I am trying to measure the charge capacity of some LTO battery cells. To do this I want to discharge each battery cell from 2.5 V to 1.7 V. I am going to measure and log the current over the discharge cycle and then integrate the obtained function over the testing time. Of course I would not have to do this if I had a constant current load, but I do not.

The cells are 20 Ah and I want the test to be quick. Thus to make the test take only an hour, the cell should be discharged at an average of 1 C, 20 amps. The cell is rated for up to 10 C so it is not a problem to go as much as 25 or 30 A if my set up can handle it.

Is the following set up feasible? One LTO cell charged to 2.5 amps connected to 10 AWG wire connected to a uni-direction Hall effect current sensor. The current sensor which is rated for up to 50 amps is then connected to 10 parallel 1 ohm resistors rated at 10 watts each. The other end of the resistors must then be somehow connected back to 10 AWG wire to connect back to the negative battery terminal. Each resistor will carry at most 2.5 amps at 2.5 volts for a maximum wattage of 6.25 Watts, well under the 10 watt threshold. The resistors are just there to dissipate the energy.

I am most concerned with how I could safely connect the 1 ohm resistors, which are rated for 10 watts but have 20 AWG wire leads, to the current sensor that is going to output up to 25 amps. Additionally will the heat generated by the resistors be a problem?

  • \$\begingroup\$ 22 AWG can handle 8A for reference, and 22 AWG has wire insulation that can melt. Component leads don't. If you bunch the resistor together it will still be fine because they are so far below their power rating that airflow is not as important. Just don't go overboard. \$\endgroup\$
    – DKNguyen
    Apr 17, 2019 at 18:07
  • \$\begingroup\$ Would you like get a power transistor and Op Amp or 2nd transistor on a breadboard to make a CC sink? \$\endgroup\$ Apr 17, 2019 at 18:18
  • \$\begingroup\$ I would also pulse the current every minute off to measure ESR from voltage rise/current drop but don’t go below 10% SoC \$\endgroup\$ Apr 17, 2019 at 19:41

2 Answers 2


There are several methods that can be used to connect your load resistors to your lead wires.

If this were me, I would simply strip about 6 inches of insulation from your 10 AWG conductors. Stretch them straight and temporarily fasten the wires to a board. Space the wires apart far enough that the leads from the resistors overhang the wires about 1/4" on each end of the resistors.

Now simply wrap the resistor lead one full turn around the 10 AWG conductor. Do one resistor right next to where the wire insulation starts. Get the resistor straight, then solder the resistor wire to the 10 AWG conductor.

You will need either a decent soldering iron or a soldering gun to properly make the solder connection.

Now space the remaining resistors along the bare conductors. When you are happy with the spacing, repeat the above process.

You can then remove the wires and resistors from the board. Hang the 10 AWG wires with resistors in the air and do your testing.

But there are other methods that work equally as well. You can get punched copper strip (looks like muffler strapping but made of copper) and mount that on wooden spacer blocks. Then wrap the resistor leads around screws that go through the holes in the copper strap into the wood blocks.

Whatever technique that you use, ensure that the resistors are able to have air flow around them for cooling.


I am assuming you are using cement/Ceramic resistors such as this, and you have to be concerned about the temperature of the resistor.
These can run at temperatures of 70-100degC, but since you've underpowered the units you might expect to see only 50-60degC.

The 10AWG wire is appropriate for your current flow, but I'd suggest that you use screwed connections to the resistors rather than soldered.
Something like this may be appropriate:

enter image description here

These come with a nice shorting strip and are more than capable of supporting your current flow. Screw four of these to a chipboard square and you can use your resistors with full length leads between the connections. Having all the resistors in close proximity will raise the temperature, but it should be quite manageable.

  • 1
    \$\begingroup\$ Why do you suggest screw connections instead of solder connections? \$\endgroup\$ Apr 17, 2019 at 19:05
  • \$\begingroup\$ What I suggested were screw clamps. They are a much more reliable connection than soldered connections where heat is involved. Solder is already a brittle connection, heat makes it worse. Think AC wiring where soldered connections are banned. You use either crimp, screw or high pressure push connections to avoid problems. \$\endgroup\$ Apr 17, 2019 at 21:44
  • \$\begingroup\$ Do you have suggestions for non-permanent connections between the wire and battery terminals? I ended up buying 8 AWG. And these are the cells scib.jp/en/product/cell.htm. They recommend spot welding but I don't want to do that. @JackCreasey \$\endgroup\$
    – jRRRR
    Apr 17, 2019 at 21:54
  • \$\begingroup\$ @jRRRR I'd suggest at 25A a simple pressure contact would be sufficient for your testing purposes. It's not clear from the datasheet what they suggest, but I'd start with no more than about 2-3lbs pressure using a simple copper tape such as this: amazon.com/Copper-Strip-043-Thickness-Shipping/dp/B07CQPTKDL \$\endgroup\$ Apr 17, 2019 at 22:26

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