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The above is an excerpt from the book "Crystal Oscillator Circuits". It said it is meaningless to measure voltage across a crystal. I can not understand why. When crystal is operating in parallel resonance region, it appears inductive. And everyday voltmeter can not calculate $$\frac {dI}{dt}$$ Which is necessary to calculate the voltage across inductor. Is this the reason why the excerpt above claimed that it is meaningless to calculate voltage across a crystal?

  • 3
    \$\begingroup\$ The voltage measured across a crystal is not meaningless in general - that quote reads "is meaningless as a measure of current through the crystal" - probably because you don't know the impedance of the crystal when it's slightly off resonance, so can't use I=V/Z to calculate current from the measured voltage. \$\endgroup\$
    – brhans
    Nov 6 '17 at 16:09

Yes, that's correct. You can't easily distinguish between the voltage drop created by the internal resistance and the drop across the (equivalent) inductor/capacitor, you only see the sum of the two drops.

You could of course calculate the voltage drop across the inductor using dI/dt or the phase of the current, but in that case you already have the current anyway so there's no point in measuring the voltage to get the power dissipated in the crystal.

Edit: As brhans noted, the voltage isn't meaningless in general however. For example, crystals have a dielectric breakdown voltage that must not be exceeded. It's just not useful for calculating power dissipation.

  • 1
    \$\begingroup\$ @niki_t1: Jonathan's answer is good but we'd recommend that you wait until the Earth rotates at least once to give everyone on the planet a chance to answer. You might encourage a better answer or one which gives some other insights - even if not as good as Jonathan's. Then come back and make your choice. \$\endgroup\$
    – Transistor
    Nov 6 '17 at 16:27

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