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I have a little problem: as I said I want to build kind of a big flashlight. With LED's and stuff. I already know some things I need, LED's of course, a battery case, a switch and stuff.. but my problem are the more special parts, what other things do I need? The first I can think of is a resistance, but which one do I need? I suppose I will power it with a 9V-Battery (The LED will have orange light, so its about 2-2,3V if im right)

The other problem i have is about the LED itself. Does it simply have + and - or is it more complicated?

I mean I could just try, but I dont want to make any capital mistakes ^^'

I hope somebody here knows what I try to do and is able to help me a little.

Thanks to anybody who tries :D

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    \$\begingroup\$ You may find 9V (PP3) batteries to not have a stellar capacity for high loads. You may wish to utilise a few D-cell batteries which are at a more appropriate voltage (i.e. 2x series = ~3 volts, 2x series + 2x parallel = higher capacity) \$\endgroup\$
    – Transient
    Feb 13, 2012 at 0:39
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    \$\begingroup\$ Assuming the LEDs will run on 30mA each, the resistor (assuming 3V not 9V), Vdrop of an Orange LED typically 2, will dissipate 1V / 39 Ohms = 0.026W (less than 1/4W) which is not bad at all. It is amplified to 260mW if you use ten LEDs in parallel, which is fairly significant for battery life. A constant current source is somewhat a lot more simple than Olin's example (no board required, just hand wiring). There may be premade ones, but they are specific and may only be for certain LED's (i.e. Cree) \$\endgroup\$
    – Transient
    Feb 13, 2012 at 1:12
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    \$\begingroup\$ Well, if I understood correctly, i can build this without too much complicated boards etc.? The values you gave me should work for my project then? (3V D-Cell, 39 Ohms resistor etc.) Then i have another question: to prevent that one broken LED will make the whole system useless, can i simply link them together as a parallel cirquit? Well, would you mind looking over a cirquit scheme i made? i can upload it tomorrow. A working scheme would really help me a lot and i could learn a little from it i suppose. (and dont worry, i know all the symbols i think i need :) ) \$\endgroup\$ Feb 13, 2012 at 2:18
  • \$\begingroup\$ absolutely upload it. I will write my previous comments as an answer as to not convolute this. Feel free to post a comment with your schematic on there when I have written this as an answer. \$\endgroup\$
    – Transient
    Feb 13, 2012 at 3:39

2 Answers 2

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LEDs differ from incandescent bulbs such as used in legacy flashlights in that you can't just connect a voltage source to a LED and expect something good to happen. LEDs want to be driven with current. Batteries are generally voltage sources and usually don't come in convenient voltages to run LEDs directly anyway. Incandescent bulbs could be made for specific voltages over a wide range by changing the fillament length and diameter. However, the voltage drop of a LED is dictated by semiconductor physics and isn't something the designer has much control over.

Fortunately, with modern electronics it is relatively easy to make a constant current source from a battery. I have done exactly that for a head mounted LED light. Follow the link and you will not only see a description of the project, but links to the schematic, board drawing, and BOM.

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  • \$\begingroup\$ Okay, I think I get the system. Now my problem is, that I don't really understand the board, because I do not know that much about electronics. Are those circuits also available in electronic-stores? \$\endgroup\$ Feb 12, 2012 at 22:03
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    \$\begingroup\$ user: Yes, you can buy LED flashlights of various sizes. They are quite common now. I thought you were interested in building your own. From a pure economics standpoint, it will be better to just buy one someone else has already designed, debugged, and mass produced. @Russell has done that and can probably give you a lot of background on available products. I created the KnurdLight largely for my own entertainment and as a demonstration design, including the firmware. \$\endgroup\$ Feb 12, 2012 at 22:19
  • \$\begingroup\$ No, you got me wrong. I want to make one myself, because the shape of a normal flashlight does not fit my plans with it, lets say its too small. I meant the device, on which the batteries are mounted to create electrical current. I won't have problems to connect all those parts, I just cannot learn the most complicated things in time, I suppose. \$\endgroup\$ Feb 12, 2012 at 22:32
  • \$\begingroup\$ @user: "Big" is no information at all. If you want more specifics, provide real specs, like light output, physical size, what kind of batteries, etc. \$\endgroup\$ Feb 12, 2012 at 22:44
  • \$\begingroup\$ Well, the whole thing consists of a 100mm ⌀ pipe. I want to embed the whole system into that pipe, so that it works like a big flashlight. The switch will be on the one side, the lights on the other, quite like one of those little flashlights. The problem now only is the power supply, which seems to be much more complicated as i expected. So i just wanted to know now, if there is a finished power-device (like the one you used with your cap-light) which is available in stores and simple needs to be wired to the rest of the construction. \$\endgroup\$ Feb 12, 2012 at 23:07
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Resistors: (as a complementary answer on them)
You may want to choose resistors if you are looking for simple results without efficiency a priority.

Note that these leds may not be be flashlight material, you can run these at a higher current at your own risk (may ruin them quickly) near maximum or purchase a higher powered LED.

Calculations:

Assuming you are using a typical orange LED with 2 volt drop and 25 mA typical forward current (check datasheets* if you have them on your LED) you should be able to calculate a series resistor value with the following:

$$\frac{V_{in} - V_{forward}}{I} = \frac{3-2}{0.025} = 40\Omega$$

Any voltage higher than the forward voltage will require a series resistor, this is why we subtract the forward voltage. The closest resistor value that you may find is 47 Ohms, these are usually rated at +/- 5% so it is close enough.

Circuit:

Then i have another question: to prevent that one broken LED will make the whole system useless, can i simply link them together as a parallel cirquit?

Yes, you may have something such as this in the end: parallel led

If one LED blows it will not generally cause the others to become brighter, and so you are safe. LEDs should not blow when run under proper conditions (temperatures, etc.) and should outlive the rest of the components at the very least.

You can have a single series resistor to do all of this in fact: series-parallel circuit

There are two major issues:

  • Resistance must be inversely proportional to the number of LEDs. 120 / 5 = 24 ohms for R1 here, and R1 will dissipate E^2/R = 0.375 Watts! It must be rated above this, normal resistors may smoke up.
  • If one LED does blow, it will in fact let the others take its current in its place. This can cause a surge in current to one if the others were to fail somehow.

I only mention this for completeness.

Feel free to upload a schematic along with chosen resistor values (i.e. if you wish to try it at 30mA, modify resistor values to compensate for that.)

As well measure the maximum current draw (i.e. 200mA, 500mA) as to determine the appropriate amount of batteries required, D-cell batteries can usually supply these currents without too much issue.

*An example LED datasheet: 5mm orange LED

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  • \$\begingroup\$ Hm, i guess the last image covered about everything i would put into my own schematic (maybe a switch, but that wouldnt change anything about its function..) I still have problems understanding the resistance.. does it need to be 8 Ohms now, if i use 5 LEDs or does it effectively use 8 Ohms for every LED while having a total of 40 Ohms? It should simply be able to take the electrical power. \$\endgroup\$ Feb 13, 2012 at 5:53
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    \$\begingroup\$ Slight error, I had meant 120 Ohms / 5 branches = 24 Ohms beforehand. This would allow 3 / 24 = 125mA to pass through, which split equally, allows 125 / 5 = 25mA per LED with the correct power. The resistor would have to be rated at 1/2 Watt or more (3V^2/24 Ohms=0.375W). This still has the pitfalls mentioned above. To add one new LED a smaller value resistor (=more heat on one package) would have to be used each time. I would recommend just parallel resistors as up there first, resistors are CHEAP. \$\endgroup\$
    – Transient
    Feb 13, 2012 at 7:38
  • \$\begingroup\$ Okay, thank you a lot, i guess i know how to do it now :) Then it now is building time. \$\endgroup\$ Feb 13, 2012 at 12:43
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    \$\begingroup\$ @user1016675 feel free to accept an answer that you find the most relevant to answering your question. You are welcome. \$\endgroup\$
    – Transient
    Feb 14, 2012 at 4:33

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