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I would the following:

When the LM358 non-inverting pin voltage greater than 1.00V then the batteries are empty and the output go high. I would like to turn the P-MOSFET off, but its not nowrking now, because the LM358 can't go as high as the source of the mosfet.

How can I modify it, when the battery voltage drops below 10V, turn the P-MOSFET off?

enter image description here

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  • \$\begingroup\$ Either make your circuit high-side referenced (ie, make a rail that is 5V lower than the input and use that as ground, instead of making a voltage rail that is 5V above negative and use that as supply) or switch the a low-side switching circuit with an NMOS (or PMOS, as long as you can turn it of high enough). Additionally, I hope your load takes a lot of current since now you will be drawing current with your 100k devider for the reference and your opamp - so even on a low battery you will still be draining it some more. \$\endgroup\$
    – Joren Vaes
    Aug 22, 2017 at 7:22
  • \$\begingroup\$ Perhaps even better would to implement a circuit that does the opposite: use very little current and only turn the system on if the battery is high enough. This is how most ICs meant for low-battery protection tend to work, if my memory serves me righ \$\endgroup\$
    – Joren Vaes
    Aug 22, 2017 at 7:23
  • \$\begingroup\$ Can you help me with a schematic? \$\endgroup\$
    – Lobi
    Aug 22, 2017 at 7:26
  • \$\begingroup\$ @Lobi, in addition to the good answers here, you might find this useful. It's an ON Semi note about this exact application. Figure 3 is basically Bimpelrekkie's answer and the "Efficiency Considerations" section is also pretty important depending on what your load is. At the end there's a good table of MOSFETs with various "on-resistances" if you haven't identified one already. Your Vgs is roughly 11V with Bimpelrekkie's solution, so you'll have to look at datasheets to figure out your rds but you can conservatively assume half the rds @ 4.5V. onsemi.com/pub/Collateral/AND9093-D.PDF \$\endgroup\$
    – jalalipop
    Aug 22, 2017 at 11:53

3 Answers 3

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Change the comparator and voltage reference with a single IC: High voltage comparator with reference, hysteresis . See one example in the picture.

enter image description here

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You might want to consider the following combined switch and low voltage cutoff

schematic

simulate this circuit – Schematic created using CircuitLab

When 'off', it consumes no power.

M1 and IC1 form a self-latching 'thyristor'. The TLV431 is a precision shunt regulator, that can also be used in other ways. Here, it is being used as if it was an NPN transistor with a precise 1.24v base threshold voltage and a very high current gain. When the output voltage falls such the IC1's input is less than 1.24v, the latch trips out. Set R4 to adjust the trip-out voltage, or leave out R4 and calculate R2 and R3 exactly.

You can use either the on/off switches, or the on/off toggle, or open collector logic to turn the power on and off. If you connect a capacitor across the on switch, you can make the circuit default to powered 'on' when it's first connected.

Beware that the TLV431 is an unusual device in that different voltage ranges are available in the same part number. The TI device only goes to 6v. You need the On Semi, Diodes Inc or the Zetec devices for a 16v version.

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A possible solution is this:

schematic

simulate this circuit – Schematic created using CircuitLab

The PMOS is now switched OFF when Q1 is OFF. Then R2 then shorts the gate-source of the PMOS and it will be OFF.

Note that the NPN will be ON when the opamp's output is HIGH and that will also turn ON the PMOS.

This is different from your solution where the PMOS is supposed to be ON when the opamp's output is LOW.

So for my solution, swap the inputs of the opamp to make the opamp's output go HIGH to enable the PMOS.

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