# Problem with Op amp and zener diode circuit

I have the following problem.

Consider the circuit below:

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.

• 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). Commented Mar 13, 2020 at 16:42
• 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) Commented Mar 13, 2020 at 16:43
• 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?
– Carl
Commented Mar 13, 2020 at 16:51
• You should be able to calculate the voltage at the output if the diode wouldn't be there. Commented Mar 13, 2020 at 16:58
• 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. Commented Mar 13, 2020 at 16:58

I will assume $$\R_1\$$ to be resistor connected to $$\V_{\text{in}}\$$ and $$\R_f\$$ to be feedback resistor.
Since the inverting and non-inverting pins are shorted, you have 0 V at one end and -3 V at another end of resistor $$\R_1\$$. So the current of 3 mA 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 $$\V_o\$$ pin to inverting pin. But this creates a voltage drop of 3V across the resistor and Zener diode.