bright LEDs...

You definitly need the resistor. The spec sheet for the LEDs will tell you the current you want and the typical voltage drop. R =3D (Vbat

- Vdiode)/Idiode... just doing the math for you....

George H.

On Thu, 15 Apr 2010 11:36:13 -0700, larya wibbled:

Yes - you will be wanting a resistor, unless the LED has a built in current limiter (rare but sometimes)

Ohms law time :)

Decide what current you will drive them at (if in doubt 10mA will fire most devices well enough, but if you are testing retina destroying Luxeons then you might want 1/2A!). This is I

Look up in the LED data sheet the forward voltage drop (ideally at the test current but there may be a generic figure. This is Vdrop

If you using a battery with nominal voltage Vsupp, then:

V=IR

So R=(Vsupp-Vdrop)/I

(using volts and amps *not* mA)

eg generic led with a forward voltage drop of 2.1V supplied at 20mA from

4.5V dry cells (use 3.6V if using rechareables):

R=(4.5-2.1)/0.02 R=120 Ohms

If in doubt 100 odd ohms is a pretty safe bet for most LEDs.

That'll be a reasonable R value to use. It ignores the internal resistance of the battery for high current draws, but it will be a good starting point that will work for most "normal" LEDs.

3 AA cells is a good choice. Some LEDs have quite moderate forward voltage drops and you need to comfortably overcome this. 2 AA cells would work but only just.

And the AA cells can supply decent small to mid currents for decent amounts of time (compared to say a 9V PP3)

--
Tim Watts

Managers, politicians and environmentalists: Nature's carbon buffer.

Actually it doesn't. I have a cheap flashlight that uses 3 1.5V coin cell batteries and one white LED. There is no resistance involved. it was drawing

10mA. I put in a 3V lithium in series with one of the 1.5V to brighten it up. It drew 50mA.

The problem, of course, without the current limiting resistor is that you can easily burn up the LED. Without the resistor your guessing at the operating current and any variations can easily push you out of spec and destroy the LED.

For example, even if you find the right voltage for your operating current even modest fluxations in the voltage supply could burn the LED up.

Having said that, if you know the risks and know the application then it is possible, but generally not recomended, to skip the current limiting resistor.

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More of the story..... a ham friend of mine gave me three strings of Christmas ourdoor lighting.. In this string is a solar panel and electronics and battery, 3 nimh batteries (aa).. and the string of bright LEDS.. I have no idea of the specs of the LEDS.. I am considering snipping the end LED and creating a test circuit of a battery and led and resistor and see what happens.. In total I may have 60 keds... Larry ve3fxq

It appears to me that something like 99.9% of LEDs can be tested and operated in prolonged manner with a 9V "rectangular" battery, that used to be known as a "transistor radio battery", along with a 330 ohm 1/2 watt resistor in series with the LED.

Should the tolerance of the resistor come into question, any that I have ever seen for 330 ohm 1/2 watt resistors having should be OK.

If the LED is a "high power" one rated to use with 350 mA or higher current, then you may want to web search for how to power it and how to heatsink it if you want to treat it as a high power LED.

- Don Klipstein ( snipped-for-privacy@misty.com)

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Larry, sounds like you need to study/read some more. Do you have any test equipment? DMM's? Batteries are pretty safe, just don't plug anything into the wall socket.

George H.

You are giving rather missleading advice here. Your cheap flashlight relies on the INTERNAL resistance of the coin cells to limit the current. If yo used 4.5 volts from a Bigger cell or power supply it would burn the LED immediately.

John G.

In this case the battery internal resistance will act as the current limiter. Using a battery will a smaller internal resistance (such as

3xAA or 3xD) will drive a larger current through the LED and most likely evaporate it.

If some flash light without a current limiting resistor is designed to work with NiMh, is operated with NiCd cells (with usually lower internal resistance) it may destroy the LED.

The use of an external current limiting resistor will reduce the LED current variations caused by battery internal resistance as well as open voltage (alcaline/NiMh) variations.

Yes - - I totally agree. A few years ago I made a simple LED tester with a Lithium coin cell and a resistor. It was made using double-sided copperclad so that the led leads could straddle the edge of the board to make connection. Of course, other parts of the tester had the test terminals separated only by a gap on the copper surface for testing surface mount or devices whose leads could be bent down to the surface. It needed to be protected with fish paper to keep it from draining the expensive battery in a pocket. The actual test current was only approximately known since it depended on LED voltage drop, and some LEDs required so much voltage they could hardly be tested.

So more recently I decided to make an improved tester with a 9 volt battery and a tiny LM317MB surface mount regulator connected as a current regulator with a 3-position toggle switch selecting 2, 10 or 50 mA. The 9 volt battery has enough voltage for 2 LEDs in series so that they both conduct the same selected current for brightness comparison. An old 16-pin IC socket was cut down using only one 8-pin side in a cutout in the board edge with 2 center pins for single LEDs or the 2 pins at each end in series for a pair of LEDs. Of course it also retained the edge and gap straddle connection ideas of the original design. With the surface mount components nestled close to the battery and the switch always returned to the 2 mA setting between tests, it could be carried in the pocket with little concern for draining the battery. Overall size was 3/4" by 2" by 2 1/2".