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gsingh1234
modified 6 years ago

Resonator

1
3
159
01:25:23
This circuit demonstrates how a resonator can be used practically , so it includes a coupling cap to feed the input signal with some source resistance , the series resistance of the inductor and loading of the measurement setup , Anyone using this should do an AC analysis from 2 to 8 MHz and see the effect of varying the coupling cap and series resistance of inductor at the output , for typical cases coupling cap with a tenth of the value of the resonator cap works best, the inductor is assumed to be air core so high frequency loss in magnetic cores is not modeled. This type of resonator can work without voltage source being part of the circuit, and therefore achieves galvanic isolation, practically the ground of the voltage source and the resonator can be very different and only limited by the breakdown voltage of coupling capacitor.
published 6 years ago
fatcat2
6 years ago
Change the capacitor value from 100pF to 10nF; it degrades the performance.
gsingh1234
6 years ago
Yes it surely will because the coupling cap will then heavily load the resonator because 10nf has very low impedance at the resonant frequency
fatcat2
6 years ago
Treat the voltage source withseries resistance as a current source with parallel resistance. Make it 10nF and convert it into Norton's equivalent. Check the bode plot. It'll be just fine.

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