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08-23-2017, 03:31 PM | #1 |
Gone Wild
Join Date: Jun 2015
Location: St. Augustine, FL
Posts: 328
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What size Power Resistor do I need to pull down lithium cells?
I need to get 14 sets of lithium cells pulled down to the same voltage, to bottom balance my pack before charging it up. The voltages currently range from a low of 3.28 to a high of 3.87.
I want to bring them all to 3.28, and am looking at various power resistors online, but not sure what values I need. I know the max voltage for the cells is 4V, and I know they can dump a huge amperage if called upon. So, how do I calculate a reasonable value for a power resistor to drain them? I see ranges from 100W to 250W (in my price range), and values from .5 to 500 Ohm. I tried working the Ohm's Law formulas, but I don't really know the answer I am looking for so it makes it hard to calculate. Any suggestions or education would be appreciated. |
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08-23-2017, 08:25 PM | #2 |
Gone Wild
Join Date: Jan 2013
Location: Saint Petersburg, FL
Posts: 2,089
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Re: What size Power Resistor do I need to pull down lithium cells?
If you are using the resistor to make all the cells the same voltage then your voltage diff is 0.59v between the highly charged cell and the low. So a 0.5 ohm resistor should pull 1.2 amps which is about 3/4ths of a watt (minimum).
If you are using the resistor to bring a bank of batteries, all in parallel after you've balanced them, down to a lower voltage then 3.6v at 1 ohm gives 3.6 amps which is about 13 watts. If you use an online calculator then you can play with the numbers until you're happy. http://rapidtables.com/calc/electric...calculator.htm |
08-23-2017, 09:23 PM | #3 |
Over This Interview Is...
Join Date: Jun 2012
Location: AZ
Posts: 17,449
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Re: What size Power Resistor do I need to pull down lithium cells?
It also depends on how long you are willing to wait for them to drain. A higher ohm value resistor will drain more slowly, but the benefit is it makes less "heat per hour" , allowing you to use a cheaper, lower watt resistor.
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08-24-2017, 06:39 AM | #4 |
Gone Wild
Join Date: Sep 2012
Location: Charlotte, NC
Posts: 9,329
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Re: What size Power Resistor do I need to pull down lithium cells?
JaxPilot, the concept of "bottom balancing" is to bring all batteries down to 0% SOC, not just the same voltage.
The voltage by itself cannot tell you how much energy (AH) is stored in each individual battery. The common analogy is that you have a bunch of water buckets of different sizes with an unknown amount of water in each. You would empty all buckets (% SOC) and then add the same amount of water to each bucket. Since the cells will be charged in Series without a BMS, they will all get the same amount of energy (AH). Keep in mind without a BMS, they may/will not all end up with the same final voltage after the pack reaches your cutoff voltage, but that is OK since you know that you can remove that same amount of energy without bringing any cell below the initial 0% SOC. You would probably stay around the 10%-90% SOC range anyway. You could use a single 0.1Ω x 250w resistor, but they are around $45. You could also use cheaper 0.5Ω x 50w (4 or 5 in parallel), keep in mind all of those resistors need to be mounted to a "heat sink" plate for the rated wattage. http://www.mouser.com/ProductDetail/...9lJNubsVfuY%3d The typical process I have seen uses a JLD5740 meter with relay triggers: http://www.lightobject.com/Programma...stem-P408.aspx The idea is that you would wire the resistor in series with a solenoid/contactor and place it across a cell positive and negative terminals. The Solenoid coil gets wired in series with the meter N.O. relay terminal. The cell being discharged would also be connected to the meter 3 decimal place 10v input. You would set the meter to turn the relay ON at 3.450v and OFF at 3.250v in order to control the load placed on the cell. Let that setup run until it no longer cycles as the cells no longer rebound above 3.450v, you can then narrow the range until all cells settle near the bottom voltage. You need to choose the cutoff voltages that work best for your specific batteries. |
08-24-2017, 10:30 AM | #5 |
Gone Wild
Join Date: Jan 2014
Location: South Georgia
Posts: 1,120
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Re: What size Power Resistor do I need to pull down lithium cells?
A 100 feet roll of awg#14 copper house wire will give you a .516 ohm load between the white and black wires when black and white wires spliced together at the far end. 100 feet of #12 will give you .324 ohms under the same conditions.
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08-24-2017, 07:21 PM | #6 |
Gone Wild
Join Date: Sep 2010
Location: Erie, Pennsylvania
Posts: 989
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Re: What size Power Resistor do I need to pull down lithium cells?
Just a thought here, once you get all cell voltages close, maybe you could connect all cells in parallel (with inexpensive current limiting resistors to each cell) and then let the pack settle overnight to the lowest voltage. - RAY
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08-24-2017, 11:53 PM | #7 | ||||
Gone Wild
Join Date: Jun 2015
Location: St. Augustine, FL
Posts: 328
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Re: What size Power Resistor do I need to pull down lithium cells?
Sergio,
Thanks for taking the time to help me out, and more importantly, trying to teach this old dog some electrical tricks. I am so far out of my league when it comes to this stuff! Just ask PingEye3, I've been killing him offline! I was a bit incomplete in my original post, I apologize. I do have a BMS that will be attached to the cells, but the balancing current is only 50-60 mA. THAT would take forever to balance with a .6V difference, which is why I wanted to pull all the cells down to the same starting voltage, then attach the BMS and charger to bring them up more closely in balance. Quote:
Quote:
Quote:
Quote:
I am thinking I want to install a JDL404AH and shunt to keep track of AH used as another safety measure. |
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08-25-2017, 01:42 AM | #8 | |
Gone Wild
Join Date: Jun 2015
Location: St. Augustine, FL
Posts: 328
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Re: What size Power Resistor do I need to pull down lithium cells?
Quote:
3.89V starting 3.28V ending 123 AH capacity 2 each of .5Ω x 50W resistors Total R = R/2 =.5Ω/2 = .25Ω I = V/R = 3.89/.25 = 15.56A P = I^2*R = 242.11/.25 = 60.53W So I will be pulling ~15.5A/60.5W, decreasing to ~13A/44.1W as voltage drops. Here is where I go off the rails, because I can't figure out how to calculate how much voltage drop will occur over time with a given load. I read a JohnnieB post that said the Peukert factor was close to 1.0 so that part is negligible. Not that it would be usable to me anyway. So I went with this. If the normal capacity is 123 AH @ C1, then AH / A = Run Time, so 123AH/15A = 8.2 hours. So I'm guessing 8.2 hours to fully drain the cells. Only I don't want them fully drained. I need to lose 85% SOC, so that should be 8.2*.85 = ~7 hours. Anyway, that's my guess on how long per 3P cell it will take to drain with what I have on hand. I may need more power resistors, since I have 10 cells I need to bring down of the 14!! |
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08-25-2017, 06:07 AM | #9 |
Gone Wild
Join Date: Jan 2013
Location: Saint Petersburg, FL
Posts: 2,089
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Re: What size Power Resistor do I need to pull down lithium cells?
Also remember the AH rating is different if drained at a different rate. 100AH lithium cells are measured at C1. If you drain them at C5 they may only hold 85AH. I would assume draining them at C0.1 would mean they hold some higher AH value, maybe 115AH.
So don't be surprised if it takes longer than expected. |
08-25-2017, 07:02 AM | #10 |
Gone Wild
Join Date: Sep 2012
Location: Charlotte, NC
Posts: 9,329
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Re: What size Power Resistor do I need to pull down lithium cells?
bronsonj, unlike an FLA battery, a lithium battery will deliver the AH stored regardless the discharge rate.
JaxPilot, Your math is correct, just make sure you mount those resistors to a heat-sink surface or keep a fan running on them. The only thing that will affect the discharge time is the AH left in each battery, but since the voltage on a lithium cell is about the same from 20% to 80% SOC, it is difficult to get an estimate of the remaining capacity based on a voltage reading. This is one of the reasons you want to discharge them to some voltage pass that "knee" in the graph when the voltage decreases sharply. I think an automated way to add/remove the resistor load is about the only feasible way of doing it, otherwise You will need to spend a lot of time looking at the meter and run the risk of over-discharging them. Since You are only dealing with about 16A, a simple bosch style 12v/40A automotive relay will work for switching the load. The JLD404AH will work, it is just not quite as precise as the JLD5740 with its 10v range setting. Make sure you set the JLD404AH to 3 decimal places. |
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