Low Voltage Cutout

I'd go for a circuit with an adjustable undervoltage trip point. Once power returns,the charging voltage is sensed and the circuit returns to normal.

I propose one comparator set up with hysteresis, one V reference and a transistor to drive the relay..

Alternatively a DTL circuit could be made up. Extra relay contacts can sometimes put to use for circuit function.

D from BC myrealaddress(at)comic(dot)com British Columbia Canada

If the load circuit is sure to pull the voltage down to zero quickly, you can use the voltage on the load side to do the positive feedback. The most extreme version would be to have a push button you need to press to turn the system back on.

If it is a 12V battery, 9V is too low. You want to disconnect the battery at somewhere between 10.5 and 11 volts - higher is better for the battery.

To prevent the "bounce syndrome", hold the relay energized through its own open point. Something like this:

N/O Relay Contact Battery(+) ---+---o--- | ^---o---+----> to load k| | [D1] [Zd] 11 volts | | +--------+------+ | + | |k Charger [Relay] [D3] | - | | Battery(-) ---+--------+------+

When the battery V drops below 11 volts, the relay drops and the contact opens, isolating the battery from the load and the zener. D1 prevents the relay from "seeing" the battery voltage, and D3 absorbs the inductive kick from the relay dropping out. When the power is restored, the charger turns on and energizes the relay, reconnecting the load.

Use a Schottky for D1, because it has a lower voltage drop than a regular diode. If that still presents an unacceptable voltage drop, use a double pole relay and replace the diode with the second normally open contact.

You can use a differnt zener or a TL431 to set the drop out voltage, but unless your battery spec says you can safely discharge it below 10.5 volts, I recommend you use the

Ed