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Need cut-off circuit for my emergency light

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HELP! :cry:
My battery charger output is about 13.8V which drives a 12VDC lamp. I have a 3.5Ah lead-acid battery connected to it which should work as back-up in case of power failure. I want to connect a cut-off circuit so that when the battery voltage drops to 9Volts, the DC lamp would switch off to avoid deep-discharge of the battery.
What components should I use? Thanks!
 

Use a low power comparator and a series transistor. A low side MOS transistor will work. There is a similar question on the site. Do a search for my user name and the word "hysteresis" to find it. The person for that question says that he will report back now the suggested circuit works.
 

Firstly, take a look at the low-power C-mos version of the 555 astabile timer:
**broken link removed**
Secondly, go to this site:
**broken link removed**
Basically a 555 works as follows:
When the voltage on pin 2 is less than 1/3 of the IC's supply voltage (Vcc), the RS FF is set and the output (pin 3) goes high (almost up to Vcc).
When the voltage on pin 6 is more than 2/3rds of the supply voltage, the RS FF is reset and the output pin goes low (grounded to the negative supply line).
Click to expand...

Principle of the operation you can see on the picture below ..

All what you will need to do is to re-design the output stage, which may consist of a P-channel MOSFET (or PNP power transistor), a zener diode (to shift levels) and a resistor connected between D and S (P-MOSFET) or B and E (PNP) .. directly driven from 555 ..

Regards,
IanP
 

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