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.