I have a bike flash light which is consisted of 5 LEDS (red). Which looks like this.

enter image description here

I unsoldered three of them and put it in a series circuit using two AAA bateries. The Leds did not turn on. I assumed that something was wrong with my circuit connections. So I just disconnected the two leds and kept only one of them with one battery( later i tried with two). The led did not turn on either. I changed the polarity but still nothing. So what am i doing wrong? does the led get destroyed when i unsoldered from the circuit board?

The batteries were not empty. The leds were operating normally while they were on the board.

Thanks in advance guys I am a total newbie.. :)

EDIT: The polarity was tried both ways so i did not add + or - in the picture. enter image description here

  • \$\begingroup\$ Please provide a schematic of what you are doing. \$\endgroup\$ – Dwayne Reid Feb 20 '15 at 20:51
  • \$\begingroup\$ edited check the circuit \$\endgroup\$ – Dionysis Nt. Feb 20 '15 at 20:56

I'd say there are two possibilities.

First, please check the original unit. In addition to the batteries and the LEDs, there ought to be a resistor. This will limit the LED current to safe levels. Without any resistor at all, it's entirely possible that you've simply destroyed one or more LEDs by hooking them directly to the battery.

The second possibility has to do with your soldering technique. What size soldering iron do you have and how long did it take you to remove and resolder the LEDs? It is entirely possible that, if you used a high-wattage soldering iron, and heated the LED leads for too long that you have just cooked one or more LED.

ETA - About choosing resistors. You did not identify the batteries you use, so there is no way to know what voltage they provide. In fact, there is no guarantee that the five LEDs are not connected in parallel.

But here's the general procedure for choosing resistors. Find the battery voltage. Then figure out the operating voltage and current requirements for the LEDs. In the case of red LEDs, 1.5 volts is a good start, and 10 to 20 mA for the current. When the LED is operating, the battery voltage will be split between the resistor and the LED. Let's say that the battery puts out 2 volts, and the LED needs 1.5. Then the difference (2 - 1.5) will appear across the resistor. The relationship between voltage and current in a resistor is called Ohm's Law, and is V = iR, where V is voltage, I is current (in amps) and R is resistance in ohms. In this case, assuming 20 mA for the LED current, the current through the resistor will also be 20 mA, since the two are in series. Then 0.5 = .020 x R, and R = .5 / .020, or 25 ohms.

  • \$\begingroup\$ I would expect the missing resistor to be a problem. It would be great if you added some theory how to choose a resistor to drive LED(s). \$\endgroup\$ – Nazar Feb 20 '15 at 21:19

My best guess is that in the original unit, the LEDs were all connected in parallel. Although it is bad practice to run the LEDs from batteries without a current limit resistor, Asian manufacturers do it all the time.

In the schematic that you show, you have connected the LEDs in series. Assuming that these are Red LEDs, you would need at least (3 * 1.7V) = 5.1V before the LEDs even begin to turn on.

The original manufacturer got away without using a series resistor to drive all those LEDs because they have 5 of them in parallel. The internal series resistance of the crappy batteries that they supply plus the resistance of the PCB traces serves to limit the total current.

If you are planning on using only a few LEDs, you had better plan to add an appropriate series resistor. You will most likely burn the LED out quickly if you don't.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.