I am trying to calculate the transfer function of a wien bridge oscillator but there is something wrong in my calculations.
In the picture below there are my results from this calcultation. I am using signal flow graphs
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1\$\begingroup\$ You never define \$z_1\$ and \$z_2\$, so I can't sort out what you're trying to do. In general, using signal flow graphs (or block diagrams) to analyze circuits needs to be approached very carefully because signal flow graphs assume that signals flow in one direction only -- in real circuits, each node is loaded by the components attached to it; that loading becomes a two-way communication that must be accommodated in the graph or block diagram. \$\endgroup\$– TimWescottCommented Sep 8, 2019 at 17:15
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\$\begingroup\$ A transfer function requires an input signal (\$TF=\large \frac{output}{input}\$). Normally, oscillators don't have input signals. \$\endgroup\$– ChuCommented Sep 8, 2019 at 18:40
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\$\begingroup\$ in the link from maxim they calculated it \$\endgroup\$– tairitCommented Sep 8, 2019 at 18:41
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\$\begingroup\$ They must be interpreting TF loosely. \$\endgroup\$– ChuCommented Sep 8, 2019 at 18:45
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1\$\begingroup\$ In a circuit like this one, you feed the non-inverting input with a portion of the output. You have to calculate the attenuation at the 0° phase frequency to exactly compensate for the attenuation and maintain oscillations. So yes, there is a transfer function linking the stimulus (the op-amp output) to the response (the voltage at NINV pin) that I calculated in the answer. \$\endgroup\$– Verbal KintCommented Sep 8, 2019 at 20:12
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1 Answer
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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 : )
answered Sep 8, 2019 at 17:20