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Author Topic: Lithium cells / batteries. More efficient to step-up or step-down  (Read 753 times)
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Hey guys,

I'm building a battery powered LED project.
I need a maximum of 1A @ 5v. Current draw will typically be less.
I want to use "14500" lithium cells.
I plan to use a separate regulator board (to meet my current demands) and bypass the onboard regulator of the arduino.
Size/weight are issues. Through-hole components preferred for ease of prototyping.

1. Is it more efficient to step up a single 3.7v cell or step down two cells in series (7.4v)?
2. Will I run into issues using 14500 cells in parallel to increase my capacity?
3. Will I run into issues using 14500 cells in series? (They have onboard protection circuits)



Possible equipment options:
Batteries: (upto 2C discharge rate) http://dx.com/p/trustfire-protected-14500-3-7v-900mah-rechargeable-lithium-batteries-2-pack-26124
Voltage Regulator:
Both: http://dx.com/p/mini-dc-dc-voltage-stabilizer-regulator-module-red-126106
Step up (TDFN package): http://www.maximintegrated.com/datasheet/index.mvp/id/5770
Step up (TDFN package): http://www.maximintegrated.com/datasheet/index.mvp/id/5127
Step up (SOT23 package): http://www.maximintegrated.com/datasheet/index.mvp/id/2451



« Last Edit: December 19, 2013, 04:38:57 am by Panici » Logged

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My gut reaction is that having a single cell is going to simplify things(*) even if it doesn't
bring any efficiency improvement - at the expense of less endurance.

(*) no charge balancing needed, fewer mechanical connections, simpler to monitor
one cell.
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My gut reaction is that having a single cell is going to simplify things(*) even if it doesn't
bring any efficiency improvement - at the expense of less endurance.

(*) no charge balancing needed, fewer mechanical connections, simpler to monitor
one cell.

Cells are being charged in a balanced charger anyways. None of your other points are concerns of mine. Mostly looking at efficiency here to prolong usable battery life.
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For what it's worth, here is my opinion.

 A switching regulator in a step down configuration is usually 80-90% efficient in most cases.  You will not get that kind of efficiency from a step up or boost scenario.  In simplest terms, if you boost the voltage, you get lower instantaneous current.
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2. Will I run into issues using 14500 cells in parallel to increase my capacity?

Yes.

3. Will I run into issues using 14500 cells in series? (They have onboard protection circuits)

You can discharge them in series but you can't charge them that way.
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Take a look at this boost regulator from Pololu. The efficiency is over 80% at up to 700 mA output, with 3.3 V input. The efficiency will be a bit lower as the voltage drops below that, but the typical behavior is well documented on this page http://www.pololu.com/product/2115
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If you google "switching regulator efficiency", you will find lots written about this subject, for example http://cds.linear.com/docs/en/application-note/an46.pdf.

What sort of LEDs will you be using, and how many of them? if you are planning to run the LEDs from the 5V supply with a series resistor, then that is a simple but inefficient solution. A more efficient solution would be to run the entire system from 3.7V. However, if your LEDs have about 3V forward voltage, then you may need to use constant-current drivers with them to avoid too much loss of brightness as the battery discharges.
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More sophisticated boost regulators can get >95% efficiency, it just depends how many
discrete MOSFETs you are prepared to use, though I suspect there are some one-chip
boost regulators now that use synchronous switching rather than a schottky diode
(a major source of loss for low voltage converters).

The awesome LT3791 claims 98.5% boost or buck efficiency at 100W, but needs 4
external MOSFETs and lots of other components, this would be overkill for a small
DC-DC converter, but the principles are sound - live in hope of finding a one-chip
solution like this for all low power buck/boost needs!
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One other factor is the same one that makes transmission lines run at very high voltages. Stepping down will draw lower current from the battery and incur less I-squared-R loss.
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