# Wien-Bridge Oscillator - gain of RC network

I'm just wondering, I'm trying to find the overall gain of I understand where the gain of the non-inverting op amp config comes from. But where does the gain of the RC network come from? Since there is technically no input in this case?

EDIT: I tried the suggestion of V+ being the input, however I still can't seem to get the same answer that was given before in my lecture notes (that I've written out above): • Gain is Vou/V"+". The input for a RC network is opamp output. And the RC network output is at the point where we connect the noninverting input. – G36 May 5 '18 at 15:25
• THis is duplicated a hundred times already but positive feedback gain must be slightly more than negative feedback gain. electronics.stackexchange.com/search?q=user%3Ame+wien compute DC gain and ignore Caps for stability then choose C.. or read my 2nd answer in this link and see how to make it linear ( with soft limiting) and startup fast – Sunnyskyguy EE75 May 5 '18 at 15:40
• @G36 See the edit above. – AlfroJang80 May 5 '18 at 17:16
• @AlfroJang80 Vo is the input and V+ is the output so, the gain is Vo/V+ – G36 May 5 '18 at 21:40

## 1 Answer

For such a circuit (with positive and negative feedback) the oscillation condition is: Positive feedback factor kp (for the desired oscillation frequency) must be equal to the negative feedback factor kn.

The positive feedback factor for two equal R values and to equal C values is kp=1/3 for w=1/RC . Hence, for the circuit to oscillate we require also kn=1/3.

In reality (for a safe start of oscillation) it is common practice to use kn<1/3 (R2>2R1).

Comment: The IDEAL oscillation condition kp=kn results from the fact that - for linear opamp operation - the signal voltages at the positive and negative input terminals must be equal (opamp ideal). In practice, however (due to real opamp properties), we choose kp slightly larger than kn.

Hence, the oscillation amplitude (overfullfillment of the Barkhausen condition) will rise until it is limited (clipped) because of the finite supply voltages - unless an additional amplituide control mechanism is included (antiparallel diodes, for example).