Finding the loop gain of an oscillator

[Sinister]

Homework Statement

Okay, so I know that I have to find the gain of the negative feedback part (1+ R2/R1).

But then to find the transfer function of the bottom part of the oscillator, would the resistor and capacitor that are attached to the '+' terminal of the op amp be considered in parallel?

I know that the resistor and capacitor connected directly to the output are in parallel.

Homework Equations

L(S)=A(S)*B(S)

A(S)= (1+R2/R1)

The Attempt at a Solution

 

[aralbrec]

That looks like Sedra & Smith :)

The feedback is simply the transfer function Va/Vo where Va is the voltage at the + terminal.

If you are looking at the current heading into the feedback circuit from Vo, you have (R+C)||R and that impedance is fed by the capacitor directly connected to Vo.

This is voltage / voltage feedback where the 'input' is assumed to be in series with the + lead into the op amp but is zero.

 

[Sinister]

Son what will be the loop gain :s still confused if the capacitor and resistor on left side in series or parallel and how would you find B(s)

 

[aralbrec]

Son what will be the loop gain :s still confused if the capacitor and resistor on left side in series or parallel and how would you find B(s)

The loop gain will be (1+R2/R1)*(Va/Vo) where A=(1+R2/R1) and β=(Va/Vo)

If the feedback is zero, the output is simply A*Vi. Zero the voltage at Va to zero the feedback and imagine Vi at the Va terminal. Then your open loop gain is A=(1+R2/R1).

The feedback is voltage/voltage, meaning the output voltage is sensed and the feedback signal is a voltage subtracted from the input voltage. If the input is Vi and is in series just before the + terminal of the opamp, it is being added to the feedback signal β (so watch the sign and positive feedback condition). Then set Vi=0 for this circuit.


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Edit: Adding a diagram. Circuit on left, standard feedback diagram on right. You have to match the circuit to the standard diagram.

With no feedback (Va=0), the output is Vo = A*Vi

The feedback β feeds a fraction of the output Vo to the summer (right diagram). The 'summer' (left diagram) is the series connection of Vi and Va. Note that an *addition* is happening, not a subtraction so your condition on the loop gain Aβ for positive feedback will be slightly different (ie not phase = 180 degrees)

I placed Vi like that so it wouldn't affect any part of the circuit operation. If Vi were attached at the ground end of the capacitor at Va, the feedback β would not be easily separated in the circuit. Similarly if Vi were attached at the ground end of the A part of the circuit, the feedback and open loop gain would not be easily found either.

 

[rezeew]

Yes, you are correct in considering the resistor and capacitor attached to the '+' terminal of the op amp in parallel. This is because they both contribute to the feedback loop and affect the gain of the oscillator. The resistor and capacitor connected directly to the output are also in parallel, but they are not part of the feedback loop and therefore do not contribute to the loop gain.

To find the loop gain, you would need to multiply the gain of the negative feedback part (1+R2/R1) by the transfer function of the bottom part of the oscillator, which includes the parallel combination of the resistor and capacitor connected to the '+' terminal.

In summary, to find the loop gain of an oscillator, you would need to consider all components that contribute to the feedback loop and affect the gain, including the resistor and capacitor attached to the '+' terminal of the op amp.