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.

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You don't need a quadraphonic Blaupunkt -- you need a curve
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ahh i see i see..thank you very much.
so the conditions for oscillation are...A=3 and w=1/RC..

however, when u said the poles are on the imaginary axis, all in understand is that when w=1/RC the Real part gets cancelled out leaving the imaginary part behind..


however, i do not understand from this part onwards..
do you mean the right half of the cartesian graph? and if it is on the right side..why does positive feedback increase?

thank 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.

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You don't need a quadraphonic Blaupunkt -- you need a curve
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yes that guess would be correct...they do not teach the s plane at all..to be honest with you ive never heard of it *sigh*...im really tired of not being able to understand a concept to the fullest of its essence...been trying my best to self-learn though ..altou it takes a longer time...but its definitely catalyzed by people like you=) thank you so much for your help=)

however i have just one last little question...if the loop back gain is >1 that means it is on the right side of the s plane deeming it unstable..

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.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
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i shall keep a mental note of that for future study...
once again thank you so much for everything BrownOut =)

 

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.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

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

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

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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.

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By Brownout:
Hey thanks =)


By audioguru:

is it correct to say when the signals are small, the diodes are not conducting and the gain is higher than 3..when the signals are big, the diodes conduct (the resistance parallel to the diode bridge is reduced)..thus making the gain exactly 3...so is the purpose of the diode bridge to reduce the gain back to 3 after oscillations have occured to ensure a smooth sine wave?

based on that, should the values of the resistances at the voltage divider resistor network be set to the values as attached in the circuit? (R3,R4 and R6)..the parallel resistance between the diodes is the extra resistance that increases the gain above 3?

however when i conducted simulation tests ...
R3=10k R4=20k R6=5k
gives me a sine wave with an amplitude of +/- 15V (My Vcc and -Vcc are +-15V) and the edges of the sine wave are clipped (but not as distorted as when no diodes are used)
(Attached as clipped.png)


but when
R3=10k R4=15k and R6=10k
i get a sine wave of reduced amplitude at +- 3.3V ..however the edges of the sine wave are NOT clipped.
(attached as unclipped.png)


ive got a feeling that the first configuration R3=10k R4=20k R6=5k is the correct way but would like to understand why the voltage amplitude is reduced so low when the 2nd configuration of R3=10k R4=15k and R6=10k is used.

Attached Thumbnails

     

 

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.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

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.

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I will NOT answer PM's requesting technical help, please use the Forum
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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=)

Attached Thumbnails

 

 

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

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Eric " Good enough is Perfect "
I will NOT answer PM's requesting technical help, please use the Forum
PIC tutorials: Nigel's www.winpicprog.co.uk/ Bill's: www.blueroomelectronics.com/

 

It seems that you don't know how to properly "set" the simulation.
Here you have two examples:







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