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I'm building a battery to drive a small 12V motor which is rated to draw max 7A.

The battery has the following stats:

  • Nominal voltage: 14.4V
  • Capacity: 20.4Ah

So I calculated that the energy should be 293.76Wh.

So I understood the discharge time should be calcuated as:

Energy/(Voltage * Current)

Since the motor is rated for max 15V, which is below the 16.8V max of the battery, I will use a buck converter to bring the voltage down to 12V.

So my question is, would the calculation then be:

293.76Wh / (12V * 7A)

Or, do I use the nominal voltage of the battery in the calculation?, i.e.

293.76Wh / (14.4V * 7A)
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  • \$\begingroup\$ What does this mean: Since the battery is rated for max 15V, which is below the 16.8V max of the battery $$$$ What does this mean: I will use a buck converter to bring the current down to 12V.. $$$$ Sounds like gobbledygook to me. Available useful ampere hours will be about 50% those quoted by the battery manufacturer. \$\endgroup\$ – Andy aka Apr 18 at 10:58
  • \$\begingroup\$ You need the datasheet showing you the discharge curves, then use those formulas, because the battery will have a fast initial discharge, a flat(-tish, plateau) region, then it will go downhill fast. Even then you'll still be only approximate. \$\endgroup\$ – a concerned citizen Apr 18 at 11:03
  • \$\begingroup\$ This was a typo - the motor is rated for 15V max according to the data sheet \$\endgroup\$ – sak Apr 18 at 11:04
  • \$\begingroup\$ The battery will only have its full Ah capacity at some arbitrary and possibly small load. Capacity will be much lower at higher currents. \$\endgroup\$ – K H Apr 18 at 11:07
  • \$\begingroup\$ If you start motor often, that draws 10x approx the max rated load initially. So derate accordingly. \$\endgroup\$ – Tony Stewart EE75 Apr 18 at 13:50
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You can use:

293.76Wh / (12V * 7A)

Or:

293.76Wh / (14.4V * 7A*(12V/14.4))

Equals to:

293.76Wh / (14.4V * 5.83A)

In any case, you need to take current and voltage from the same part of the circuit. My first option is if power is measured near the motor. Second-third is if power is measured near the battery. Result is the same. Also it is a good idea to account for battery inefficiency at high current, it could be about 20% loss. And converter loss. Also could be about 20%.

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Your assumption is to design a Buck converter and then you ask to compute power .

This is the wrong assumption.

DC motors have a DC resistance or DCR that is about 10% or the rated load impedance and thus starts with 10x the rated current at rated voltage. Buck converters do not have this excess capacity beyond the millisecond storage time in the capacitors.

However ESC motor controllers are designed for this transient current and heat and is also limited by your acceleration controls which reduces the current by PWM thru the motor RL impedance starting up.

Thus if you check ESC motor capacity specs rated for a 100W 12 Motor, once expects the rated current to be <8A and the surge currents to be ~ 80A peak and thus require heat sinks to dissipate the Ron load loss.

For cooler operation derate the ESC capacity and choose a motor controller that suits your interface needs for 15V. It is also likely to be more efficient in power conversion. Assume >90% but verify.

The Li Ion battery specs are usually:
Nominal: 14.4V;
Charge Cut-off: 16.8V; Discharge Cut-off: 10V

The cut-off quickly drops to the 15V on initial load or after a long settling time.

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