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Given this circuit

schematic

simulate this circuit – Schematic created using CircuitLab

I am trying to prove that the opamp works in saturation region as instructed in the first answer to this question : How are positive and negative feedback of opamps so different? How to analyse a circuit where both are present?

So, we have

$$ V^- = V_{in} $$ $$ V^+ = \dfrac{R_1}{R_1+R_2}V_{out} $$ $$ V_{out} = A_v(V^+ − V^-) $$ $$ V_{out} = A_v(\dfrac{R_1}{R_1+R_2} V_{out} − V_{in}) $$ $$ \lim_{A_v\to\infty}\frac{V_{out}}{V_{in}} = \lim_{A_v\to\infty}\dfrac{Av}{Av \frac{R_1}{R_1+R_2} - 1} $$ $$ \lim_{A_v\to\infty}\frac{V_{out}}{V_{in}} = 1 + \frac{R_2}{R_1} $$ \$\frac{Vout}{Vin}\$ is finite, even though the feedback is positive! Why is the circuit working in saturation region instead of the linear one?

Am I missing something here?!!

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  • \$\begingroup\$ Could you please explain how you got the following equation: $$\lim_{A_v\to\infty}\frac{V_{out}}{V_{in}} = \lim_{A_v\to\infty}\dfrac{Av}{Av \frac{R_1}{R_1+R_2} - 1}$$ \$\endgroup\$ – Vivek Subramanian Jul 17 '18 at 17:06
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When you solve positive feedback circuits like this, you need some initial values.

We can say that \$V_{sat+}\$ as the upper limit to what the opamp can drive to and \$V_{sat-}\$ as the lower limit.

If we make an initial assumption that \$V_{out} = V_{sat+}\$ then you will get $$ V_+ = V_{sat+}\dfrac {R_1}{R_1+R_2} $$ $$ V_{out} = A_v(\dfrac{R_1}{R_1+R_2} V_{sat+} − V_{in})$$

When \$V_{in} < \dfrac{R_1}{R_1+R_2} V_{sat+}\$ the output will be \$V_{sat+}\$ When \$V_{in} > \dfrac{R_1}{R_1+R_2} V_{sat+}\$ the output will be \$V_{sat-}\$

You would do the exact same procedure with an initial assumption that \$V_{out} = V_{sat-}\$ to see when you would see a transition going in the opposite direction.

Check out page 7 of Opamp circuits - Comparitors and Positive Feedback

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    \$\begingroup\$ Yes you're right, and I am familiar with this way of seeing things.. What I am wondering is, what is wrong with the equations above? \$\endgroup\$ – ielyamani Dec 5 '14 at 4:09
  • \$\begingroup\$ Bumped to top, I'm interested in knowing what's wrong too as I thought this method worked everytime because the equation was actually describing the internal working of the amplifier (happily surprised to see my initial post was useful to you as well ;) ). \$\endgroup\$ – Mister Mystère Dec 13 '14 at 20:35
  • \$\begingroup\$ It must have something to do with Vout/Vin being correct only IF Vout is not restricted to +Vsat/-Vsat (which is impossible) but I'd like to see a proper demonstration because it's only intuitive for now. \$\endgroup\$ – Mister Mystère Dec 13 '14 at 20:36
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The answer is positive feedback and the noise always tends to force the amplifier into saturation. Assume \$V_{in} = 0\$, at power on, your output \$V_{out}\$ is zero. Any input disturbance that might try to force \$V_{out}\$ away from zero will elicit opposite response. The positive feedback is in the same direction as the perturbation, tending to reinforce it. This will drive the amplifier into saturation.


Update:

Actually, there are some problem in your third equation. You've assumed the amplifier working in the "linear region" already.

If

$$ V_{out} = A_v(V^+ − V^-) $$

When \$A_{v}\$ goes to \$\infty\$, by positive feedback, the input should to \$\infty\$ too, \$V_{out}\$ should be infinite then, then you'll blow out the universe. If you want the output finite, and have a infinite \$A_{v}\$, your input should tend to zero, this is just negative feedback does.

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  • \$\begingroup\$ The first paragraph in your much appreciated answer means that the amplifier would start working in one of the saturation regions. As for the third equation, I believe it's always true: if V+ is different from V- then the opamp works in saturation, if not then it's a constant(infinity multiplied by zero) \$\endgroup\$ – ielyamani Dec 5 '14 at 4:34
  • \$\begingroup\$ Working not mean there must exist the linear relation between input and output. \$\endgroup\$ – diverger Dec 5 '14 at 4:37
  • \$\begingroup\$ I am afraid the third equation is always true for an ideal op amp \$\endgroup\$ – ielyamani Dec 5 '14 at 4:46

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