Even though this question may be solved, I ran into this sort of problem when trying to simulate the Wien bridge oscillator.
Using LTSPICE:
- The voltage gain of the amplifier circuit MUST be equal too or greater than 3 for oscillations to start because the input is 1/3 of the output. This value, ( Av ≥ 3 ) is set by the feedback resistor network.
- Since oscillators start up from thermal noise you may need to provide an initial pulse input for the simulation to get going.
The following steps enabled me to successfully simulate this oscillator without an initial pulse input on Proteus Software.
Using Proteus:
- Step 1 above applies.
- Proteus simulation software does not require this initial 'push' and thus it may be easier to use Proteus software.
- Make sure to use active capacitors while simulating, otherwise nothing will happen.
The gain you measure depends on the amplitude of the signal in this circuit. This is the function of the diodes.
For example, if the output signal amplitude is small (say less than 0.6 V), then neither of the diodes across the resistor (R6 ? -- it's hidden) will conduct, and your gain will be 4x (with the 10k resistors). When the amplitude exceeds about 2.4 V peak, then at the extremes of the sinusoidal output, the diodes will conduct, and effectively limit the voltage across R6 to about 0.6 V. This reduces the gain beyond this amplitude to 3x (the other resistors control it).
If you were to draw a graph, the gain (slope of VOUT vs. VIN) would be 4x until the peaks reached 2.4 V, then would reduce to 3x for the portion of signal beyond that. Because gain changes with signal amplitude, you would get distortion.
This is the exact purpose of the diodes in this circuit -- they allow the gain to change, and if you get the resistors (and other components) right, for small signal amplitudes the overall lop gain will be just greater than 1, and oscillations will build up. However, when the amplitude reaches about 2.4 V, the gain will fall, and the amplitude will stabilize around where the 'average' gain is 1.0
Best Answer
To determine the transfer function of this Wien-bridge oscillator, you can try the fast analytical circuits techniques or FACTs. This is the documented problem number 9 actually.
The principle is quite simple: you consider the transfer function denominator as a combination of the circuit time constants determined when the stimulus is turned off. Basically, the exercise consists of temporarily disconnected a capacitor (or an inductor) and "look" through its connecting terminals to determine the resistance driving the capacitor. For a circuit like this one, you can determine the denominator and the numerator without writing a single line of algebra. The basic circuit to look at is this one:
The stimulus in your circuit is the op-amp output while the response is the voltage across the grounded capacitor. You will then calculate the necessary gain the non-inverting gain will have to exhibit to exactly compensate the attenuation of the filter at the oscillation frequency.
If you do the maths ok, then you should end-up with a low-entropy transfer function arranged in the following way:
\$H(s)=H_{res}\frac{1}{1+Q(\frac{\omega_0}{s}+\frac{s}{\omega_0})}\$
which is the transfer function of a band-pass filter. In this expression, \$H_{res}\$ is the attenuation to be compensated by \$R_f\$ and \$R_i\$ to ensure sustained oscillations.
The complete Mathcad sheet is given below and shows how the expression perfectly matches the one obtained from brute-force algebra:
You can look at an introductory seminar taught at APEC 2016 which smoothly shows how FACTs work. When you've tried them, there is no turning back : )