Current flowing through a PCB trace heats it. The heat escapes into the board and the air, and the trace settles at a temperature above ambient that rises with current. Exceed the laminate’s limits and you risk delamination, damaged solder joints or a trace that acts as a fuse – so designers size power traces for an acceptable temperature rise, commonly 10–20 °C.
The classic sizing rule comes from IPC-2221 (formerly IPC-D-275), which fitted curves to tests from the 1950s: I = k × ΔT^0.44 × A^0.725, with the cross-section A in square mils. External traces get k = 0.048 because they shed heat to the air; internal traces get k = 0.024, half the current for the same size. This calculator solves that equation for the current, and also reports the trace’s resistance, voltage drop and power loss for a given length.
Its successor, IPC-2152 (2009), measured many more configurations and found that internal traces are not nearly as bad as the old charts suggest, while board thickness, copper planes and nearby traces matter a lot. The IPC-2221 values are therefore conservative for internal layers and remain a safe, widely accepted first estimate; use IPC-2152 charts or thermal simulation for critical high-current designs.
How to use the PCB Trace Current Calculator
- Trace width
- Copper thickness — 1 oz/ft² = 1.378 mil = 35 µm; 2 oz = 70 µm.
- Maximum temperature rise
- Trace location
- Ambient temperature
- Trace length
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Enter the trace width, copper thickness, allowed temperature rise and location, plus the ambient temperature and length. The calculator computes the cross-section, applies the IPC-2221 formula for current, then uses copper resistivity at the trace temperature for resistance, voltage drop and power.
Worked example
A 10 mil wide outer-layer trace in 1 oz copper (1.378 mil) has a cross-section of 13.78 mil². For a 10 °C rise: I = 0.048 × 10^0.44 × 13.78^0.725 = 0.048 × 2.754 × 6.70 = 0.886 A. An inner-layer trace carries half: 0.443 A. At 35 °C a 5 cm length has a resistance of 0.102 Ω, so at 0.886 A it drops 91 mV and dissipates 80 mW.
Current for 1 oz copper, 10 °C rise (IPC-2221)
| Width | External | Internal |
|---|---|---|
| 10 mil (0.25 mm) | 0.89 A | 0.44 A |
| 20 mil (0.51 mm) | 1.46 A | 0.73 A |
| 50 mil (1.27 mm) | 2.83 A | 1.41 A |
| 100 mil (2.54 mm) | 4.70 A | 2.35 A |
Inputs explained
| Field | What to enter |
|---|---|
| Trace width | Units: mil, mm, µm, in |
| Copper thickness | Units: oz/ft², mil, µm, mm. 1 oz/ft² = 1.378 mil = 35 µm; 2 oz = 70 µm. |
| Maximum temperature rise | Units: °C, °F, K |
| Trace location | Choose one: External (outer layer), Internal (inner layer) |
| Ambient temperature | Units: °C, °F, K |
| Trace length | Units: cm, mm, in, mil |
What each result means
- Cross-sectional area = [Trace width]×[Copper thickness]
- Maximum current = [k constant]×[Maximum temperature rise]^0.44×[Cross-sectional area]^0.725
- Trace temperature = [Ambient temperature] + [Maximum temperature rise]
- Resistance at that temperature = 1.72e-6×[Trace length]÷([Cross-sectional area]×6.4516e-6)×(1 + 0.00393×([Trace temperature] – 20))
- Voltage drop at maximum current = [Maximum current]×[Resistance at that temperature]
- Power dissipation = [Maximum current]^2×[Resistance at that temperature]
- Current density = [Maximum current]÷([Cross-sectional area]×0.00064516)
Formula
I = k × ΔT^0.44 × A^0.725 (A in mil², ΔT in °C, k = 0.048 external, 0.024 internal) · A = W × T · R = ρL ÷ A × [1 + α(T_trace − 20 °C)] · V = I × R · P = I² × R
Worked example
Inputs — Trace width: 10 mil; Copper thickness: 1 oz/ft²; Maximum temperature rise: 10 °C; Trace location: External (outer layer); Ambient temperature: 25 °C; Trace length: 5 cm.
Results — Cross-sectional area = 13.78 mil²; Maximum current = 0.8855 A; Trace temperature = 35 °C; Resistance at that temperature = 102.4371 mΩ; Voltage drop at maximum current = 90.7098 mV; Power dissipation = 80.3251 mW; Current density = 99.6 A/mm²; IPC-2221 chart range = Within the IPC-2221 chart range.
Key terms
- Temperature rise (ΔT)
- How much hotter the trace runs than its surroundings.
- Copper weight (oz/ft²)
- Copper thickness expressed as mass per area: 1 oz ≈ 35 µm.
- Cross-section
- Trace width × thickness, in mil² for the IPC formula.
- Current density
- Current per cross-sectional area, in A/mm².
Frequently asked questions
How much current can a 10 mil trace carry?
About 0.89 A on an outer layer with 1 oz copper and a 10 °C rise, or 0.44 A on an inner layer (IPC-2221).
Why are internal traces rated lower?
IPC-2221 assumes they cannot shed heat to air; IPC-2152 tests show they actually do nearly as well, so the old rule is conservative.
What temperature rise should I allow?
10 °C is a conservative default; 20–30 °C is common for power traces if the board and parts tolerate it.
Does doubling the width double the current?
No – current grows with area to the power 0.725, so twice the width gives about 1.65 times the current.
What about short traces?
Very short traces near large pads or planes run cooler than the formula predicts, because copper conducts heat away.
Should I use 2 oz copper?
Yes for high currents – it halves the width needed for the same cross-section.