# How can I set a frequency in a Wien bridge oscillator

Discussion in 'General Electronics Discussion' started by tburn, Apr 30, 2014.

1. ### tburn

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Apr 30, 2014
Hi all,
Im new here So Im designing a simple Wien bridge oscillator, and need it to be set to a specific frequency.

This is my circuit: Where C1 = C and C2 = C
and R3 = R and R4 = R

Im using ω=1/(C2R2)
where C = my capacitor values and R = my Resistor values.

My problem is, however, how do I get it to equal ω?
Should I just set either C or R to a random value, and then work out the other?

Many thanks,
Tommy

2. ### Ratch

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Mar 10, 2013
Nope, you cannot design a circuit helter-skelter like that. Since you have a inverting amplifier, you have to figure out what frequency the positive terminal of the op amp will be at 180° with respect to Vout. The total of the phase shift of 180° from the op amp inversion and the 180° from the phase shift = 360° and the circuit will oscillate if the gain is high enough. Gain is set by R1 and R2. Can you figure that out? Did you Google for Wien bridges before you asked here?

Ratch

3. ### tburn

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Apr 30, 2014
Ah I see. I forgot about the phase difference. My problem is that all the sources assume that you know what values you are using for the RC network, and show you how to find the frequency. Im having difficulty reverse-engineering this.
Thanks again,
Tommy

4. ### duke37

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Jan 9, 2011
C1, C2 and R3, R4 set the frequency.
To vary the frequency two components need to be altered. Normally this is done with a dual potentiometer varying R3 and R4 together.

5. ### LvW

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Apr 12, 2014
The WIEN oscillator consists of a NON-INVERTING amplifier and a simple RC-bandpass.
It is easy to show that the center frequency of the bandpass (with zero phase shift) is at wo=1/RC.
This zero phase shift is the reason we need a non-inv. amplifier with a nominal gain of "3" because the attenuation of the bandpass at wo is 1/3.
Thus, we arrive at a (nominal) loop gain of unity with zero phase shift (oscillation condition).
In practice, for a safe start of oscillation, we need a gain slightly larger than "3" (e.g. 3.2 or so).

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6. ### tburn

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Apr 30, 2014
Thanks, That makes sense. But what Im having trouble with is setting the values of the RC circuit so i gane achieve a particular frequency.
Many thanks.

7. ### LvW

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Apr 12, 2014
What is the problem?
When 2*Pi*Fo=1/RC you can choose either R or C as you like and compute the remaining value correspondingly.
However, as mentioned already by duke37, the disadvantage of the WIEN-type oscillator is that you have to vary two elements (R or C) at the same time (because the equations as given apply for R1=R2=R and C1=C2=C only).

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8. ### duke37

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769
Jan 9, 2011
You have given the equation 2*pi*f = 1/(C2.R2)
So f = 1/(2*pi*C2*R2)
Chose convenient capacitors and calculate the resistors. Resistors between 10k and 100k would be best.

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9. ### Ratch

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Mar 10, 2013
If you use two op amps, then only one variable resistor is necessary.

Ratch

10. ### tburn

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Apr 30, 2014
So I do just pick one at random and then calculate the other? because Ratch (2nd post) said that this wasnt the way to do it...
Many Thanks

11. ### Ratch

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Mar 10, 2013
I can see why you don't want to listen to old tired burn-outs like us. So why don't you let a sweet young thing like Jeri tell you how to do it without using any lame parts you might have.

Ratch

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12. ### tburn

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Apr 30, 2014
Thanks lol! I appreciate all the help. Youve all put me on the right track and it's working beautifully now I did wonder though, I presumed I'd be able to put 2 resistors in series to get a better approximation of the required value from the available store of esistors as they dont affect the phase, but i wanted 20.2k resistors so I used a 20k and a 200ohm resistor in series for each of the R3 and R4 values but it seems to kill the amplitude. Am I missing something here?
thanks! :-D

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13. ### Ratch

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Mar 10, 2013
Those resistors DO affect the phase, otherwise they would not be in the frequency equation. The gain requirements are different if you don't use the same values of resistors in the phase shift network. Did you Google and did you read post #9?

Ratch

14. ### davennModerator

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Sep 5, 2009
ahhh yes
I learnt a long time ago that a thermistor or light bulb was needed in the R2 position

Dave

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15. ### tburn

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Apr 30, 2014
Thanks Ratch, I did, but what Im saying is that Im using resistors in series to make up the required frequency. (I need 22K2 so Im using 22k in series with 200 ohms) I actually just realised that The simulator seems to give me a low gain but if I detach it, short circuit the lower value resistor and then re-attach it it seems to work... It might just need made in practise to see if it'll actually work.

16. ### Ratch

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Mar 10, 2013
Match the resistors and caps in the phase shift network as best you can. Then adjust the feedback R1 and R2 to get it oscillating. Then adjust R3 and R4 together keeping the same value to get the right freqency.

Ratch

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17. ### duke37

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Jan 9, 2011
In one simulator that I have used, (forget which one), the oscillator had to be given a kick to start it.
You need a gain of three, less than this and nothing happens. More than this and you get distortion hence Dave's suggestion of gain control with a variable component depending on signal amplitude.

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18. ### tburn

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Apr 30, 2014
When you say match the resistors and capacitors what do you mean? as in, I have 20.2k resistors and 3.3nF capacitors. This gives me the correct frequency. Would I be better off with different values that give the same frequency?

19. ### Ratch

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Mar 10, 2013
No, you appear to be OK. I thought you said something was not working right. You do need to set up an automatic gain control (AGC) like a light bulb so you don't have to constantly fiddle with the feedback resistors.

Ratch

20. ### tburn

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Apr 30, 2014
Ah ok Then That's all my questions answered. Thank you All for all your help. Youve been brilliant!! :-D
Really appreciate it1
Tommy  