Calculate the minimum trace width required for a given current while maintaining a specified temperature rise.
0.5 oz
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1 (Single Layer)
2 (Internal Layer)
Required Trace Width
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mm
Understanding PCB Trace Width and Current Carrying Capacity
Printed Circuit Boards (PCBs) are the backbone of modern electronics. Traces on a PCB act as electrical conductors, similar to wires. The width of these traces is crucial, especially when they need to carry significant electrical current. Insufficient trace width can lead to overheating, signal degradation, and in severe cases, board failure.
This calculator helps determine the necessary trace width to safely conduct a given current, considering factors like desired temperature rise, copper weight, trace length, PCB thickness, ambient temperature, and whether the trace is on an external or internal layer.
The Science Behind the Calculation
The calculation is primarily based on empirical formulas derived from standards like IPC-2152, which relates trace width, current, and temperature rise. A simplified approach often involves calculating the trace resistance and then determining the width that keeps the power dissipation (and thus temperature rise) within acceptable limits.
The core idea is to manage the power loss, P = I²R, where 'I' is the current and 'R' is the resistance of the trace. This power loss manifests as heat. The trace's ability to dissipate this heat depends on its geometry (width, length, thickness), copper weight, and the PCB's thermal properties.
A common empirical relationship used for calculations is derived from studies by Brooks, which consider the cross-sectional area of the trace and its thermal resistance. The formula often looks something like this (simplified):
Width = (Current / (K * (TempRise ^ B))) ^ (1/C)
Where:
Current is the expected current flow in Amps.
TempRise is the maximum allowable temperature increase of the trace above ambient, in degrees Celsius.
K, B, and C are empirical constants that depend on copper weight, PCB material, and sometimes trace geometry (like length and thickness).
The IPC-2152 standard provides charts and more sophisticated formulas that account for various factors like:
Copper Weight: Thicker copper (higher oz) has lower resistance for the same width.
Layer Type: External layers dissipate heat more effectively than internal layers.
Ambient Temperature: A higher ambient temperature means less allowable temperature rise for the same heat generated.
Trace Length and Thickness: These influence the overall thermal resistance and resistance of the trace.
This calculator aims to provide a practical estimate using a common approximation derived from these principles, often related to the IEC-60364-5-52 standard or similar empirical models for PCB traces.
Use Cases:
Power Delivery: Designing traces that supply power to components without overheating.
High Current Applications: In motor control, power supplies, and battery charging circuits.
Signal Integrity: While primarily for current, ensuring traces are adequately sized can prevent unintended resistance changes affecting signals.
Thermal Management: Understanding the thermal impact of current on PCB traces.
Disclaimer: This calculator provides an estimate. For critical applications, always consult official IPC standards (like IPC-2152) and perform thorough thermal simulations or testing. The actual temperature rise can be influenced by many factors not included in this simplified model, such as airflow, proximity to other heat sources, and specific PCB material properties.
function calculateTraceWidth() {
var currentAmps = parseFloat(document.getElementById("currentAmps").value);
var tempRiseCelsius = parseFloat(document.getElementById("tempRiseCelsius").value);
var copperWeightOz = parseFloat(document.getElementById("copperWeightOz").value);
var traceLengthMeters = parseFloat(document.getElementById("traceLengthMeters").value);
var pcbThicknessMm = parseFloat(document.getElementById("pcbThicknessMm").value);
var ambientTempCelsius = parseFloat(document.getElementById("ambientTempCelsius").value);
var layerCount = parseInt(document.getElementById("layerCount").value);
var resultElement = document.getElementById("calculatedWidth");
var unitElement = document.getElementById("unit");
// Validate inputs
if (isNaN(currentAmps) || currentAmps <= 0 ||
isNaN(tempRiseCelsius) || tempRiseCelsius <= 0 ||
isNaN(copperWeightOz) || copperWeightOz <= 0 ||
isNaN(traceLengthMeters) || traceLengthMeters < 0 || // Length can be 0
isNaN(pcbThicknessMm) || pcbThicknessMm <= 0 ||
isNaN(ambientTempCelsius) || ambientTempCelsius < 0 || // Ambient temp can be 0
isNaN(layerCount) || layerCount higher factor
// Consider trace length – longer traces have higher resistance for the same width.
// This effect is usually more pronounced for signal integrity than DC current capacity, but we can include a factor.
var lengthFactor = 1.0;
if (traceLengthMeters > 0.1) { // Only apply if trace is somewhat long
lengthFactor = 1.0 + (traceLengthMeters / 1.0); // Simple linear increase, max factor of ~2 for 1m trace
}
// Consider layer type – internal layers are worse for heat dissipation
var layerFactor = 1.0;
if (layerCount === 2) { // Internal layer
layerFactor = 1.5; // Internal layers need ~1.5x wider trace (very rough estimate)
}
// Calculate Trace Width in mm
// This formula is an empirical approximation derived from fitting data.
// It's meant to provide a reasonable estimate for common scenarios.
var calculatedWidthMm = adjustedWidthFactor * Math.pow(currentAmps, exponent_I) * Math.pow(effectiveTempRise, exponent_T) * lengthFactor / layerFactor;
// Ensure minimum width if calculation yields very small numbers
if (calculatedWidthMm < 0.1) {
calculatedWidthMm = 0.1; // Set a practical minimum width
}
resultElement.innerText = calculatedWidthMm.toFixed(2);
unitElement.innerText = "mm";
}