We have the following scenario: over-voltage protection IC (MAX18590/86) is placed in front of a buck regulator in order to protect (obviously) against excess-voltage at the input. Normally the over-voltage threshold (say 15V) is set by a resistor divider. When an over-voltage is detected the output is disconnected from the buck regulator. When the over-voltage condition is removed, normal operation is resumed after a 15ms blanking period.
A second fault condition is a failure of the buck regulator, most notably a short of the pass transistor. This would expose the remainder of the circuit to the unregulated input voltage (here 15V max).
I am wondering whether I could use a second voltage divider and two diodes to let the MAX18590 also disconnect the supply when the regulator output exceeds a certain level (say 3.6V) as shown in the schematic.
Of course, when the regulator shorts and the supply is disconnected the fault condition vanishes and the supply is reconnected after the blanking period (15ms). Since a short of the pass-transistor is likely to persist, the protective device will toggle on and off forever with a period of 15ms.
The maximum leakage current into the OVLO pin is given as 100nA which should not load the resistive dividers significantly.
The resistor values given in the schematic are calculated without taking the diode forward drop into account. Is there actually much of a forward drop when only 100nA of forward current are present?
How much of an effect does the reverse-current of the diodes have?
Please comment on the idea in general and on the issue of diode drop in particular.