I am trying to figure out a simple way when my battery drops to around 6.2V to disconnect the battery. I have a LP2950ACZ-5.0 (5V regulator) and an LP2950ACZ-3.0 (3V regulator) feeding my 16F628A processor along with an rf module,which is why I have the 3.3 regulator. Since the modules are not cheap to replace, at least not to me, and neither are 9V batteries I am trying to figure out how to cutoff the 9V battery when the voltage drops to around 6.5, 6.2 but that part of the circuit would have to draw very little current. I have been experimenting with my other post http://forum.allaboutcircuits.com/showthread.php?t=16232 but that circuit alone wants to drain the battery. It has been suggested to use the comparator modules on the processor, but even if I do that the circuit would have to allow the processor time to power up, compare the voltage, then break the circuit. Seems like it would keep doing that indefinately until the battery is dead.

Anyhow thanks for any suggestions on this pipedream

 

Why are you using two regulators, 3.3V and 5.0V? The PIC will run fine on the 3.3V regulator.

To answer your question, the best way would be to use the comparator to sense the supply voltage. Set up an interrupt or poll the comparator occasionally. When the voltage drops, put the PIC into SLEEP mode where it will consume only a few microamps worst case. You may want to wire the 3.3V RF module to the PIC so you can turn it off, or somehow send a command to it to go into standby mode.

When all the power consuming micros and modules are shut down, the current draw will be a few microamps in total. Your batteries will go for years on this drain; in fact, the self discharge rate would probably exceed this for some batteries.

 

Interesting thought, hmmm...I will try that but still am worried about if low battery voltage will damage my regulator.

Thank you very much for the suggestion.

 

Your regulator will not be harmed by low voltage.

 

Thank you again for the help.