Voltage detector Question

Thread Starter

NM2008

Joined Feb 9, 2008
135
Hi,
Would anybody have any ideas for a relatively simple/basic voltage detector?

I have a 12v sealed lead acid battery, which powers four circuit boards,
I do not want the battery to discharge to below 10v.

Would someone have circuit/diagram, of the following?
Once the 12v battery drops to 10v, it will trigger a relay and shut off power to the circuit boards.

Thanks for your help.
Regards NM
 
There are a lot ways to do this, my suggestion:

Have the battery feed a voltage regulator like a 7805 (+5V regulator), then use the +5V to power a comparator. Have one input of the comparator connected to the +5V, while the other input of the comparator is fed from the center of a voltage divider made of two equal resistors (say two 10k Ohm). The output of the comparator will be biased to drive the control relay.
This way the compare voltages will always be stable as long as the battery voltage is above 6.5V. As soon as the battery voltage drops just slightly below the 10V, the comparator will sense this and the output should then go low, thus deactivating the relay that is providing battery power to your other circuits.

Have fun!
 

CDRIVE

Joined Jul 1, 2008
2,219
This Spiced with an adjustable range (P1) of +9V to +10V. The circuit is turned on by pressing S1, which will close K1 contacts until the bat voltage drops below 10V with P1 is at center.
 

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SgtWookie

Joined Jul 17, 2007
22,230
Relays are mighty power-hungry. Better to avoid them if you can.

Not sure what you're powering from the battery, but here's an idea I came up with for another poster with a similar situation back last November. The voltages are a bit different, but no big deal; just replace the 6.2v micropower Zener diode with one rated 10v.

The circuit basically disconnects the negative lead (ground) from the load when the voltage drops below the Zener diode's breakdown voltage; Q1 stops conducting (which was keeping MOSFET U1's gate high) so R1 pulls the MOSFET's gate low.

Since Q1 is then turned off, there is only a very small amount of leakage current through Q1 left to drain the battery, which would not discharge it much faster than self-discharge.

U1 could be replaced with a more robust MOSFET, such as an IRFZ24 or the like; in that case C1 would need to be increased.
 

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CDRIVE

Joined Jul 1, 2008
2,219
Relays are mighty power-hungry. Better to avoid them if you can.
Sgt, I must confess that I didn't give this issue the attention that it deserved and you're dead on correct on that account. I like the simplicity of your circuit but I don't see any hysteresis in there. Because of this I would expect the battery voltage to immediately increase when this circuit triggers off, thus causing it to re-trigger on. This would happen more rapidly (to the point of buzzing) as the battery charge decayed.

Am I missing something? :confused:
 

tkng211

Joined Jan 4, 2008
65
The attached circuit is a simple one using the adjustable shunt regulator TL431/LM431 (TO92 package). Hope it can meet your requirements.
 

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Last edited:

SgtWookie

Joined Jul 17, 2007
22,230
Sgt, I must confess that I didn't give this issue the attention that it deserved and you're dead on correct on that account. I like the simplicity of your circuit but I don't see any hysteresis in there. Because of this I would expect the battery voltage to immediately increase when this circuit triggers off, thus causing it to re-trigger on. This would happen more rapidly (to the point of buzzing) as the battery charge decayed.

Am I missing something? :confused:
When the MOSFET turns off, the path to ground for D1 is removed, so Q1's base is held at the emitter's potential by the R2/R3 network. The battery is removed for charging. When the charged battery is reconnected, C1 bootstraps the MOSFET gate.

C1 could be replaced by a NO momentary switch to provide the bootstrap turn-on function; this would be arguably more reliable than relying on the capacitor, as if the battery connection "bounced", the MOSFET gate may discharge via the 47k resistor. As things are, the only way to recover from this now is to remove the battery for a period of time until C1 fully discharges.
 
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