PCB Trace Width Calculator
IPC-style baseline current sizing
Required Parameters
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Quick Answer
Use this to calculate how wide a copper trace must be to safely carry a specific current without overheating (based on IPC standards).
PCB Trace Width Calculator
Calculate the minimum trace width required for a given current on a printed circuit board, based on the IPC-2221 standard.
IPC-2221 Formula
The trace width depends on current, temperature rise above ambient, copper thickness, and layer position (internal vs. external).
Design Tips
- Use wider traces than minimum for reliability
- Internal layers need wider traces due to less cooling
- Use copper pours for high-current paths
- Consider via thermal relief for power planes
- Always verify with a thermal simulation for critical designs
Related Tools
- Trace Impedance Calculator
- Wire Size Calculator
- Ohm's Law Calculator
Design Notes
The IPC-2221 standard equations used here provide a safe baseline, but modern high-density boards often rely on the newer IPC-2152 standard. External layers dissipate heat into the air efficiently, meaning they can be thinner than internal layers for the same current. If routing power planes or high-current motor drivers, consider using thicker copper (e.g., 2oz) or removing solder mask to apply solder directly over the trace to lower resistance.
Common Mistakes
- 1
Using internal layer calculations for external traces (results in overly wide traces).
- 2
Ignoring the effect of via bottlenecks: A 50-mil trace won't handle 5A if it connects to the next layer through a single 0.3mm via. Use via arrays.
- 3
Assuming 1oz copper is exactly 1oz thick everywhere: Plating processes can vary copper thickness by up to 20%.
Engineering Handbox
1. Find Cross Section: A = (3 / (0.048 × 10^0.44))^(1/0.725) ≈ 39.5 mil² 2. Given 1oz = 1.378 mils thick 3. Width = 39.5 / 1.378 = 28.6 mils (0.726 mm)
Knowledge Base
What is the IPC-2221 trace width formula?
The IPC-2221 formula is I = k × ΔT^b × A^c, where I = current (A), ΔT = temperature rise (°C), A = cross-sectional area (mil²), and k/b/c are empirical constants. For external layers: k=0.048, b=0.44, c=0.725. For internal layers: k=0.024.
How wide should a PCB trace be for 1 amp?
For 1A on a 1oz copper external layer with 10°C temperature rise: approximately 10 mils (0.25 mm). For an internal layer, double that to about 20 mils (0.5 mm). Always round up to add a safety margin.
What is the difference between internal and external layer trace width?
Internal layers require approximately 2× wider traces than external layers for the same current. This is because external traces dissipate heat via air convection, while internal traces are insulated by FR4 dielectric material on both sides.
How does copper weight affect trace width?
Copper weight determines trace thickness: 1 oz = 1.378 mils (35 µm), 2 oz = 2.756 mils (70 µm). Doubling the copper weight halves the required trace width for the same current, because the cross-sectional area doubles at the same width.
What temperature rise should I use for PCB trace calculations?
Common values are 10°C for standard designs and 20-30°C for power boards. The temperature rise is above ambient — so a 10°C rise in a 50°C enclosure means the trace reaches 60°C. FR4 glass transition temperature (Tg) is typically 130-170°C.
How do I calculate voltage drop across a PCB trace?
V_drop = I × R_trace, where R_trace = ρ × Length / (Width × Thickness). For 1oz copper: resistivity ≈ 0.49 mΩ/square. A 100mm trace at 1oz copper, 0.5mm wide, carrying 3A drops about 100mV.
Can a via handle the same current as a trace?
No. A single standard via (0.3mm drill, 1oz plating) can handle roughly 0.5-1A. For high-current connections, use via arrays — typically 4-8 vias in parallel to match the trace's current capacity when transitioning between layers.
What is the difference between IPC-2221 and IPC-2152?
IPC-2221 (1998) is based on 1950s military test data and provides conservative estimates. IPC-2152 (2009) uses modern test data and accounts for board thickness, thermal conductivity, and copper plane proximity. Use IPC-2152 for precision high-current designs.
Does solder mask affect trace current capacity?
Yes. Solder mask acts as a thermal insulator, slightly reducing heat dissipation from external traces. Removing solder mask and tinning the trace with solder effectively increases trace thickness by 25-50 µm, improving current capacity. This is common practice for high-current paths.
How wide should a USB 5V power trace be?
For USB 2.0 (500 mA): minimum 10 mils (0.25 mm) on 1oz copper external. For USB 3.0 (900 mA): minimum 15 mils (0.38 mm). For USB-C Power Delivery (3A at 5V): minimum 30 mils (0.76 mm). Always verify with this calculator for your specific stackup.