You just routed a 2-amp power trace on your PCB. It looks fine in the layout. But will it actually deliver the voltage your IC needs? That depends on trace resistance — and if you've never calculated it, you're guessing.
Trace resistance causes voltage drop. Too much drop and your 3.3V rail shows up as 3.0V at the load. Brownouts, data corruption, mysterious resets. Not fun. The good news: calculating PCB trace resistance is straightforward once you know the formula and have the right numbers.
The Basic Formula
PCB trace resistance follows the same physics as any conductor:
R = ρ × L / A
Where:
- R = resistance (Ω)
- ρ = resistivity of copper = 1.72 × 10⁻⁸ Ω·m (at 20°C for pure copper)
- L = trace length (m)
- A = cross-sectional area (m²)
For PCB traces, it's more convenient to work in mils (thousandths of an inch) and ounces of copper. Here's the practical version:
R = (0.000552 × L) / (W × T)
Where:
- R = resistance in ohms
- L = trace length in mils
- W = trace width in mils
- T = trace thickness in mils
The constant 0.000552 Ω·mil comes from converting copper's bulk resistivity into these units.
Copper Weight and Thickness
Before you can calculate anything, you need to know how thick your copper is. PCB copper is specified in "weight" — ounces per square foot.
| Copper Weight | Thickness (mils) | Thickness (µm) | Typical Use |
|---|---|---|---|
| 0.5 oz | 0.7 | 17.5 | Signal layers, fine-pitch |
| 1 oz | 1.4 | 35 | Standard inner/outer layers |
| 2 oz | 2.8 | 70 | Power traces, high current |
| 3 oz | 4.2 | 105 | Bus bars, high-current power |
| 4 oz | 5.6 | 140 | Extreme current, bus bars |
Most standard PCBs use 1 oz copper on outer layers and 0.5 oz or 1 oz on inner layers. If you're doing power distribution, you can specify 2 oz for specific layers — it costs more but dramatically reduces resistance.
Important: these are nominal thicknesses. Actual thickness varies by ±10% from the PCB fab. Plating adds another 0.5–1.0 mils on outer layers.
Worked Example: Power Distribution Trace
Let's say you have a 3.3V regulator at one end of a PCB and a microcontroller drawing 500mA at the other end. The trace is 4 inches long and 20 mils wide on a 1 oz copper layer.
Step 1: Convert to consistent units
- Length: 4 inches = 4,000 mils
- Width: 20 mils
- Thickness: 1.4 mils (1 oz copper)
Step 2: Calculate resistance
R = (0.000552 × 4000) / (20 × 1.4)
R = 2.208 / 28
R = 0.0789 Ω
Step 3: Calculate voltage drop
Vdrop = I × R = 0.5 × 0.0789 = 0.039V (39mV)
Is 39mV acceptable? For a 3.3V rail, that's about 1.2%. Most ICs tolerate ±5% on their supply, so you're fine — for one load.
But what if you have five loads, each drawing 500mA, all connected along this trace?
Step 4: Distributed load analysis
The worst case is the load furthest from the regulator. For distributed loads, the effective resistance to the farthest point is the same 0.0789Ω, but the current isn't all at the end — it's spread along the trace.
For a rough estimate with uniformly distributed loads, the effective voltage drop is about half of what you'd calculate with all current at the end:
Vdrop ≈ (Itotal × R) / 2 = (2.5 × 0.0789) / 2 = 0.099V (99mV)
That's still fine — about 3% drop. But you can see how quickly this adds up.
The Temperature Factor
Copper resistance increases with temperature. The temperature coefficient of resistance (TCR) for copper is about +0.393% per °C.
R(T) = R(20°C) × [1 + α × (T - 20)]
Where α = 0.00393 per °C.
At 85°C (common industrial spec):
R(85°C) = R(20°C) × [1 + 0.00393 × 65]
R(85°C) = R(20°C) × 1.256
That's a 25.6% increase in resistance. Our 39mV drop at room temperature becomes 49mV at 85°C. Still fine, but the margin shrinks.
At 125°C (automotive under-hood):
R(125°C) = R(20°C) × [1 + 0.00393 × 105]
R(125°C) = R(20°C) × 1.413
41% increase. Now that 39mV becomes 55mV. Add in the current draw increasing because the IC's quiescent current goes up at temperature, and you might be closer to the edge than you think.
Quick Reference: Resistance per Inch for Common Traces
Here's a lookup table so you don't have to run the formula every time.
| Width (mils) | 0.5 oz (mΩ/in) | 1 oz (mΩ/in) | 2 oz (mΩ/in) |
|---|---|---|---|
| 10 | 78.9 | 39.4 | 19.7 |
| 20 | 39.4 | 19.7 | 9.9 |
| 50 | 15.8 | 7.9 | 3.9 |
| 100 | 7.9 | 3.9 | 2.0 |
| 200 | 3.9 | 2.0 | 1.0 |
To get total resistance, multiply by trace length in inches. For example, a 6-inch, 50-mil trace on 1 oz copper: 7.9 × 6 = 47.4 mΩ.
Practical Design Tips
Use copper pours for power distribution
Instead of routing narrow traces, use large copper pours (planes) for power. A 1 oz copper pour that's 500 mils wide has a resistance of about 0.79 mΩ/in. Over 4 inches, that's only 3.2 mΩ. At 2 amps, the drop is 6.4 mV — negligible.
Don't forget via resistance
Each via adds resistance too. A standard via (10 mil drill, 1 oz plating) has about 0.5–1.0 mΩ of resistance. That's small, but if you're dropping 3.3V through multiple vias to an internal layer, the drops add up.
For high-current paths, use multiple vias in parallel. Four vias in parallel reduces via resistance by 4x.
Consider the return path
Current flows in a loop. The resistance of your ground return path matters just as much as the power path. If your ground is a solid plane, the return resistance is negligible. If your ground is a routed trace, calculate its resistance too.
Check critical sensitive nodes
Some circuits are especially sensitive to voltage drop:
- ADC reference inputs — A 10mV drop on a 3.3V reference gives you a 0.3% error on every measurement
- Sensor excitation — Resistive sensors (strain gauges, RTDs) are ratiometric; supply drop directly affects accuracy
- Clock power pins — Voltage drop on oscillator supply pins can shift frequency
- LED drivers — A small drop in forward voltage means a noticeable change in brightness
IPC-2221 vs IPC-2152
IPC-2221 gives you the basic trace resistance formula we used above. IPC-2152 is the newer standard that accounts for more variables — trace geometry, board thickness, copper planes adjacent to the trace, and airflow.
For most voltage drop calculations, the simple formula from IPC-2221 is accurate enough. IPC-2152 matters more when you're calculating current-carrying capacity and temperature rise, which is a related but different concern.
A Real-World Problem: The 1.8V Core Rail
Here's a scenario I've seen cause field failures.
A design has a 1.8V core rail feeding an FPGA. The regulator is on one corner of the board, the FPGA is 5 inches away. The power trace is 25 mils wide on a 1 oz inner layer.
R = (0.000552 × 5000) / (25 × 1.4) = 0.0789 Ω
The FPGA draws 1.5A during heavy processing:
Vdrop = 1.5 × 0.0789 = 0.118V
Voltage at the FPGA: 1.8 - 0.118 = 1.682V
The FPGA's minimum core voltage spec is 1.71V. The design fails at room temperature. At 70°C, it fails harder.
The fix: Route the 1.8V as a copper pour (plane) instead of a 25-mil trace. A 300-mil wide pour on the same layer:
R = (0.000552 × 5000) / (300 × 1.4) = 0.0066 Ω
Vdrop = 1.5 × 0.0066 = 0.010V (10mV)
Now the FPGA sees 1.79V. Problem solved, and you probably improved EMC performance too by having a low-impedance power distribution.
Tools for the Job
- IPC-2221 — Generic standard for PCB design, includes trace resistance formulas
- IPC-2152 — More detailed thermal analysis standard
- Saturn PCB Toolkit — Free Windows tool for trace resistance, impedance, and current capacity
- KiCad Calculator — Built-in trace resistance calculator in the KiCad PCB tool
The Checklist
- Calculate trace resistance for every power distribution path
- Account for temperature (use worst-case operating temperature)
- Verify voltage drop is within IC tolerance at maximum current
- Check both power and ground return paths
- Use copper pours for anything over a few hundred milliamps
- Add parallel vias when transitioning power between layers
- Re-check after any layout changes that move components or extend traces
Skip the spreadsheet and use the PCB Trace Calculator — enter your trace width, length, and copper weight, and it gives you resistance, voltage drop, and power dissipation instantly. Supports both IPC-2221 and IPC-2152 methods.