Making a resistive load to test battery capacity

Thread Starter

RogueRose

Joined Oct 10, 2014
376
I have a bunch of 18-18.5v batteries I need to test to see how much capacity they have and I figured the best way is to put a load on it and test, but I don't have anything that will work as a test load but I have a lot of ceramic/cemet resistors (100's) from 2w up to 20w. I just don't know what resistance I should put on the battery nor how many watts I should use.

I do have a couple 12v LED's that are "3w" but are more like 1/2w that I could put inline if that would help at all.

I know that if I put resistors in parallel they 1/2 the resistance (if same ohm rating) and double if put in series, but do the wattages stay the same, IIRC, they do.

So what ohm rating and what wattage should I be looking for, or what combination should I make to test these properly. I plan on taking the V reading then running the test for X amount of time and checking the V. I may take a reading every minute or so and keep track that way as well.

Any suggestions on this? Thank you for any help you can suggest!
 

Tonyr1084

Joined Sep 24, 2015
3,584
Get a couple automotive headlights and put them in series so that they won't burn out. Their wattage is rated at 55 watts at 12 volts (actually 13.8 volts). If you can measure the current then you can discern the drain on the batteries. If you're drawing (lets say) 5 amps at 18 volts, 18 x 5 = 72 watts. As you monitor the battery voltage and the amperage you can get a profile of how the battery is performing. Without any of us knowing the specs on the batteries you are testing we can't begin to guess at what you can expect out of your batteries. You COULD test them for one hour and measure the amount of voltage and current drop at the end and extrapolate how long the batteries should last under such a load.

Batteries are not only rated in volts they're also typically rated in Amp Hour. Meaning how many amps you can expect to draw for an hour. I'm sure that's not completely accurate and someone here will correct my statement, but the way I think of it is that a 20 amp hour battery can deliver 20 amps for one hour or 1 amp for 20 hours. I'm sure that's not completely accurate, but that's just how I look at it.
 

dendad

Joined Feb 20, 2016
3,025
You could make a constant current load with an under voltage cutout. Years ago a friend of mine did that, and used a battery clock to indicate the time.
He removed the battery and supplied 1.5V from across the test load. Setting the clock to 12:00, then starting the test. The clock stops when the battery disconnects at under volts. The time it took can now be read from the clock.
 

Thread Starter

RogueRose

Joined Oct 10, 2014
376
What if I set up this:

run a series of 7 20w @ 2.5 ohm to get 17.5 ohms and then a 10w 1 ohm or . 47 ohm that should give me 18.5 ohm or just under 18ohm. That should draw 1 amp from a 18v battery, correct? I figure with a fan blowing on these that should be enough to dissipate some heat.

I didn't remember Ohm's law was as simple as it is.
 

dendad

Joined Feb 20, 2016
3,025
What if I set up this:

run a series of 7 20w @ 2.5 ohm to get 17.5 ohms and then a 10w 1 ohm or . 47 ohm that should give me 18.5 ohm or just under 18ohm. That should draw 1 amp from a 18v battery, correct? I figure with a fan blowing on these that should be enough to dissipate some heat.

I didn't remember Ohm's law was as simple as it is.
It is probably a better idea to have a constant current load then the figures have more meaning. Otherwise, you have a varying current and voltage as the batteries discharge.
For the power level you are looking at, an LM317 connected as a constant current supply should work on as long as you have it on a heatsink, and probably with a fan.
 
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