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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.

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  • \$\begingroup\$ buck-boost converters are the most efficient way (it can reach up to 90+% efficiency, which is great...) \$\endgroup\$ – Eugene Sh. Mar 19 '15 at 16:09
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    \$\begingroup\$ If you find one that steps up voltage and current at the same time, send me some, I want to chain them and power my house from a pair of AA batteries... \$\endgroup\$ – PlasmaHH Mar 19 '15 at 16:10
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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

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What you want is a type of switching power supply called a boost converter. These convert a low DC voltage at high current into a high DC voltage at low current. In theory with ideal parts, they are 100% efficient. For what you want to do, real efficiency of 80% would be relatively easy. Over 90% is doable, but requires tricks like synchronous rectification (a method of working around the fact that the diode is not ideal and has a small voltage drop across it when conducting).

Here is the basic concept of a boost converter:

When the switch is closed, current builds up linearly with time thru the inductor, storing energy in the inductor. When the switch is opened, the current must continue flowing somewhere in the short term, which is thru the diode onto the output. Now the voltage across the inductor is reversed and the current decreases linearly with time. When it reaches 0, the diode stops conducting and everything is off until the next time the switch is closed again. This is the discontinuous mode operation case. In practise, the switch is often closed and opened so fast that the current doesn't die down to 0 during the brief time it is open. That is called continuous mode operation.

In any case, a control system watches Vout and adjusts the timing of the switch being opened and closed to dump more or less current onto Vout depending on whether it is low or high.

Boost converters are quite common, and there are many chips out there that integrate a lot of the function of them. For your voltages and currents, you should be easily able to find a chip that needs little more than the inductor and the output cap externally. I would start by checking the offerings from Microchip, ST, and TI. In the unlikely event I don't find something suitable there, I'd check Linear and others that are generally more pricey.

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