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I am a little confused, shouldn't I take into account the maximum battery voltage when trying to add-up batteries in order to achieve certain voltage?

For instance if I want to drive a 48V 1000W DC Motor with li-ions, I need 11 li-ions to approximate 48V (46.2) with probably a voltage regulator or a step up module. If I make my calculations with 3.6V I will need 3-4 more batteries and a step-down module.

So what should I watch when making such calculations, maximum or nominal voltage or it doesn't matter?

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    \$\begingroup\$ Does your motor require exactly 48V or will it work over a range of voltages? How long do you want to run the motor from a set of fully charged batteries? \$\endgroup\$ Commented Oct 19, 2018 at 19:55
  • \$\begingroup\$ @ElliotAlderson No I don't think the motor needs exactly 48V, it's just a motor. But if the voltage is higher it will wear out quicker. If I supply lower voltage it will be slower, but I don't think it will overdraw current to make-up for the loose. 1 hour is a good start I think. \$\endgroup\$
    – Edenia
    Commented Oct 19, 2018 at 19:58
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    \$\begingroup\$ If you use a step-up voltage converter to bump it up to exactly 48V, then you won't have to worry about the voltage fluctuations on the batteries (until they are dead). That way you can do what you said, "take into account the maximum battery voltage when trying to add-up batteries". \$\endgroup\$
    – mith
    Commented Oct 19, 2018 at 19:59
  • \$\begingroup\$ Also each battery is rated 20A which means it can supply the needed current, but it will probably overheat them quickly too, without BMS. \$\endgroup\$
    – Edenia
    Commented Oct 19, 2018 at 20:01
  • \$\begingroup\$ Do you have a constant speed requirement? Can you accept the efficiency loss from a power converter? \$\endgroup\$
    – scorpdaddy
    Commented Oct 19, 2018 at 20:05

1 Answer 1

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If you don't mind a variable speed from your motor, then powering it with a variable voltage from a battery of Li-ions is OK. However, you need to consider what happens at the maximum and minimum voltage of the cells. Your particular motor application (which you don't give us the detail of) will control what's important.

There are two issues with providing too high a voltage.

(1) The motor goes faster than you want. This will mean it runs less efficiently, and wears out more quickly.

(2) Motors are usually rated to be able to switch on, without the current surge they take when stationary being so large as to demagnetise (read 'destroy') the motor. Applying a higher voltage than rated could do this, unless you also use active current limiting.

Too low a voltage will mean the motor may go too slow, or not be able to draw enough current to develop the required torque.

If you need to provide a precise 48v to your motor, then there are two easy ways to organise your cells.

(a) Make the highest voltage from the battery < 48v, and use a simple boost converter. Uses fewer cells. Results in higher currents in the battery wiring for the same motor power.

(b) Make the lowest voltage from your battery > 48v, and use a simple buck converter. This will also allow you to control the motor down to zero speed, and to limit the current easily avoiding the switch on surge. These two factors together with the lower battery current will often make this the preferred choice, if you're happy with the larger battery.

(or c) Have the voltage range from your battery overlap that of the motor, and use a buck-boost or other converter that allows you to convert up or down in voltage.

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  • \$\begingroup\$ You need to teach people how to construct an answer. \$\endgroup\$
    – Edenia
    Commented Oct 20, 2018 at 19:25

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