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Microstrip impedance

Microstrip impedance: the example drawn by the calculatorw = 3 mmh = 1.6 mmεr = 4.4ground planeZ0 ≈ 50.2 Ω
50.2 Ω for 3 mm over 1.6 mm of εr 4.4

A fast signal sees a line, not a wire: width, height and material set its impedance.

Opens in the app on 1 December, with Circuit Notebook 1.0 See it worked out

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The example, worked out

Given

I want
Z0 of a width
Trace width
3 mm
Height to the ground plane
1.6 mm
Permittivity εr
4.4
Copper thickness
35 µm
Frequency (for the lengths)
1 GHz

Working

  1. the field lives partly in the board, partly in the air
  2. Hammerstad's form for w/h > 1, the copper thickness folded into the width (3.061 mm)
  3. slower than light, so shorter waves

Results

Characteristic impedance Z050.21 Ω
Effective permittivity3.33
Velocity factor0.548
Delay6.09 ps/mm
Wavelength at 1 GHz164 mm
Quarter wave41.1 mm
w / h1.88

The formulas

  1. εeff = (εr + 1) / 2 + (εr − 1) / 2 × (1 + 12 h / w)^−1/2the field lives partly in the board, partly in the air
  2. Z0 = 120π / (√εeff (w/h + 1.393 + 0.667 ln(w/h + 1.444)))for w/h above 1
  3. v = c / √εeffslower than light, so shorter waves

The usual traps

  • FR-4 is not one material: εr runs from 4.2 to 4.8.
  • A slot in the ground plane under the trace breaks the line.

Questions

How wide is a 50 Ω microstrip trace?

εeff = (εr + 1) / 2 + (εr − 1) / 2 × (1 + 12 h / w)^−1/2: the field lives partly in the board, partly in the air. Z0 = 120π / (√εeff (w/h + 1.393 + 0.667 ln(w/h + 1.444))): for w/h above 1. v = c / √εeff: slower than light, so shorter waves.

What does it give for typical values?

50.2 Ω for 3 mm over 1.6 mm of εr 4.4. Characteristic impedance Z0: 50.21 Ω.

What are the usual mistakes?

FR-4 is not one material: εr runs from 4.2 to 4.8. A slot in the ground plane under the trace breaks the line.

In the app

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Opens in the app on 1 December, with Circuit Notebook 1.0

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