I am hoping someone out there can help me with my Low Voltage Drop Out Circuit. I have an application where I want to protect a 12v Lead Acid Battery from over draining and thereby shortening it's life. It needs to isolate the load at about 11.6v. The circuit needs to draw very little current so a relay is out of the question. The circuit shown is the one I am trying to use but has a major flaw. When the battery voltage drops it can reach a point where instead of isolating the load fully, it just drops the voltage and the FET gets hot and if left would, I am sure burn out. In testing with a power supply, dropping the battery voltage still further finaly turns off the FET entirely but it is this mid point where thermal runaway would occur and is anyway unuseable. Any ideas?

 

Is the purpose of the 1M resistor to add positive feedback to latch the circuit? I think it is wrong or not doing that job. I did not try this. Just in my head. LOL You might have to have a load applied for it to work. Even a small load.

 

Try reducing the gate divider by 10 times:
change 10k to 1k, and 2.2k to 220.
It should provide a sharper threshold.
The TL431 needs at least 1mA cathode current but I'd set it to 10mA.

 

When the battery voltage drops it can reach a point where instead of isolating the load fully, it just drops the voltage and the FET gets hot

That's because the 1meg resistor in your circuit is providing negative feedback not positive, which puts the circuit in a partial linear mode, and thus does not switch sharply when it reaches the transition point (sim of this below).


Below is the LTspice sim of post #2 circuit with slight changes to the configuration to provide the needed operating current for the TL431.
The positive feedback from R4 gives about 200mV of hysteresis to provide a sharp, complete transition, and minimize any oscillation (chatter) at the trip point.