Given this circuit


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?!!

  • \$\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

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

  • 1
    \$\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

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.


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


$$ V_{out} = A_v(V^+ − V^-) $$

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.

  • \$\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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