PCB Example Walkthrough: Calculating Trace Width and Voltage Drop

The PCB Example Problem Statement

Method Selection and the Hidden Trap

To solve this PCB example, we must apply two distinct physical principles: the IPC-2221B standard for thermal management and Ohm's Law for voltage drop. The IPC-2221B standard provides empirical formulas derived from MIL-STD-2118 to determine the cross-sectional area of a copper trace required to carry a specific current without exceeding a target temperature rise.

The Trap: Most junior engineers and hobbyists fail this specific problem in two ways. First, they blindly use the IPC constants for an internal layer. Internal layers rely entirely on conduction through the FR4 dielectric to dissipate heat, whereas external layers benefit from convective air cooling. Using internal constants for an external trace will result in an unnecessarily wide, space-wasting trace. Second, and more dangerously, they calculate the thermal width, assume the job is done, and ignore the voltage drop constraint. A trace might run perfectly cool while dropping 200mV, starving a downstream microcontroller of its required voltage headroom.

Step-by-Step Algebraic Solution

We must calculate the width required for the thermal constraint, then the width required for the voltage drop constraint. The larger of the two wins.

Part 1: Thermal Constraint (IPC-2221B)

The governing IPC-2221B formula for current is:

I = k * (ΔT^b) * (A^c)

Where:

• I = Current (5A)
• k, b, c = Constants for external layers: k = 0.048, b = 0.44, c = 0.725
• ΔT = Temperature rise (20°C)
• A = Cross-sectional area in mils²

Step 1: Isolate A.

A = [ I / (k * ΔT^b) ] ^ (1/c)

Step 2: Plug in the values.

ΔT^b = 20^0.44 ≈ 3.7334

k * ΔT^b = 0.048 * 3.7334 ≈ 0.1792

I / 0.1792 = 5 / 0.1792 ≈ 27.902

A = 27.902 ^ (1 / 0.725) = 27.902 ^ 1.3793 ≈ 97.86 mils²

Step 3: Convert Area to Width.

For 1 oz/ft² copper, the thickness (T) is 1.37 mils.

Width_thermal = A / T = 97.86 / 1.37 ≈ 71.43 mils

Part 2: Voltage Drop Constraint (Ohm's Law)

The governing formulas are V = I * R and R = (ρ * L) / A.

Where:

• V = Max voltage drop (0.050V)
• I = Current (5A)
• ρ = Resistivity of copper (1.724 × 10⁻⁸ Ω·m)
• L = Length (4 inches = 0.1016 meters)

Step 4: Find Maximum Allowable Resistance.

R_max = V / I = 0.050V / 5A = 0.010 Ω (10 mΩ)

Step 5: Solve for Required Area in Metric, then Convert.

A_metric = (ρ * L) / R_max

A_metric = (1.724e-8 * 0.1016) / 0.010 = 1.7515e-7 m²

Convert square meters to square mils (1 m² = 1.55 × 10⁹ mils²):

A_mils = 1.7515e-7 * 1.55e9 ≈ 271.48 mils²

Step 6: Convert Area to Width.

Width_voltage = A_mils / T = 271.48 / 1.37 ≈ 198.16 mils

Sanity Check and Independent Verification

Order of Magnitude Check: A 5A trace on an external layer typically requires between 50 and 100 mils for thermal management. Our thermal result of 71.43 mils sits perfectly in this expected range. The voltage drop width of ~198 mils is roughly three times wider, which makes physical sense: forcing a 50mV limit over a relatively long 4-inch run requires a massive cross-section to reduce resistance.

Unit Check: We converted inches to meters for the resistivity calculation, and square meters to square mils for the final width division. The units cancel cleanly to yield mils for the final width dimension.

Decision Tree: Thermal vs. Voltage Drop Constraints

When routing power rails, you must evaluate both constraints and let the physics dictate the final geometry. Use this decision matrix to finalize your CAD parameters.

By selecting 200 mils, you guarantee the voltage drop remains under 50mV (actual drop will be ~49.5mV), and because the trace is nearly three times wider than the thermal requirement, the actual temperature rise will be negligible (roughly 3°C above ambient).

Frequently Asked Questions

What if my PCB fab house cannot manufacture a 200-mil trace due to space constraints?

If board real estate is too tight for a 200-mil trace, you cannot simply shrink the width without violating the 50mV drop constraint. Your concrete alternatives are: (1) Increase the copper weight to 2 oz/ft² (which halves the required width to ~100 mils), (2) Shorten the physical routing path by moving the power supply closer to the load, or (3) Solder a bare copper wire or busbar directly over the trace to handle the bulk current.

Does the IPC-2221B standard apply to high-frequency RF signals?

No. IPC-2221B is strictly for DC and low-frequency AC thermal/current capacity calculations. For RF signals (e.g., 2.4 GHz WiFi or 5G antennas), trace width is dictated by controlled impedance requirements (typically 50Ω microstrip or stripline), which depend on the FR4 dielectric constant (Dk), trace height above the reference plane, and operating frequency. Refer to Altium's impedance routing guidelines for RF-specific width calculations.

Why do we use 1.37 mils for 1 oz copper instead of exactly 1 mil?

'1 oz copper' refers to the weight of the copper (1 avoirdupois ounce) spread evenly over a 1-square-foot area. When you calculate the volume and divide by the area, the resulting physical thickness is approximately 1.37 mils (34.8 µm). Using '1 mil' in your algebra will result in a 27% error in your cross-sectional area, leading to a trace that overheats or drops excessive voltage.