Electrical – Wien Bridge Oscillator Resonant Frequency cannot reached!

Even though this question may be solved, I ran into this sort of problem when trying to simulate the Wien bridge oscillator.

Using LTSPICE:

1. 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.
2. 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:

1. Step 1 above applies.
2. Proteus simulation software does not require this initial 'push' and thus it may be easier to use Proteus software.
3. Make sure to use active capacitors while simulating, otherwise nothing will happen.

There seems to be some confusion about this relatively straight forward circuit: -

The components C1, R1, C2 and R2 are not really a filter that let one frequency through. They don't selectively filter a single frequency and attenuate all the others. Here's the response of the network. I've called the input to the non-inverting input "Vin" and the output from the op-amp "Vout": -

\$\dfrac{V_{IN}}{V_{OUT}} = \dfrac{sC_1 R_2}{s^2(C_1 C_2 R_1 R_2) + s(C_1 R_2 + C_1 R_1 + C_2 R_2)+1}\$

In other words it's a bandpass filter with not very high Q so, in fact, it lets a load of frequencies thru around it's optimal tuning point and that optimal tuning point is when the \$s^2\$ terms = the terms not associated with s i.e.

\$\omega_0 = \dfrac{1}{\sqrt{C_1 C_2 R_1 R_2}}\$

At this point, there is no phase shift between output and input of the op-amp - this is the strictest condition for oscillation to both start and remain stable. This is the only frequency that the Wein bridge will oscillate at because phase shift, front-to-back has to be zero (not 1 degree or 0.1 degree but zero degrees, always). Why is there no phase shift? Because the \$s^2\$ terms are cancelled by the terms not associated with s leaving: -

\$\dfrac{V_{IN}}{V_{OUT}} = \dfrac{sC_1 R_2}{s(C_1 R_2 + C_1 R_1 + C_2 R_2)}\$ - clearly s cancels top and bottom and the phase shift is zero.

What about attenuation? Usually both Cs are made equal and ditto the R terms. This leaves: -

\$\dfrac{V_{IN}}{V_{OUT}} = \dfrac{C R}{(C R + C R + C R)}\$ = 0.3333. In other words a 9.5dB attenuation.

This is countered by R4 and R3 providing 9.5dB of gain. That's the theory but in fact you need a bit more gain to get this circuit kick-started into action then you get the problem that the gain is too high. To counter this non-linear elements are used to control the gain like a lamp:L -

On power-up the lamp is cold and therefore its resistance is lower then when it starts to warm. This gives the circuit the gain lift it needs and the output from the op-amp starts to produce a rapidly rising sinewave. This in turn, warms the small lamp and its resistance increases stabilizing the gain at just the right amount for controlled amplitude oscillation.