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Few days ago I was trying to recharge a bank of ten batteries connected in parallel. I connected the charger directly to the battery terminals at 13.75 V as specified in battery manufacturer datasheet.

The problem: the batteries were about 11.4 V at that time and my DC supply transformer was supplying a the maximum current of 50 A. Well ... 50 A for 10 \$\times\$ 9 Ah batteries is way too much, not to mention that the wires started to melt due to high current so I realised that I need something to limit the current and quickly figured it out that I need a resistor.

I've seen numerous youtube videos and documentation about how a resistor works and it's limitations, I ended up using this calculator.

13.75 V at 4 A = 3.4375 \$\Omega\$ resistor, but I found a 3.3 \$\Omega\$ resistor which (based on the calculator) should be capable to limit current at 4.16 A, dissipating with 57.29 W of heat.

Big surprise when I connected it on positive terminal (tried on negative also) and only about 0.6 A of current was flowing to the batteries, despite the fact that the batteries were discharged and without any kind of resistor they alone would draw > 30 A.

So my question is: why is that? Why does the 3.3 \$\Omega\$ resistor limit the current to 0.6 A and not to 4.16 A as calculated. Is there a possible difference in resistance which doesn't allow the current to flow at max capacity?

I've also tested the resistor with a multimeter and the resistance is fine. I've also connected it to positive and negative of a single battery and about 4.3 A of current was flowing - so it works as it should but not when connected with the charger.

I have absolutely no clue about it, could someone please help me to figure it out?

Thank you!

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  • \$\begingroup\$ NB: Make sure each of the batteries voltage levels are about the same, or you'll have possibly high currents running between the batteries themselves, as well! \$\endgroup\$ – Steve Mar 15 '17 at 18:40
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You have to use the voltage difference to calculate the resistor. $$13.75~V - 11.4~V = 2.35~V$$

If you measure the voltage across your resistor, this is also approximately what you will read.

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  • \$\begingroup\$ So ... if I have: Power source: 13.8V / Minimum battery voltage: 11V - 13.8 - 11 = 2.8V - rapidtables.com/calc/electric/watt-volt-amp-calculator.htm - 2.8V at 10A = 0.28ohms resistor ? \$\endgroup\$ – user3405598 Mar 15 '17 at 9:26
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    \$\begingroup\$ Yes, and the resistor will dissipate 28W, in other words you need a pretty beefy resistor with either a big heatsink or active cooling of some sort. \$\endgroup\$ – Dampmaskin Mar 15 '17 at 10:06
  • \$\begingroup\$ Thank you very much! Didn't knew that ... it's logical but not for a newbie like me. \$\endgroup\$ – user3405598 Mar 15 '17 at 15:15
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    \$\begingroup\$ Current limiting with resistors is inefficient, and best suited for low voltages and/or low currents. When you're dealing with several amps, it's generally better to use a power supply or charger with current limiting functionality. \$\endgroup\$ – Dampmaskin Mar 15 '17 at 16:57

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