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(Very Basic) Low Voltage Detector

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Menticol

Active Member
Hello guys!

I'm working on a very basic circuit to prevent battery over-discharge. It's a 741 op amp working as a comparator, and the results are... well, very predictable.

The Battery is a 12V, 7 Ah SLA, powering a home made emergency light. I think 10 volts is low enough to turn off the light.

The problem is, how can I power the circuit with the same battery wich the circuit is designed to protect?

Other schematics show a 5.1V zener diode to adress this problem, but the weather is too bad to go out to buy them.

I have 7805, 7809 voltage regulators in stock, but I guess they are unuseful for the purpose: Vcc must be higher than the voltage the circuit is intended to detect, right?

Thank you very much in advance
 

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MikeMl

Well-Known Member
Most Helpful Member
You need a voltage reference which stays fixed even as the battery discharges. The opamp doesn't care what its supply voltage is; it only amplifies the difference between it's inputs. You make one input a fixed voltage derived from a zener or other voltage reference, while the other input is fed from a voltage divider off the battery.

I posted a much better circuit for a low-voltage disconnect based on a LM431 on these forums recently. Read this thread.
 
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Boncuk

New Member
I think 10 volts is low enough to turn off the light.
A lead battery is considered to be depleted at 1.8V per cell.

That makes 10.8V for an exhausted battery of nominal 12V. (max 14.4V before gasing)

That gives 3.6V to play with or still 3.0V if the battery is charged to 13.8V (standard).

Be careful discharging the battery below that voltage. It will be junk in no time.

Boncuk
 
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