Message	Date	From	Subject
2589	2009-01-11 06:10:33	joop_l	Loaded Q and the amount of negative resistance
Hi, I have been playing with a crystal oscillator in LT-SPICE. It seems that it is possible to "measure" the negative resistance and, when changing some parameters, show oscillation occurs when it is at least equal to the negative of the crystal Rs. But I am not sure what the final loaded Q of the crystal is. Can I safely assume the Q can be calculated with: Reff = ABS(Rs + Rneg) so e.g. Rs = 5 Ohm, Rneg = -12 Ohm => Reff = 7 Ohm and use this in Q = 1/R * SQRT(L/C) Because then theoretically the loaded Q could be even higher than the unloaded Q, provided the oscillator will still start. Giving the simulation a "jumpstart" shows that it is possible to maintain oscillation with Rneg only slightly bigger than the crystal Rs (e.g. Reff = ABS(5-8) = 3). Having a higher loaded Q than unloaded is also suggested in this article (which I cannot access, anybody?..): http://www3.interscience.wiley.com/journal/114198913/abstract?CRETRY=1&SRETRY=0 Cheers, Joop - pe1cqp
2590	2009-01-11 09:19:48	Rick	Re: Loaded Q and the amount of negative resistance
Increasing the Q of a resonator by supplying a little external energy at the correct phase is the fundamental concept behind pendulum clocks, children on swings, regenerative receivers, etc. When there is an external source of energy, it is fine for loaded Q to be higher than unloaded Q. I've had modest success modeling crystal oscillators in LTspice and predicting oscillation with various measured crystal parameters. But my primary tool for improving oscillator designs is the scientific method (think, experiment, think about the measured results, design a new experiment, repeat). With oscillators I always discover something important in the measured results that the simulator doesn't predict and I didn't think about ahead of time. Interesting work--have fun, Rick
2591	2009-01-11 14:48:08	joop_l	Re: Loaded Q and the amount of negative resistance

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2589

2009-01-11 06:10:33

joop_l

Loaded Q and the amount of negative resistance

Hi, I have been playing with a crystal oscillator in LT-SPICE. It
seems that it is possible to "measure" the negative resistance and,
when changing some parameters, show oscillation occurs when it is at
least equal to the negative of the crystal Rs.

But I am not sure what the final loaded Q of the crystal is. Can I
safely assume the Q can be calculated with:

Reff = ABS(Rs + Rneg)

so e.g. Rs = 5 Ohm, Rneg = -12 Ohm => Reff = 7 Ohm

and use this in Q = 1/R * SQRT(L/C)

Because then theoretically the loaded Q could be even higher than the
unloaded Q, provided the oscillator will still start. Giving the
simulation a "jumpstart" shows that it is possible to maintain
oscillation with Rneg only slightly bigger than the crystal Rs (e.g.
Reff = ABS(5-8) = 3).

Having a higher loaded Q than unloaded is also suggested in this
article (which I cannot access, anybody?..):
http://www3.interscience.wiley.com/journal/114198913/abstract?CRETRY=1&SRETRY=0

Cheers,

Joop - pe1cqp

2590

2009-01-11 09:19:48

Rick

Re: Loaded Q and the amount of negative resistance

Increasing the Q of a resonator by supplying a little external energy at the correct phase is
the fundamental concept behind pendulum clocks, children on swings, regenerative
receivers, etc. When there is an external source of energy, it is fine for loaded Q to be higher
than unloaded Q.

I've had modest success modeling crystal oscillators in LTspice and predicting oscillation with
various measured crystal parameters. But my primary tool for improving oscillator designs is
the scientific method (think, experiment, think about the measured results, design a new
experiment, repeat). With oscillators I always discover something important in the measured
results that the simulator doesn't predict and I didn't think about ahead of time.

Interesting work--have fun,

Rick

2591

2009-01-11 14:48:08

joop_l

Re: Loaded Q and the amount of negative resistance