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I'm ordering a 2-pin crystal (8.86723 MHz) and need to specify the load.

A little help calculating this, please?

Thanks!

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I'm pretty sure that the circuit does not function as a crystal oscillator. Is it an own invention?

For a crystal, you should decide if the circuit needs the series or parallel resonance. There are both, pretty near each other. For details, get e.g. the ARRL Handbook and read about crystal oscillator basics.

```--

Tauno Voipio```
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I double-checked the board layout; it is as I describe.

The circuit is part of an existing video driver board that outputs ASCII text to a monochrome CRT monitor. I have substituted a sine wave generator in place of the crystal and the board works.

The original crystal is gone (missing when I obtained the equipment) else I would have looked up the numbers on the original.

I'm not asking for evaluation of the circuit's function -- that is confirmed

-- only to help calculate the load on the crystal.

Thanks, Dave

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Can you draw the circuit around the crystal and ask the crystal company for their help?

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Yup, makes no sense, DC or AC.

```--
John Larkin         Highland Technology, Inc

jlarkin at highlandtechnology dot com ```
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Hmm... maybe this does? ::

Sorry for the crummy first attempt...

Thanks!

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Looks like Series, Generally CL can be ignored. It would be PCB related. If you specify it, you can always ad it to the circuit. The 10pf compensates for the propagation delay in the gates.

Cheers

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I see you're having difficulties in determining the load..

If you are looking for an exact figure, I think you need to use your signal generator via 100 Ohm R for example and a scope to measure the drop in the circuit.

First, test the scope probe by measuring the drop after the R only, to make sure you know the exact cap value in your probe.

Feed the circuit with this signal via the 100 ohm R, measure the drop. Calculate the load and remove the scope probe load from the results.

The net results should give you a load that is going to be close enough.. You may want to operate the circuit for a bit before taking final values. The logic chips are going to shift a little.

I'm guessing you'll end up with an approximate value that equals 8 pf.

In the capacitor manufacture world, the common practice was to zero beat a tuned circuit with a fixed frequency. You attach a test subject to the post which were part of this tuned circuit. You then moved the calibrated dial which was nothing more than a capacitor, to make it zero beat again. A scope was used with the X,Y inputs for that nice lissajous circle or spiral curves.

In any case, this dial would give you the exact capacitance load.

Jamie

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Isn't load capacity for the crystal a straightforward calculation?

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On a sunny day (Wed, 13 Feb 2013 12:15:49 -0800) it happened DaveC wrote in :

Fir * sake use a trimmer. Those 8.8.. 2xFc crystals are not MENT to be exactly on frequency. They are normally used in a PLL, possibly with varicap or some other reactance, to lock to incoming color burst.

If you do not do such a genlock, and want it free running, then you need a trimmer, zero tc components (caps), and for sure not a 2 inverter LSTTL oscillator, but a real one, and maybe even an oven.

JFET makes nice oscillator.

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By load they mean the total capacitance in parallel with the crystal. Some of this will be the obvious circuit, and the rest will be the inherent capacitance of wiring to the crystal.

Just take a reasonable stab by quoting say 20pf, and if you need precision, then use a trimmer capacitor across the crystal to bring it to the required frequency.

peter

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Load capacitance applies when the crystal is operated at its parallel resonance. Your circuit operates at the series resonant frequency. Tell the crystal manufacturer you want it to be series-resonant at 8.86723 MHz.

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