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I have 3 18650 Li-ion cells connected in series. Each have a capacity of 2200 mAh and can discharge at 10C. First of all , tell the maximum amps I can draw from them. If I am not wrong, I can draw 22 A. Now, I have a setup as follows- I have a A2212 2200kv brushless motor and 30 A SimonK ESC connected and powered by the battery setup as descried earlier. I have a Arduino to control the speed of the motor from 0% to 100%.I connect the motor to a 1045 propeller and run the motor in the loop as Arduino is programmed. The motor speeds up until 50% throttle, but post that it starts slowing down and starts running steadily and I can make out by the sound that the motor is spinning at lesser than 50% throttle. I tested the circuit with a multimeter and with some research I understood that the voltage as the motor reaches 50% throttle reduces to 9 V which triggers the ESC's low-voltage protection system. I felt that this was an issue with the propeller and switched to a 7038 (7 inch propeller) but it ended with similar results. I also found out that the motor was only drawing 4 A of current by 50% throttle just before the cut-off was triggered. Can someone please help me figure out better Li-ion battery specs which can help eliminate this issue and prevent the voltage from drooping so low, or any circuit or module which can help reduce if not eliminate this issue maybe like using capacitors, buck converters etc. ESC - https://robu.in/product/standard-30a-bldc-esc-electronic-speed-controller/ BLDC MOTOR - https://robu.in/product/a2212-10t-13t-2200kv-brushless-motor-with-soldered-connector/ LI-ION CELLS - https://robu.in/product/orange-isr-18650-2200mah-10c-lithium-ion-battery/ BATTERY PACK - https://robu.in/product/black-plastic-storage-box-case-holder-battery-3-x-18650-cell-box-without-cover/

Just another point, I have fit the cells in the battery holder and connected the thin wires from the holder directly to the ESC's power wires, just saying if this might provide more context!

Appreciate any relevant help!

Regards

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  • \$\begingroup\$ After reading the one answer you have right now, I looked. Here's a 10C battery and it specifies \$\le 25\:\text{m}\Omega\$. That feels reasonable to me. So, assuming the worst case there, this works out to about \$1.65\:\text{V}\$ drop, under peak specified load. (They won't last long, though. 6 minutes?) Perhaps the batteries you have are incorrectly rated (to sell them.) Or you are measuring them when they are closer to their discharged state? \$\endgroup\$
    – periblepsis
    Commented Oct 2, 2023 at 12:18
  • \$\begingroup\$ I am pretty sure that the batteries are correctly rated as I could retrieve the expected number of amps. I have charged the batteries to full and tried as well but the only change here is that it can go to maximum to 55% throttle compared to the usual 50% throttle. The motor with 7038 propeller was drawing 4A at 50% throttle. Hope this helps you navigate and thanks for the reply! \$\endgroup\$
    – S.A
    Commented Oct 2, 2023 at 12:35
  • \$\begingroup\$ Sorry to overlook your first statement, but there are 3 batteries are connected in series so the drop must add up and be around 3-4 V which is what is happening(from 12.6V to 9 V), I am a beginner in this so please correct me if I am wrong. \$\endgroup\$
    – S.A
    Commented Oct 2, 2023 at 12:43
  • \$\begingroup\$ No. If there is \$25\:\text{m}\Omega\$, worst case, for each battery... then for all three in series there is worst case \$75\:\text{m}\Omega\$. At \$22\:\text{A}\$ this will drop at most \$1.65\:\text{V}\$. \$\endgroup\$
    – periblepsis
    Commented Oct 2, 2023 at 12:45
  • \$\begingroup\$ Ok, highlighting some points from my question, the current draw by the motor connected to the 7038 prop was around 4A just before the low-voltage protection was triggered. The Voltage drop of the entire pack was from 12V to 9V. Now, I am plugging in the values to V= IR, for each cell- 1 = 4*R -> R = 0.25 Ω , Is there an issue with the values or the calculation, if not what can I do to fix the overall issue ? \$\endgroup\$
    – S.A
    Commented Oct 2, 2023 at 12:54

1 Answer 1

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I figured out the answer. I was using individual Lithium-Ion cells all connected to each other using a temporary battery holder. Because of the huge resistance between the parts of the holder, there was a significant voltage sag in the Pack's Voltage not the Cell's Voltage. I bought a new spot-welded and factory made pack which works perfectly fine now.

Thanks to -

Finbarr, Audioguru and periblepsis for their quick and useful responses

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