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I'm connecting 6 LEDs serially with two 12volt batteries. Each LED needs 3.2 volt and takes 20mA of current.

I'm using two alkaline max 12 volt batteries(A23 LR23) and ATC is their manufacturer.

I tried to calculate the batteries' lifetime like below(taken from a website):

To find out how much energy I'm going to use I multiplied the Watts times the number of hours to get Watt-hours. So 12v battery rated(?google search) 1.2Ampere-hour then the energy is:

12 volt * 1.2 Ampere-hour(battery) * 2 batteries = 28.8 Watt-hours

The LEDs run at 3.2v and uses 20mA each. So

3.2 volt * 6 leds(each needs 3.2 volt and current 20mA) * (20 / 1000)Amp = 0.384 Watt

And finally the battery lasts:

28.8 watt-hour/0.384 watt = 37.5 hours

Is this calculation correct? If not is it possible to help me correct my mistake?

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  • \$\begingroup\$ Found the solution at this website: oznium.com/12v-led-battery-holder Sorry for not researching properly. \$\endgroup\$ – mvr950 Jul 2 '12 at 5:58
  • \$\begingroup\$ LR23 capacity is two magnitude order less than 1.2Ah. If you take 40mAh out of them, you're lucky. Your led string will last less than 2 hours \$\endgroup\$ – Axeman Jul 2 '12 at 5:58
  • \$\begingroup\$ You're not mentioning series resistors for the LED's at all. Are you aware you need them in most any setup and certainly this one? \$\endgroup\$ – jippie Jul 2 '12 at 7:31
  • \$\begingroup\$ I was wondering about using resistors for this setup. Now I've to rethink the setup. Thanks for suggestion. \$\endgroup\$ – mvr950 Jul 3 '12 at 4:02
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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.

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  • \$\begingroup\$ And moreover I have some doubt on the internal impedence of those batteries... will they be able to supply 40mA? \$\endgroup\$ – Axeman Jul 2 '12 at 7:32
  • \$\begingroup\$ Good point about serialization. Won't a converter be able to discharge the batteries much deeper and therefore increase efficiency? I doubt that in a setup like this you'll be able to draw the full 1200mAh from the batteries, but then again I never tried it. \$\endgroup\$ – jippie Jul 2 '12 at 7:34
  • \$\begingroup\$ @jippie - Yes, the converter will get more juice out it, also because the current will already drop early on with just the resistors. But you'll have to place the batteries in series to allow them to go lower, and then switching from 24 V to 9.6 V is again less efficient than from 12 V to 9.6 V. \$\endgroup\$ – stevenvh Jul 2 '12 at 7:41
  • \$\begingroup\$ Thanks stevenvh for comprehensive answer. I'll follow your advice. \$\endgroup\$ – mvr950 Jul 3 '12 at 4:21
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Those who are interested: First we need to know the milli-ampere-hour of battery pack.

Because we know that:

mAh of battery pack / current draw of LEDs = approximately how long the batteries will last

It is known that: Pack of 8 x AA size batteries: about 2000 mAh

1 x 23A size battery: about 50 mAh

1 x 27A size battery: about 20 mAh

mAh of 2 12Volt A23 batteries is = 50mAh * 2 = 100mAh

current draw of 6 LEDs is = 20mA * 6 = 120mA

So the batteries will last approximately:

100 mAh/ 120mA = 0.833 hour or (0.833 * 60 minutes) = 49.98 minutes(approx.)

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    \$\begingroup\$ No, the 0.833 hour is approximately. If you calculate 100/120 * 60 you get 50 exact. And even if a calculation result is 49.9814574 you don't say 49.98 approximately, you say 50 approximately. \$\endgroup\$ – stevenvh Jul 2 '12 at 6:42

protected by W5VO Feb 15 '13 at 14:01

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