Newbie to PCB design so the question may be very basic for experts here.
I am designing a 4 Layer RF board - just a break out board for an RF Amplifier chip based on their recommended design. Board size is 5cmx5cm
I need the RF track to be 50Ω I am trying to use the PCB calculator but I need some help to determine what should be the thickness I should enter.
PCB vendor says Dielectric constant is 4.3 - Er
copper thickness on all layers is 35µm - T
Now comes H - the RF track is on the Top layer - rest of the layers are flooded ground planes.
Board stack up is as below
![](//kicad-info.s3.dualstack.us-west-2.amazonaws.com/original/2X/0/0fd528d4d6866d8b723af5955c257703ec15612c.jpg)
What should be the value of H, is it 0.2 i.e thickness between track layer and immediate adjacent layer below itself or is it the thickness of the core.
I have included a sample track.
The track runs between copper flood which is ground.
For this scenario, should I choose micro strip line calculator or co-planar wave guide with ground plane to calculate the track thickness.
I need 50Ω impedance what should be the angle value I should choose to get the W parameter - if I add 0 to angle then L value in the calculator becomes 0 - what is the correct way to get a 50Ω trace width using the calculator
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H would be top to the first ground plane (0.2 mm in your case).
W shouldn’t depend on angle, only L. That’s just for the trace delay.
Coplanar or microstrip should be reasonable in this case.
I’d expect the track to be around 0.36 mm wide.
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Many thanks for your reply, As you are more experienced could you point out how do I handle this scenario
the pad width is only .23mm and the pitch is 0.5mm
When I use a 0.36mm wide track I get DRC problems and the impedance is 61 if I reduce the width of the trace
The length of the track connecting the PAD to a connector is only 3.43mm
The frequency of the RF signal is 1575.42MHz
Any suggestions
Route a short track out of the pad at the pad width, then go up to the 50 ohm width.
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Yes GPS is 15cm wavelength, maybe 10cm on a PCB dielectric, so any track segment less than 10mm is negligible
Could you please explain how is the 10mm that you quote arrived at?
I read lengths less than λ/4 are shorter lengths and doesn’t impact the impedance from the source’s perspective. Is there a different rule of thumb
λ = 300/1575.42 = 0.19m
λ/4 =0.047m = 4.76cm = 48mm
1/10 has always been an engineering rule of thumb for negligible, so λ/10 is a sensible upper limit.
λ/4 is not nearly short enough - it is the standard length for an impedance transformer and might even maximise reflection problems, see https://en.wikipedia.org/wiki/Quarter-wave_impedance_transformer
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I came across this document
http://www.cypress.com/file/136236/download
page 45 where it states that its not a good idea to change RF track width on a continuous strip
Ideally you taper at width changes