Trace resistance, heating, inductance and parasitic capacitance

Trace resistance, trace heating, trace inductance and parasitic capacitance. Copper on a board has resistance and inductance like any other conductor. It matters on high-current traces, and on high-speed signals where inductance is a concern. Typical copper thickness: 35 µm on the outer layer of a rigid board, 12 µm on a flex. The parasitic capacitance figure is for a microstrip line — a trace on an outer layer.

Trace width

Length

Copper thickness

Current

SI prefixes accepted (4k7 / 1M / 10m / 220). Upper-case M = mega, lower-case m = milli

Trace resistance

33.142857mΩ

Trace inductance10.3745 nH

Trace dissipation33.142857 mW

Parasitic capacitance0.571 pF

Advanced — Dielectric thickness, PCB relative permittivity

Dielectric thickness

PCB relative permittivity

εr

Derived values

Voltage drop: 33.143 mV; Current × trace resistance

History

Press Enter in any field to keep a snapshot here.

Formula

Trace resistance: R = 1.74×10⁻² × L / (W × T) [mΩ]
Inductance: from an approximation
Dissipation: P = R × I² [mW]

Design notes

Copper traces carry parasitics of their own — resistance, inductance and capacitance. On high-current rails and fast signals you cannot ignore them.

What to check:
• High-current traces: is the drop, R × I, within budget?
• Fast signals: inductance is what turns into ringing
• Heating: P = R × I² — is the rise acceptable?

Standard copper thickness is 35 µm (1 oz) on an outer layer and 12–18 µm on a flex. For heavy currents consider 2 oz (70 µm) or more.

When you need this

For sizing power traces and answering "can this trace carry 3A?". Alongside resistance and dissipation it reports inductance and parasitic capacitance, so it serves both power and high-speed work.

What the four outputs mean

Output	Meaning	Where it matters
Trace resistance	R = ρL/(w·t), copper resistivity 1.74×10⁻⁸ Ω·m	IR drop and heating
Trace inductance	Self-inductance of the trace	Supply ripple, high-speed return paths
Trace dissipation	P = R·I², i.e. the heat generated	Temperature rise
Parasitic capacitance	Capacitance to the reference plane	Capacitive loading, sensitive nodes

Copper weight

"oz" is the weight of copper per square foot, converted to thickness.
Copper	Thickness	Sheet resistance	Notes
0.5 oz	17.5 µm	about 0.99 mΩ/sq	Common on inner signal layers
1 oz	35 µm	about 0.49 mΩ/sq	The usual default
2 oz	70 µm	about 0.25 mΩ/sq	Power boards; fine traces get harder
3 oz	105 µm	about 0.17 mΩ/sq	High current; watch cost and minimum width

Sheet resistance makes this quick to estimate: multiply length÷width by 0.49 for 1oz copper to get milliohms. A 1mm wide, 100mm long trace is 100 × 0.49 = 49mΩ.

Resistance rises with temperature

Copper has a temperature coefficient of about 0.393%/K. Referenced to 20°C, a trace at 60°C is already about 16% more resistive; at 100°C, about 31%.

On traces that dissipate power this is a positive feedback loop — more resistance, more heat. Check IR drop and dissipation at the highest temperature you expect. This calculator reports the 20°C value, so apply the factor by hand.

Worked example: a 3.3V / 2A rail

1mm wide, 100mm long, 35µm (1oz) copper, 2A.
Resistance 49.7mΩ, dissipation 198.9mW.
IR drop is 2A × 49.7mΩ = 99mV, or 3.0% of 3.3V — leaving only 2% of a ±5% budget.
At 60°C the resistance rises 16% to 57.5mΩ and the drop becomes 115mV (3.5%).
Doubling the width to 2mm halves it to 24.9mΩ and 50mV (1.5%). Take that.

Common mistakes

Forgetting via resistance. A φ0.3mm via in a 1.6mm board is roughly 0.5–1mΩ. It adds up every time a rail changes layer, so always use several vias in parallel for power.
Judging current capacity from resistance. What actually sets it is temperature rise; use the IPC-2152 curves. Inner layers dissipate heat poorly and need to be wider than outer layers for the same current.
Missing the narrowest point. Resistance is dominated by necks around pads, thermal reliefs and pinch points under components. Enter the narrowest section.
Forgetting the return path. IR drop happens on the round trip. A thin ground trace adds its drop directly to your supply error.

Frequently asked questions

How many amps can this trace carry?

This calculator gives resistance and heating, not a current rating. Current capacity follows from allowable temperature rise — see IPC-2152. As a rough guide, an outer-layer 1oz trace with a 10K rise carries about 1.5–2A per millimetre of width; on inner layers, roughly 60% of that.

Why do inner layers carry less current?

Heat has nowhere to go. An outer trace loses heat directly to the air; an inner one is surrounded by resin. For the same temperature rise, the inner trace must carry less.

What does "per square" mean in sheet resistance?

The resistance of one square of the conductor. A square of foil has the same resistance regardless of its size, so counting how many squares fit along the trace (length ÷ width) and multiplying by sheet resistance gives the total. Traces of varying width can be handled section by section.

When does trace capacitance matter?

On high-impedance nodes: feedback dividers, high-resistance analogue inputs, crystal circuits. A few picofarads change the behaviour there. On power rails the same capacitance is helpful, acting as extra decoupling.

Standards and references

IPC-2152 — Current capacity of conductors; temperature rise versus width, copper weight and layer. The current standard.
IPC-2221B — General design standard, including the older current capacity charts.

Last updated: 2026-08-29