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Low Voltage protection (cut at 33v - Full charged 44v)

Hi All,

My first post here...
I am at a loose end...

I have a battery pack 33-44v (low to fully charged)...

I have a 6-8 amp load (10 to be safe) and I want to hard CUT the voltage when the voltage gets to to 33 and below....

I need help with starting this circuit or a link to something that has already been done.

Cheers
 
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P.S. I already turn the lights on using 2v to a 20 amp transistor, so all i need to do is cut the 2v supply to stop it from running....
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Here is an example, however the constant current draw to operate the relay may be an issue. (and you'll have to select components and values to suit the voltage and current).

Nevertheless, this circuit does have some useful properties that you'll need:

1) it has hysteresis. The voltage will rise once the load has been removed, and the last thing you want is the load turning on and off and on and of and on and off... when the battery gets close to flat. You want it to turn off and stay off until the battery is significantly charged (so the voltage which turns it on needs to be somewhat higher than the voltage which turns it off.)

2) it switches rapidly. A 10A load from a 20V battery is 200W. Unless the switching device turns on and off quickly you will release lots of heat and possibly destroy the switching device (many low voltage cut-out circuits do NOT switch rapidly and they would likely fail in your application). Having said that, this circuit relies on the relay dropping out, and that's not going to provide an especially fast turn on or off so the relay contacts need to be rated for a significant current.

3) it needs to be able to withstand 44V (or even higher). Many electronic devices used in similar circuits will have a maximum voltage well under 45V.

Given your load is several amps, the current required to operate the relay may be an acceptably small additional current. You probably need a 24 or 36 volt relay and an 18 to 20 volt zener.
 

KrisBlueNZ

Sadly passed away in 2015
That's an interesting page you found there Steve. Despite the claims in the text, that circuit doesn't have hysteresis. There is an emitter degeneration resistor that provides negative feedback, but no positive feedback. Just another example of circuits that don't really work, presented on the internet :-(

In this case, I would use a MOSFET and add some positive feedback for hysteresis.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
I believe it has hysteresis due to the pull-in voltage being higher than the drop-out voltage.
 
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