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In the following circuit i want to find the output voltage Vo related to the input Voltage Vi. Diodes D1 and D2 are the same and have a 0,7V drop when they conduct. Op-amp is ideal too. For the zener diode we know that: $$Vzener=8,6V$$

Well, when our input is positive (Vi>0) , the output Vo is negative(Vo<0), so Zener diode conducts but both D1 and D2 are reversed biased,so the circuit works as an inverting operational amplifer with Gain=-Rf/Rin=-10K/1K=-10. So Vo=-10Vi.

I am a bit confused with the other case about the negative input(Vi<0). Then the output will be positive(Vi>0). Will this make the zener diode conduct in the opposite direction and D1,D2 forward biased as well, so that we have another path from the output to the inverting input?

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When the input is negative (and small) the diodes and zener can be ignored.

As the input signal gets bigger negatively you will reach a point when the output becomes +(8.6 V + 2*0.7V = 10 volts). Prior to this you have a gain of -10k/1k and at (or above) this point you have full negative feedback and the circuit has zero gain hence, the output limits at 10 volts. It can do nothing else theoretically.

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  • \$\begingroup\$ I'm not sure if i got it. Well i guess my thought about the positive input is correct. I mean that when input is positive(Vin>0), then output is negative(Vo<0), so Diode zener is forward biased but D1 and D2 are reversed biased, am i right? So we ignore this path. On the other hand, for small values of negative input(before Vo reaches 10V) none of the three diodes conduct. But after Vo gets bigger negatively than -1V, all 3 diodes conduct with a total voltage drop of 10 Volts, which means that now current will flow in this new path instead of the path with 10k resistance? \$\endgroup\$ – MJ13 Apr 30 '18 at 16:02
  • \$\begingroup\$ And if so why output remains at 10V after that? \$\endgroup\$ – MJ13 Apr 30 '18 at 16:03
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    \$\begingroup\$ The output will remain at 10 volts yes. Remember, that with negative feedback, the opamp is always trying to make both its inputs have the same voltage and 10.01 volts on the output would have to put 0.01 volts at one of the inputs. Negative feedback fights to solve that problem and keep the output at bang on 10 volts. For non ideal diodes and zener there will of course be a rounded knee effect. \$\endgroup\$ – Andy aka Apr 30 '18 at 16:15
  • \$\begingroup\$ @So we can say that an easy way to think about this problem for the negative inputs is to imagine as all the three diodes consist a new diode D' that has the same polarity with the D1,D2 diodes and conducts above 10V(which is the total voltage drop of all of them) ? \$\endgroup\$ – MJ13 Apr 30 '18 at 16:21
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    \$\begingroup\$ Yes, just one diode with a breakdown of 10 volts. \$\endgroup\$ – Andy aka Apr 30 '18 at 17:28

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