simulate this circuit – Schematic created using CircuitLab

Basically I have an issue where a security manufacturers power supply has built in eol resistors for monitoring a battery or mains fault.

The problem is that these resistors are the wrong value for an install and can't be changed, they're built in and won't be able to be modified as it will void warranty.

The circuit is quite simple. There is 2 4K7 resistors in series connected to 0v, these are then connected to an input pin with a pullup resistor to 5V. There is a transistor in parallel with one resistor that essentially shorts it out if no faults.

So on an input there is a path to 0v with either 4K7 or 9K4 resistance.

Now how do I go about having these 2 states drive a FET or transistor ?

Sorry for the stupid question however while I understand some basics on how they work I have never designed anything like this.

Is it as simple as a resistor voltage divider say with a FET where 1 resistance value is within gate threshold voltage and the other not or would I need something like a comparator or a zener clamp ?


Have added a schematic, what is is in the box on the left is built into the supply and cannot be altered. The terminal essentially then connects straight to an input but the problem is that the resistance isn't correct for this security panel.

Based on what MikeP has said on the right is what I believe may vaguely work given a gate threshold of 2V.

Unfortunately do not have the 5VDC rail available to play with either. Only the 13.7VDC rail from the supply itself, I may have to play around with this as essentially it is monitoring itself and if running down on battery that voltage rail will sag so my external circuit may switch off before the power supplies monitoring determines batteries are too low.

Next questions would be does that FET require any sort of protection or pull-up/downs and also is there any way to easily invert it so it turns on or off the same as the built in transistor rather than opposite ?

Depletion mode seems hard to find...

Thanks so far.

  • 2
    \$\begingroup\$ We love schematics. There's an easy to use schematic editor button on the toolbar. Otherwise it is difficult to figure out what you are talking about. Drag components onto the page, double click to edit properties. Drag from nodes to create connections. Make it read left to right with positive rail at the top and negative at the bottom. \$\endgroup\$
    – Transistor
    Commented Aug 5, 2016 at 14:15
  • 1
    \$\begingroup\$ Yeah, I need to see a schematic showing where the resistors are and where the circuit you are designing is. Is the node where the two resistors are connected together available for inspection? \$\endgroup\$
    – user57037
    Commented Aug 5, 2016 at 16:06
  • \$\begingroup\$ Thanks, I've added a schematic, was on mobile before which doesn't seem to work. \$\endgroup\$
    – D-on
    Commented Aug 6, 2016 at 3:48
  • \$\begingroup\$ I don't understand... so in a fault condition, the monitor output from the supplied power supply is 0V, and when it's working, it's 13.7V through 4.7k? \$\endgroup\$
    – Daniel
    Commented Aug 6, 2016 at 5:05
  • \$\begingroup\$ No, sorry if it's not clear, what's within the box is the monitoring circuit for the supply. No fault condition is transitor on and therefore 4.7K to 0v, fault condition transistor off and is 9.4K to 0v. \$\endgroup\$
    – D-on
    Commented Aug 6, 2016 at 5:15

2 Answers 2


I think this is what you need. When R4 is 4.7K, FAULT will be low. When R4 is 9.4K, FAULT will be high. You can swich the logic around by swapping inverting and non-inverting connections.


simulate this circuit – Schematic created using CircuitLab

Good luck.

  • \$\begingroup\$ I think this and MikeP's should be enough to start breadboarding some circuits. Thanks very much. \$\endgroup\$
    – D-on
    Commented Aug 6, 2016 at 6:38
  • \$\begingroup\$ After some testing this is pretty much perfect. Breadboarded with an LM393 and 2N7000 driving an LED for visual feedback with a trimpot for R4 and works well. Also tested the 13.7VDC rail dropping as batteries deplete and no issue with the comparator circuit as the reference is also dropping, working well beyond the point a low battery fault will trigger. Thanks again. \$\endgroup\$
    – D-on
    Commented Aug 9, 2016 at 15:11
  • \$\begingroup\$ Thanks for the feedback. Provides much more motivation than points! \$\endgroup\$
    – user57037
    Commented Aug 10, 2016 at 3:05

suppose that your FET Vgs(th) is 2 V In that case, pull the pin up through a 9 kohm resistor to 5 V based on the voltage divider, you'll get 1.72 V or 2.55 V at the pin which should turn the FET off and on.


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