[SOLVED] Building LC Meter Circuit

I have one more question to ask that is occupying my mind.





the circuit given above again. What I don't get is how the resonance starts. In my opinion, When the output of the IC becomes high, thanks to voltage divider network, PIN2 will be equal to Vout/2. And then I guess it will charge the LC tank circuit which will trigger resonance. I don't know if I am right till this point. Then resonance wave is observed at PIN2. But how come it doesn't have DC Offset, because of that Vout/2 I explained above. Thanks in advance.

Aydin Ozcan

Simulators rarely demonstrate oscillators starting up. The reason is that in order for the oscillation to start there has to be some change in the feedback signal to be amplified. In real life, that comes from noise generated in the components but simulators tend to ignore that and only show DC conditions at start up. For the simulation to work you have to 'kick' the circuit into life by briefly injecting a change in voltage.

Brian.

betwixt said:

The result is as expected but for the wrong reasons, I suspect it you short out C6 completely it will still show a similar frequency.

What is happening is the dominant frequency selecting component is now C4. Normally, it's value is very large compared to the LC parts you are measuring and it becomes almost invisible to the signal leaving the L and C to decide the frequency. When you make C6 large as well, the inductor starts to look like a short circuit across it and it plays little part in deciding the frequency. What you are left with is an RC oscillator using the comparators inversion and feedback through R4 to charge and discharge C4.

The LC components will still play some role in setting the frequency but not enough to be able to use it for measurement purposes.

Brian.

Click to expand...

Just saw your previous answer. Can't hold back to mention that the explanation sounds plausible but is wrong, though. I didn't post my answer without checking before that the oscillation is actually set by the LC resonance. The surprising feature of this oscillator is to work with an extremely wide characteristic impedance range.

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I don't read post #21 specifically simulation related. It's asking about the oscillator operation in general.

A prerequisite for stoking up oscillations is an operation point in the linear amplifier range. DC feedback is the usual means to achieve it, in the present circuit the -1 DC feedback to inverting input pin is out competing the +0.5 feedback to n.i. input. In the oscillation frequency range, positive feedback is dominant, allowing an increasing wave at the LC resonance. Magnitude is limited by output saturation. Initial deviation from steady state is achieved by power-up transients or even circuit noise (the latter not modelled by a simulater).