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What I understand of an astable multivibrator so far is this:

  • The output saturates to one of two extremes, which drives the capacitor in the negative feedback loop to either charge or discharge.
  • Once there is a discrepancy between the inverting and non inverting terminal voltages (|v+| < |v-|), the positive feedback amplifies that difference and pushes the output to the opposite of the two extremes mentioned above.

Astable Multivibrator using an op amp

For example, if the output was initially at +Vsat, the capacitor would charge until it got to +beta * Vsat. Once it exceeded this upper threshold voltage, however, v- would be greater than v+, which would drive the op amp to negative saturation (-Vsat). The process would repeat as the capacitor discharges from +beta * Vsat to -beta * Vsat; beta here is R1/(R1+R2).

My question is as follows. During the portion of the time that |v-| > |v+| (either during capacitor charge or discharge), why is the negative feedback not coming into play and quickly equalizing the two voltages? Can we assume that the positive feedback of the schmitt trigger works faster than the negative feedback involving a resistor and a capacitor? Is there some sort of a "battle" between the two feedbacks and one ends up "winning" eventually?

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    \$\begingroup\$ Exactly! You have found the answer to your question yourself. All you have to do is find out why the positive feedback is faster than the negative one ... or, in other words, what slows down the negative feedback... \$\endgroup\$ May 1 at 6:59
  • \$\begingroup\$ There is a real "battle" between the two feedbacks in the weird circuit of negative impedance converter (NIC) where the negative feedback always "wins". But there they are both fast. See my Wikibooks story. \$\endgroup\$ May 1 at 8:44
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why is the negative feedback not coming into play and quickly equalizing the two voltages? Can we assume that the positive feedback of the schmitt trigger works faster than the negative feedback involving a resistor and a capacitor?

Yes.

The voltage at (-) terminal can rise and fall only at a speed with which the capacitor can charge/discharge. That rate is limited by the opamp saturation voltage and the resistor Rf. It has time constant Rf x C.

The positive feedback through the two resistors R1 and R2 are nearly instantaneous.

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  • \$\begingroup\$ This makes so much sense, thanks a lot! \$\endgroup\$ May 2 at 12:32
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In the simplest analysis consider this: -

enter image description here

Vn is the average of Vp because Rf and C form a low pass filter.

Not until Vn catches up with Vp can negative feedback have any instantaneous effect. Then, when that occurs, the output flips and Vn is playing catch up with Vp once more but, with the polarities reversed.

Negative feedback ensures that the average value of Vn equals the average value of Vp. That's all negative feedback does here.

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  • \$\begingroup\$ Right observation... This is how all bistable (temperature, pressure, etc.) regulators work - the controlled quantity wiggles between the two thresholds so its average value is equal to the middle value of the hysteresis region. \$\endgroup\$ May 1 at 10:52

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