i dont understand how the gain, A is connected into the transfer function equation as well..

You've already written the equation: Av(closed loop) = A/1 + A*B. I just showed the math for the denominator of the equation, because that determines the poles, and I was hoping that would give you some insight into why the oscillator works. Sorry if I confused you.

I'm guessing you haven't studied s-plane graphs? Basically, a feedback system is stable if the poles are in the left half plane, and unstable in the right-half plane. Without that background, it suffices to say that the loop gain of the oscillator must be >= 1. And I think you already know how to determine that.

If you've studied 2nd order differential equations, you can apply that to see that the solution is a growing exponential function when the real parts of the poles are positive. That would be identical to what I've been saying about right half plane.

Originally Posted by qtommer

therefore is it correct to say that if the loop gain is less than 1, it would be on the left side of the s plane and becomes a stable op amp thus resulting in attenuations in the circuit?

The position of the roots on the left side affect the response of the amplifier. You can have a damped system or undamped, depending on exactly where the poles lie. However, you're right that the system would be stable.

Thanks for your questions. You have me thinking about electronics again. I haven't had much opportunity to do that since I was laid off some months ago.

Hey qtommer, check this out: Exploring the s-Plane

Play with the interactive s-plane. It's pretty cool

The lowpass filter reduces the signal to 1/1.5 times plus the highpass filter reduces the signal to 1/1.5 times so the total loss is 1/3 times. Then the opamp needs a gain of slightly more than 3 times to oscillate. At the oscillation frequency the phase differences of the filters cancel and the opamp also has no phase shift.

A diode clipper clips the signal to limit the amplitude. A light bulb (it changes its resistance with the amplitude of the signal across it) or a FET amplitude stabilizer circuit do not clip the signal.

The gain is always 3. It takes a little time for the oscillations to build up to the clipping value, but it takes a gain of 3 or more to even get the oscillations started. The reason your circuit worked with R3 = 10k, R4= 15K and R6=5K is because the gain equation for your circuit is:

1 + (R4 + R6/R3); = 3. Remember as oscillator is an infinite gain amplifier ( remember how the denominator of the loop gain disappeared; meaning infinite gain ) So, something has to intervene to reduce the gain or else the signal will continue growing and growing and growing...

In this case, when the signal reaches a cetain value, the diodes short out R6, and now the gain is:

1 + (15K + 0/10K) = 2.5. that will stop the signal from growing. In that way, the tops of the signal aren't clipped by hitting the power rail.

Originally Posted by qtommer

Does this mean that the purpose of amplitude stabilization is NOT to ensure that the sine wave edges are distorted (clipped) but it IS to ensure that a sine wave with constant amplitude is generated right from the start without waiting for it to have to grow?

Thank you all so much

hi,
The purpose of the amp stabilisation is to keep the gain such the sinewave output is as 'pure' as possible [ look up form factor]

The sinewave will increase in amplitude as the gain increases after switch ON, then the amp stab should start to work to maintain a good sine wave.

If you have a clipped sinewave output, the gain stab components need adjustment in order to bring the gain down by a small amount.

Originally Posted by qtommer

thank you so much Brownout and eric=)

ive a better understanding now thank you so much...

now im simulating the wien bridge without amplitude stabilization (ive removed the diodes).. when i set the gain to exactly 3, i know that there will be no oscillatons and there will be a flat line at 0V.

however, when i zoomed into the flat line i noticed a rather odd repetitive waveform (as attached) is there any reason why the shape is so?

thank you all once again=)

hi,
Again that looks like sim artefact, the period is about 1nSec.!!!