Free Engineering Tool
PCB Trace Width Calculator
Size copper traces to the IPC-2221 ampacity standard. Enter your current, copper weight and allowable temperature rise to get the minimum trace width — plus cross-sectional area, resistance, voltage drop and power loss. Live results, external or internal layers.
Trace Width Calculator
Results update as you type. Defaults: 1 A, 1 oz copper, 10 °C rise, external layer.
Design Inputs
Length powers resistance, voltage drop and power loss. Leave blank to size width only.
Results
Minimum Trace Width
at 1 oz (1.378 mil) copper, external layer
Cross-Sectional Area
16.3 mil²·0.0105 mm²
Resistance
170.42 mΩ
Voltage Drop
170.4 mV
Power Loss
170.4 mW
Per IPC-2221 for uniform copper on a single layer. Figures are a starting point — add margin for vias, connectors, ambient temperature and manufacturing tolerance, and have the stackup reviewed for high-current or safety-critical designs.
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Get Instant QuoteThe Method
How trace width is calculated
This tool uses the trace-width model from IPC-2221, the IPC generic standard for printed board design. The standard relates the current a conductor can carry to its copper cross-sectional area and the temperature rise that current produces. The calculation runs in two steps.
1. Required cross-section
The copper area needed to carry current I at an allowable temperature rise ΔT:
A(mil²) = ( I / (k · ΔT^0.44) )^(1/0.725)- k = 0.048 — external (surface) layers
- k = 0.024 — internal (buried) layers
- ΔT — temperature rise above ambient, in °C
2. Convert area to width
Divide the required area by the copper thickness. One ounce of copper spread over a square foot is about 1.378 mil (35 µm) thick:
thickness(mil) = oz × 1.378
width(mil) = A(mil²) / thickness(mil)Resistance, voltage drop (V = I·R) and power loss (P = I²·R) then follow from copper resistivity, corrected to the trace's operating temperature.
Reference: IPC-2221 Generic Standard on Printed Board Design (IPC, Bannockburn, IL). The equation is an empirical fit for uniform copper in still air; for high-current or safety-critical work, cross-check against IPC-2152 and thermal testing.
The Variables
What affects the required width
Copper weight
The cross-section you need is fixed by current and temperature rise, so thicker copper means a narrower trace. Going from 1 oz to 2 oz roughly halves the width for the same current — the reason heavy-copper boards pack high currents into compact routing.
Temperature rise
ΔT is how much hotter the trace may run above ambient. Allowing a larger rise lets a thinner trace carry more current, but the peak temperature (ambient + rise) must stay safely below the laminate Tg and component ratings. 10 °C is a common conservative default.
Internal vs external
Surface traces shed heat to the air; buried traces are wrapped in dielectric and run hotter. IPC-2221 halves the constant for internal layers (k = 0.024 vs 0.048), so an inner-layer trace needs roughly double the width for the same current.
Quick Reference
Common current → trace width
External (surface) layer, 1 oz copper, 10 °C temperature rise, per IPC-2221. For internal layers, roughly double the width. Use the calculator above for other copper weights and temperature rises.
| Current | Cross-section (mil²) | Width (mil) | Width (mm) |
|---|---|---|---|
| 0.5 A | 6.3 | 4.5 | 0.12 |
| 1 A | 16.3 | 11.8 | 0.30 |
| 2 A | 42.4 | 30.8 | 0.78 |
| 3 A | 74.2 | 53.8 | 1.37 |
| 4 A | 110.3 | 80.0 | 2.03 |
| 5 A | 150.0 | 108.9 | 2.77 |
| 7 A | 238.6 | 173.2 | 4.40 |
| 10 A | 390.3 | 283.2 | 7.19 |
Values are minimum widths to hold the stated temperature rise; add margin for vias, connectors, ambient temperature and manufacturing tolerance.
FAQ
Trace width, answered
How is PCB trace width calculated?
From the IPC-2221 model: A(mil²) = (I / (k · ΔT^0.44))^(1/0.725), with k = 0.048 external and k = 0.024 internal. Divide that area by copper thickness (1 oz ≈ 1.378 mil) to get width. Example: 1 A, 10 °C rise, 1 oz external ≈ 11.8 mil (0.30 mm).
Why do internal traces need to be wider than external traces?
Buried layers are wrapped in dielectric and cool poorly, so IPC-2221 uses a lower constant (k = 0.024 vs 0.048). That means an internal trace needs roughly 2× the cross-section — and about 2× the width — to carry the same current at the same temperature rise.
How much does copper weight change the width?
Required cross-section is set by current and ΔT, so width scales inversely with thickness. Doubling copper from 1 oz to 2 oz roughly halves the width for the same current — which is why power and automotive boards specify heavy copper.
Is the IPC-2221 formula conservative enough for my design?
It is a solid starting point but assumes uniform copper in still air with no nearby heat sources. It ignores vias, connectors, adjacent high-current traces and elevated ambient. For high-current or safety-critical boards, add margin and validate with IPC-2152, thermal simulation or a temperature-rise test.
What temperature rise (ΔT) should I use?
ΔT is heating above ambient, not absolute temperature. 10 °C is a conservative default; 20 °C is common where extra heating is acceptable. Keep peak temperature (ambient + rise) safely below the laminate Tg and component ratings — for standard FR-4, well under 130 °C.
Related: controlled-impedance PCB, custom PCB manufacturing, Megtron low-loss laminates, and the instant PCB quote tool.
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