# Noninverting schmitt trigger circuit LTP, UTP calculation

I knew very well how to work negative feedback using virtual short but this is the first time encountering positive feedback. I think the feedback voltage here is $$\\frac{R_1}{R_1+R_2}V_{out}\$$. If so, this must be the reference voltage too. How is my textbook getting just $$\\frac{R_1}{R_2}\$$ ? • Ask yourself, a question for which Vin the voltage at the non-inverting input is equal to 0V. Consider two cases Vout at +Vsat and -Vsat – G36 Mar 31 '20 at 16:29
• What about Vin? electronics.stackexchange.com/questions/430912/… – G36 Mar 31 '20 at 16:36
• And your task is to find Vin at wich the voltage at non-inverting input is equal to 0V. Can you do it? – G36 Mar 31 '20 at 16:37
• Simply, the Vout will be fixed at +Vsat or -Vsat until the voltage at non-inverting input reaches 0V. So you need to find Vin that "gives" 0V at non-inverting input when Vout is at +Vsat and -Vsat. – G36 Mar 31 '20 at 16:39
• Yes, you can use the superposition. – G36 Mar 31 '20 at 16:41

## 2 Answers

The voltage at the $$\V^+\$$ is actually,

$$V^+=V_i\dfrac{R_2}{R_1+R_2}+V_o\dfrac{R_1}{R_1+R_2}$$

You can find that using superposition, for example. So if $$\V_i\$$ is indeed 0, you end up with what you've shown so far.

Now here is the thing, if doesn't take much to saturate the OP-Amp output to either +VSAT or -VSAT. Under negative feedback, the OP-Amp is forced to work under the linear region, not the case under positive feedback. So if $$\V^+ > V^-\$$ , the output is +VSAT, if $$\V^+ < V^-\$$, the output is -VSAT.

The trick here is that you have to assume an intial state for the output, either +VSAT or -VSAT. Say, it's +VSAT, then at the $$\V^+\$$ node, you'll have:

$$V^+=V_i\dfrac{R_2}{R_1+R_2}+V_{SAT}\dfrac{R_1}{R_1+R_2}$$

Since the $$\V^-\$$ node is fixed at ground, the OP-Amp output will remain at +VSAT so long as:

$$V_i\dfrac{R_2}{R_1+R_2}+V_{SAT}\dfrac{R_1}{R_1+R_2}>0$$

$$V_i\dfrac{R_2}{R_1+R_2}>-V_{SAT}\dfrac{R_1}{R_1+R_2}$$ $$V_iR_2>-V_{SAT}R_1$$ $$V_i>-V_{SAT}\dfrac{R_1}{R_2}$$

So, as long as $$\V_i\$$ stays above that, the output will remain at +VSAT. If $$\V_i\$$ starts to decrease such that it doesn't meet the condition above, then it means that $$\V^+ and the output will switch over to -VSAT. You can follow the same procedure as above to find what the threshold would be to get it back to +VSAT.

• So clear and elegantly explained love how you applied the superposition here to work the combined voltage at $+$ terminal. Thank you so so much:)) – across Mar 31 '20 at 16:58

Notable things for the lower threshold point: -

• The current through $$\R_1\$$ has to equal the current through $$\R_2\$$ (always)
• The actual threshold point is 0 volts because $$\v_{in-}\$$ is at 0 volts
• Current through $$\R_1\$$ at the actual threshold point is $$\V_{IN}/R_1\$$
• Current through $$\R_2\$$ at the actual threshold point is $$\-V_{SAT}/R_2\$$

Hence, because these two currents are equal, $$\V_{IN} = -V_{SAT}\dfrac{R_1}{R_2}\$$

• Oh I can use most of the negative feedback tricks here too! XD Also I'm pretty sure at second point you mean $V_{IN\color{red}{-}}$ is at $0$ volts – across Mar 31 '20 at 17:00
• @beccaboo oops I did! Fixed. – Andy aka Mar 31 '20 at 18:17