Electronic – Extending battery life: Calculating advantage of adding more batteries

The setup in your answer is a bad one.

First you parallel two batteries, which you shouldn't, because their voltages are never exactly the same. Their low internal resistance will cause a current from one battery to the other. So feed three LEDs from 1 battery, and the other three from the other.

Then you place all LEDs parallel, which means that you lose (12 V - 3.2 V) x 20 mA = 176 mW per LED in its series resistor, while the LED itself uses only 64 mW. Total power loss is more than 1 W. That's because the large voltage difference between battery and LED. The best way to get a longer endurance is to keep losses in the series resistors as low as possible. So better place two times three LEDs in series, so that their total current is 40 mA instead of 120 mA. The power loss in the series resistors is then (12 V - 3 x 3.2 V) x 20 mA = 45 mW per 3 LEDs, or 96 mW in total. That's less than 10 % of the power loss for all LEDs in parallel.

Then your batteries will last 100 mAh / 40 mA = 2.5 hours or 150 minutes. This is pretty optimal. The batteries' capacity is 1200 mWh, and the LEDs consume 384 mW, so with an ideal conversion you can get a little over 3 hours out of them. But the most efficient conversion using a switching current regulator will get you maybe 85 % efficiency, and then you only gain 9 extra minutes.

edit re comments
An alkaline battery's voltage quickly drops by 10-15 %, and then remains more constant for a great part of the discharge cycle. So either you calculate the resistors for a larger current at the start, and 20 mA for the rest, or for 20 mA at the start, and a lower current later on. The latter solution will give you a longer battery life, but a bit less brightness.

jippie suggests to use a switcher anyway to get more out of the batteries, and it's a thought. You'll have to place the batteries in series to get 24 V to allow a voltage drop as high as possible. The larger Vin/Vout ratio of the switcher will make it less efficient, but overall you should get some extra time from the batteries.

(1) Probable issue is attempt to massively over drive LEDs - see below.
Series LED resistors will be needed.

(2) Boost converter MAY be not working properly - see below for testing method.

LED datasheet here
TPS61201 boost converter datasheet here

Two x AA alkaline with provide a voltage between 3.2V and about 2V.
A larger value of C1 on Vin will do no harm and will help very low voltage/bad battery startup.

The TPS61201 boost converter will happily start and run on this voltage range.

The LEDs are NOT rated at 20 mA continuous - see data sheet. LEDS are rated at

• 10 mA continuous

• 20mA peak at 10% duty cycle at 0.1 ms pulse width !!!.

Also thermal limitations of IC must be observed.
Running LEDs directly off IC pins risks LED damage and possibly IC malfunction.

What is design requirement?:

Say 10 mA/LED and all on.
12 LEDS x 10 mA = 120 mA.
120 MA x 3V3 =~ 400 mW.
Efficiency at 120 mA out and 2V Vin ~= 75% - see fig4 from datasheet below.
So 400 mW/75% = 560 mW into converter.
At 2V Vin Iin = 560 mW/2V = 280 mA Iin.
This is well within IC capability.

So - IC is capable of providing requerement IF LEDs are correctly driven.
Problem may be excess LED drive OR bad converter components.

Test: Provide a 120 mA resistor load to converter.
R = V/I = 3.3 V / 120 mA = 27 Ohm.
Will converter supply 3V3 to 27 Ohm with 2V supply including startup?
Use lower R's for higher load current if desired.

If converter will not support desired load current then an inadequately rated inductor is the most likely problem.

Your Cout = 22 uF = 2 x data sheet value - should not be a problem. Sometimes high Cout can cause startup problems but 22 uF should be fine.

Most likely problem is massively high LED currents.
Add series resistors to set currents to 10 m max. Note that Vf LED varies with colour - see datasheet.

ADDED:

New information:
LEDs are not as shown on diagram.
Assume max per LED current is 20 mA.

Actual LEDs are Dialight 5988710307F.
These LEDs are rated at 20 mA ABS MAX so you could run them at 20 mA, perhaps.
[Are you feeling lucky, punk?]
If so then double figures I supplied above to 1
20 mA x 12 = 240 mA.

This is still on the curve in Fig 4 above with 1.8V Vin so the converter can handle it.

TRY MY RESISTIVE LOAD TEST with R to suit real load.
If this passes OK then converter is OK.
If this fails then fix it first - chasing driver problems when the power supply is failing is liable to be unproductive :-).

ADDED:

A deleted answer suggested that rising battery impedance would mean that AA batteries could not be used in this application and that C or D cells were needed.
IMHO this is not true.
While AA cells will probably not reach full discharge potential due to falling current capability, they should work reasonably well.

The converter has a performance curve in fig 4 of the data sheet which shows operation at 1.8V = 0.9V/cell. As long as the batteries will provide the required load current at this voltage OR HIGHER the system will work OK.

At I_LED = 10 mA and all segments lit Ibattery at Vbattery=2V will be ABOUT 280 mA (see above) and at I_LED = 20 mA I_Battery at Vbattery = 2V will be about 550 mA (efficiency slightly HIGHER at higher load - see graph).
IF the battery is capable of providing about 500 mA at 2V+ then it will work. This is getting extreme for AA Alkaline but the battery will provide more power than this for much of its discharge life.
My simple resistor loading test will show whether the converter is OK at any given battery state.
Note that input capacitor matter muchly when batteries are near end of life. A large capacitor greatly reduces battery effective impedance on current peaks. Mean ESR is not altered but failures usually occur when current peaks occur during the boost cycle.