# Why does a Schmitt trigger work in saturation region?

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?!!

• Could you please explain how you got the following equation: $$\lim_{A_v\to\infty}\frac{V_{out}}{V_{in}} = \lim_{A_v\to\infty}\dfrac{Av}{Av \frac{R_1}{R_1+R_2} - 1}$$ – 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

• 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? – ielyamani Dec 5 '14 at 4:09
• 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 ;) ). – Mister Mystère Dec 13 '14 at 20:35
• 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. – 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.

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.

• 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) – ielyamani Dec 5 '14 at 4:34
• Working not mean there must exist the linear relation between input and output. – diverger Dec 5 '14 at 4:37
• I am afraid the third equation is always true for an ideal op amp – ielyamani Dec 5 '14 at 4:46