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I have a circuit that is supposed to run four 12 V DC motors with built-in motor drivers off of one LiPo battery. The LiPo battery (single cell 3.7 V) is boosted to 12 V (the converter circuit is rated for 1.75 A) and the 12 V line is connected to four separate motors on the PCB.

When I flip the power switch with more than one motor connected, the power LED immediately turns off and the battery only outputs 2.5 V (the battery is fully charged and outputs 4.2 V with the circuit off). When connecting a single motor by itself, there are no problems, the motor runs perfectly.

The battery is a 6000 mAh battery, with a supposed .5C continuous discharge current allowed with a maximum of 1C discharge current. I measured the current that one motor by itself takes when hooked up to a 12 V power supply, and I found a single motor draws 70 mA of current. In my eyes, four motors should not be causing an overcurrent issue with the battery, if that is their current draw.

What could be the cause of the battery shut-off? I have included the useful parts of my schematic below.

battery and 12 V converter

Battery to 12 V converter

12 V motor connections (right) and load switches to control motor power on/off (left)

12 V to load switches that control motor power on/off and the four motor pin headers

UPDATE: If I turn on the circuit with one motor connected and then connect the other three the battery never turns off and all four motors work at full speed. The turning-off behavior only happens when all four motors are connected before the power is turned on.

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  • \$\begingroup\$ What is the maximum current output characteristic of the battery? \$\endgroup\$
    – Ilya
    Mar 21, 2022 at 17:06
  • \$\begingroup\$ You are drawing to much current. When you boost the voltage you also boost the current. Assuming a 200mA load at 12V and a battery voltage of 3V you need 800mA * 1.1 assuming a 90% efficiency of the boost converter leaving you at 214 mA current on the battery. \$\endgroup\$
    – Gil
    Mar 21, 2022 at 17:12
  • \$\begingroup\$ @Gil You are saying that if the motors need a 200mA load the battery would see a ~880mA load in your example? Is that because the voltage is being boosted by ~300%? Can you explain the math you mention? \$\endgroup\$
    – Dodger8188
    Mar 21, 2022 at 17:44
  • \$\begingroup\$ @ilya I mentioned it in the post, but the battery can sustain 3A output and has a max output of 6A. \$\endgroup\$
    – Dodger8188
    Mar 21, 2022 at 17:45
  • \$\begingroup\$ @gil did you do 12v/4.2v * 70mA/motor * 4motors = 800mA? \$\endgroup\$
    – Dodger8188
    Mar 21, 2022 at 18:17

1 Answer 1

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Never design a 3.6 to 12V DCDC converter for driving motors. You will never get an economical low driver impedance.

The static DCR resistance of the motor will short circuit your converter unless it is rated for 100x the max rated current of 1 motor. The idle current of your motor is irrelevant if it is a light load. No load can be 1% of the start current unless you are using speed and acceleration control to limit current.

Consider the max rated current and multiply by 10 to determine the R=V/I or just measure the DC resistance (DCR) Then transform that impedance to primary by \$R_{load}={(V_i/V_o)}^2 * DCR * 0.9\$ for 90% efficiency .

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