Finding the loop gain of an oscillator

Sinister

1. The problem statement, all variables and given/known data
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.



2. Relevant equations

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

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

3. 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

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.

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