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Suppose I had a unit that could accept DC power between 6 and 14 volts, and the choice of using either a 7.4V 1000 mAh or 11.1V 1000 mAh battery.

Do either of these choices offer a clear advantage? What are the tradeoffs involved?

update: I have not selected the device yet. It will be a small monitor (3-5 inches) to display telemetry data from an RC plane (which explains the battery voltages -- it's convenient to use either a 2S or 3S LiPo pack). Now I've got a better understanding of the principles involved, and know the main thing I need to check; whether it has a linear or switching power supply.

Mark Harrison
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    Can you give us more information, will the device pull the same power regardless of voltage? What is the device doing with the input power, is it a device that will improve performance with increased power input or a device that will degrade? – Kortuk Sep 20 '11 at 08:25
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    Yeah, the correct answer depends heavily on the type of load the units presents. (For example, if it uses a linear regulator to drop to 5V, the answer is probably the 7.4V battery. If it uses a switching regulator, the answer is probably the 11.1V battery.) – David Schwartz Sep 20 '11 at 10:51
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    "Now I know the main thing I need to check; whether it has a linear or switching power supply." -- Easiest to check with an ammeter. If the current goes down as the voltage goes up, you have a switcher. If it stays put, you have a linear. If it goes up, you have a resistor. – Mike DeSimone Sep 21 '11 at 03:48
  • @Mike DeSimone - Nicely put summary of test for Switcher / linear / resistor feed. Note that "resistor" can include "shunt regulator which "burns off" excess energy rather than "blocking it" as a linear regulator does or converting it as a switcher does. – Russell McMahon Sep 22 '11 at 10:27

3 Answers3

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Work at the lowest possible voltage. If you draw \$100mA\$ at \$7.4V\$ that's \$740mW\$, at \$11.1V\$ that's \$1100mW\$, or 50% more. The capacity of the \$11.1V\$ is also 50% higher, so there's no difference there, apart from the \$11.1V\$ battery being bulkier and heavier and more expensive.

But if the load is resistive there will be indeed a difference. Suppose the resistance is \$74\Omega\$, then the power drawn from the 7.4V battery is

\$P = \dfrac{V^2}{R} = \dfrac{(7.4V)^2}{74\Omega} = 740mW\$

while the power drawn from an 11.1V battery is

\$P = \dfrac{V^2}{R} = \dfrac{(11.1V)^2}{74\Omega} = 1665mW\$

or 2.25 times as much. Since the \$11.1V\$ battery's capacity is only 1.5 times as high it will drain faster.

edit
Kevin wants to read about linear and maybe other regulators
If the battery feeds a regulator then the circuit's power will be independent of the battery's voltage. Say you regulate to 5V and draw 100mA from that, then the circuit will always consume 500mW, no matter what the battery's voltage is, because the 5V is fixed. If it's a linear regulator then the voltage difference between battery and regulator output will cause dissipation in the regulator. For the 7.4V battery:

\$P = \Delta V \cdot I = (7.4V - 5V) \cdot 100mA = 240mW\$

If the battery's voltage is the 11.1V however this becomes

\$P = \Delta V \cdot I = (11.1V - 5V) \cdot 100mA = 610mW\$

so your power loss in the regulator will be higher for the higher voltage battery. But just like in the first example the battery's longevity will be the same:

Battery life = \$ \dfrac{1000mAh}{100mA} = 10h \$

since the 11.1V battery has a higher capacity. It's just a pity to waste that capacity instead of using it to get a 50% longer battery life.
A switcher (SMPS, or Switch-Mode Power Supply) solves this. A switcher allows high efficiency power conversion. Let's assume an ideal switcher with a 100% efficiency. If you draw 100mA from the 5V output this will represent only 68mA from the 7.4V battery, or 45mA from the 11.1V battery. That's because the power remains the same:

\$P = V \cdot I = 5V \cdot 100mA = 7.4V \cdot 68mA = 11.1V \cdot 45mA \$

A 1000mAh battery will then last

\$ \dfrac{1000mAh}{68mA} = 14.8h (7.4V) \$ or \$ \dfrac{1000mAh}{45mA} = 22.2h (11.1V) \$

so here we see that we have a 50% longer battery life for the bigger battery, since we don't have conversion losses. Remember that we calculated for a 100% efficient switcher. We could recalculate for a more realistic 85% conversion efficiency, and find the same ratio. I'll leave that as an exercise for the interested reader :-).

stevenvh
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  • Good analysis, but it should be noted that this is only accurate if the device has a linear regulator. I agree that the 6-14V input range can be reasonably assumed to be a 5V linear regulator with a power maximum at 14V, but it's not stated (yet...). – Kevin Vermeer Sep 20 '11 at 10:28
  • @Kevin - added to my answer. Happy now? ;-) – stevenvh Sep 20 '11 at 11:17
  • It is an improvement, but I was happy to leave it as a comment. I upvoted you before you made the addition. – Kevin Vermeer Sep 20 '11 at 11:48
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You don't mention what kind of unit you have, especially what kind of power supply circuit it has. You need to know what current it draws at those two voltages. You did not tell us, so for all we know it could be drawing 0 mA at both voltages. That would give you infinite lifetime with either battery, which might be what you want (or not, you did not tell! but this is a reasonable assumption).

This is just speculation:

  • Your gadget could have a linear regulator, which essentially draws constant current. In this case both batteries will roughly last equally long. But the 11.1V battery will probably be heavier and more expensive.

  • Your gadget could have a switch mode regulator, which essentially draws constant power. In this case the 11.1V battery stores more energy, so if you are after the longest lifetime, that one would be preferred.

There are other (less likely) situations, for instance: your gadget is a heater with a simple resistive load.

Matt B.
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Wouter van Ooijen
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2

Summary:

  • With a linear regulator use 2 cells

  • With a switching regulator, which you should use, 3 cells provide about 50% more energy storage than 2 cells.


As others have said, a better explanation of the requirement I assume that is 2 x or 3 x LiIon cells (as 3.7V nominal each.)
You say 100 mAh so maybe they are AA (14500) cells or maybe LiPo flat pack.

The 2 cell solution can go as low as 6/3 = 3V/cell to give 6 Volts. This is at the comfortable lower limit of LiIon so is acceptable.

LiIon will be 4.2V when fully charged = 8.4V/2 cells or 12.6V/3 cells - still inside your spec.

As Steven says - IF you are using a linear regulator you will waste more energy with the 3 cell solution. BUT if you use a switching regulator (SR) (as you really really really should) then energy efficiency will be similar in each case.

With a SR the 3 cell solution gives you about 50% more energy storage.


Added. Catalysed by Mike's comment. (@Mike DeSimone)

What sort of internal regulator?

You'd hope that anything made to run from 6 to 14 V would use a switching power supply inside, but maybe not. So, to determine what sort of regulator is inside a device, increase supply voltage and watch what the current does. Then -

  • A switching power supply draws approximately constant energy.

    Vin x In x Efficinecy = Vout x Iout.

    So, as Vin rises, Iin falls.
    Doubling Vin about halves Iin.

    A switching regulator is usually the most efficient solution except when Vin is close to Vout, but is noisier and usually somewhat more expensive and complex than a linear regulator.

  • A linear regulator passes the input current through to the output and drops any excess voltage across the regulator. It also needs a (usually) relatively small amount of current for its own use. So

    Iin ~= Iout

    When increasing Vin , Ion should stay relatively constant.

    A linear regulator should be used when Vin is not massively higher than Vout.
    Power_dissipated = (Vin-Vout) x Iload.

  • A shunt regulator or a zener supply or a resistive dropper to an unregulated suppl (very bad) maintains output voltage and current and as Vin is increased it draws increasing current over and above what is needed and dissipates the energy as (Vin-Vout_internal) x Iin

    So as Vin is increased Iin increases.

    It is possible to write equations for what you can expect to see but there are a surprising number of variable factors which make it not worth doing. A shnt regulator is usually very cheap but should usually only be used for low power and when Vin is not vastly more than Vout_internal,

  • -
Russell McMahon
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