Microstrip impedance: what Z₀ means, the Hammerstad–Jensen formulas, substrates, and a worked example
A trace over a ground plane is a transmission line. Its characteristic impedance, the ratio of voltage to current in the wave travelling along it, depends on the trace's width, the dielectric's thickness and its dielectric constant: wider or closer to the plane means more capacitance and a lower Z₀. Fast edges and RF signals need the line's Z₀ to match the source and the load, 50 Ω as a rule, or energy reflects. This calculator gives Z₀ and the effective dielectric constant from the geometry using Hammerstad and Jensen's formulas with a copper-thickness correction, the wave's speed and delay, the capacitance and inductance per length, the width for a target impedance, the wavelength at a frequency and the delay of a run, with the cross-section and its field drawn to scale.
How to use the microstrip calculator
- Type the trace width and the dielectric height: the distance from the trace down to its reference plane, not the whole board.
- Pick the substrate in the menu row (it fills in εr; edit it to the laminate's data-sheet value) and the copper weight.
- Add a target Z₀ for the width that gives it, a frequency for the wavelength and a length for the delay. The Width table tab sweeps the width.
What the characteristic impedance means
Every length of the line has some inductance and some capacitance to the plane, and a wave moving along it sees them as a resistance Z₀ = √(L/C), typically 20–120 Ω for a PCB trace. If the far end is terminated in Z₀ the wave is absorbed; otherwise part reflects, and on a fast signal the reflection shows as ringing and overshoot. A line matters only when it is long compared with the signal: a run whose delay exceeds about a third of the rise time, or a sixth of a wavelength at RF. For short, slow runs Z₀ is irrelevant.
The formulas
εeff = (εr+1)/2 + (εr−1)/2 · (1 + 10h/w)−ab Z₀ = (η₀ / 2π√εeff) · ln(F·h/w + √(1 + (2h/w)²))
Hammerstad and Jensen published these closed forms in 1980 as a fit to exact field solutions; they are within 0.2 % for any practical width ratio with a vanishingly thin strip. a, b and F are fitted functions of w/h and εr. The copper's thickness is allowed for by treating the strip as slightly wider (Schneider's correction, the approach IPC-2141 takes). The formulas ignore dispersion, the slow rise of εeff with frequency, which is a percent or two on FR-4 at a few gigahertz, and they assume a plane much wider than the strip and no other copper nearby.
Substrates and stacks
FR-4 has εr around 4.3 at low frequencies, falling to about 4.0 above a gigahertz and varying by batch; a 50 Ω strip on a 1.6 mm two-layer board is 3 mm wide, awkward to route, which is why RF and high-speed boards use four layers with a thin prepreg under the outer layer: on 0.2 mm the same 50 Ω needs about 0.35 mm. Rogers laminates (εr 3.5–3.7) and PTFE (2.2) have a stable εr and far lower loss for RF. For a serious design get the stack-up from the fab: they know the real prepreg thickness after pressing and the resin's εr, and will tune the width to hit the impedance you specify.
When the impedance matters
Antenna feeds, RF inputs, USB, HDMI, Ethernet, LVDS, DDR memory, PCIe and any clock faster than a few tens of megahertz with a run longer than a few centimetres. For a 1 ns edge on FR-4 the critical length is about 50 mm; for a 100 ps edge it is 5 mm. Differential pairs use the same physics with a coupled-line formula, and inner-layer traces between two planes are striplines with a different formula; both are coming to this suite.
Your microstrip, step by step
- Geometry: w/h = 1.875, effective width 3.06 mm with the copper; εeff = 3.263.
- Impedance: Z₀ = 50.5 Ω (51.1 Ω for a thin strip); 6.03 ps/mm, C 0.12 pF/mm, L 304 pH/mm.
- For 50.0 Ω: 3.05 mm wide.
- At 2.4 GHz: λ = 69.2 mm, λ/4 = 17.3 mm; 25.0 mm is 130°, 151 ps.
Worked example: a 3 mm trace on 1.6 mm FR-4
w/h = 1.875, εr = 4.3, 1 oz copper. The copper widens the strip to 3.06 mm; Hammerstad–Jensen give εeff = 3.26 and Z₀ = 50.5 Ω, which is why 3 mm is the textbook 50 Ω width on a two-layer board. The wave travels at 0.55 c, 6.03 ps per millimetre, so at 2.4 GHz the wavelength on the strip is 69.2 mm and a quarter-wave section is 17.3 mm; a 25 mm feed is 130° long, well past the point where it must be matched. For exactly 50 Ω the width would be 3.05 mm. On a four-layer board with a 0.2 mm prepreg the same 50 Ω needs only 0.37 mm.
Questions
Why 50 Ω?
A historical compromise for coaxial cable between lowest loss (about 77 Ω) and highest power handling (about 30 Ω); the test equipment, connectors and chips all settled on it. Video uses 75 Ω, and differential standards 90–120 Ω between the pair.
Does the solder mask change Z₀?
A little: the mask over the strip raises εeff slightly and lowers Z₀ by one or two ohms. Fabs account for it when they tune controlled-impedance widths.
What if there is copper beside the trace?
A ground pour within about three dielectric heights of the strip makes it a coplanar waveguide with a lower Z₀. Keep pours back, or use a coplanar formula.
How accurate is this on a real board?
The formulas are within 1 %; the laminate's εr tolerance and the etch are worse, typically ±5–10 % on Z₀ unless the fab controls it. For critical lines, specify the impedance and let the fab adjust the width to their stack.