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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}\$ ?

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across
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    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
  • If Vout is +Vsat, then the noninverting input will see \$\frac{R_1}{R_1+R_2}V_{sat}\$ when \$V_{in}=0\$ – across Mar 31 '20 at 16:35
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    What about Vin? https://electronics.stackexchange.com/questions/430912/what-is-the-purpose-of-a-resistor-in-parallel-with-a-buffer-gate/431076#431076 – G36 Mar 31 '20 at 16:36
  • That positive feedback is throwing me off. I'm feeling helpless as I can't use the virtual ground thingy here :( – across Mar 31 '20 at 16:37
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    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
  • I'll give it a try, but really I'm a bit clueless as my textbook just gave formula without explanation – across Mar 31 '20 at 16:37
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    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
  • Ah ok I think I can use superposition here. The voltage seen at the \$+\$ terminal of opamp due to \$V_{in}\$ would be \$\frac{R_2}{R_1+R_2}V_{in}\$ – across Mar 31 '20 at 16:41
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    Yes, you can use the superposition. – G36 Mar 31 '20 at 16:41
  • Let's say to start with \$V_{out} = -V_{sat}\$, then the feedback voltage at \$+\$ terminal is \$\frac{R_1}{R_1+R_2}(-V_{sat})\$. When the input voltage component at \$+\$ terminal equals \$-ve\$ of this feeback voltage, the output changes its state. That's it? XD – across Mar 31 '20 at 16:43
  • \$\frac{R_2}{R_1+R_2}V_{in} = -\frac{R_1}{R_1+R_2}(-V_{sat}) \Rightarrow V_{in} = +\frac{R_1}{R_2}V_{sat}\$ – across Mar 31 '20 at 16:45
  • Looks I get itXD you're awesome and an excellant teacher! Thank you so much:) – across Mar 31 '20 at 16:46
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    I'm glad that I could help. – G36 Mar 31 '20 at 16:48
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    And remember that the transitions will occur when V"+" = "-"V. https://electronics.stackexchange.com/questions/465430/waveform-at-the-negative-terminal-of-an-opamp/465585#465585 – G36 Mar 31 '20 at 16:50

2 Answers2

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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.

Big6
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  • 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
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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}\$

Andy aka
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