wein bridge oscillator

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

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

Hey thanks =)


By audioguru:
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.
Click to expand...


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.
 

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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.
 
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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.
 
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=)
 

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hi,
Again that looks like sim artefact, the period is about 1nSec.!!!
 
The diode clipper clips the signal the same as the same circuit in a guitar fuzz effects circuit. But if there is a resistor in series with the diodes then the clipping distortion is reduced.
 
It seems that you don't know how to properly "set" the simulation.
Here you have two examples:
Click to expand...

thank you for your reply Jony130 as well as helping teach me simulate. if you are referring to the previous post where i had the gain set at 3 for the circuit without amp stab, i purposely set it to those parameters to have a closer view..under normal parameters, i get an overall flat line..which is correctly represented when the gain is exactly 3 returning no oscillations. am i having the correct concept? i stand corrected..other than that the parameters on my time transient analysis are identical to yours.=)

referring to your second circuit with the diode bridge. i would like to ask why resistances of R3,R4 and R6 are assigned those values of 10k, 4.7k and 2.35k (VR set to 0.5) respectively.From the gain formula 1+R2/R1..after the diodes get turned on and R5 is bypassed, wouldnt the gain drop to 2.41K? or is there some little resistance still available from the diode bridge resistor(R5)?
 
with those values of resistances, i just realized that the pk to pk voltage of that circuit is reduced to approx +- 3V although the power rail is +-15V..ive asked that question too earlier ...is the lowering of amplitude down to +-3V something intentional? ( i do get a really nice sine wave) or should the output sine wave oscillate from pk-pk according to the power rail values?
 
You don't want rail-to-rail oscillations. You're getting a good sine wave because you're intentionally not going to the rail.
 
You don't want rail-to-rail oscillations. You're getting a good sine wave because you're intentionally not going to the rail.
Click to expand...

ahh i see i see...thank you so much!!!

as for the clipping form 15V to appprox 3V .how does the diodes clip so much voltage when it only has 0.7V drop?
when the diodes start conducting, is the resistance parallel to it still "resisting" or is it totally bypassed?
 
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.
Click to expand...

sorry for the oversight.
when the gain becomes 2.5. the gain is no longer unity thus not providing oscillatory conditions..however, the loop gain becomes less than 1 (2.5*(1/3))
=0.83



so when the signal starts to get big and the diodes short out the resistor, it becomes a damped system making the amplitudes to be clipped down?
 
=D
i love those too! haha...once again im indebted to you.=)
cant believe that someone with your expertise got laid off...anyway its their loss..
thank you once again!!!=)
 
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