Device or circuit to cut off power when voltage drops from 12 VDC down to 9VDC. from 12 VDC

I have a 12V battery pack. When the voltage drops to 9 volts, I need to
shut off the power. Is there a device to do this? A wiring diagram
would help as I am weak in electronics.

 

-------------
Wire an open-collector op-amp comparator with 100 ohm/5.1V zener-diode
defined reference voltage to turn on and off a latch that runs a
transistor. Use a 10K pot to set the 9V set-point, and then replace
with two resistors if you like.

If you want a diagram send me $10. Or you can look it all up using
those words and phrases in Google. Clues: LM339, 74LS74, 2N3055.

-Steve

 

+----/ o-----+-------------------- +12 output
| |
+ ---+---o o-----+-----RELAY-----+
PB | |
R1 \ -----
/ / \ TL431
|<------------/ \ or
/ /_____\ LM431
R2 \ |
| |
- ----------------+---------------+

You need 2 resistors, a relay and a TL431, plus a momentary
pushbutton switch. MPB - 1B for the switch, and RLY-428 for
the relay are good choices, available from Allelectronics
http://www.allelectronics.com/
The specified relay will handle 8 amps and can be driven
directly from the TL431. If you use a different relay,
you need to be sure that the TL431 can drive it directly.
Make sure that the relay's coil resistance is over 120
ohms. The higher the resistance of the relay coil, the
better. It will be tough to find a better relay choice than
the one specified. R1 is 2200 ohms. R2 is 352 ohms and can
be made with a 330 ohm and a 22 ohm resistor in series.

When you press the pushbutton, the relay will energize,
provided the battery voltage is over about 9.1 volts.
The relay contact, shown above the PB (pushbutton)
symbol will close conducting the 12 volts to the output
and to the TL431 (or LM431 - same device) through the
relay coil. The 431 will conduct as long as the voltage
from the battery is over about 9.1 volts.

The circuit shown will draw about .025 amps from your
battery pack. When the voltage drops to close to below
about 9.1 volts the relay will de-energize, and no
current will flow either to the load or to the circuit.

Ed

 

Here's a corrected version - I forgot to show a diode:

+----/ o-----+-------------------- +12 output
| | D1
| | +---|<--+
| | | |
+ ---+---o o-----+---+-RELAY-+---+
PB | |
R1 \ -----
/ / \ TL431
|<------------/ \ or
/ /_____\ LM431
R2 \ |
| |
- ----------------+---------------+


Note the added (D1) 1N4001 diode


Ed

 

You forgot the positive feedback. Also, using a P-MOSFET instead of a
relay will save alot of power.

The circuit is as follows:

P-MOSFET
Battery+ -------+------+---S+--. -D------.
| | | ^ | |
[620k] [22k] -- -- -- |
| | --------
| | G |
| +----' |
o------(---------[2.2MEG]--o
| | |
| --- |
o-----/ \ TL431 |
| ----- |
| | [Rload]
[100k] | |
| | |
Battery- --------o------o-------------------'

The TL431 will sink current when it's control lead is > 2.5V. That will
happen when the output is high, and battery gets down to 9V. It won't
turn back on until the battery gets back up to 10V. Thus, it may not
chatter as the battery gets close to 9V like the relay one could easily
do, depending on the load.

You actually need a suprisingly small MOSFET, because it'll always
either be off or completely on, so it won't need to deal with large
amounts of power. When the circuit is off, the simulator claims it
should only draw about 50uA.

I have not built this circuit, only simulated it, so you should
obviously experiment before committing to building 1000 of them.

This circuit has crappy temperature compensation, so it may not work
outside a reasonable temp range.