# Impedance track

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

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

2 Likes

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.

1 Like

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.

2 Likes

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

1 Like

I came across this document