Help Please: low voltage auto shut off.

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Projek_01

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Hi, i am looking to build somthing that will shut off automaticly when my voltage drops below around 4.6v but have no ideal how to make it. May somone please help me.
Thanks
 
Could you use a variable resistor to adjust the voltage to a relay? While the relay is closed, the circuit works, but when the voltage goes below 4.6v, the relay switches off and the circuit opens. Use the variable resistor to adjust at what voltage the relay turns off.
Use your meter and a varible resistor (different one that you remove later) to adjust the voltage to 4.6v. Make sure the circuit is on then turn the variable resistor for the relay slowly till it switches off, then turn it back so its on. Remove the second veriable resistor and connect back to main power. When the voltage goes below 4.6, the relay should shut off.
~Mike
 
You don't say how much (or little) amount of current you want to shut off.
A big relay can handle switching tens of amps, but would waste a lot of current being powered all the time in a low-current system. A low-power opamp can be used to switch a transistor to turn off the power and the opamp would be wired as a comparator.
 
Or use industry reset circuit (TC1272, TS809CLEUfor example) with transistor...

Pavel.
 
You might want to look into detail at the requirements here.
A battery drained to 4.6v will go above 4.6v immediately once the load is shut off. This could lead to on/off oscillation which may be undesirable. Ideally it should go into a latching shutdown mode and require 5v nominal or something like that to reset the latch.
 
Good point, Oznog! :lol:
 

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Electric powered radio control model airplanes frequently use a low-voltage shutdown for the motor so that the receiver still works for a dead engine controlled landing.
 
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