Introduction
When designing printed circuit boards (PCBs), the width and thickness of copper traces impact how much current they can safely carry without overheating. Traces must be appropriately sized based on expected current levels. Copper weight, trace width, and current capacity have a direct mathematical relationship. This article provides an in-depth examination of these parameters and their correlation in PCB design.
Copper Weight
Copper weight refers to the thickness of the copper foil used to form PCB traces, pads, and planes. The most common weights are:
- 1 oz – 1 ounce per square foot, equivalent to a thickness of 1.4 mils (34 μm)
- 2 oz – 2 ounce per square foot, equivalent to 2.8 mils (68 μm)
Heavier copper foil allows for higher current capacity. But it costs more and can complicate fine-pitch PCB fabrication.
Trace Width
Trace width is the manufactured width of a PCB track, typically measured in mils (1 mil = 0.001 inches). Wider traces can handle more current due to reduced resistance. Minimum widths are dictated by current levels.
Current Carrying Capacity
The current carrying capacity defines how much continuous DC or RMS AC current a trace can conduct without exceeding temperature limits, usually 10-30°C above ambient. Excess current causes overheating damage.
Factors Affecting Current Capacity
Current capacity depends on:
- Copper weight – Heavier copper has lower resistance
- Trace width – Wider traces have lower resistance
- Temperature rise – Allowable increase over ambient
- Environment – Operating temperature influences limits
- Heat sinking – Thermal dissipation enables higher current
Appropriately sizing traces for expected currents prevents overheating while minimizing unnecessary PCB space and cost.
Copper Weight and Resistance
The primary factor relating copper weight to current capacity is the change in electrical resistance:
- Heavier copper has lower resistance
- Lower resistance results in less heating from a given current
- Reduced heating allows higher current capacity
For example, the table below shows typical per-length resistances relative to common copper weights:
| Copper Weight | Resistance (ohms/mm) |
|---|---|
| 1/2 oz | 0.0048 |
| 1 oz | 0.0029 |
| 2 oz | 0.0016 |
The resistance drops as copper weight increases, enabling higher current capacity.
Calculating Resistance from Weight
The resistance through a length of conductor is calculated using:
Where:
- ρ is the resistivity of copper (1.678 x 10<sup>-8</sup> Ωm)
- L is the length (m)
- A is the cross-sectional area (m<sup>2</sup>)
For a rectangular PCB trace, the cross-sectional area is:
Where:
- W is trace width (m)
- T is copper thickness (m)
Combining the equations allows resistance calculation based on trace dimensions and copper weight.
Trace Resistance Example
For a 50 mm long, 0.5 mm wide trace in 1 oz (34 μm) foil:
Increasing to 2 oz (68 μm) thickness halves the resistance:
Heavier copper foil significantly reduces electrical resistance due to the larger cross-sectional area.
Lower Resistance Increases Current
The power dissipated as heat in a conductor is:
Where I is the current and R is the resistance.
For a given temperature rise, higher current is possible with lower resistance before reaching power dissipation limits. The reduced resistance of thicker copper enables higher current capacity.
Trace Width and Resistance
In addition to copper weight, trace width also impacts resistance:
- Wider traces have a larger cross-sectional area
- Larger area produces lower resistance
- Lower resistance allows higher current capacity
For example, a 100 mm long trace with 0.25 mm width has 4X the resistance of a 0.5 mm wide trace in the same 1 oz copper:
Wider traces reduce resistance and enable increased current carrying capacity.
Combining Weight and Width
The effects of copper weight and trace width are multiplicative. For example, the combination of:
- Doubling copper weight from 1 oz to 2 oz (halves resistance)
- Doubling trace width from 0.25 mm to 0.5 mm (halves resistance again)
Decreases resistance to 1/4 of the original, increasing current capacity by a factor of 4X.
Optimizing both copper weight and trace width provides the maximum current capacity for a given PCB area.
Trace Temperature Rise
While lower resistance allows more current, we must also consider the resulting temperature rise. Power dissipated as heat raises trace temperature:
Where:
- ΔT is temperature rise
- P is power dissipation
- Rθ is thermal resistance
Rθ depends on trace size, environment, and heat sinking. Allowable ΔT determines current capacity.
Calculating Current Capacity
An analysis combining electrical and thermal considerations calculates current capacity:
- Start with fixed constraints:
- Target temperature rise ΔT
- Ambient temperature Tambient
- Max allowable temperature Tmax
- Determine acceptable power dissipation:
- P = ΔT / Rθ
- Use Rθ for given construction
- Use Ohm’s law to find current at target power:
- I = (P / R)**0.5
- Resulting I is the current capacity for the constraints
More thorough calculations maximize accuracy but often use assumed standard conditions for simplicity.
IPC-2152 Current Capacity Tables
IPC-2152 provides current capacity tables based on:
- Copper weight
- Trace width
- Assumed temperature rise and conditions
The tables relate width and weight to maximum current for common PCB parameters. An excerpt is shown below:
| Width (mm) | 1 oz Current (A) | 2 oz Current (A) |
|---|---|---|
| 0.25 | 1.4 | 2.5 |
| 0.5 | 2.2 | 3.9 |
| 0.75 | 3.1 | 5.9 |
| 1.0 | 3.9 | 7.2 |
This simplifies current capacity estimates based on standard assumptions.
Current Density Rule of Thumb
For approximating current capacity, a general rule of thumb is:
Maximum current (A) = Current density (A/mm2) x Cross-sectional area (mm2)
Where the current density is:
- 0.8 to 1 A/mm2 for external traces without heat sinking
- 1.8 to 2 A/mm2 for external traces with heat sinking
- 3 to 4 A/mm2 for internal plane layers
The cross-sectional area is calculated from trace width and copper thickness.
Heat Sinking Effects
Heat sinking to nearby plane layers enables narrower trace widths and higher current density, increasing capacity for a given area.
For example, with 2 oz copper:
- External trace, 0.5 mm wide -> 3.9 A capacity
- Internal trace, 0.25 mm wide -> 5 A capacity
The thinner internal trace matches the capacity of the thicker external trace by utilizing heat sinking.
Estimating Required Width
To estimate the trace width needed for a target current:
- Select an appropriate current density based on heat sinking
- Calculate the required cross-sectional area:
- Area = Target current / Current density
- Use area and copper weight to get minimum width:
- Width = Area / Copper thickness
Then verify capacity using IPC-2152 tables or more detailed analysis.
Trace Width Design Factors
- Match trace widths to expected currents
- Ensure high-current traces meet minimum width needs
- Use larger widths than required when possible
- Maximize heat sinking from ground planes
- Confirm key traces with thermal modeling
- Document assumptions and design rules used
Careful trace sizing optimizes cost, reliability, and PCB performance.
Case Study: USB 3.0 Cable
As a case study, we can examine PCB trace sizes for a USB 3.0 cable.
Key parameters:
- 5V supply current: 0.9 A
- Data pairs carry 1.0 A per pair, 8 pairs total = 8 A
- Target ΔT = 20°C ambient, 60°C max temperature
- 1 oz external traces with ground plane heat sinking
Using IPC-2152:
- 5V trace: 0.25 mm width
- Data traces: 0.5 mm width
This case study illustrates appropriate trace sizing for standard USB currents.
Summary
- Heavier copper weight reduces electrical resistance
- Lower resistance allows increased current capacity
- Wider traces also decrease resistance due to larger area
- Trace width must be sized based on target current
- IPC-2152 tables relate width and weight to current capacity
- Heat sinking improves capacity for a given trace size
- Matching trace size to current prevents overheating damage
Correctly correlating copper weight, trace width, and current carrying capacity ensures safe and reliable PCB performance under expected current loads.
Frequently Asked Questions
How accurate must current capacity calculations be?
Rough estimations are often sufficient early in design to determine minimum widths. More detailed analysis may be warranted for high-power or long-life applications.
What copper weight should be used?
1 oz copper offers the best balance of cost, manufacturability, and performance for most applications. 2 oz provides higher capacity for high-power boards.
Is it always better to use thicker copper?
Not always – thicker copper increases material and fabrication costs. Use the minimum weight that satisfies capacity needs. Excessive thickness can also lead to thermal stresses.
How much margin should be added to current capacity?
A 10-20% margin above calculated capacity is recommended to account for analysis inaccuracies and environmental variations during operation.
Can vias decrease current capacity?
Yes, narrower vias can create bottlenecks increasing resistance and heating. Size vias at least as wide as connected traces to prevent reductions in capacity.
