I am studying Wien Bridge Oscillator. The part I do not understand is how the other frequencies die out. For example, let us say initial disturbance was not of the resonant frequency but a higher frequency, then what mechanism ensures it dies out and similarly what is the explanation for lower frequency?


2 Answers 2


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

enter image description here

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 -

enter image description here

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.

  • \$\begingroup\$ Normally, we select (without any restrictions) R1=R2 and C1=C2.In this case, the RC bandpass attenuates the center frequency with a factor "1/3". Therefore, the opamp gain must be somewhat larger than "3". Thus, the loop gain is slightly larger than unity (Barkhausen´s oscillation condition). \$\endgroup\$
    – LvW
    Apr 20, 2014 at 20:52
  • \$\begingroup\$ @LvW I think you might find the attenuation is 2:1 not 3:1 - see my revised answer. \$\endgroup\$
    – Andy aka
    Apr 20, 2014 at 21:01
  • \$\begingroup\$ Andy - an error has occured in your calculation. The denominator must contain the expression s*(R1C1+R2C2+R1C2). Consequently, for equal components the attenuation is 1/3 and the required gain of the opamp is nominal "3" (in practice slightly larger,e.g. "3.2"). By the way, The gain of the circuit in your drawing has the correct value of "3". \$\endgroup\$
    – LvW
    Apr 21, 2014 at 8:31
  • \$\begingroup\$ @lvw darn it, I'll check the math and thanks for your diligence dude. \$\endgroup\$
    – Andy aka
    Apr 21, 2014 at 9:46
  • \$\begingroup\$ @LvW OK I get s(C1R1 + C1R2 + C2R2) which is slightly different to yours but thanks anyway. \$\endgroup\$
    – Andy aka
    Apr 21, 2014 at 10:31

Wien bridge oscillator basically works like this:

  • you have a somewhat wide band amplifier
  • you have a filter with phase response that zeroes at \$f_W\$
  • you have a circuit that can measure the output amplitude and tune the amplifier gain

You connect the filter input to the amplifier output, and the filter output to the amplifier input.

When you power the circuit the amplifier output ideally is zero, so the filter output is also zero and nothing happens. Actually this is a stable condition, i.e. if the output would be truly zero and noise would not exist, then your circuit would rest. But unfortunately noise exist, and in particular white thermal noise. Its power spectral density is quite flat from dc to very high frequencies (let's leave flicker out of this), so what happens? The filter "shapes" the noise it gets in its input, so at the filter output you will still have noise but phase will be zero only at \$f_W\$, so you get positive reaction only for that frequency. It will then amplify it, feed it to the filter that would be more than happy to feed it back to the amplifier, and so on. That should answer your question.

But that filter is not enough for your circuit to work. You need to respect the Barkhausen stability criterion:

  • the loop gain of the oscillator must be equal to 1 in absolute magnitude, \$|\beta A|=1\$
  • the phase shift must be zero or an integer multiple of \$2\pi\$

These condition must be met at your chosen frequency \$f_W\$.

The first one is met in this way: the system is designed to have a gain greater than one, usually three or something, and then the "gain control circuit" kicks in, limiting the gain when the circuit actually oscillates.

The second condition is met through the filter: at \$f_W\$ phase shift must be exactly \$n2\pi\$.

  • \$\begingroup\$ my question is how do other frequencies get attenuated \$\endgroup\$
    – happymath
    Apr 20, 2014 at 17:44
  • 1
    \$\begingroup\$ Well, the very very selective filter attenuates them... I don't really know how to put it in a simpler way. You might want to check how a filter works but I am sure you know that, you might want to solve the circuit of the feedback network of a wien oscillator and see that it actually filters out all the frequencies except a very narrow set. \$\endgroup\$ Apr 20, 2014 at 17:47
  • 1
    \$\begingroup\$ The WIEN bandpass is NOT a very selective filter. Just the opposte is true. However, that is not a problem at all because there is only one single frequency with zero phase shift - and that is the center frequency of the RC bandpass. That means: There is only one frequency which can fulfill Barkhausen´s condition. The loop gain at this frequency must be somewhat larger than unity at t=0 (switch-on). That means, the gain of the opamp must be somewhat larger than "3" (perhaps 3.2). Therefore, the oscillator can safely start oscillations. \$\endgroup\$
    – LvW
    Apr 20, 2014 at 20:45
  • \$\begingroup\$ @LvW thanks for pointing it out, you are perfectly right. \$\endgroup\$ Apr 20, 2014 at 21:05

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