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I have the following problem.

Consider the circuit below:

enter image description here

The resistors have the resistance \$R=1000 \Omega\$.

The zener-diode has the breakdown voltage \$V_{break}=2\mbox{V}\$

When \$V_{in}=-3 \mbox{V}\$ what is \$V_{o}\$?

Okay, so here is my thought process:

Since \$V_{in}\$ is negative, the zener-diode cannot be in forward bias mode. If \$V_o\$ is less than \$\mbox{2V}\$, then there is no feedback loop, thanks to the zener-diode's breakdown voltage.

When \$V_{o}=2\mbox{V}\$ the breakdown voltage is reached, and a feedback loop is in place. And since this is negative feedback, the system will stabilize.

So my answer is that \$V_{o}=2\mbox{V}\$.

Now, this solution required me to "think a little", which is not a bad thing. However, in an exam situation I maybe won't be as calm or relaxed, to allow me to think like this.

So my question is: Is my solution even correct (if not then where did I go wrong?) and secondly, is there a solution that I can do in "auto-pilot" so to speak. Like node-voltage analysis or another way that doesn't need clever thinking.

I hope someone can help me with this.

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    \$\begingroup\$ Isn't it fun to think about circuits in "clever" way? I always try to think in "clever" way when I find these kinds of circuits instead of going through all those equations involving solutions (if any). \$\endgroup\$ – G-aura-V Mar 13 '20 at 16:42
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    \$\begingroup\$ Assuming ideal components your answer is fine, but your reasoning is not. Even if the diode is not in breakdown mode, there is a feedback loop with the two resistors. Maybe think along the lines that the voltage difference between the inputs of the opamp is zero (that's what it's trying to achieve) \$\endgroup\$ – Arsenal Mar 13 '20 at 16:43
  • \$\begingroup\$ Hmm, now I'm a little confused. If there is a feedback loop even before the breakdown voltage is reached, then doesn't the system stabilize at a lower output voltage, before \$V_{o}\$ reaches the breakdown voltage? \$\endgroup\$ – Carl Mar 13 '20 at 16:51
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    \$\begingroup\$ You should be able to calculate the voltage at the output if the diode wouldn't be there. \$\endgroup\$ – Arsenal Mar 13 '20 at 16:58
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    \$\begingroup\$ The system stabilizes only when the inverting and non-inverting pins are at equal potential. We do our calculations assuming that the system is stable whenever there is a negative feedback. \$\endgroup\$ – G-aura-V Mar 13 '20 at 16:58
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I assume you know the basic properties of an op-amp.

I will assume R1 to be resistor connected to Vin and Rf to be feedback resistor.

Since the inverting and non-inverting pins are shorted, you have 0V at one end and -3V at another end of resistor R1. So the current of 3mA flows from inverting pin towards the supply voltage. But where does this current come from? This is where the feedback resistor comes in action. The feedback resistor allow 3mA of current to pass from Vo pin to inverting pin. But this creates a voltage drop of 3V across the resistor and zener diode.

As you have told, the zener voltage is 2V which implies that the voltage across the resistor can only be 2V. So, the output can no longer inncrease the voltage. Then how is the current balanced? Well, it is the duty of Zener diode to supply the current when it has crossed its reverse breakdown voltage.

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You're correct. A useful concept here is a virtual ground...when connected in a negative feedback situation, the output does whatever it can to make the negative input look exactly like the positive one. Since the positive input is grounded (and I'm assuming that's within its common mode voltage range) the output will increase with decreasing Vin until it reaches the zener breakdown, after which it can supply all the current it needs with a voltage of 2V.

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