5A Trace Width Calculator

Understanding 5A Trace Width Requirements

When designing a PCB that needs to carry 5 amperes of current, proper trace width calculation is critical for thermal management and reliability. Based on the IPC-2221 standard, a 0.5oz copper internal layer trace requires approximately 14.47mm (569.8 mil) width to safely handle this current with a 10°C temperature rise. This calculation ensures your printed circuit board maintains optimal performance without excessive heating, which could lead to solder joint failure, delamination, or fire hazards in extreme cases.

Key Parameters for 5A Design

• Current Capacity: 5A (amperes) - the maximum continuous current this trace is designed to carry
• Copper Weight: 0.5 oz/ft² (18μm) - determines the conductor cross-sectional area
• Layer Type: Internal ( sandwiched between FR4 dielectric material)
• Temperature Rise: 10°C above ambient (IPC-2221 conservative design guideline)
• Cross-Section Area: 390.29 sq mil (0.2518 mm²)

Why Layer Type Matters for 5A Current

Internal layers are sandwiched between FR4 dielectric material, which has relatively poor thermal conductivity (~0.3 W/m·K). For 5A on 0.5oz copper internal layer, the trace must be wider to compensate for reduced heat dissipation. The calculated width of 14.47mm ensures safe operation even with limited thermal pathways.

IPC-2221 Standard Formula

The IPC-2221B standard provides the industry-accepted formula for PCB trace width calculation: I = k × ΔT^b × A^c, where I is current in amperes, ΔT is temperature rise in °C, A is cross-sectional area in square mils, and k, b, c are empirical constants specific to internal layers. This formula has been validated through extensive testing and is used by PCB designers worldwide for 5A and higher current applications.