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From: Fred Bloggs
User-Agent: Mozilla/5.0 (Windows; U; Win 9x 4.90; en-US; rv:1.0.1) Gecko/20020823 Netscape/7.0
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Subject: Re: Querry, Kick Starting Crystal Oscilator.
References: <3DCA905A.59FD928B@mmm.com.DELETETHIS> <3DCC9A71.7DF3C3A9@mfi.net>
Date: Sun, 10 Nov 2002 18:17:46 GMT
NNTP-Posting-Date: Sun, 10 Nov 2002 10:17:46 PST
Organization: EarthLink Inc. -- http://www.EarthLink.net
Michael Stein wrote:
>> Like most crystals, those have a Q of 60,000 to 90,000 which
>> means they'll take to 2 to 3 seconds to reach 50% of the full
>> oscillation energy. That's the way it goes; the energy cannot
>> simply be kick-started into the crystal.
> Picture a parallel LC with a constant current (DC) being forced through
> the L. The voltage across the L and C is constant. Now suddenly
> disconnect the current source.
> Why should there be a delay in the start of ringing and why should
> this be dependent on the Q of the LC? And if the ring doesn't
> start at full amplitude, then where does the energy from the L go?
First of all the crystal is actually a series L-C-R and then only when
it goes into oscillation. The motional impedance does not actually exist
until the motion a.k.a. oscillation gets started- and that is why
consideration of this impedance at start-up is nearly worthless. The
32KHz tuning-fork style of crystal typically shows ~50K of motional
resistance in steady state , but at startup this may be on the order of
500K ohm , tapering to 50K as a complicated function of power
dissipation. The simplistic linear model only comes into play at the
first low-level oscillation which is a complicated proposition to bound.
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