ESP32 Calcs

PCB Trace Width Calculator — IPC-2221, External and Internal

Who this is for: For anyone routing a board with power on it: the difference between a trace that's just a wire and a trace that's a fuse you accidentally designed.

IPC-2221 trace width for your current, temperature rise and copper weight — external or internal layer — plus resistance and voltage drop when you enter the trace length.

Quick answer: The calculator applies the IPC-2221 chart formula: cross-section area (mil²) = (current ÷ (k × ΔT^0.44))^(1 ÷ 0.725), with k = 0.048 for external layers and 0.024 for internal, then width = area ÷ copper thickness (1 oz = 1.378 mil). A 1 A trace with a 10 °C rise allowance on 1 oz external copper: ≈ 0.30 mm minimum (11.8 mil). Same current on an internal layer: ≈ 0.78 mm — internal traces run about 2.6× wider because FR-4 traps the heat.

10 °C is the conservative default; 20–30 °C for short bursts

Adds resistance, voltage drop and power loss

Copper weight
Minimum trace width
0.3 mm
In mils
11.8 mil
Use at least
0.35 mm
Voltage drop (add length)
—
CurrentWidth (mm)Width (mil)Use at least
0.5 A0.124.50.15 mm
1 A0.311.80.35 mm
2 A0.7830.80.8 mm
3 A1.3753.81.4 mm
Widths follow the IPC-2221 chart formula (external conductors k=0.048, internal k=0.024, 1 oz copper = 1.378 mil). Internal layers run roughly 2.6× wider for the same current because FR-4 traps the heat. Two habits worth keeping: add vias and copper pour to help the trace shed heat in real boards, and remember this is a thermal calculation — if your trace also feeds an analog sensor, size it for voltage drop (enter the length) rather than temperature. For a typical ESP32 project, 0.5 A on a 1 oz external layer lands near 0.15 mm — surprisingly thin — which is why powering a 5 V LED strip through the board still wants 1 mm-class pours and connectors, not a trace.
Core facts
StandardIPC-2221 chart formula
External k factor0.048 · internal 0.024
1 oz copper35 µm / 1.378 mil thick
Reference (verified in tests)1 A · 10 °C · 1 oz external ≈ 0.30 mm
Also outputsresistance · voltage drop · power loss (with length)

How the IPC-2221 trace width calculation works

Trace width is a thermal question, not an electrical one. A trace has some resistance, current through it makes heat, and the copper-to-air (or copper-to-FR-4) surface has to shed that heat without the trace exceeding the temperature rise you allow. The IPC-2221 chart — the industry standard since 1998 — encodes the empirically-measured relationship, and this calculator implements its formula directly with unit tests asserting the published reference values (1 A / 10 °C / 1 oz external ≈ 11.8 mil).

The two decisions that change the answer most: the layer (external traces cool through air and solder mask; internal traces are sandwiched in plastic and need roughly 2.6× the width for the same current) and the allowed temperature rise (10 °C is the conservative default; 20–30 °C is defensible for short inrush pulses but a bad idea for a continuous supply rail).

Width is only half the job — the calculator also returns the trace's resistance when you enter its length, and therefore the voltage drop and power loss. For a sensor's supply line, a 0.2 V drop across a thin trace corrupts measurements long before any heat problem appears; that's why the length field exists. For real boards, the standard practice stack is: size the trace by this formula, then add copper pours, thermal relief and stitching vias around power paths so the real board runs cooler than the calculation assumes.

Common uses

  • Sizing the 5 V rail trace that feeds a 2 A LED strip header
  • Checking a motor driver's output traces on 2 oz copper
  • Verifying a sensor supply line won't drop enough volts to matter
  • Deciding between 1 oz and 2 oz copper for a small power board
  • Estimating the voltage drop on a long, thin analog trace before respinning the board

When to ignore the formula

Three cases: (1) very short traces can carry much more than the chart says because the board itself is the heatsink — a 1 mm stub to a via is not a thermal problem; (2) traces that are actually fuses on purpose (should never exist for users), and (3) high-frequency signals, where width is set by impedance (50 Ω microstrip) not current. For everything else — power rails, motor lines, LED feeds — the IPC formula with a 10 °C rise is the right default and errs conservative.

Copper weight in one line

1 oz copper = 35 µm thick = 1.378 mil. Cheapest fab default is 1 oz; 2 oz roughly halves the width needed for the same current and costs a little more. For a 5 V 8 A LED strip project, 2 oz external copper at 10 °C gives ≈ 1.9 mm — still a wide bus, which is why high-current boards use pours and multiple layers rather than wider and wider single traces.

Frequently Asked Questions

How wide should a PCB trace be for 1 amp?
With the conservative IPC-2221 defaults (10 °C rise, 1 oz copper): 0.30 mm (11.8 mil) minimum on an external layer, or 0.78 mm on an internal layer. In practice most designers round up to 0.4 mm for 1 A on 1 oz external copper for margin, and 0.25 mm is common for 0.5 A.
Why do internal traces need to be much wider?
Internal traces are laminated inside FR-4, which is a thermal insulator — heat can only escape through the copper to the vias and edges. The IPC-2221 k factor (0.048 external vs 0.024 internal) captures it: for the same current and temperature rise, internal traces need about 2.6× the width. If a design has room, routing power on outer layers keeps the copper budget small.
What temperature rise should I use?
10 °C for anything continuously powered, 20–30 °C only for short pulses or when board space forces it (and check what that heat does to nearby parts). The rise limit isn't about the copper surviving — it's about how hot you're willing to let the board get. A 30 °C rise over a 40 °C ambient board makes 70 °C copper, warm enough to matter for reliability and for the plastic parts touching it.
Does trace length matter for the width calculation?
Not for heat (the formula assumes steady state over the trace's area) but absolutely for voltage drop: resistance = ρ × length ÷ area. A 100 mm, 0.3 mm wide, 1 oz trace is ≈ 0.19 Ω — at 1 A that's 190 mV lost and 190 mW heating the board elsewhere. Enter the length and check the drop before finalizing anything that also carries measurements.
Can I just use a copper pour instead of wide traces?
Yes — pours (or simply very wide polygons) beat the formula because they spread heat into more copper. The formula still governs: the current has to enter and leave the pour somewhere, and the narrow neck at a pad or via is where the heat concentrates. Design the neck with the formula, let the pour handle the rest.
How many vias do I need to carry power between layers?
Roughly: a standard 0.3 mm drill / 0.6 mm pad via handles about 1 A with moderate heating — but via current capacity is its own calculation (plating thickness dominates). The common safe practice is one via per amp for a given net, doubled for reliability, or simply pour both layers and stitch with a grid so no single via is critical.

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