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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}\$ ?

enter image description here

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    \$\begingroup\$ 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 \$\endgroup\$ – G36 Mar 31 at 16:29
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    \$\begingroup\$ What about Vin? electronics.stackexchange.com/questions/430912/… \$\endgroup\$ – G36 Mar 31 at 16:36
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    \$\begingroup\$ And your task is to find Vin at wich the voltage at non-inverting input is equal to 0V. Can you do it? \$\endgroup\$ – G36 Mar 31 at 16:37
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    \$\begingroup\$ 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. \$\endgroup\$ – G36 Mar 31 at 16:39
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    \$\begingroup\$ Yes, you can use the superposition. \$\endgroup\$ – G36 Mar 31 at 16:41
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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^+ <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.

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

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  • \$\begingroup\$ 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 \$\endgroup\$ – beccaboo Mar 31 at 17:00
  • \$\begingroup\$ @beccaboo oops I did! Fixed. \$\endgroup\$ – Andy aka Mar 31 at 18:17

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