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I've been trying to design a simple low voltage cut-off circuit for my dual supply +-9V circuit. The positive side works as intended with the LM431 and a PMOS, but the negative side not so much.

The idea behind the circuit is that if the voltage of the battery starts decreasing than at a certain point (about 7.5V to 8V) then the circuit will cut-off the supply entirely to protect the Load. This needs to happen for both the +9V side and the -9V side (-7.5V to -8V).

One of the important things is that the start-up voltage should be higher than the cut-off voltage, which is true for the positive side of my circuit (start-up is at about 8.7V and cut-off at about 8.1V with the resistors used).

My problem is that the LM431 doesn't seem to work for the negative side and i cant really find an alternative. Is there any similar device as the LM431 just for negative low voltage cut-off or did i just mess up the circuit entirely?

I did try to make the circuit symmetrical, but it seems to me that the LM431 is just the wrong component. That of i just have to make a completely new circuit, but I'd like to keep it as symmetrical as possible. Maybe any of you can help me with this problems as google doesn't really have much in regards to negative low cut-off structures.

If you are wondering about the resistance values of R17 to R19, i did try around some different values to see the change in the diagramm, but it doesn't really change all that much aside from the voltage starting to oscillate at certain values.

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    \$\begingroup\$ There's a body diode on M7 from drain to source that is keeping the negative rail always on. \$\endgroup\$
    – td127
    Apr 14, 2022 at 21:09
  • \$\begingroup\$ M7 is an N-channel MOS with its body diode having the cathode on drain. Since the V_out- has a higher potential than V_in- the body diode should be turned off. This can be seen in the schematics too as V_out- only starts falling to V_in- when the NMOS turns on a bit. The problem i am facing is the weird turn-on behaviour of the MOS that i cant really adjust like in the positive part \$\endgroup\$
    – Kaiser F
    Apr 15, 2022 at 9:07
  • \$\begingroup\$ Sorry, you're right - I was looking at the Si743DP MOSFET and scandalously assumed they were the same. \$\endgroup\$
    – td127
    Apr 15, 2022 at 18:49
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    \$\begingroup\$ You're using the LM431 as a comparator and that works in the top half because with its anode firmly held at ground LM431 will conduct as soon as its reference crosses 2.5V. But the bottom's LM431 anode isn't fixed, so when negative rail starts falling both the anode and reference fall together, 2.5V apart, keeping the LM431 conducting. \$\endgroup\$
    – td127
    Apr 16, 2022 at 3:25
  • \$\begingroup\$ "the circuit will cut-off the supply entirely to protect the Load." - Why do you need to do this? Shouldn't it cut both supplies if one goes down? \$\endgroup\$ Apr 16, 2022 at 22:00

1 Answer 1

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Here's your circuit using comparators. The comparators are across the entire voltage range. The threshold voltage of 5V is set with a voltage source in the simulation but in real life can be a 5V regulator with very low quiescent current. This circuit will turn off both supplies if either battery falls, as opposed to the original circuit that only turned of the rail associated with the failing battery.

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  • \$\begingroup\$ I really like this design and the way it works. The only real downside is that i make this project for a college class and i am not allowed to use comparators since everything should be build using discrete components such as R,L,C, MOS, BJT and maybe some small references like the TL431 etc. I found a rather easy solution using only a zener diode, two resistors and a MOS. The only thing i wanna try now is to make an interlocking between the two circuits. \$\endgroup\$
    – Kaiser F
    Apr 18, 2022 at 8:21
  • \$\begingroup\$ I tried using a zener but zeners require a fair amount of current to get into their specified range so could be a significant power suck on batteries. And to that end, all those 10K resistors in my schematic should be 100K to limit quiescent current draw. \$\endgroup\$
    – td127
    Apr 18, 2022 at 16:11

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