Trace Current Capacity (IPC-2221)
I = k × ΔT^0.44 × A^0.725
Quick answer
This calculator solves I = k × ΔT^0.44 × A^0.725 for I. With Constant k 0.5, Temperature Rise (°C) 50 °C, Cross-Section (mil²) 1000 mil², the result is 418.3 A.
Calculator
Formula
Description
The IPC-2221 standard provides an empirical formula for estimating the current-carrying capacity of PCB copper traces based on allowable temperature rise and cross-sectional area. The constant k differs for external (outer layer) and internal (inner layer) traces because external traces benefit from better convective cooling. Cross-sectional area is the product of trace width and copper thickness (typically 1 oz/ft² = 1.4 mil = 35μm). The formula is conservative and assumes standard FR-4 substrate with natural convection cooling.
Variables
- I — Maximum current (A)
- k — Layer constant: 0.048 for external, 0.024 for internal traces
- ΔT — Allowable temperature rise above ambient (°C)
- A — Trace cross-sectional area (mil²)
Practical Notes
For 1 oz copper (1.4 mil thick): a 10 mil wide trace has A = 14 mil², a 50 mil trace has A = 70 mil². At ΔT = 10°C on an external layer: 14 mil² carries about 0.5A, 70 mil² carries about 1.5A. For higher currents, use wider traces, heavier copper (2 oz = 2.8 mil), or polygon fills. The IPC-2152 standard provides updated, more accurate charts.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| k | Constant k | — | — |
| dT | Temperature Rise (°C) | °C | °C |
| A | Cross-Section (mil²) | mil² | mil² |
Step by step
- 1. What you know: Constant k 0.5, Temperature Rise (°C) 50 °C, Cross-Section (mil²) 1000 mil²
- 2. The relation: I = k × ΔT^0.44 × A^0.725
- 3. Substitute: k = 0.5, dT = 50 °C, A = 1000 mil²
- 4. Result: I = 418.3 A
Worked examples
Minimum range
Given: Constant k 0.025, Temperature Rise (°C) 2.5 °C, Cross-Section (mil²) 10 mil²
Result: I = 0.1986 A
Low range
Given: Constant k 0.25, Temperature Rise (°C) 25 °C, Cross-Section (mil²) 100 mil²
Result: I = 29.04 A
Typical
Given: Constant k 0.5, Temperature Rise (°C) 50 °C, Cross-Section (mil²) 1000 mil²
Result: I = 418.3 A
High range
Given: Constant k 0.8, Temperature Rise (°C) 85 °C, Cross-Section (mil²) 10000 mil²
Result: I = 4488 A
Maximum range
Given: Constant k 8, Temperature Rise (°C) 850 °C, Cross-Section (mil²) 100000 mil²
Result: I = 656200 A
How I changes with Constant k
| Constant k (input) | I (result) |
|---|---|
| 0.05 | 41.83 A |
| 0.125 | 104.6 A |
| 0.25 | 209.2 A |
| 0.375 | 313.7 A |
| 0.5 | 418.3 A |
| 0.75 | 627.5 A |
| 1 | 836.7 A |
| 2 | 1673 A |
| 3.5 | 2928 A |
| 5 | 4183 A |
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<p><a href="https://vbatpower.com/formulas/trace-current/">Trace Current Capacity (IPC-2221)</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate Trace Current Capacity (IPC-2221)?
Use I = k × ΔT^0.44 × A^0.725. With Constant k 0.5, Temperature Rise (°C) 50 °C, Cross-Section (mil²) 1000 mil², I works out to 418.3 A.
How do you calculate current capacity of 50mil 2oz trace?
Rearrange for I: with Constant k 0.25, Temperature Rise (°C) 25 °C, Cross-Section (mil²) 100 mil², I is 29.04 A.
What is the formula for Trace Current Capacity (IPC-2221)?
The formula is I = k × ΔT^0.44 × A^0.725. The IPC-2221 standard provides an empirical formula for estimating the current-carrying capacity of PCB copper traces based on allowable temperature rise and cross-sectional area.
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