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Discharging a 12V Lead Acid Battery

Posted on 2014-02-06
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Last Modified: 2014-02-13
Heyas,

Can anyone recommend equipment to discharge a 12 volt lead acid battery down to specified level, as I don't want to have to use a series of light blubs :).

I realize this is isn't strictly IT related, but I thought I give it a shot.

Thank you.
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Question by:Zack
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8 Comments
 
LVL 19

Expert Comment

by:Miguel Angel Perez Muñoz
ID: 39838807
There are some chargers with discharge capacity, I have one used to charge Lipo batteries but can charge/discharge lead/nimh/nicd/lipo/life batteries: http://www.hobbyking.com/hobbyking/store/__6478__imax_b6_ac_charger_discharger_1_6_cells_genuine_.html
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LVL 44

Expert Comment

by:Darr247
ID: 39839141
I would use a battery load tester.
e.g. http://www.oreillyauto.com/site/c/detail/SS02/44200/N0366.oap
Give it a minute or so to cool off between cycles. Typically the load is a high-wattage resistor with a heat sink on it.
Here's a cheaper one - http://www.harborfreight.com/100-amp-6-volt12-volt-battery-load-tester-69888-9191.html
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Author Comment

by:Zack
ID: 39840568
Thanks for the info.

Are there any Battery Testers out there that will continuously discharge a 12 volt lead acid battery down to specified level.

Thank you.
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LVL 44

Expert Comment

by:Darr247
ID: 39840669
Well, if you discharge Lead-Acid batteries much below 10 volts, the risk of killing a cell rises rapidly. At 10 volts there is still plenty of amperage left in it.
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LVL 19

Expert Comment

by:Miguel Angel Perez Muñoz
ID: 39841207
I don´t know about battery testers that do this, but voltage may not indicate how many change remains on battery, best way is use density meter like this: http://www.coleparmer.co.uk/Category/Battery_Test_Digital_Hydrometer_Density_Meter/53465
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LVL 25

Accepted Solution

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kode99 earned 250 total points
ID: 39843851
Simple solution is a inverter which would let you use anything you want to drain the battery.  Combine this with a adjustable low voltage disconnect (LVD) to shut it down when you get to the point you want.

Here's a LVD that can handle 35 amps and can be set to a range of voltages,

http://www.powerwerx.com/batteries-chargers/dc-low-voltage-disconnect-lvd.html

Here's a bunch of inverters at Canadian tire,
http://www.canadiantire.ca/en/search-results.html?searchByTerm=true&q=inverter

Running a electric heater is probably the most consistent way to do it (other than light bulbs).  Just get a inverter that can handle the wattage of the heater.  Lower amperage LVD's and inverters are fairly cheap.  A LVD is also not that hard to build from scratch.

Keep in mind that if you heavily load the battery the voltage will drop but will go back up when the load is taken off.  Depending on exactly what the purpose is it may be better to do a slower discharge for more accurate results.

If you are looking to ensure that your voltage level is below a set number you could just use a voltage regulator to guarantee a specific output voltage no matter what the actual battery voltage is.  Alot better than messing around discharging batteries.

We use moderately expensive battery testers that can tell us in a few seconds how far gone a lead acid battery is.   On a lead acid the voltage drop under a set load for a set time period is a pretty good indicator or how much charge is left.
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LVL 44

Assisted Solution

by:Darr247
Darr247 earned 250 total points
ID: 39844328
Put eight 3.3 ohm 50 watt resistors connected in parallel across the output of that LVD from powerwerx.com and you should have a shunt that switches itself in and out until the battery will no longer recover to 12.8V...

Resistances in parallel equal the reciprocal of the sum of the reciprocals
1/3.3 + 1/3.3 + 1/3.3 + 1/3.3 + 1/3.3 + 1/3.3 + 1/3.3 + 1/3.3 = 1/2.4242...
1/2.4242... = 0.4125 ohm

The power rating is the sum of the ratings when equal resistances are paralleled.
So eight 3.3 ohm 50W resistors in parallel = 0.4125 ohm @400W

R = Resistance (in ohms)
I = Intensity (aka amperage)
P = Power (in watts)
E = Electromotive force (aka voltage)

E/R = I
12.8/0.4125 = 31.03A (that load will shrink as voltage drops)

E*I = P (true for DC only... AC also uses inductance, capacitance and frequency)
12.8 * 31.03 = 397W (that power draw will shrink as voltage drops)

So that would be about $124, plus sundries like wire (I would go with 16 pieces of #18) to attach the resistors to the LVD and the LVD to the battery (2 pieces of #8 or larger) and terminals (even if it comes with the yellow terminals shown in the picture, they would not be large enough for 8 #18's nor a #8 or larger) and solder (to attach the #18 to the resistors) and clamps to attach to the battery terminals...  and a DC-rated switch in one of the wires to the battery would probably be a good idea so you can turn it off while making the connection to the battery.  And, say, a 2' piece of 1x10 or 3/4'' plywood on which to mount it all.

You could built ~20 of those rigs for the price of one of those specific gravity processors. :)
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Author Closing Comment

by:Zack
ID: 39855677
Thanks the info I am buying those products as we speak :).
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