# 9v Battery 2D cell torch conversion

Discussion in 'The Projects Forum' started by Will, Dec 11, 2006.

1. ### Will Thread Starter New Member

Dec 11, 2006
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Hello! I am new here... just needed some help and advice!

I recently purchased a 2D cell 10LED Torch from my local Tesco, and would like to convert it to a standard 9V battery. It currently runs at 2.0v, however, I would like to increase this. Obviously NOT to 9V! As such, I devided to wire in a 9V battery adapter and pop in a resistor. I was probably going to go for 3V, so I was wondering how I would calculate what kind of resistor I need? And indeed, what resistor DO I need?

It seems that it currently runs at 2V with the use of capacitors wired into the LED's circuit (I will get some pictures of this if it helps) - as D cells are 1.5, and it runs at 5, there is a capactior between the live wire and the LED's. So should I go for 2.5 from the battery so it is overall 3, or, would 3.5 probably be fine?

They are ultrabright LED's, and it seems that, looking online, most will take up to 3.5v.

2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
282
Hi,

Not sure why you want to increase the supply voltage. LED's do take a certain amount of voltage to go into conduction, but after that it's all current. The conduction voltage is determined by the nature of the p-n junctions that go into the structire of the LED. It cannot be changed.

Your two D cells put out roughly 3 volts, so with the white LED's needing two of then to conduct, the current pushed by the remaining volt is limited by a resistor. It can't be a capacitor in the flashlight, as capacitors block DC current.

10 LED's have to be arranged in parallel for two D cells to run them (10 in series would need 20 volts), so current in the flashlight might be up there in the range of 100 ma (10 ma per LED). That would mean the resistor is on the order of 10 ohms.

9 volt batteries have much less ampacity compared with D cells. My guess is that even with a larger resistor for the extra voltage (70 ohms), the 9 volt battery would have a useful life of less than one minute. Their discharge curve is based on a current draw of no more than 15 ma. Realistic current out of 9 volt batteries is more like 3 ma.

3. ### Will Thread Starter New Member

Dec 11, 2006
3
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Hi! I am pretty sure they are aranged in series actually... all wired in a big circle shape, one leg in postivie, other in negative. I am also pretty sure they ARE two capactiors wired in there. One is blank, the other says 200 uf.

I have a few pics now, sorry about qaulity, phone cam used.

4. ### beenthere Retired Moderator

Apr 20, 2004
15,815
282
Hi,

It looks as if they are in parallel, with legs across the circular foils. The outer foil might be positive, and the inner negative. Those are capacitors. I can't make out the other device to the right of the top capacitor in the first pic. Seems elaborate for a flashlight. Still don't think you can pull enough current out of a 9 volt battery.

5. ### Søren Senior Member

Sep 2, 2006
472
28
Hi,

They're in parallel!
The unmarked "capacitor" is an inductance (a coil) and the remaining components *seems* to be an MKT capacitor and a diode, but better photos would help in determining this.

I wonder if it could be a crude switcher to boost the voltage, since white LED's needs a higher voltage then 2 cells provide. If it's not a swither, i don't see the purpose of using a coil, a diode and caps either.

Anyhow, forget about running this from a PP3 (9V) battery !!

May 16, 2005
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7. ### Søren Senior Member

Sep 2, 2006
472
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Hi,

Not driven from a PP3 !

There is several switch mode IC's specifically made for running white LED's, most of them up switchers though, but I fear the OP would shy away from anything that involved.

Best way to drive LED's from switchers is using current mode anyway.

8. ### Will Thread Starter New Member

Dec 11, 2006
3
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Using current... could I pop a capacitor into the system between the batteries and the LED circuit? Or maybe replace the original with a higer rated one? The small thing you said might be a diode is labelled JO2A (maybe J02A) 6003.

Also, what aobut the method you said with using an IC? I am not too good with electronics knowledge wise, but am interested and willing to learn! I would have a better idea, but my Systems and Control teacher didn't actually turn up for two years at school.