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Crystals question

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MikeMl

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Note that when is says 9 to 12pF load capacitance, that means the effective capacitance looking back at the oscillator circuit as measured between the two pins of the crystal.

To find the effective capacitance, you would have to consider the two external capacitors on the uC pin, shunted by the capacitance of the uC pins themselves(~2-3pF), shunted by the stray capacitance of the PCB traces (~2pF). Say you use external capacitors of 15pf. That would mean that there is 15+3+2 = 20pF effective shunt capacitance on each end of the crystal to ground. Now the two 20pF shunts are effectively in-series when considering how they effect the crystal, so 20*20/(20+20) = 10pF is the "load" capacitance seen by the crystal...
 

MrAl

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Hello there,

For parallel cut crystals the load capacitors go from the crystal leads to ground, one cap per lead. The total load capacitance to match the data sheet is then the series combination of the two caps plus about 5pf as an estimate. For example, two 12pf caps in series would equal 6pf, then adding the 5pf we come up to 11pf total so the load capacitance would be 11pf and this would have to match the data sheet to get the actual crystal package stamped frequency. If you do not match this value the frequency will be slightly off of the package stamped value. Remember that one 12pf cap connects to each lead of the crystal so that's two caps needed, so for that example two 12pf caps would be needed.
If both caps are the same (they usually are) then the formula is simple:
CLoad=C/2+5pf
where
CLoad is the load capacitance from the data sheet, and
C is the capacitance of each capacitor in pf.
Thus for your crystal you might be able to use two 8pf caps to get a total of 9pf load capacitance:
CLoad=8/2+5
in picofarads.

Another consideration with 'chip' crystals is their low drive requirement. This often requires a series resistor, in series with the capacitor (but the two load caps stay connected directly to the crystal pins to ground). The initial test value would be equal to 5 times the crystals internal series resistance. Thus for a crystal with 60 ohms internal resistance the resistor would be 300 ohms. Be aware that overdriving the crystal could damage it permanently.
 
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