mcp · tool name ohmforge_pcb_via_estimator
Example request
curl -s -X POST https://ohmforgepcbviaestimator.usefulweb.app/api/v1/calculate \
  -H 'content-type: application/json' \
  -d '{"finishedHoleMm":0.4,"platingMm":0.035,"lengthMm":1.6,"currentA":1}'
Feedback · ideas & bug reports welcome