Why does a Schmitt trigger work in saturation region?

Question

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?!!

Answer 1

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

Answer 2

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.