100nF

PCB trace width calculator

A 3 A trace on a 1 oz outer layer needs 1.65 mm for a 10 °C rise. That is the IPC-2152 figure; the older IPC-2221 curve gives 1.37 mm. Current, allowed rise and copper weight set the width; a plane within 1 mm roughly halves the rise. Both assume steady DC in still air; short pulses and hot neighbours break that.

Width, current or temperature rise — whichever one you are missing — from the IPC-2221 curves or the IPC-2152 universal chart, plus what the trace drops and dissipates at that current.

W = 11.8 milsT = 1.38 milouter layer — open to air
Fig 1 — Conductor cross-section, 16.3 mils² of copper.
Width
11.8 mils · 0.300 mm
Current
1.00 A
Temperature rise
10.0 °C
Cross-section
16.3 mils² · 0.0105 mm²
Copper temperature
35 °C
Resistance, 100 mm
169 mΩ
Resistance per metre
1.69 Ω/m
Voltage drop
169 mV
Power in the trace
169 mW

How this is calculated

Standard: IPC-2221A Section 6.2 (conductor sizing charts); IPC-2152 universal chart

I=k⋅ΔT0.44⋅A0.725I = k \cdot \Delta T^{0.44} \cdot A^{0.725}
IPC-2221, with k = 0.048 on external layers and 0.024 on internal layers.
I=0.0454⋅ΔT0.5453⋅A0.6533I = 0.0454 \cdot \Delta T^{0.5453} \cdot A^{0.6533}
IPC-2152, a curve fit of the universal chart to within ±10 %.
I×20.5453I \times 2^{0.5453}
The IPC-2152 correction for a plane within 1 mm, which halves the temperature rise.
width=Aoz⋅1.378\text{width} = \frac{A}{\text{oz} \cdot 1.378}
Copper weight in oz, cross-section in mils², width in mils.
R=ρ⋅LA⋅(1+0.00393⋅(T−20))R = \frac{\rho \cdot L}{A} \cdot \bigl(1 + 0.00393 \cdot (T - 20)\bigr)
Copper resistivity is 1.68e-8 Ω·m, and T is the ambient temperature plus ΔT.
V=I⋅R,P=I2⋅RV = I \cdot R, \quad P = I^2 \cdot R
Voltage dropped along the trace and the power it dissipates.

Assumptions

What sets a trace's width for a given current

How wide a copper trace has to be so that a given DC current heats it by no more than a chosen amount above ambient. Two data sets are on offer, and they do not agree.

IPC-2221 (Generic Standard on Printed Board Design, Section 6.2, conductor sizing charts) is the one every older calculator uses. The curves were drawn in the 1950s from a small set of measurements; the closed form everyone fits to them is the first expression under How this is calculated above, with current in amps, ΔT in °C and cross-section in mils². Solving it for area rather than current is a matter of raising both sides to 1/0.725, and the width follows from the copper weight: 1 oz/ft² is 1.378 mil, or 34.79 µm.

IPC-2152 (Standard for Determining Current Carrying Capacity in Printed Board Design, 2009) replaced those charts with a much larger measured data set. It is published as charts, not equations. The calculator uses the common curve fit of the universal chart — 1 oz external copper, still air, 70 mil board, no plane nearby — which tracks the chart to within about ±10 % — the second expression above, in the same units, inverted the same way through its own exponent.

IPC-2152's most useful finding is that a copper plane within about 1 mm of the trace roughly halves the temperature rise for the same current. The optional plane correction applies exactly that: the predicted rise is halved, which is the same as solving the fit with 2·ΔT. It is a rule of thumb lifted from the standard's correction charts, not a separate fit.

Any of the three variables can be the unknown — width, current or temperature rise — the equation is just rearranged. Given a trace length the tool also reports DC resistance, voltage drop and dissipated power at the operating temperature, which is the ambient plus the rise — the last two expressions above.

IPC-2221 trace width chart, 10 °C rise

Minimum trace width for a 10 °C temperature rise, computed by the calculator above from the IPC-2221 formula for the two copper weights a board house quotes by default. Internal layers get half the current of external ones at the same width, because the standard assumed they cool only through the laminate.

Current1 oz external1 oz internal2 oz external2 oz internal
0.5 A0.12 mm (5 mil)0.30 mm (12 mil)0.06 mm (2 mil)0.15 mm (6 mil)
1 A0.30 mm (12 mil)0.78 mm (31 mil)0.15 mm (6 mil)0.39 mm (15 mil)
2 A0.78 mm (31 mil)2.03 mm (80 mil)0.39 mm (15 mil)1.02 mm (40 mil)
3 A1.37 mm (54 mil)3.56 mm (140 mil)0.68 mm (27 mil)1.78 mm (70 mil)
5 A2.77 mm (109 mil)7.19 mm (283 mil)1.38 mm (54 mil)3.60 mm (142 mil)
10 A7.19 mm (283 mil)18.71 mm (737 mil)3.60 mm (142 mil)9.36 mm (368 mil)

IPC-2152 trace width chart, 10 °C rise

The same currents through the IPC-2152 fit. The standard measured internal traces within a few per cent of external ones on the same board, so the layer column is gone; what replaced it is the copper plane, which the standard found roughly halves the rise when it sits within about a millimetre of the trace.

Current1 oz, no plane1 oz, plane within 1 mm2 oz, no plane2 oz, plane within 1 mm
0.5 A0.11 mm (4 mil)0.06 mm (2 mil)0.05 mm (2 mil)0.03 mm (1 mil)
1 A0.31 mm (12 mil)0.17 mm (7 mil)0.15 mm (6 mil)0.09 mm (3 mil)
2 A0.89 mm (35 mil)0.50 mm (20 mil)0.44 mm (17 mil)0.25 mm (10 mil)
3 A1.65 mm (65 mil)0.92 mm (36 mil)0.82 mm (32 mil)0.46 mm (18 mil)
5 A3.60 mm (142 mil)2.02 mm (79 mil)1.80 mm (71 mil)1.01 mm (40 mil)
10 A10.40 mm (410 mil)5.83 mm (230 mil)5.20 mm (205 mil)2.92 mm (115 mil)

Worked example: 3 A at a 10 °C rise in 1 oz copper

3 A, 10 °C rise, 1 oz copper, external layer, 100 mm long, 20 °C ambient.

IPC-2221 external
  ΔT^0.44      = 10^0.44        = 2.754
  k · ΔT^0.44  = 0.048 · 2.754  = 0.1322
  I / that     = 3 / 0.1322     = 22.70
  A            = 22.70^(1/0.725) = 22.70^1.379 = 74.2 mil²
  W            = 74.2 / 1.378   = 53.8 mil = 1.37 mm

IPC-2221 internal (k = 0.024)
  A = (3 / (0.024 · 2.754))^1.379 = 45.40^1.379 = 192.9 mil²
  W = 192.9 / 1.378 = 140.0 mil = 3.56 mm

IPC-2152 universal chart, no plane
  ΔT^0.5453        = 10^0.5453        = 3.510
  0.0454 · 3.510   = 0.1594
  I / that         = 3 / 0.1594       = 18.83
  A                = 18.83^(1/0.6533) = 18.83^1.531 = 89.4 mil²
  W                = 89.4 / 1.378     = 64.9 mil = 1.65 mm

Resistance of the 1.65 mm trace at 30 °C
  W · T = 1.6476e-3 m · 34.79e-6 m = 5.732e-8 m²
  R20   = 1.68e-8 · 0.1 / 5.732e-8  = 29.3 mΩ
  R30   = 29.3 · (1 + 0.00393 · 10) = 30.5 mΩ
  V     = 3 · 0.0305 = 91 mV
  P     = 9 · 0.0305 = 0.27 W

So IPC-2221 external says 1.37 mm, IPC-2152 says 1.65 mm, and IPC-2221 internal says 3.56 mm. Read that carefully: at this current the 2152 universal chart is slightly more conservative than 2221 external, and 2221 internal is more than twice as wide as anything 2152 measured. Put a plane 0.2 mm under the trace and 2152 with the plane correction gives 3 A at 10 °C on 0.92 mm (solve with ΔT = 20 °C: A = 50.1 mil², W = 36.4 mil).

Where IPC-2221 and IPC-2152 stop being valid

Common trace width mistakes

Further reading