I am looking at using this crystal for a new circuit design. It will provide the main oscillator for my PIC application.

Here is the part on digikey.

The datasheet indicates that it has a load capacitance of 8pF. This seems very low compared to most. I have read that the load capacitance should be equal to:

$$ \frac{(C_1\times C_2)}{(C_1+C_2)} +C_{stray}$$

where Cstray is the capacitance of the PCB traces which are typically between 5 and 15 pF.

This equation is provided in a Microchip FAQ sheet. If this is correct my PCB trace capacitance is likely already equal to the load capacitance of the oscillator. So does this mean that I really don't need any capacitors connected between the crystal and ground?


2 Answers 2


A good tip for getting optimum cap values is that an oscillator like this is happiest when the amplitudes of the signals at each side of the crystal are similar. PCB capacitance will be a factor, so this needs to be tweaked on the actual board, and using a x100 scope probe to avoid the probe affecting the reading. A 10M resistor right on the tip of a standard probe is an alternative if you don't have a x100 - accuracy doesn't matter as you're just looking at relative amplitudes.

  • 4
    \$\begingroup\$ Gotcha: don't get trapped into thinking you don't need the load caps when you see a happy oscillation, though, as your probes load the lines with 5pF+ (20pF for me). \$\endgroup\$
    – tyblu
    Commented Dec 13, 2010 at 14:02

A standard 20 pF crystal would be more suitable, and make the design much easier.


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