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PCB Trace Width Calculator (IPC-2221 / IPC-2152)

Calculate minimum PCB trace width for current capacity per IPC-2221 and IPC-2152. Get resistance, voltage drop, and power dissipation. Free, instant results.

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Formula

A=(Ik⋅ΔTb)1/cA = \left(\frac{I}{k \cdot \Delta T^b}\right)^{1/c}

Reference: IPC-2221B Section 6.2; IPC-2152

A— Cross-sectional area (mil²) (mil²)
I— Current (A)
ΔT— Temperature rise above ambient (°C)
k,b,c— IPC-2221 empirical coefficients

How It Works

The PCB Trace Width Calculator determines minimum conductor width to safely carry DC/AC current without excessive temperature rise — essential for power distribution networks, motor drivers, and LED circuits. Power electronics engineers use this to prevent trace burnout that occurs when current density exceeds 30-50 A/mm2 on standard FR4.

Per IPC-2152 (supersedes IPC-2221), trace current capacity follows I = k x deltaT^0.44 x A^0.725, where k is a constant (0.048 for external layers, 0.024 for internal), deltaT is temperature rise in Celsius, and A is cross-sectional area in mils squared. A 1oz copper (35um) external trace carrying 3A requires 1.0-1.5mm width for 10C rise; internal layers need 2mm+ due to 50% reduced cooling.

Temperature rise compounds with ambient: a 20C rise design at 25C ambient reaches 45C, but at 55C ambient (automotive) reaches 75C — approaching FR4's Tg of 130-170C. Per IPC-2152 Table 5-1, current capacity derates 15% at 50C ambient versus 25C baseline.

Copper weight directly impacts capacity: 2oz copper (70um) carries 40% more current than 1oz at the same temperature rise because cross-sectional area doubles while surface cooling area only increases by trace width. For high-current designs (>5A), 2oz copper on external layers is standard practice per IPC-2221B Section 6.2.

Worked Example

Problem: Size a trace for 5A continuous on 4-layer FR4 (1oz external copper, 55C ambient, 20C maximum temperature rise, external layer).

Solution per IPC-2152:

  1. Target: I = 5A, deltaT = 20C, k = 0.048 (external)
  2. Required area: A = (I / (k x deltaT^0.44))^(1/0.725) = (5 / (0.048 x 20^0.44))^1.38 = (5 / 0.195)^1.38 = 25.6^1.38 = 89.4 mils2
  3. Convert to mm: 89.4 mils2 = 57.7 mm2 ... wait, units: 89.4 mils2 with 1.4 mil (35um) thickness gives W = 89.4/1.4 = 64 mils = 1.63mm
  4. Add 25% margin for ambient: W = 1.63 x 1.25 = 2.0mm
  5. Verify resistance: R = 1.724e-8 x 0.1m / (0.002 x 35e-6) = 24.6 mohm; P = 5^2 x 0.0246 = 0.62W
Result: Use 2mm (80 mil) trace width. Voltage drop over 100mm = 123mV (2.5% of 5V supply — acceptable for most designs).

Practical Tips

  • ✓Use 2oz copper for power traces >3A — doubles current capacity for only 15% cost increase per IPC-2152 Table 6-1 recommendation.
  • ✓Add copper pours around power traces — adjacent copper increases heat spreading and improves effective current capacity by 10-20% per thermal simulation studies.
  • ✓For motor/LED drivers: size for peak current (often 2-3x continuous) with 30C rise limit, not average current — prevents thermal cycling fatigue per IPC-9701A.

Common Mistakes

  • ✗Using IPC-2221 charts instead of IPC-2152 — older charts underestimate current capacity by 20-40% due to conservative 1950s data. IPC-2152 (2009) uses modern thermal modeling.
  • ✗Ignoring via resistance in current path — a 0.3mm via adds 1-3 mohm; 10 vias in series can add 30 mohm, causing 150mV drop at 5A that exceeds typical regulator accuracy.
  • ✗Calculating at 25C ambient when product operates at 55-85C — per IPC-2152, derate current capacity 3% per 10C ambient increase above 25C baseline.

Frequently Asked Questions

IPC-2152 (2009) provides trace current capacity charts based on controlled thermal testing, replacing IPC-2221's outdated 1950s data. Key finding: external traces carry 40-60% more current than internal at same temperature rise due to convective cooling. The standard covers 0.5oz to 3oz copper, 5C to 100C temperature rise, and includes derating for elevated ambient.
Current capacity scales with area^0.725 per IPC-2152. Doubling copper (1oz to 2oz) increases capacity by 2^0.725 = 65%, not 100%, because thermal resistance also changes. Practical impact: 2oz external copper carries 3A in 0.8mm width versus 1.5mm for 1oz — critical for space-constrained designs like smartphone PCBs.
Temperature rise adds to ambient: 20C rise at 25C ambient = 45C trace temperature; at 85C automotive ambient = 105C — risking solder joint reliability (IPC-J-STD-001 limits at 125C) and approaching FR4 Tg. Additionally, copper resistivity increases 0.4%/C, so a hot trace has 8% higher resistance at 20C rise, creating thermal runaway risk at high currents.
Yes for FR4/CEM-3 standard materials with copper traces. For aluminum-core PCBs (LED applications), current capacity is 2-3x higher due to metal substrate heat spreading — use manufacturer's thermal data. For flex circuits, derate 20-30% due to reduced heat dissipation per IPC-2223. HDI boards with thin dielectrics may allow higher capacity due to closer ground plane cooling.
Per IPC-2152 for 10C temperature rise: 1oz external copper needs 1.0-1.2mm width; 2oz external needs 0.6-0.7mm; 1oz internal needs 1.8-2.2mm. For 20C rise (less conservative): reduce widths by 30%. Always add 20% margin for manufacturing tolerance and ambient temperature variation. Internal layers need 80-100% wider traces than external at same current.
Common causes per IPC-2152 troubleshooting: (1) No copper pour nearby — heat spreading loss increases temperature 15-25%. (2) Solder mask traps heat — adds 5-10C versus bare copper. (3) Vias in path — each adds 1-3 mohm resistance and local heating. (4) Actual copper thickness is 20% below nominal after etching. (5) Ambient higher than design assumption. Solution: add 30-50% margin for reliability.
R = rho x L / (W x T). For copper (rho = 1.724e-8 ohm-m), 1oz (35um), 1mm wide, 100mm long: R = 1.724e-8 x 0.1 / (0.001 x 35e-6) = 49 mohm. At 25C. At 85C trace temperature: R increases 24% to 61 mohm. For 5A current: voltage drop = 305mV, power dissipation = 1.5W over 100mm — likely needs wider trace or 2oz copper.

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