Once a signal's rise time is short compared with how long it takes to travel down a trace, the trace stops being a simple wire and starts behaving like a transmission line with a characteristic impedance. If that impedance does not match the driver and load, energy reflects back and corrupts the signal. A microstrip — a trace on an outer layer above a ground plane — has an impedance set by just three things: how wide the trace is, how far it sits above the ground plane, and the dielectric constant of the board between them. This calculator solves the standard Hammerstad-Wheeler equations so you can hit a target like 50 Ω.
Reviewed: June 19, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: Hammerstad & Jensen / IPC-2141 transmission-line guidance.
The microstrip equations
The effective permittivity captures the fact that the field is shared between the board and the air above it, so it always sits between 1 and εr. Z₀ then scales inversely with the square root of εeff and falls as the trace widens relative to the dielectric height. To raise impedance, narrow the trace, thicken the dielectric, or move to a lower-εr substrate.
Worked example — a 50 Ω RF trace on FR4
Scenario: A 2.4 GHz antenna feed on 1.6 mm FR4 (εr 4.4). What trace width gives 50 Ω?
3 mm lands almost exactly on 50 Ω, so it is the go-to width for 1.6 mm FR4. Nudging to 3.1 mm trims it slightly under 50 Ω. On a thin 0.2 mm inner layer the same 50 Ω target needs only ~0.35 mm of width — which is why high-speed inner-layer routing uses much finer traces.
Frequently Asked Questions
A flat trace over a ground plane forms a transmission line; Z₀ is its voltage-to-current ratio, set by width, dielectric height and εr. RF and high-speed design usually targets 50 Ω.
On 1.6 mm FR4 (εr 4.4), ~3 mm width gives 50 Ω. On a 0.2 mm layer it's ~0.35 mm. Compute with your real stack-up for accuracy.
The field sits partly in the board and partly in air, so the wave sees a blended εeff between 1 and εr. It sets propagation speed and electrical length.
Yes, slightly — thicker copper lowers Z₀ a few ohms. These thin-conductor equations are good to a couple of ohms for 1 oz copper; use a field solver for tight specs.
Microstrip is on an outer layer with one ground plane (field partly in air); stripline is buried between two planes. Stripline shields better; microstrip is easier to route and probe.