Electronic – How to efficiently boost a low voltage high current input to a high voltage low current output

batterieslithium ionpowerpower supplyswitch-mode-power-supply

I have a 3.7V 6600 mAh lithium-ion battery that I'd like to use to power some small 6V motors that only require 500 mA. However, I need the battery supply to last as long as possible (which is why I'm using a large mAh rated battery).

How would I convert the lithium's output to 6V @ 500mA as efficiently as possible using an off-the-shelf component?

I see a ton of cheap Chinese buck-boost dc-to-dc converters with switching regulators that claim to do this, but are they the most efficient method? I'm worried these are designed to step up both the voltage and current, not invert them as I need, and so will still be wasting a lot of watts.

Best Answer

The most efficient solution widely available to convert a voltage into a higher voltage is a Boost Converter.

A Boost converter is efficient because it is a DC-DC Power Converter. Other methods such as charge pumps are typically voltage converters. The distinction is important because by transferring power from the input to the output it's theoretically (but not practically), 100% efficient.

Going by your numbers and a 100% efficiency your 6V, 500mA (3W) output would draw 3W from the input (i.e. 3.7V, 811mA).

More practically, a 90% efficiency gives an input power of 3W/0.9 = 3.33W = 3.7V, 900mA