PCB Trace Width vs Current: The IPC-2221 Method With a Worked Example
A copper track has resistance, so current flowing through it produces heat. If the track is too narrow, it warms up, its resistance rises, the voltage drop increases and in the worst case the copper lifts or fuses like a fuse wire. Choosing a track width is therefore not a matter of taste for power nets; it is a calculation.
The most widely used starting point is the chart from the old IPC-2221 design standard, which can be expressed as a simple formula. This article explains the formula, works through examples by hand, and shows where KiCad's built-in calculator fits.
The IPC-2221 Formula
IPC-2221 relates the current a track can carry to its cross-sectional area and the temperature rise you are willing to accept above ambient.
I = k x dT^0.44 x A^0.725 I = current in amperes dT = allowed temperature rise in degrees C A = cross-sectional area in square mils k = 0.048 for external (outer) layers k = 0.024 for internal layers Rearranged for area: A = ( I / (k x dT^0.44) ) ^ (1 / 0.725) Then width from area: Width (mil) = A / thickness (mil) Thickness for 1 oz copper = about 1.378 mil (35 um)
Internal layers get half the constant because they are buried in FR-4, which conducts heat poorly, while outer tracks can shed heat to the air. The formula is empirical, based on old test data, and is intentionally conservative for most practical boards.
Choosing the Temperature Rise
The temperature rise is your design decision, not a property of the board. A rise of 10 degrees C is a common, comfortable default for general electronics. If the board sits inside a warm enclosure, or next to a hot regulator, the track temperature is ambient plus self-heating plus whatever nearby parts contribute, so a smaller rise is wiser. A 20 or 30 degree C rise gives narrower tracks and can be acceptable for short runs in a well ventilated product, but remember that FR-4, solder joints and nearby components all age faster when hot.
Also decide which current to design for. Use the highest continuous current the track will see, not the typical value. For loads with short peaks, such as a motor starting or a capacitor bank charging at power-on, the brief peak usually does not heat the copper much, but the voltage drop during the peak can still reset a microcontroller sharing the same rail.
Worked Example: 2 A on an Outer Layer
Suppose a motor driver board needs a track carrying 2 A on the top layer, using 1 oz copper, and you accept a 10 degree C rise.
Step 1: dT^0.44 = 10^0.44 = 2.754 Step 2: k x dT^0.44 = 0.048 x 2.754 = 0.1322 Step 3: I / 0.1322 = 2 / 0.1322 = 15.13 Step 4: A = 15.13^(1/0.725) = 15.13^1.379 = about 42.4 sq mil Step 5: Width = 42.4 / 1.378 = about 30.8 mil 30.8 mil x 0.0254 = about 0.78 mm Round up: use a 0.8 mm (or wider) track
Repeat the calculation for other currents with the same assumptions and you get a handy reference table. The values below are calculated with the formula above for a 10 degree C rise and are rounded.
| Current | Outer layer, 1 oz | Inner layer, 1 oz | Outer layer, 2 oz |
|---|---|---|---|
| 1 A | about 0.30 mm (11.8 mil) | about 0.78 mm | about 0.15 mm |
| 2 A | about 0.78 mm (30.8 mil) | about 2.0 mm | about 0.39 mm |
| 3 A | about 1.37 mm (53.8 mil) | about 3.6 mm | about 0.68 mm |
Do Not Forget Voltage Drop
Temperature rise is only one limit. A track that stays cool can still drop too much voltage over a long run, which matters for low-voltage rails. Use the resistance formula with copper resistivity.
R = rho x L / (W x T) rho (copper, 20 C) = 1.72 x 10^-8 ohm.m Example: 0.8 mm wide, 35 um thick, 100 mm long R = 1.72e-8 x 0.1 / (0.8e-3 x 35e-6) = 1.72e-9 / 2.8e-8 = about 0.061 ohm At 2 A: V drop = 2 x 0.061 = about 0.12 V Power lost = I^2 x R = 4 x 0.061 = about 0.25 W
A 0.12 V drop is fine on a 12 V motor supply but significant on a 3.3 V logic rail. If the drop matters, widen the track, use a copper pour, or move the regulator closer to the load.
Using KiCad's Calculator and Net Classes
- Open Calculator Tools from the KiCad project manager and select the Track Width tab.
- Enter the current, temperature rise, conductor length and copper thickness. The tool reports the required width for external and internal layers, plus resistance and voltage drop, using the IPC-2221 formula.
- In the PCB Editor, open File > Board Setup > Design Rules > Net Classes and create a class such as
Powerwith the computed track width. - Assign power nets to the class, either in the schematic or with pattern matching in the Net Classes page, so the router uses the right width automatically.
Newer guidance in IPC-2152 is based on more recent measurements and accounts for board thickness and nearby planes. For most hobby and student boards, IPC-2221 results with a sensible safety margin are adequate.
Common Mistakes
- Sizing a track for the average current when the load has large peaks, such as a motor at start-up.
- Forgetting that a wide track narrows at a pad or a via neck; the narrowest point limits the current.
- Using a single small via to carry amps between layers. Use several vias in parallel for high current paths.
- Mixing mm and mil. The formula uses mils; convert at the end with 1 mil = 0.0254 mm.
Summary
IPC-2221 gives the cross-section needed for a current and temperature rise, and dividing by copper thickness gives width. For 1 oz outer copper and 10 degree C rise, roughly 0.3 mm handles 1 A and 0.8 mm handles 2 A. Check voltage drop separately, use net classes so power widths are applied automatically, and add margin for peak currents.