# newbie, 5 volt led run on 12 volt

Discussion in 'The Projects Forum' started by bigdodge, Jan 25, 2011.

1. ### bigdodge Thread Starter New Member

Jan 25, 2011
8
0
Hello to all.

My name is Dustin. I work as a mechanic for INDOT. One of the guys I work with is building a 32 ford hot rod. I'm helping the guy with all the wiring on the car, and we have just started on whole project.

My problem is with leds. The owner of the car is using 2 green leds for turn signals and a blue one for the bright headlights. We were able to find two greens at radioshack that are 12 volt, so they will work with no problem.

The question I have is with the blue led. The only one we could find is a 5 volt 30ma led. I'm sure that there is a way to make this work using a resistor. I am not able to figure it out. Also witch one of the leads is the positive, I thank its the longer one but not shure.

Any help would be greatly embraced. Thanks in advance for helping a fella in need.

2. ### wayneh Expert

Sep 9, 2010
12,365
3,220
If you're really sure it's a 5v LED, then you need to drop about 10v across a resistor, so that when the system is at 15v, the LED is properly powered. I'd shoot for less than the max rated current, eg. 20mA instead of 30mA. It'll last longer and be less likely to burn out. If it's not bright enough and you're confident how it's all working, you could then go to a lower resistance and higher current.

Anyway, Ohms law is your friend. V = I*R Rearrange to solve for R = V/I.

So for your LED, R = 15v/0.02A = 750 ohm. By the same calc, you'd use a 500 ohm for 30mA. But again, I'd start high in ohms and work down if needed. The LEDs may be plenty bright at 20mA.

3. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
It'll probably be TOO bright at night. Be aware that super-bright LEDs available nowdays can cause permanent damage to your eyes if you stare at them. They will burn dead spots in your retina. One way that you can diffuse the light somewhat is to roughen up the front of the LED with some fine sandpaper. Otherwise, the narrow focus and brightness of most LEDs will just be too distracting - and even harmful to your eyesight.

4. ### bigdodge Thread Starter New Member

Jan 25, 2011
8
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wayneh,

What watt would that be. Mabey a 1/4 watt 750 ohm.

Also witch lead is the positive. And where would I put the resistor, on the negative or positive side.

5. ### bigdodge Thread Starter New Member

Jan 25, 2011
8
0
SgtWookie,

Thanks.

The pachage says that it has 300mcd, I take it this is the brightness of the led. Is this very bright?

6. ### ian_gregg New Member

Jan 19, 2011
21
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As the LED is still basically a diode you can try it both ways and see which one works. Convention is that longer lead is positive.
If you have 12V supply and 5V across the LED, you want 7V across the resistor.
7V/20mA = 350R, go 330R, power = 7V*20mA = 140mW, 1/4 Watt is safe
7V/30mA = 233R, go up to next standard value, 270R, power = 7V*30mA = 210mW, 1/2 Watt would be on the safe side

7. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Raw LEDs, before they are mounted in a fancy case, do not have a set voltage, as they are current controlled devices, the exact voltage they operate is set by current limiting resistor.

LEDs, 555s, Flashers, and Light Chasers

8. ### bigdodge Thread Starter New Member

Jan 25, 2011
8
0
The pachage says it a 5 vdc 30 ma.

I gess I'm kinda confused. Would a 1/4 watt resistor with 750 ohms work.

9. ### wayneh Expert

Sep 9, 2010
12,365
3,220
Power = current ^2 * resistance Watts=Amps^2*ohms

(^2 = squared)

10. ### bdd4 New Member

Jan 27, 2011
3
0
1. Do not trust the specifications on Radio Shack packages. They are not always accurate.
2. You are interested in the Vf (Forward voltage drop) of the LED. This is the amount of voltage that will appear across the LED when it is on. Likely it is less than 5 volts say 3.6 to 4 volts
3. A LED data sheet will give you Vf as follows: Minimum, Typical and Maximum. You can design for the Typical and if adjust the dropping resistor value for the desired brightness.
4. The calculated values you were given are fine.
5. The long lead is likely the anode (+). and the short lead is the cathode (-). The cathode is identified by a "flat" on the side of the LED at the bottom. Look at the bottom of the LED toward the leads and you will see the "flat" next to the cathode. You can connect the current limiting resistor either from the anode (+) to the battery + or from the cathode lead (-) to ground, battery negative. For your design, same results.

11. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
You can add a resistor to the existing indicator to make it 12V. Since there is 7V difference between the two set points (5V indicator, 12V supply) and a known current you can calculate it as follows.

7VDC ÷ 0.03A = 233.3Ω ≈ 240Ω @ ½W 5%

12. ### bigdodge Thread Starter New Member

Jan 25, 2011
8
0
bdd4 thanks,

I'd like to say thanks for everybodys help. If I ever need any further advice I will ask again.

I'm gonna try the 1/4 watt 750 ohm resistor and see how brite that is. If it turns out to be to brite I should be able to go up with the ohms, right. If I went down that would give me a brighter led?

13. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Yes, and I calculated the exact resistor you would need to do the job right the first time.