Peter L. Peres email (remove spam text)
On Mon, 26 Oct 1998, John Payson wrote:
> sumption, reduce frequency accuracy, and lead to increased wear on
> the crystal.
There is no 'wear' on the crystal when operated in its SOA. There is
ageing but that has almost nothing to do with the oscillation.
> The PIC will try to operate as a fairly linear amplifying
> inverter. As the amplitude increases, it's necessary that
> the PIC's gain decrease (since if the net gain around the
> loop were greater than one, the oscillations would grow
> exponentially); this is the cause of the slight "flat top"
> observed on the PIC's output.
The PIC is nowhere 'fairly linear'. You can expect 5% and worse distortion
from it, in the 'linear' operating area of the oscillator gate.
> Because the PIC's output will try to match the input (but
> inverted) the two signals should be precisely 180 degrees
> out of phase. If the phase difference is greater or less,
> then the R in the circuit (real or inherent) will have to
> absorb the difference between what the PIC's output is try-
> ing to do and what that end of the crystal is actually do-
No, this would be true for a series crystal. The crystal used with a PIC
is parallel so you can treat it as a very narrow bandpass filter. Whatever
the signal at its output, if it contains even a little component on its
center frequency, it will select it, flip it 180 degrees and send it back
to the amplifier input with some loss.
This is also how it starts up. Noise at the output of the amplifier
contains energy on all the frequencies in the GBW band of the amplifier.
The tiny part in it that falls on the crystal center frequency is selected
and fed back to the amplifier and oscillation starts.
The startup for a given gain and noise depends directly on the Q of the
crystal (or rather, on the bandwidth of the filter ;). Low Q oscillators
start faster. Ceramic and SAW filter oscillators start very fast vs.
> scope inputs will be affected equally. If the caps have
> been chosen correctly, the two signals should appear 180
> degrees out of phase. If this is achieved, circuit power
The signals will practically never be 180 degrees out of phase. The extra
gain in the amplifier sees to that. You can only get 180 degrees clean if
the gain of the amplifier is controlled down in an active loop to stay in
the linear area. I'll keep saying this, sorry.
> consumption will be minimized, and so will EMI emissions
> (since the two caps will be trying to feed roughly equal
> and opposite currents through their ground returns).
You can never rely on this. The caps and the chip grounds must be as
intimate as possible.
> The biggest question I'd still have, then, would be how to
> select the best value for the caps from among all those that
> produce the perfect 180 degree phase shift. Would the ideal
> value be one where the ratio of the caps roughly matched the
> ratio of the signal strengths (so as to best balance out the
> ground currents)?
If you want to be picky, you can use the actual crystal parameters, the
equivalent circuit of the amplifier input, and the equivalent circuit of
the amplifier output, and calculate Cout so the impedance at the output
matches the generator Q*Zo and Cin to the impedance at the input so it
matches Q*Zin. The actual calculus can be obtained from a crystal filter
design chapter in a book. These are just scribbled equations here, they
have nothing to do with reality. One interesting thing is, that you will
most likely end up with totally impossible part values for the circuit,
will give up, and will use the crystal manufacturer's specced capacitors
instead, so you can phone him and complain if it does not work ;)
In reply to: <01BE00CE.2CB84790@JOHN_WKST1>
See also: www.piclist.com/techref/index.htm?key=crystal+tutorial
You must be a member of the
piclist mailing list
(not only a www.piclist.com member) to post to the