# Should each trace carrying RF be 50Ohm in characteristic impedance? How?

I have to translate a VHF (160MHz) receiver schematic into a PCB. After looking here and there, I am a bit confused.

It seems like the main issues with RF are

1. to avoid stray inductors and capacitors, by avoiding close tracks (capacity up), wide tracks (capacitor with ground plane underneath), and long tracks (inductance up)
2. to avoid signal reflections by avoiding sudden changes in "characteristic impedance".

[Please do tell me if I missed others]

I only have a vague idea of what characteristic impedance is (this wonderful video helped a lot though), but it sounds like it's the impedance of the equivalent RLC circuit.

1. It should be dependent on the length and the frequency of the signal, how come it isn't?
2. Intuitively I should calculate the characteristic impedance of each pad-to-pad trace and make sure it is always 50Ohm. Is that the case?

An online calculator gives (for 18um thick copper, 4.7 of permittivity, 0.5mm thick substrate) me 0.9mm of width to get 50Ohms. Does that mean I should route all traces at this width, keeping them short but without having them too close to eachother, and then I have nothing to worry about?

I only have a vague idea of what characteristic impedance

Characteristic impedance is the ratio of voltage to current (thus, an impedance) for signals propagating along the trace, which is determined by the balance of capacitance and inductance along the trace.

It should be dependent on the length and the frequency, how come it isn't?

Characteristic impedance depends on the ratio of inductance to capacitance. Since both inductance and capacitance increase linearly when the trace length increases, their ratio doesn't depend on the trace length.

Also, within limits, these parameters also don't change much with frequency, so again the ratio doesn't depend on frequency and the characteristic impedance doesn't depend on frequency.

Intuitively I should calculate the characteristic impedance of each pad-to-pad trace and make sure it is always 50Ohm. Is that the case?

If the driving circuits are designed to drive 50 ohm loads, then generally yes. You also want to provide matched termination at at least one end of the trace, and possibly both, depending on the details of your circuit.

Generally you don't have to make a separate calculation for each connection. You just look at your board stack-up and find a trace width that achieves 50-ohm characteristic impedance, and make all of your traces that width. You might use microstrip, stripline, or coplanar waveguide geometry depending on the circumstances of your layout. You would do a separate calculation for each signal layer on your PCB, and maybe for the different types of geometry (microstrip and coplanar, single-ended and differential) if you need to use all those combinations.

If the trace length is less than about 1/10 of a wavelength at your operating frequency, then you can often get away with using an unmatched trace.

• Thanks a lot. When to use microstrips, striplines and waveguides? Also, I'm confused now about how reflections change the voltage. COuld you have a look at my edit? – user42875 Apr 16 '15 at 11:47
• Here in fact: electronics.stackexchange.com/questions/165099/… – user42875 Apr 16 '15 at 12:39

It seems to me you summed up pretty much all what I would have summed up, so I'll get into the (easy) math that answers your question(s).

Check this out. I'll rewrite it down here:

$$Z_\mathrm{in} (\ell)=Z_0 \frac{Z_L + jZ_0\tan(\beta \ell)}{Z_0 + jZ_L\tan(\beta \ell)}$$

The above formula allows you to compute the input impedance of a lossless transmission line if you know the characteristic impedance $Z_0$, the load impedance $Z_L$ and the wavenumber $\beta=\frac{2\pi}{\lambda}$, where $\lambda$ is the wavelength in the transmission line.

Now that seems to be a complicated formula, what it tells you is that the input impedance is a mess.

There are two ways to improve this "mess":

Let's examine the second situation. If $Z_L=Z_0$:

$$Z_\mathrm{in} (\ell)=Z_0 \frac{Z_0 + jZ_0\tan(\beta \ell)}{Z_0 + jZ_0\tan(\beta \ell)}=Z_0$$

And that's where the magic happens. The input impedance does not depend on the trace length and that's great since you don't usually want to care how long transmission lines are when they are used to transmit: think of a poor technician who needs to cut a coaxial cable for some mm long wave, maybe over 10m of cable... Good luck with that.

What you usually do then is make devices so that all the impedances at their ports are known so that the PCB designer (you!) can easily size the tracks. It happens that $50\Omega$ is a very widely used value, I imagine because coaxial cables happen to have an inherent (as in size and materials inherent) $50\Omega$ impedance. Your ICs probably have 50 ohms output and input ports so using 50 ohms traces is exactly what you want to do.

As per your other questions well, reducing cross talk, parasitic capacitance or inductance and whatever non ideality comes to mind is always a Good Thing$^{TM}$ so do your best to keep your tracks short and far.

• Apparently 50Ohm is used because it's a good compromise between power transfer and attenuation... Not entirely clear how those work though. THanks for this! – user42875 Apr 16 '15 at 11:26
• I've updated my answer to try and understand how reflections work, care to have a look at it? – user42875 Apr 16 '15 at 11:46