Wein Bridge Oscillator

[Hero999]

Yes, but that could be useful for high gains at higher frequencies it could safely omitted.

 

[audioguru]

The old uA709 opamp needed two extenal compensation capacitors and a resistor. With a gain of 1 its output was a triangle wave at 28V p-p above only 3kHz. An old 741 opamp makes a good sine-wave up to only 9kHz and a newer TL07x opamp goes up to 100kHz.

 

[Hero999]

That's only when compensated, 28p-p is possible upto 200kHz when using a small compensation capacitors.
**broken link removed**

Uncompensated op-amps are still usefull on occasions and they're still made too.
https://focus.ti.com/lit/ds/symlink/tl070.pdf

 

[bananasiong]

Hi,
I couldn't understand, why the transfer function is 1/3?

Because the op amp's output impedance is low and its impedance at both inputs is very high, deriving the transfer function of the Wien network (the resistive divider made up of R1 and C1 on the top and R2 and C2 on the bottom) is relatively straightforward. Thus, the transfer function is: **broken link removed**
Where j represents a 90-degree phase shift in the transfer function (either positive or negative). The other (non "j") terms represent zero phase shift in the transfer function. As the magnitude of the terms change, the resultant phase shift also changes. Equation 1 shows that the "non-j" terms in the denominator equate to zero, leaving only j terms in the numerator and denominator.

How come the numerator is C1R2?
For the denomenator, 1 + ... is because of close loop gain?

Thanks

 

[audioguru]

The signal from the output of the opamp to its non-inverting input is reduced to 1/3rd by the cascaded highpass and lowpass filters. Then the opamp needs a closed loop gain of slightly more than 3 for the circuit to oscillate.

 

[bananasiong]

Yes, I got it. It is just mathematical calculation.
The voltage devider of R1C1 and R2C2 right?

Thanks

 

[bananasiong]

Hi,
Now I'm having problem to understand the Wien Bridge Oscillator with amplitude control. I've made the circuit as shown in the attachment. The 10.5kOhm resistor is a variable resistor to control the amplitude. At this resistance, the amplitude of the sinewave is very small.
My question is, before approaching 0.7V, the diodes are not conducted and hence, total feedback resistor is 20.5kOhm and the gain is more than 2. But when the amplitude of the waveform is larger than 0.7V, the diode in forward biased is conducted, so the total feedback resistor is only 10.5, the gain is less than 2. I thought it will stop oscillating.

Thanks

 

[audioguru]

A conducting diode has resistance. Its voltage drop is nearly always the same so its resistance changes with the amount of current flowing in it. With low current then its resistance is high so the gain of the opamp is high. With higher current then its resistance is lower so it clamps the max voltage swing of the output of the opamp.

The frequency-determining parts are resonant so the output level takes time to change. When it starts to become too small then the resistance of the diodes increase and causes the gain to increase so that the output begins increasing in level. The opposite occurs if the output level begins to increase too much.

The diodes cause the sine-wave output to have a flat top and a flat bottom. This creates harmonic distortion. Other methods of level control don't produce so much distortion.

 

[bananasiong]

The waveform is distorted only when the total feedback resistance is more than 28kOhm (my simulation result). When it is less than 28kOhm, it gives nice sinewave.

I need some time to digest the resistive of the diode..

 

[audioguru]

In a real circuit either the diodes do nothing and the sine-wave fades away to zero, or the sine-wave gets higher and higher until the sine-wave has a flat spot on its top and bottom. You won't be able to adjust it with no distortion and still allow enough gain for it to always start.

 

[Nigel Goodwin]

Use a small lighbulb to stabilise the oscillator, a common VERY high quality method, used in some VERY expensive commercial gear.

 

[audioguru]

Use a small lighbulb to stabilise the oscillator.

Yeah.
Bounce, bounce, bounce, bounce, bounce, bounce, bounce, bounce, bounce, ...

 

[Nigel Goodwin]

Yeah.
Bounce, bounce, bounce, bounce, bounce, bounce, bounce, bounce, bounce, ...

But VERY low distortion! - and other low distortion methods bounce as well, but cost a lot more to do.

 

[audioguru]

My Wien Bridge tone generator uses the best of both:
1) A light bulb for low distortion level control.
2) Some back-to-back LEDs in series across the feedback resistor to clamp the very high swings of the bouncing.
Only the first few cycles at a low frequency are clipped then it quickly settles to a fixed level.

I tried a FET transistor automatic gain control and even tried an optically isolated-FET but the distortion was too high or the output level was too low.

 

[bananasiong]

What is 'bounce' in electronics?
2 back to back LED's in series across the feedback resistor? Then the LED's will not be conducted? I can't figure out how are them connected.

Thanks

 

[audioguru]

A light bulb changes its resistance when its current changes. When it is in the negative feedback loop of the opamp of an oscillator then it controls the amplitude. But its reation time is slow (which allows very low distortion) so the amplitude bounces too high, then too low, then high again then low again for many bounces.

In my Wien Bridge oscillator I used LEDs instead of your diodes and I used two in series for a higher diode voltage. The LEDs stop the output amplitude from going too high during the bouncing so it doesn't bounce as much.

 

[bananasiong]

The resistance of the bulb keeps changing due to the AC signal?

 

[Nigel Goodwin]

The resistance of the bulb keeps changing due to the AC signal?

Yes, but only until it stabilises, this happens when you change it's frequency - but once it's stabilised it's a very low distortion oscillator.

 

[audioguru]

The resistance of the light bulb changes with changes in the average current in it from the signal. It changes slowly because it takes time for it to heat up and to cool down. If it changed quicker then it would cancel the signal at low frequencies.

 

[bananasiong]

The frequency is determined by the frequency selective network right? Then how to change the frequency?