Electrical

PCB Trace Width Calculator

Find the PCB trace width for a current, the current a trace can carry, or its temperature rise, using the IPC-2221 formula.

Formula W = [ I / (k ΔT^0.44) ]^(1/0.725) / tReviewed Sep 20, 2026

A copper trace heats up when current flows through it, and a trace that is too narrow can overheat, lift off the board or burn open. This PCB trace width calculator uses the IPC-2221 current-capacity formula to find the width a trace needs for a given current and allowed temperature rise. You choose the copper weight and whether the trace is on an outer or inner layer. It can also work backwards to the maximum current for a width, or the temperature rise for a given current, and it estimates the trace resistance, voltage drop and power loss when you enter a length. It covers current-carrying width only. Controlled impedance is a separate calculation that depends on the dielectric and stack-up.

Calculation Bench

Solve for

Copper Weight

Layer

01Current

I · Steady (DC or RMS) current

02Temperature Rise (°C)

ΔT · Allowed above ambient; 10 is conservative

03Trace Length (optional)

L · For resistance, drop and power

04Ambient Temperature (°C, optional)

Ta · Defaults to 25 °C

Show Result In

Solution

Enter the current, temperature rise and copper weight to get the trace width.

W = [ I / (k ΔT^0.44) ]^(1/0.725) / t

Formula Sheet

W=\dfrac{1}{t}\left(\dfrac{I}{k\,\Delta T^{0.44}}\right)^{1/0.725}

I=k\,\Delta T^{0.44}(W t)^{0.725}

\Delta T=\left(\dfrac{I}{k\,(W t)^{0.725}}\right)^{1/0.44}

WTrace Width
tCopper Thickness
ICurrent
ΔTTemperature Rise
kLayer Constant
ACross-Section Area

Variables & Units

Symbol	Variable	Description	Common Units
W	Trace Width	Width of the copper trace.	mil, mm
t	Copper Thickness	Thickness of the copper layer; 1.378 mil per oz/ft².	mil, µm
I	Current	Continuous current through the trace.	A
ΔT	Temperature Rise	Trace temperature above ambient.	°C
k	Layer Constant	0.048 for external layers, 0.024 for internal layers.
A	Cross-Section Area	Width times thickness, in square mils.	mil²

How to Use This Calculator

01Choose what to solve: the trace width for a current, the maximum current for a width, or the temperature rise for a width and current.
02Pick the copper weight in ounces per square foot. Most boards use 1 oz on the outer layers; heavy-current boards use 2 oz or more.
03Pick the layer: external for traces on the top or bottom surface, internal for traces buried between layers.
04Enter the current and the temperature rise you can accept above ambient. 10 °C is a common conservative choice.
05Optionally enter the trace length and the ambient temperature to get resistance, voltage drop, power loss and the operating temperature.
06Read the width in mil and mm, then round up to a width your board maker can produce and leave some margin.

How the Formula Works

IPC-2221 gives the current a trace can carry as I = k × ΔT^0.44 × A^0.725, where I is in amperes, ΔT is the temperature rise above ambient in °C, and A is the copper cross-section in square mils. The constant k is 0.048 for external layers and 0.024 for internal layers.

The cross-section is the width times the copper thickness. One ounce of copper per square foot is 1.378 mil (about 35 µm) thick, so 2 oz is 2.756 mil. Rearranged, the required area is A = (I / (k × ΔT^0.44))^(1/0.725), and the width is the area divided by the thickness.

The area exponent is 0.725, not 1, so doubling the current needs about 2.6 times the area, not twice. That is why wide traces for high current grow quickly.

External traces can shed heat to the air, so they use the larger constant. IPC-2221 halves the constant for internal traces, so an internal trace needs about 2.6 times the area of an external one for the same current and rise.

Resistance uses R = ρ L / A with copper resistivity 1.724 × 10⁻⁸ Ω·m at 20 °C, raised by 0.393% per °C to the operating temperature (ambient plus temperature rise).

Worked Example 01

5 A on an outer layer, 1 oz copper, 10 °C rise

Known

Current: 5 A
Temperature rise: 10 °C
Copper: 1 oz/ft² (1.378 mil), external layer

Formula

W = [ I / (k ΔT^0.44) ]^(1/0.725) / t

Substitution

A = [5 / (0.048 × 10^0.44)]^(1/0.725) = 150.0 mil²; W = 150.0 / 1.378 = 108.9 mil

Result

108.9 mil (2.77 mm)

The 5 A trace needs about 109 mil, close to 2.8 mm. Over a 100 mm run its resistance is 0.0189 Ω at 35 °C, giving a 94 mV drop and 0.47 W of heat.

Worked Example 02

3 A on an inner layer, 2 oz copper, 20 °C rise

Known

Current: 3 A
Temperature rise: 20 °C
Copper: 2 oz/ft² (2.756 mil), internal layer

Formula

W = [ I / (k ΔT^0.44) ]^(1/0.725) / t

Substitution

A = [3 / (0.024 × 20^0.44)]^(1/0.725) = 126.7 mil²; W = 126.7 / 2.756 = 46.0 mil

Result

46.0 mil (1.17 mm)

Internal layers use half the constant, which costs 2.6 times the area. Using 2 oz copper and allowing 20 °C keeps the width to about 1.2 mm.

Worked Example 03

Temperature rise of a 10 mil trace carrying 2 A

Known

Width: 10 mil
Current: 2 A
Copper: 1 oz/ft² (1.378 mil), external layer

Formula

ΔT = [ I / (k (W t)^0.725) ]^(1/0.44)

Substitution

A = 10 × 1.378 = 13.78 mil²; ΔT = [2 / (0.048 × 13.78^0.725)]^(1/0.44) = 63.7 °C

Result

63.7 °C rise (88.7 °C at 25 °C ambient)

A 10 mil trace carries only 0.89 A at a 10 °C rise. At 2 A it runs about 64 °C hotter than the board, which is too hot for most designs. Widen the trace or use thicker copper.

Applications

01Sizing power and motor-driver traces on a new board
02Checking whether an existing trace can carry a higher current
03Estimating the temperature rise of a narrow trace before layout is final
04Estimating the voltage drop and power loss of a long power trace
05Choosing between 1 oz and 2 oz copper for a high-current board

Trace Width for a Given Current at 10 °C Rise, mil (mm)

Current (A)	1 oz external	1 oz internal	2 oz external	2 oz internal
0.5	4.5 (0.12)	11.8 (0.30)	2.3 (0.06)	5.9 (0.15)
1	11.8 (0.30)	30.8 (0.78)	5.9 (0.15)	15.4 (0.39)
2	30.8 (0.78)	80.0 (2.03)	15.4 (0.39)	40.0 (1.02)
3	53.8 (1.37)	140.0 (3.56)	26.9 (0.68)	70.0 (1.78)
5	108.9 (2.77)	283.2 (7.19)	54.4 (1.38)	141.6 (3.60)
10	283.2 (7.19)	736.8 (18.71)	141.6 (3.60)	368.4 (9.36)

Copper Weight and Thickness

Copper weight	Thickness (µm)	Thickness (mil)
0.5 oz/ft²	17.5	0.69
1 oz/ft²	35	1.38
2 oz/ft²	70	2.76
3 oz/ft²	105	4.13

Assumptions

01The current is steady (DC or RMS) and flows long enough for the trace to reach a steady temperature.
02The board is a standard epoxy-glass laminate in still air at room conditions, with no copper planes or pours nearby to spread heat.
03The trace has a uniform width and thickness along its length.
04One ounce of copper per square foot is 1.378 mil (35 µm). Plated outer layers can end up thicker than the base copper.
05Copper resistivity is 1.724 × 10⁻⁸ Ω·m at 20 °C with a temperature coefficient of 0.393% per °C.

Where This Model Stops

01The formula is a curve fit to the IPC-2221 charts, which are commonly quoted for currents up to about 35 A, temperature rises of about 10 to 100 °C and widths up to about 400 mil. The calculator warns when a result falls outside that range.
02IPC-2221 assumes a lone trace on a thick board. It ignores nearby planes, board thickness, and thermal vias, so it is usually conservative. IPC-2152 accounts for these and is the better reference for high-reliability designs.
03Does not cover controlled impedance, high-frequency skin effect, or signal integrity. Those depend on the dielectric and stack-up.
04Does not size vias, fusing current, short pulses, or connector and pad current limits. A trace is only as good as its weakest neck-down.
05Does not replace the design rules, minimum width and spacing limits, or current ratings from your board manufacturer and applicable safety standards.

References

[1]
IPC-2221B Generic Standard on Printed Board Design
IPC
Source of the conductor current-capacity charts that the k = 0.048 and 0.024 curve fit reproduces.
[2]
IPC-2152 Standard for Determining Current-Carrying Capacity in Printed Board Design
IPC
Newer test-based standard that accounts for nearby copper and board thickness; the reference for more accurate sizing.
[3]
IPC-2221 Calculator for PCB Trace Current and Heating
Altium
Explanation of the formula, its conservatism, and the derating advice for thin boards.
[4]
International Annealed Copper Standard (IEC 60028)
IEC
Basis for the 1.724 × 10⁻⁸ Ω·m copper resistivity at 20 °C.

Frequently Asked Questions

How do you calculate PCB trace width?

Use the IPC-2221 formula I = k × ΔT^0.44 × A^0.725. Rearrange it for the cross-section area A = (I / (k × ΔT^0.44))^(1/0.725) in square mils, then divide by the copper thickness to get the width. For 5 A, a 10 °C rise and 1 oz copper on an outer layer, that is about 109 mil.

What temperature rise should I use?

10 °C is a common, conservative choice. 20 to 30 °C is often accepted when space is tight and the board has margin. Add the rise to the highest ambient temperature the board will see, and check that the result stays below the rating of the laminate and nearby parts.

Why do internal traces need to be wider?

IPC-2221 assumes an internal trace cannot shed heat to the air, so it halves the constant, which needs about 2.6 times the area. Later data in IPC-2152 shows the real penalty is smaller, because heat spreads through the surrounding laminate and copper, so the internal figure here is on the conservative side.

Should I use IPC-2221 or IPC-2152?

This calculator uses IPC-2221 because it is a simple, widely used formula. IPC-2152 is based on more test data and accounts for planes and board thickness, and it usually allows a narrower trace at the same rise. For high-current or high-reliability boards, check the result against IPC-2152 or a thermal simulation.

Does the formula work for pulsed or peak current?

No. It assumes a steady current that heats the trace to a steady temperature. A short pulse heats the copper far less, and a very large pulse can fuse it. Size for the RMS current, and treat the peak separately.

How wide should a trace be on top of what the calculator says?

Add margin. Boards vary in copper thickness, and the formula ignores hot components and enclosed air. A common practice is to derate by 10%, and by 15% on boards thinner than about 30 mil. Also round up to a width your manufacturer can etch reliably.

What if the calculated width is very thin?

Very small currents give widths under 4 mil (0.1 mm), which many manufacturers cannot etch reliably. Use the manufacturer's minimum trace width instead. The trace will then carry far more than the signal needs, which is fine.

Does this calculate impedance?

No. This tool only finds the width for current-carrying capacity. Controlled impedance depends on the dielectric thickness and constant, the copper thickness, and the trace spacing. A width chosen for current is often far too wide for a 50 ohm line.

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