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Old 18th June 2003, 01:54 AM   #1
zmoz is offline zmoz  United States
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Default Low voltage cutoff circuit?

I plan on powering something off of 4 AA rechargeable batteries. The problem is if I discharge the batteries any lower than .8v per cell it will damage them. I need to make a cutoff circuit that will turn off the batteries once they reach 3.2 volts. Anybody have any idea how?
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Old 20th June 2003, 06:50 AM   #2
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Weird idea.

Bistable relay (latching) to allow
4 AA battery power to pass.

Create a comparator circuit using
LP339 (not LM339) for low power
consumption (60uA standby?) and
trigger the relay to unlatch when
the comparator hits threshold voltage.

Normal idea.
Mosfet transistor on = 4 AA battery power to pass.

Mosfet transistor off when voltage
drops below threshold = power cutoff.
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Old 20th June 2003, 08:55 AM   #3
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Default Re: Low voltage cutoff circuit?

Quote:
Originally posted by zmoz
I plan on powering something off of 4 AA rechargeable batteries. The problem is if I discharge the batteries any lower than .8v per cell it will damage them. I need to make a cutoff circuit that will turn off the batteries once they reach 3.2 volts. Anybody have any idea how?
How much current do you need to supply to the circuit?

Cheers,

Mark
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Old 20th June 2003, 09:46 AM   #4
x-pro is offline x-pro  United Kingdom
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There are special IC's called voltage detectors to do things like that. Look for ILC5062 from Fairchild, XC61C series from Torex, some available from Microchip (TC53, TC54), Texas Instruments (TPS3836) and others. All these chips consume on average about 1 microamp, some even less than that (TI chip - 220nA) and available for a wide range of voltages in about 0.1V increments. These are not terribly expensive and usually very precise (1 or 2 % tolerance).

Cheers

Al
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