Why is the feedback of op-amp oscillators connected to inverting input instead of positive input

Question

I'm new to the subject of oscillators and learned that sustained sinusoidal oscillations are made possible by positive or regenerative feedback. Is that right? If that's true then I can't quite grasp why, especially in op-amp LC or RC oscillators, is the feedback network terminated to the inverting input while the non-inverting input is usually grounded. Doesn't it make it negative feedback? My circuit analysis ability is not yet fully developed so please bear with me.

Also, I would highly appreciate if someone patiently traces out the operation of this particular colpitts oscillator for me in relation to this question.

One more thing: does the LC network act as the feedback or it's just a tank circuit while Rf entirely serves as the feedback? I know. silly questions.

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Thanks everyone.

Answer 1

This circuit is getting positive feedback at the inverting input, because the feedback network adds 180 degrees of phase shift at a certain frequency.

I simulated the feedback network with some arbitrary values. You can see that the phase of the feedback network is 0 across all frequencies, except for the oscillation frequency.

This means at that frequency (in my sim ~191 MHz) the feedback is inverted before coming back to the amplifier. Typically, if the op amp input was a 191 MHz sine, the output would be inverted (180 deg phase shift) and added back to the input - this is negative feedback. Since the feedback network adds ANOTHER 180 deg of phase shift (360 deg total) the sine being added back is in phase, resulting in positive feedback.

Answer 2

Making this as simple as possible, when oscillating one end of the inductor will be rising in voltage while the other end is falling. The current in the inductor charges one cap positively while it charges the other negatively. As those charges accumulate the current falls in the inductor till it reverses and the whole thing repeats in the other direction. The resultant see-saw in currents is what makes the whole thing ring at it's designed frequency.

Since, as I mentioned above, the voltage polarity is reversed across the inductor the op-amp needs to be connected as an invertor, i.e. using negative feedback as shown.

Note: Given an initial stimulus the circuit will ring at the oscillation frequency on it's own, with no op-amp in the circuit, till it decays due to losses. The op-amp overcomes those losses and keeps the thing ringing as long as power is applied.

Answer 3

In a typical inverting amplifier circuit, you are correct that the output feeding back to the negative input would be a form of negative feedback. The difference here lies in the resonant LC circuit that is present. In such a parallel arrangement, the capacitors and inductor create a 180° phase shift in the signal.

This means that if you start by applying a 0° input on the negative side, the inverter op amp arrangement will produce a 180° shifted (inverted) output. This output gets shifted by another 180° by the LC circuit, and you end up with 360° or effectively, 0° again. This signal adds constructively with the input signal, and you have positive feedback.

Trevor's answer explains a little bit more about how the circuit specifically works. Hope this helps, cheers.

Answer 4

Any Colpitts oscillator uses two capacitors and an inductor AND one other extra passive component - a resistor. This may seem strange that I mention this but, with a perfect op-amp driving C1 (as per your circuit diagram) you don't quite get 180 degrees phase shift from the two capacitors and inductor. Here's the general idea: -

So now the output of the op-amp is additionally phase shifted by R3 and C1 (in my circuit) a few degrees and, this ensures that the frequency selective feedback network can produce a phase shift of 180 degrees with ease.

Your circuit doesn't show this so you may be wondering how yours can work - op-amps are not perfect and at some arbitrary high frequency there will be a delay through the op-amp and that will be equivalent to a few degrees phase shift. This ensures that the phase shift from L and C2 (maximum of 180 degrees at infinite frequency) can be used with the additonal phase shift from R3 and C1.

If you built the circuit you might want to put the extra resistor in circuit and initially choose a value of 10 ohms - if you get an overly distorted sine wave try increasing it but, be prepared, the sine quality is not that great from a basic colpitts oscillator because there is nothing that takes care of ensuring gain/amplitude stability. Nonetheless it's a fairly useful circuit and despite the plethora of information out there on how they work, hardly any of it mentions the extra resistor or the delay incurred by the op-amp in order to make it oscillate.

Answer 5

" sustained sinusoidal oscillations are made possible by positive or regenerative feedback. Is that right? "

Not always. Sustained oscillation occurs when the Barkhausen stability criterion are met, and one of the criteria is that the total phase shift through the the amplifier and its feedback loop is 0 degrees, or 360, or 720, etc. A common way to achieve this is with 180 degrees in the feedback loop and 180 degrees in the amplifier. The amplifier part is met by making using an inverting gain circuit.

https://en.wikipedia.org/wiki/Barkhausen_stability_criterion

Answer 6

As others have stated, the feedback components shift the phase of the feedback signal, so although the feedback signal is fed into the inverting input of the op-amp, the total phase shift around the entire feedback loop makes the feedback some integer multiple of 360 degrees -- i.e. positive feedback.

If the amplifier were a BJT instead of an op-amp, then R3 and C1 would be needed, for the reason that @AndyAka gives.

However, when the amplifier is a typical op-amp, R3 and C1 are not needed for obtaining sufficient phase shift to oscillate. This is because a typical op-amp has an internal "Miller capacitor" that gives the op-amp itself an approximately 90 degree phase shift for much of its frequency range. R3 and C1 do, however reduce the amount of amplification needed to initiate oscillation.

^{simulate this circuit – Schematic created using CircuitLab}

Note that an op-amp oscillator of this sort does not produce a nice clean sine wave. Once the amplitude of the wave gets big enough, the op-amp "saturates", and the tops of the sine wave are clipped. Unfortunately, not in a way that gives a nice clean square wave, either.