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.


simulate this circuit – Schematic created using CircuitLab

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.

  • \$\begingroup\$ To let that circuit "stand on its feet" you need at least a resistor from OVLO to ground, or these diodes are not properly biased. And you might want to check your linear regulator part, you say that's a 7805 but the output voltage is marked as 3V3... \$\endgroup\$ Jul 9, 2014 at 15:08
  • \$\begingroup\$ @VladimirCravero you mean in order to have some defined current flowing through the diodes all the time? \$\endgroup\$
    – Arne
    Jul 9, 2014 at 15:11
  • \$\begingroup\$ Yep, I'd say that's needed. Your idea is great though, I bet you can tune the resistors to take in account the diodes drops. I don't have time to help now but somebody will. \$\endgroup\$ Jul 9, 2014 at 15:20
  • \$\begingroup\$ Take a look at the datasheet. The forward voltage of those diodes at 100uA is about 75mV. Current 1000 times smaller has an associated drop that is negligible and not shown in the plot. Also, I have never before dealt with something like this. I understand that case where a simple reference voltage is generated for OVLO, but what exactly are the diodes doing? IN OVLO a high-impedance input? \$\endgroup\$
    – sherrellbc
    Jul 9, 2014 at 15:28
  • \$\begingroup\$ @sherrellbc the diodes are meant to "select" the higher of the two voltage dividers. \$\endgroup\$
    – Arne
    Jul 9, 2014 at 15:34

1 Answer 1


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:

  1. 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.
  2. Choose an alternative buck regulator with built in OVP and OCP. Many include this protection already.
  • \$\begingroup\$ regarding 1. given that the time-to-trip of PTCs is rather long, a suitable TVS would need a huge power rating as far as I understand. Thin implies a larger package, which I would like to avoid. Please correct me if I am wrong. \$\endgroup\$
    – Arne
    Jul 10, 2014 at 7:47
  • \$\begingroup\$ Can you suggest such a part available (in small quantities) at mouser/digikey/farnell rated for 4-12V input, 3V3 / 1A output? \$\endgroup\$
    – Arne
    Jul 10, 2014 at 7:49

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