Knowing I have only a 12V battery to supply my circuit, how I can turn off the circuit if the voltage reaches 11V? I cannot use simple comparator CI since I don't have another supply voltage. I could I do that?


Get a comparator that runs from (say) 10 volts and, feed power to it via a low drop-out regulator. The LDO regulator is 10V and this can be potted down to (say) 5V (easy). This is your measurement reference input.

This reference feeds one input of your comparator. The other input is fed from your battery via another potential divider. Set this 2nd potential divider to produce 5V when the battery is 11 volts.

The output of the comparator will switch as the battery falls below 11 volts and this switching output can drive a relay to disconnect your load.

You could also use a mosfet instead of the relay.

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    \$\begingroup\$ This is way more work than needed.... \$\endgroup\$ – ThreePhaseEel Nov 13 '15 at 0:11
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    \$\begingroup\$ @ThreePhaseEel feel free to leave a wonderful answer that makes me feel inadequate. \$\endgroup\$ – Andy aka Nov 13 '15 at 0:17
  • \$\begingroup\$ Answer improved with example circuit :) feel free to nitpick as needed. \$\endgroup\$ – ThreePhaseEel Nov 13 '15 at 1:00
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    \$\begingroup\$ The battery voltage will bounce back quickly as the load is shed, so this either needs a hysteresis, or the battery voltage going into the comparator connected behind the relay so the circuit stays off. \$\endgroup\$ – Simon Richter Nov 13 '15 at 1:21
  • \$\begingroup\$ That is an excellent point @SimonRichter \$\endgroup\$ – Daniel Nov 13 '15 at 3:23

Is a supply-voltage supervisory IC along with a FET switch not an option for you?

One example is the TL7712A, but there are many others -- 12V versions aren't as common as 5/3.3/...V counterparts, but they still exist.

An example circuit can be found below. U1 is a generic 3-pin voltage monitor/reset supervisor, designed for 12V service -- more sophisticated parts can be used, of course! M1 can be any N-ch depletion-mode power FET with suitably high ratings (Supertex/Microchip and IXYS make 'em, the DN1509 shown being the former's work), while R1 is not needed if U1 has a push-pull output.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ In practice, you would almost certainly use a purpose-designed supervisory IC like this answer suggests. \$\endgroup\$ – Daniel Nov 13 '15 at 3:25
  • \$\begingroup\$ Unless your downstream circuit has some sort of available 'disable' pin, this is the way to go. \$\endgroup\$ – Daniel Nov 13 '15 at 3:26

Yes, you can do that. You should create a reference voltage for one leg of the comparator with a zener diode or similar voltage reference device.

The other leg of the comparator has a resistor divider that divides 11V to be equal to your reference voltage.

When these two values meet, your comparator will change states. Just make sure your reference is less than 9 or 10 volts...perhaps 5v.

  • \$\begingroup\$ Sorry @Daniel, i start to write my response before and i didn't see your's. You are totally correct. \$\endgroup\$ – Gabriel Rezende Germanovix Nov 12 '15 at 23:48

I really didn't understand why you can't use the same supply for comparator? Correct me if i am wrong, but you can use a LDO regulator to regulate the voltage like 6 ou 7V and search for an Ampop that run on this voltage and generate a vref with a voltage divider and another voltage divider to compare with main voltage source 12V.


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