To calculate the width of a PCB trace, particularly when determining the required width for a specific cross-sectional area based on copper thickness, you can use the following formula provided in the reference.
Understanding PCB Trace Width
PCB trace width is a critical parameter in circuit design. It determines how much current a trace can safely carry, its resistance, and contributes to controlled impedance requirements. While trace width is most often determined by current carrying capacity using standard calculators or charts (like those based on IPC-2152), the provided formula shows how trace width, cross-sectional area, and copper thickness are mathematically related.
Calculation Using the Provided Formula
The reference provides a direct formula to calculate trace width if you know the desired cross-sectional area of the trace and the copper thickness of the PCB layer. This calculation is based on the fundamental geometric relationship: Area = Width × Height (or Thickness).
The Formula
The formula provided is:
Width[mils] = Area[mils²] / ( Thickness[oz] * 1.378[mils/oz] )This formula calculates the trace width in mils (thousandths of an inch).
Explaining the Components
Let's break down the terms used in the formula:
- Width[mils]: This is the result you are calculating – the required width of the PCB trace, expressed in mils.
- Area[mils²]: This is the desired cross-sectional area of the copper trace, expressed in square mils. The cross-sectional area is the area of the trace when viewed from its end.
- Thickness[oz]: This is the specified copper thickness of the PCB layer, commonly expressed in ounces per square foot (oz). Standard thicknesses include 0.5 oz, 1 oz, 2 oz, etc.
- 1.378[mils/oz]: This is a conversion factor. It represents the thickness in mils of a copper layer that is 1 ounce per square foot. Specifically, 1 oz/ft² of copper is approximately 1.378 mils thick. This factor converts the copper thickness from ounces to mils.
How the Formula Works
Essentially, the formula rearranges the basic area calculation (Area = Width × Thickness).
- The denominator
( Thickness[oz] * 1.378[mils/oz] )converts the standard copper thickness (in ounces) into an actual height (in mils). Let's call thisThickness[mils]. - The formula then becomes
Width[mils] = Area[mils²] / Thickness[mils]. - This is the standard formula for finding the width of a rectangle when you know its area and height (Area / Height = Width).
This formula is useful when you need a specific cross-sectional area for reasons like achieving a target trace resistance or impedance, and you know the copper thickness you are using.
Practical Example
Let's calculate the required width for a trace with a target cross-sectional area using the formula.
Example Scenario
Suppose you need a trace with a cross-sectional area of 15 square mils (mils²) on a board with 1 ounce (oz) copper thickness.
Applying the Formula
Using the formula:
Width[mils] = Area[mils²] / ( Thickness[oz] * 1.378[mils/oz] )
Substitute the values:
Width[mils] = 15 [mils²] / ( 1 [oz] * 1.378 [mils/oz] )
First, calculate the thickness in mils:
Thickness[mils] = 1 * 1.378 = 1.378 mils
Now, calculate the width:
Width[mils] = 15 [mils²] / 1.378 [mils]
Width[mils] ≈ 10.885 mils
So, to achieve a 15 mil² cross-sectional area with 1 oz copper, the trace needs to be approximately 10.885 mils wide.
Context and Other Considerations
While the provided formula accurately describes the geometric relationship between trace width, area, and thickness, it's important to remember that in practical PCB design, trace width is most frequently determined by the maximum current the trace is expected to carry without overheating. This calculation typically involves factors like current magnitude, temperature rise limits, and whether the trace is on an external or internal layer, using standards like IPC-2152.
The formula presented here is particularly relevant when the design constraint is the required cross-sectional area itself (e.g., for specific resistance values or as part of an impedance calculation), rather than directly the current capacity.
