I think the easiest method to solve such problems is to assume that the diodes are off (both, and then one of the two), compute the voltages across the diodes and see if there's a contradiction with your assumption. Let's call the top left diode \$D_1\$ and the diode in the middle \$D_2\$.
Case 1: \$D_1\$ off, \$D_2\$ off: Since \$D_1\$ is off there is no current through the top 5k resistor, and since \$D_2\$ is off, there is also no current through the bottom left 10k resistor. So \$V=0\$ and the voltage at the anode of \$D_1\$ is 15 Volts. Contradiction! (\$D_1\$ should be on).
Case 2: \$D_1\$ off, \$D_2\$ on: again no current through top 5k resistor. Voltage \$V\$ is
$$V=\frac{15V\cdot(5k||10k)}{10k+(5k||10k)}=3.75V$$
Contradiction! (Because the voltage across \$D_1\$ would be \$15V-3.75V=11.25V\$ and it should be on.)
Case 3: \$D_1\$ on, \$D_2\$ off: Voltage \$V\$ is
$$V=\frac{15V\cdot 10k}{5k+10k}=10V$$
The voltage at the anode of \$D_2\$ is \$15V\cdot 5k/15k=5V\$. This agrees with our assumption, because with these voltages \$D_2\$ must be off. So your solution is
$$I=0A,\quad V=10V$$
If your approach simulates well, then you can prototype it. It sounds like the OVP detector is some type of comparator with a line disconnect switch, which you could simulate. I'm not sure if timing is critical.
On the other hand, other approaches, which can be used together or separately:
- Vin and/or Vout has a series PTC resettable fuse followed by a an appropriate TVS shunt. This will provide its own blanking and OVP without the extra Maxim IC. The TVS is the OVP, and the resettable fuse will shutdown the power until it's hold time is cleared.
- Choose an alternative buck regulator with built in OVP and OCP. Many include this protection already.
Best Answer
Here's another bad water analogy.
simulate this circuit – Schematic created using CircuitLab
Figure 1. "I told you this was bad."
To what level do you think the tank will fill?