Battery low voltage cut-off problem

[pier1]

Hello,

I am facing a problem in my circuit where I made it for switching of a light when the battery voltage reaches 10.55 volt. The battery is 12 volt 3.4Ah SMF type.

The problem is:

When the battery reaches 10.55 volt (roughly) the battery is disconnected from the load. When the battery is disconnected from the load the battery voltage rises a bit and turns on the relay again. This repeats and repeats.

I don't know if the circuit I made is right. Could somebody help me to fix this ? 

View attachment 40642

 

[KMoffett1]

You need to add hysteresis feedback.  This will require a higher voltage than 10.55 (you select) to switch the light back.

http://www.intusoft.com/onsemipdfs/LM324-D.pdf

Look at Figure 13 for the circuit and formulas.

Ken

 

[Hero999]

Yes, there needs to be some hysteresis to prevent oscillation, just connect a 100k resistor between the op-amp's output and the + input. This will make the turn on voltage higher than the turn off voltage, see the data-sheet linked above for more information.

I wouldn't use the LM324 anyway, the LM339, LM393 or LM311 would be much better because they're specifically designed for use as comparators. If you only need one op-amp/comparator then I'd recommend the LM311 which can drive a small relay without an additional transistor.

Don't get me wrong, it'll be fine with the LM324, it just wouldn't be my first choice.