A big flashlight, nothing special

Question

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

Answer 1

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.

Answer 2

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:

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:

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