• Assuming that the (12x1.2V AA's) are at least 800mAh each, the collective mAh rating of 12x1.2V AA's would be 9,600mAh. If I take the collective 60mA from 3 LED's, that sould give me 160h of use? It seems a bit much to me, I know its simple math but wow.

Sorry, but your math here is not correct.

Let's say you had two 1.2v 800mAh batteries.
If you placed them in parallel, voltage stays the same, mAh and current capacity doubles.
If you place them in series, voltage doubles, mAh and current capacity stays the same.

 

Can you explain that a bit more Sgt? Since you have two sources of the same energy content, shouldn't the total energy double as well in both parallel/series cases?

 

Can you explain that a bit more Sgt? Since you have two sources of the same energy content, shouldn't the total energy double as well in both parallel/series cases?

Actually, I re-thought this calculation and he is absolutely right. The batteries when in series create more voltage but their capacity is the same. I must have had some serious lunch to think that. lol.

So in this case, I have higher voltabe but the same current. I would be better off using 2 (9V) cells then. They are 200mAh. Meh.

 

Except that the best AAs are over 2000mAh, so they would still beat your 9V battery hollow in a stack of equal voltage. Even the less ambitious 800mAh models are still several times the little 9V capacity.

In the end, it comes down to cost and weight. Personally I think small 9V batteries are very poor value for money. At any rate, they are quite expensive here in UK. AA cells are sold in big numbers, and you can often get them in large packs at a discount, including rechargeable types. For some reason the AAA type are often no cheaper than their bigger brothers, so they are not so good unless you are stuck for space.

 

Well, how about this:

\(
B=mAh\ \text{rating}
E=P \cdot t
=V \cdot I \cdot t
=V \cdot B
\)

For the parallel combination, the energy content is:
\(E_p=V \cdot 2 \cdot B=2 \cdot V \cdot B\)

For the series combination, the energy content is:
\(E_s=2 \cdot V \cdot B\)

Hence \(E_s=E_p\)

Is my reasoning wrong?

 

No, that sounds fine, and lines up with what SgtWookie said. The OP, on the other hand, appears to have thought at first that 12 1.2V 800mAh cells could be used in series to make a 9.6V, 9600mAh battery, which is not right.

 

Except that the best AAs are over 2000mAh, so they would still beat your 9V battery hollow in a stack of equal voltage. Even the less ambitious 800mAh models are still several times the little 9V capacity.

In the end, it comes down to cost and weight. Personally I think small 9V batteries are very poor value for money. At any rate, they are quite expensive here in UK. AA cells are sold in big numbers, and you can often get them in large packs at a discount, including rechargeable types. For some reason the AAA type are often no cheaper than their bigger brothers, so they are not so good unless you are stuck for space.

I agree that the 9v's are not worth the bang for the buck you have to spend on them. I can buy 9v's here in Canada for $1/each. Not too bad. AA's can be bought in packs of 24 or so, but if I were to buy 12x rechargable AA's they are $20.00 / pack of 4. That is $60.00 (not including taxes) to buy rechargable batteries.

The highest mAh rating on AA's I have found, just yesterday, was 2700mAh... Not bad at all. They are too pricey though.

Well, how about this:

\(
B=mAh\ \text{rating}
E=P \cdot t
=V \cdot I \cdot t
=V \cdot B
\)

For the parallel combination, the energy content is:
\(E_p=V \cdot 2 \cdot B=2 \cdot V \cdot B\)

For the series combination, the energy content is:
\(E_s=2 \cdot V \cdot B\)

Hence \(E_s=E_p\)

Is my reasoning wrong?

*Edit*

I found this Link from a website... check it out. It explain how laptop batteries use cells in series to create higher voltages, and some in parallel to get the current capacity doubled.

Seeing how that laptops get the 14.4V @ 4800mAh ratings, I may use the same idea, have 1 9V cell in series with a parallel set of 9V cells. Has anyone ever done this?

Example:
B = 9V battery assumed @ 200mAh

B1 + B2 = 9V @ 400mAh

B1B2 + B3 = 18V @ 400mAh.

Does this make sense?

 

No, that sounds fine, and lines up with what SgtWookie said. The OP, on the other hand, appears to have thought at first that 12 1.2V 800mAh cells could be used in series to make a 9.6V, 9600mAh battery, which is not right.

Ok hang on a minute, SgtWookie stated that the batteries in series double in voltage & parallel batteries double in current capacity, while Georacer stated that the E(parallel) = E(series)... If they are the derrived to be the same, then how can it be correct to what SgtWookie stated?

 

Ok hang on a minute, SgtWookie stated that the batteries in series double in voltage & parallel batteries double in current capacity, while Georacer stated that the E(parallel) = E(series)... If they are the derrived to be the same, then how can it be correct to what SgtWookie stated?

I think that Georacer means E for Energy, which is the same in each case.

Parallel cells have more Ah, series cells have more voltage, but the same energy can be stored if an equal number of cells is used.

 

Alright. Thanks for the clarification.

Back on track with the leds, I think that I will go with 3 of the leds, use 4 9V cells, 2 packs of parallel cells wired together in series, that should give me about 800mAh at 18V. Has anyone ever done this and confirmed it to work?

 

Energizer and Duracell are American name-brand batteries.
Their 9V alkaline batteries are 625mAh when the current is only 25mA and the battery is run down to only 4.8V.
Realistically they have a capacity of only 267mAh when the current is 100mA and the battery runs down to 7.2V. Then two in parallel have a capacity of 534mAh.
I am not talking about cheap Chinese batteries that cost only $1.00 at The Dollar Store.

At 100mA, four series/parallel batteries will last for about 534/20= 27 (+12% because the current slowly drops)= 30 hours and the dimming during that time will not be much.
At 20mA, four series/parallel batteries will last for about 53 hours.

 

Energizer and Duracell are American name-brand batteries.
Their 9V alkaline batteries are 625mAh when the current is only 25mA and the battery is run down to only 4.8V.
Realistically they have a capacity of only 267mAh when the current is 100mA and the battery runs down to 7.2V. Then two in parallel have a capacity of 534mAh.
I am not talking about cheap Chinese batteries that cost only $1.00 at The Dollar Store.

At 100mA, four series/parallel batteries will last for about 534/20= 27 (+12% because the current slowly drops)= 30 hours and the dimming during that time will not be much.
At 20mA, four series/parallel batteries will last for about 53 hours.

That sounds great, thank you for this info.

I have completed this project last weekend, it worked out very well! I put everything into a hand held project box with rocker switch for the on/off. 3 leads going to the difference leds were measured out to about 10" at max length & 6" at min length. I used 3 leds and only 2 9v's for now, though I wired it to use 4 as I planned from before. I will attached a picture later on showing the results in a dark room.

Thank you to all who gave me insight to finish this project, your help was very valuable to say the least.