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I am trying to understand working principle of the circuit given in the link below :

http://kripton2035.free.fr/Resources/lcmetersch.pdf

I realized that the circuit is basically relaxation oscillator with LC tank circuit attached to its non-inverting input. I know that when the output of the comparator becomes HIGH (5V) capacitor in the negative feedback path charges and and after reaching some point, voltage on the inverting input of the comparator becomes higher than the non-inverting input and the output of the comparator becomes LOW. And then the capacitor in the negative feedback path discharges and the voltage on the inverting input becomes smaller than the voltage on the non-inverting input, causing the output to go HIGH again. It goes on like that.

But the thing is I am not able to understand the effect of LC tank circuit here. I don't know how come the output frequency is equal to LC tank resonance frequency. And I am not able to understand the function of the resistance between output and non-inverting input of the comparator.

A comprehensive explanation of the circuit is the given in the link below :

http://www.vk6fh.com/vk6fh/lc_meter_vk6fh.htm

Thanks in advance!

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Basically the "classic" relaxation oscillator only works as a kick off to start the LC oscillator.

The little damped ringing on non inverting input triggers the output to swing full supplies and increases itself.
In the frequency domain you may think that parallel resonating LC tank gives zero phase and hence positive feedback at its resonance frequency only.

Now the "slow" RC network on inverting input simply sets the threshold at the average output value to ensure proper operations even with non symmetric (both in level and time) output

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  • \$\begingroup\$ I see that from the link below : en.wikipedia.org/wiki/Relaxation_oscillator if there was no LC tank circuit connected to non-inverting input, the output frequency would be equal to f=1/2In(3)RC. This is much lower than the LC tank circuit resonance frequency. I guess that is why you call it "slow". Since the frequency generated by the tank circuit is much higher than the frequency of the relaxation oscillator without tank circuit, it dictates the output frequency. Can you verify this? By the way I didnt understand entirely what you have written. What kind of threshold u talk about? \$\endgroup\$ – Aydin Özcan Dec 6 '16 at 21:58
  • \$\begingroup\$ I just spent an hour re-typing this exact question (for another iteration of this LC meter circuit)! I have now closed the tab with the question because my problem is solved by this answer and the OP's link... +1 Thanks to both of you! \$\endgroup\$ – Dmitri Oct 16 '17 at 1:07

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