# LED's too dim. Wrong resistor? [duplicate]

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?

• what is the specifications of the LED you are using?? Oct 4, 2014 at 4:18
• 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 Oct 4, 2014 at 4:37
• 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. Oct 4, 2014 at 5:09
• 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. Oct 4, 2014 at 9:43

$$R = \frac{ V_{in}-(V_{LED} \times quantity\;of\;LEDs)}{current\;of\;LED}$$
$$R = \frac{4.5<batteries> - \;quantity\;of\;LEDs}{voltage}$$