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In a cave we are exploring we are currently blocked by a small siphon, which we would like to empty into another gallery.

For this, I would like to use a 12 V pump (not selected yet). As a power source, we would like to use the 36 V Li-ion batteries from our big hammer-drill.

I see 2 options:

  1. Using a DC-DC converter to convert the about 36 V to 12 V
  2. Using a MOSFET (+ free-wheel diode) to apply PWM, with a duty cycle of about 33% (to be adapted based on measured battery voltage)

For option 1, is there any precaution to be taken because the load is inductive rather than resistive?

For option 2: is it OK to apply such a high voltage on a 12 V motor if the average voltage remains within specs? Is it OK if the motor is a brushless DC instead of a brushed one (I think not, but not sure)?

Is there a significant difference in efficiency between both methods?

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With 48 V PWM on a brushed motor you are going to get 4 times the resistive and eddy current heating you would have gotten with direct 12 V drive. That might be OK, or it might not.

With 48 V PWM on a 12 V brushless motor, it will break immediately.

I'd recommend a DC-DC converter; one with an output current limit may be able to start the pump without overloading itself, if not, you may need to turn the voltage down and gradually raise it (over 10 seconds) to get the pump started.

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For option 1, is there any precaution to be taken because the load is inductive rather than resistive?

Not really, as long as the DC-DC converter can handle the motor´s starting current which can be pretty high.

For option 2 : is it OK to apply such a high voltage on a 12V motor if the average remains voltage within specs? Is it OK if the motor is a brushless DC instead of a brushed one (I think not, but not sure).

For a brushless motor, it probably won't work. A brushless motor is really an AC motor with a controller providing DC->AC conversion. Feeding that controller with high voltage and PWM will likely not work and might burn the controller.

For a brushed motor, it shouldn't be a problem as long as the PWM is relatively high frequency. If the PWM is too low frequency it might cause excessive wear on the motor as the peak current will be very high causing stress on the brushes.

Is there a significant difference in efficiency between both methodes?

Very little difference.

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  • \$\begingroup\$ Thanks. So if I take a brushed motor, I take whatever I want. If it is a brusless, then I have to take the DC-DC option \$\endgroup\$
    – Sandro
    Jan 9, 2022 at 12:48
  • \$\begingroup\$ I'd say so, yes. \$\endgroup\$
    – Klas-Kenny
    Jan 9, 2022 at 12:56
  • \$\begingroup\$ An ESC-motor combination does act like its own DC-DC converter to some extent. If you can find an ESC rated to accept the 36 V, it should be able to run the motor at low speed, with a corresponding saving in the current it draws from the battery. \$\endgroup\$
    – Neil_UK
    Jan 9, 2022 at 13:02
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    \$\begingroup\$ have you considered a drill powered pump? \$\endgroup\$ Jan 10, 2022 at 12:39
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For brushed DC motor, yes! I tested on my boat with 12 V trolling motor powered directly using 4 100 W solar panels in series. The solar panels have open voltage of 19.2 V and MPPT voltage of 16 V. Connect 4 in series is almost 80 V, and I used 6 kHz PWM to drive the 12 V motor. The setup worked and can move a 2200 lbs, 25 feet boat at 3 km/h.

I think the efficiency could be better because the resistive heat loss should be lower.

The accepted answer is totally wrong on resistive loss.

For the same output, average current is 25 % at 48 V with PWM, how can resistive loss 4 times higher?

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    \$\begingroup\$ I think you made a wrong assumption. At low speed, the motor can be considered a linear component (inductor and resistor in series). So if you multiply instantaneous voltage by 4, you also multiply instantaneous current by 4. As you use 25% PWM, the average current remains therefore the same. However, resistive losses are R*I², so when using 4 time the voltage/current, you get 16 times the power in the resistor. As you use 25% PWM, you get 4 times more power losses in the resistor on average. The accepted answer still seems correct to me \$\endgroup\$
    – Sandro
    Sep 28, 2022 at 18:12
  • \$\begingroup\$ @sabdro say the total output is constant 480 W, 48 V , average current is 10 A, 12 V , average current is 40 A, I am not talking about 25% PWM, I am talking about average current is 25 % \$\endgroup\$
    – mikewen
    Oct 26, 2022 at 21:31

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