At 03:08 PM 7/14/2005 -0400, you wrote:
>Spehro Pefhany wrote:
>>The output acts like a *current* source.
>
>That's the same as saying it has inifinite impedence.  As I said, I agree
>that the capacitors are in series for calculating the load capacitance in
>this case.
>
>However I don't agree that the driver has infinite impedence.  It has CMOS
>high/low output drive transistors, and will look roughly resistive.  And, I
>think there is a good chance that this resistance is about the same or lower
>than the impedence of the capacitor on the pin at that frequency.
>
>Do you at least agree that if the driver has finite impedence, then just
>adding the two capacitors is not valid?

Sorry, I don't agree, at least prima facie, that some resistance would
affect the validity of the equations-- I think the effect would be to
lower the Q somewhat, not to change the resonant frequency directly.
The output capacitance is in parallel with one of the load capacitors so it
just adds to it.

Probably the best bet to investigate in depth would be to find some very
good MOSFET models (including subthreshold) and crystal models and model
the steady-state (large signal) operation of an oscillator (and compare
with reality in terms of the external voltage swing and so on). The
theoretical analysis usually assume linear operation, which is not really
valid. I asked Microchip to consider specifying the minimum and typical gm
(transconductance) for their clock oscillator amplifiers, since it can be
used directly to predict whether the oscillator will start or not, but
they didn't seem to think that was worthwhile (I guess nobody else is
doing it...). It's easy to see that if you lower the *impedance* at
the operating frequency at the input and/or the output, that a higher
gm is required in order for the circuit to oscillate (they are roughly
in parallel for that purpose, ignoring the crystal impedance).

BTW, speaking of deviations from theory, I should add that the
equation I gave for drive power assumed worst-case voltage swing across
the crystal, which is going to be on the pessimistic side. If the actual
swing is known (say from a very high impedance probe), it can be substituted
for Vdd in the equation.

Your thesis could also be checked empirically with a frequency counter.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
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