I have a 24 MHz, crystal on my board and datasheet of IC defines the load capacitance between 5-9pF. The reference design I follow don't use any load capacitor.

My traces look like this:

enter image description here

shorter trace is 3.3 mm long, the longer one is 8.6 mm long.

Referring to the application note here, I use a signal generator at 24 MHz. Output of the generator and a scope of my oscillator are placed to an end point of a path, the second scope is placed to the other end. But, I don't observe any voltage/waveform/phase shift changes at the oscillator. I tried it by applying ramp, square, and sine wave forms, but output is the same for all.

Could anyone explain how to measure the parasitic capacitance of the paths? What do I need to do after measuring the capacitance, say I measure 1pF at the short path and 4 pF at the long path, do I simply need to add them and then compensate the required capacitance by placing load capacitors?

  • \$\begingroup\$ I think you're focusing too much on the capacitance of these traces. I guess that they have a capacitance (to ground or to eachother) of less than 0.5 pF. The crystal's load capacitors do not need to be that accurate in value anyway. \$\endgroup\$ Commented Nov 16, 2015 at 11:32
  • \$\begingroup\$ I was referring the answer on electronics.stackexchange.com/questions/121659/… says that parastic capacitance should be between 3-5 pF, I wondered if I need to measure the capacitance in order to determine if I really need the caps or not ( other method is to try and see, but I want to learn how it is properly done) \$\endgroup\$
    – Angs
    Commented Nov 16, 2015 at 11:40
  • 1
    \$\begingroup\$ Most of that 3-5pF capacitance I mentioned is typically the input/output capacitance of the chip. \$\endgroup\$ Commented Nov 16, 2015 at 11:42
  • \$\begingroup\$ The capacitance of a trace 4 thou wide on a 4 thou core is about 1.1pF per inch. Your tracks are much shorter than that, so the parasitic capacitance is, as Spehro notes, dominated by the capacitance of the IC the crystal is attached to. \$\endgroup\$ Commented Nov 16, 2015 at 12:19
  • \$\begingroup\$ Perhaps your measurements did not work because you were not using a fifty ohm scope input? That’s what the application note uses. \$\endgroup\$
    – user69795
    Commented Jan 20 at 0:41

1 Answer 1


There is a way to measure the impedance of a trace which requires transmission line theory.

But usually is only considered if the wavelength is comparable to the board size and for 24 MHz is approximately: $$\lambda = \frac{c}{f} = \frac{3\times10^8}{24\times10^6}=12 metres$$

so any reflection of that wavelength is going to be virtually nothing on a trace that's a cm or two.

Let's first assume that your traces are lossless which is safe assumption for low power traces. So our propagation constant is given as: $$\beta=\omega\sqrt{LC}, \omega=2\pi f$$ which is related to wavelength by $$\lambda=\frac{2\pi}{\beta}$$

Anyway, by the magic of the telegrapher equations derived by Oliver Heaviside, we get the characteristic impedance of $$Z_0 = \sqrt{\frac{L}{C}}$$ Which is impedance per unit length so how do we find L and C? Answer: dimensions

Mind you this is extremely simplified and only rule of thumb.

Image below is a cross section of PCB track

trace crossection

$$L=\frac{\mu d}{w}\rightarrow\mu = 4\pi \times10^{-7}$$

$$C=\frac{\epsilon w}{d}\rightarrow\epsilon = 8.85\times10^{-12}$$

This is PER METER so multiply this by how long your track is IN METERS

$$ 8.8mm = 8.8 \times 10^{-3} $$

The 'u'-looking Greek letter mu (µ) is the permeability of free space in Henrys per meter.

The 'e'-looking Greek letter epsilon (ϵ) is the permittivity of free space in Farads per meter.


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