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I have a 4S LIPO battery 5800mAH at 14.8V and I want to use it to power some 7.2V motors I have.

The Rover 5 Chassis to be exact (https://www.sparkfun.com/products/10336), the only problem I have is I don't know the most efficient way to step down the voltage. I bought this battery to use with a quadcopter and I realy don't want to buy another one is there any way I can use this?

I know about regulators but that must be really inefficient.

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  • \$\begingroup\$ Yes, a linear regulator would be extremely inefficient, but have you considered a switch-mode buck regulator? \$\endgroup\$ – brhans Sep 21 '14 at 12:19
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As @brhans says in his comment, the answer is a buck style regulator. Those can be 90%+ efficient. They are a type of switching power supply that also uses an inductor. A buck/boost power supply can even raise the voltage, but you don't need that in your application.

Buck regulators are quite common in motor circuits, so you should be able to find a 7.2V buck regulator as an inexpensive off-the-shelf part.

In the RC world, such a supply is called an UBEC (which stands for "Universal Battery Eliminator Circuit"). A quick Google search on "7.2V UBEC" found lots of results, including this one:

7.2V "UBEC" regulated power supply

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If I'm reading the spec correctly, then you have two motors per side of the chassis, and the chassis doesn't include the motor drivers.

If you're not putting mecanum wheels on it, the two motors will be driving the tread in the same direction, so you could just run them in series.

Alternatively, decrease the PWM duty on the h-bridge driving the motors so they get a lower average voltage.

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    \$\begingroup\$ Just a precision: powering a motor with a voltage higher than its nominal one is generally ok, as long as you don't get higher torques than the max one specified by the motor manufacturer and that the motor is not heating up too much. The PWM solution proposed in the answer will work well to limit the average voltage on the motor (i.e., you can limit the PWM duty cycle so it never reaches 100%). \$\endgroup\$ – Ale Sep 21 '14 at 13:45

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