How to choose capacitors for crystal oscilator?

Hi - I was just wondering - how does one choose the capacitors that go between the leads of a crystal oscilator and ground? (and do they have a special name?)

I was thinking that you just got 2 low voltage ceramic capacitors of the value listed as "load capacitance" in the crystal datasheet, but on a project I worked on recently the reccomended caps were not the same value as the load capacitance of the crystal (32pf vs. 20pf, as I recall). So does one determine the value of these capacitors through some other means? Or is it just not very important what their value is? Or were those part reccomendations not very good? Thanks!

-Michael Noone

Reply to
Michael Noone
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In article , Terry Given wrote: [... using 2 33pF caps on a XTAL speced for 20pF ...]

Don't forget the input capacitance of the amplifying device(s). That capacitance appears in parallel with one of the two 33pF capacitors. When all the capacitances add up to the makers rating and the amplifier is fast, the crystal will oscillate at the makers specified frequency.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

[...]

Yes, this is something to really worry about in the discrete amplifier case. The output impedance of a discrete amplifier can be highish.

A logic gate's output has a lowish impedance so the capacitance of the driven stage is deleveraged quite a bit in the logic gate oscillator.

In real life, the lag in the logic gate can also effect the frequency. It lowers the frequency like someone added an extra capacitor onto the driven side of the crystal. Gate delay can also lead to operation in other modes.

Theres a funny thing going on in a logic gate oscillator. Like this:

U1 ------!>o------+--------- ! ! ! R1 ! +---///-------+ ! ! ! ! / R2 ! Y1 ! !-! ! +-----! !-------+ ! !-! ! ! ! --- --- ---C1 ---C2 ! ! GND GND

If we ignore the input capacitances, all of the AC current that flows in C1 must also flow in Y1. If the Q is high, the same current also flows in C2. For logic gate based designs to repeat well, the fundamental voltage on C2 must be almost equal to the fundamental voltage on the output of U1. When all of that is true, C1 and C2 have more control over the current in Y1 than R2 does.

Also, if you are not using an unbuffered gate, the phase shift through U1 can be over 45 degrees if you are using slow logic gates relative to the oscillation frequency. When this is the case, it starts to be a good idea to change R2 to a small capacitor.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

From the point of view of the crystal (IOW ignore the ground connection) those two capacitors are in series, so 16pF total. That capacitance is in parallel with the crystal shunt capacitance, so they add up to (hopefully) 20pF.

Cheers Terry

Reply to
Terry Given

oops. And if the gate output drives another gate, that one crops up too. At least the pcb capacitance is usually small enough to ignore.

How accurately is gate input capacitance spec'd? its typically given as about 10pF, but what in practice is it?

Actually, I just checked the Philips AHC/AHCT14 datasheet (not the right part, but it was handy), which gives Cin = 3pF, Cout = 4pF. So check for whatever chip you are using.

Normally the gate output (and load) capacitance wont show up because there is a series resistor to limit crystal drive to a suitably (and perhaps surprisingly) low level, but often there is not, although crystal failure is perhaps more important in this case than exact frequency.

Cheers Terry

Reply to
Terry Given

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