[SOLVED] Wein Bridge Oscillator Clipping

[Chewie RFC]

I've built a Wein Bridge Oscillator using a 741 Op Amp with the following parameters:

RF = 20k
RG = 10k
R = 16k
C = 10n
VCC = +5V
VEE = -5V

When I view the waveform on my scope, the positive swing of the sine wave has no noticeable distortion. The negative swing is clipped about 0.8V or so less than the positive swing. The positive swing peak is about 4V, the negative swing about -3.1 to -3.2V.

Any ideas?

 

[LvW]

Ideas about what? Why don`t you try to ask a clear question?
Are you concerned about clipping at all? Or about the unsymmetry of clipping?
(By the way: The inventor of the bandpass that gave the name to the oscillator was WIEN; but I know - even in some textbooks this name is written incorrect).

 

[Chewie RFC]

Why the unsymmetrical clipping? I know the nature of the WIEN bridge oscillator is to drive the output to the rails, or very near them, but why would the negative swing be clipped and not the positive swing? I haven't built a WIEN bridge in years and I have yet to find a good answer, or any answer at all.

 

[LvW]

At first, each RC oscillator drives the output to the rails (with clipping), unless some non-linear circuitry is provided to limit the amplitude. This is the nature of a harmonic oscillator.
The symmetry of the clipping effect is determined by two parameters:
* Bias point (ideal: zero volts for split supply),
* symmetry of the opamp dynamic range - determined by the internal opamp circuitry. Most probably, this is the reason for the observed effect (in particular for the 741 type opamp powerde by only +/-5 volts).

 

[Chewie RFC]

Ok, that makes sense to me. Thanks for the clarification, that helps.

 

[Kral]

See Wien bridge oscillator - Wikipedia, the free encyclopedia

 

[Chewie RFC]

I upped the voltage on the power supply to 12-15V and saw symmetrical clipping, which makes sense. Why I didn't do that before... I don't know. Thanks for setting me straight.