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op-amp oscillator circuit

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ElectroChas

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As suggested in another thread, I built this circuit which comes from an LM324AN datasheet. It uses a bipolar supply and takes a noninverting input of ½ Vcc as the reference via voltage division.

It was also suggested that with the single supply version of this circuit that the op-amp swings from 0v to Vcc using the ½ Vcc as the "center", but when the output is viewed with an oscilloscope the square wave has a positive and negative component (of nearly equal amplitude and time period), yet the image of the circuit shows the low as 0 and not -V.

I've been pondering this circuit for some time in an effort to get a solid mental image of all its stages and I seem to be stuck on the point regarding the negative voltage output cycle.

Any explanation would be helpful.

Thank you,
Charles
 

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  • Squarewave Oscillator.jpg
    Squarewave Oscillator.jpg
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I'll give you a clue: it's a Schmitt trigger oscillator.
 
The circuit has a voltage divider that biases its input at half the supply voltage because it uses a single positive supply, not a bipolar supply. So of course the sketch shows an output from 0V to its positive high voltage that is about 1.2V less than the positive supply voltage.

Remove the resistor that goes to the positive supply if you want to use a bipolar supply.
 
Sorry that was a typo on my part. I've been playing with two different versions of this circuit which led to the typo. I meant to say unipolar in the first paragraph, but the 2nd paragraph was correctly stated, referring to "single supply" and the image also is one that uses unipolar (single) supply.

Anyhow, my question still stands and applies to the single supply version of the circuit (the one attached), why does the image show 0v low and the oscilloscope shows a negative component?
 
Your 'scope will show a negative component if it is set to view AC instead of DC.
 
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