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I'm a halloween tinkerer, and I have these UV/Blacklight flashlights that I use in my decor, but they eat batteries like crazy. They start going dim within an hour or so.

So, since I already have been putting together my own wiring and LED spotlights using bulk LED's etc, I thought I'd experiment with converting them from battery to DC power supply.

The flashlights run on 3 AA batteries in series ( [- +}[- +}[- +} ) and I've been working with a 12v power supply. When I connected it with a resistor in-line (I believe it is a 200 ohm resistor that came with my bulk LED's) it lights up, but it's too dim compared to running on the batteries. I suspect that means I need a smaller resistor but I'm not sure?

Troy Pruyt
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    what is the specifications of the LED you are using?? – Sanjeev Kumar Oct 04 '14 at 04:18
  • Unfortunately, I don't know that. They're manufactured permanently into the flashlight head, and there's no label or anything on the flashlight. There are 51 bulbs in the head, but the most info I can find is Bulb Type: 390 to 395 (nM), but that refers to the color wavelength. I'm guessing (by appearance, but that's not always reliable) that they are similar to my bulk UV LED's which are 3.0~3.4v at 20mA max. When I connect the flashlight without a resistor in-line it lights up full brightness, but it blinks! LOL – Troy Pruyt Oct 04 '14 at 04:37
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    We have a fair number of threads for selecting resistors for LEDs: [What resistor should I use when wiring 10 LEDs in series?](http://electronics.stackexchange.com/q/7755/7036) [How to choose correct resistor with proper wattage and ohms for 6 leds connected in a serial circuit?](http://electronics.stackexchange.com/q/36880/7036) [Correct formula for LED current-limiting resistor? (single LED)](http://electronics.stackexchange.com/q/17179/7036) – Nick Alexeev Oct 04 '14 at 04:42
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    If 1 led consumes 20mA then total current required will be around 20*51mA=1020mA. I hope all the led will be in parallel so required resister value= (12-(~3))/1020=8.82Ohm. – Sanjeev Kumar Oct 04 '14 at 05:09
  • With a 12v supply that would be wasteful (and bad for the leds since they would see quite different currents). Instead, I'd check the forward voltage with a multimeter (say that gives 3.2V), then put as much as possible in series to stay below 12v (that would be 3 of them, total 9.6V drop), and add a suitable series resistor (here (12-9.6)V/0.02A=120ohm). Put as may of those (3led+1res) series as you can in parallel, and you get more efficiency and more led lifetime. – Nicolas D Oct 04 '14 at 09:43
  • Can you look inside the flashlight to see the board the leds are mounted on? If its anything like the cheapo 9 led flashlights, the leds are all in parallel, and depend on the 3AA batteries' internal EquivalentSeriesResistance to control the current. – Passerby Oct 04 '14 at 15:07
  • As for the blinking, I'm assuming that the 51 uncontrolled, parallel leds, are attempting to draw too much current, and **your power supply is not powerful enough to provide it, going into overcurrent protection mode.** – Passerby Oct 04 '14 at 15:08
  • If your power supply was powerful enough though, without a resistor, you would have a bunch of blown out leds though. – Passerby Oct 04 '14 at 15:17
  • Also related: [Why exactly can't a single resistor be used for many parallel LEDs?](http://electronics.stackexchange.com/q/22291/2028) – JYelton Oct 04 '14 at 18:47

1 Answers1

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Every LED has a current value such as yellow have 10mA and white is 20mA and so on. Each light also has a voltage. And quantity of lights is also imperative to know. Then use the formula

$$R = \frac{ V_{in}-(V_{LED} \times quantity\;of\;LEDs)}{current\;of\;LED}$$

$$R = \frac{4.5<batteries> - \;quantity\;of\;LEDs}{voltage}$$

Ricardo
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Daniel Kimbril
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    No, each LED color doesn't have a current. They have typical forward voltages, but the current is dependent on the LED's designed purpose. There are 1W white LEDs for lighting that take well over 20mA. I recently used one that had a Vf of 3V at 60mA. Your first formula makes sense, no idea what you're trying to achieve with the second. – Matt Young Jun 19 '15 at 14:52