100nF

Trace width

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 = 1.38outer 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

What it computes

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

I = k · ΔT^0.44 · A^0.725          I in A, ΔT in °C, A in mil²
k = 0.048  external layers
k = 0.024  internal layers

A = (I / (k · ΔT^0.44))^(1/0.725)
W_mil = A / (1.378 · oz)            1 oz/ft² = 1.378 mil = 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 %:

I = 0.0454 · ΔT^0.5453 · A^0.6533     same units
A = (I / (0.0454 · ΔT^0.5453))^(1/0.6533)

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:

R = ρ · L / (W · T) · (1 + α · (T_ambient + ΔT − 20 °C))
ρ_Cu = 1.68e-8 Ω·m     α = 0.00393 /°C
V = I · R      P = I² · R

Worked example

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 slightlymore 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 it stops being valid

Common mistakes

Further reading