6 volt lighting

[darrylri]

Follow the link above and go to the Parts Store page. There you'll find these entries:

07 11 9 978 006--$10.00--bulb Bilux-bulb B 6v 35/35w headlight BA20D base R26-R69S
07 11 9 978 006HC--$26.50--bulb headlight 6v halogen 30/35w H4 P43T base fits conversion headlight 63 12 1 243 541
07 11 9 978 007.1--$26.50--bulb headlight 6v halogen 35/35w with BA20D bayonette incandesent base fits original reflector & bulb holder. Original wattage in a brighter halogen bulb! -- a Bench Mark Works USA exclusive!

[bernhardt]

Another source for halogen and LED bulbs

I bought bulbs from this site, and they work well. I use a 6v dual filament halogen bulb in the tombstone tailights, but I would avoid them in a later style tailight with a plastic lens - like a 60's pan or shovel. Too much heat. The headlight bulb they offer works well also. I noticed they also have a 6v LED...

http://www.classicandvintagebulbs.com/page2.html

[awander]

To figure out the circuit, you need to have a voltage drop across your serial LEDs of 4V or less. Multiply the voltage drop by 50 to get the ohm rating of an appropriate resistor. Typically this will let you run 3 LEDs in series, and then you create as many parallel circuits as you want lights (or have the generating capacity to support).

The LEDs I used had a 1.2V drop and needed 200ma. In this condition, I had 3 LEDs after a 180 ohm resistor.

Calculating the resistor:
3 LEDs * 1.2 voltage drop = 3.6V =>
3.6 * 50 = 180 ohm resistor

Wattage required:
200ma/LED * 3 LEDs = 0.6amp draw =>
0.6 amp * 6V system = 3.6 watts

Hi Darryl:

I have to take issue witha few things in your post. First, 200mA is a LOT of current for an LED. Are you sure you didn;t mean 20mA?

Second, with a 180 ohm resistor is series with the LEDs, you will get 6V-3.6V/180 ohms = .0133A or 13.3mA. Where did you get that "multiply the voltage by 50" equation?

I'm not saying this didn;t work for you, but I think your numbers and methodology(or at least the explanation)are way off.

Also, if you wire a bunch of "series LEDs" in parallel with each other, as you suggest, you will only get part of the total current through each parallel leg. How much goes through each leg will depend on the individual tolerances of the LEDs in each leg.

Andy

Actually, i think I see what you are doijng:

Ohm's Law states that V=IR, or V/I=R, where V is volts, I is amps, and R is ohms.

Your equation V*50=R (Multiply the voltage drop by 50 to get the ohm rating of an appropriate resistor) is the same as
V/.02=R, so it will give you the resistor required to limit current to 20mA, NOT 200mA. But that is assuming the 3.6V is dropped across the resistor, while in your example, the 3.6V is dropped across the LEDs, and you will have, instead, 6V-3.6V=2.4V across the resistor. If you assume the charging system actually keeps the battery at 7V, you would then have 3.4V across the resistor.

So with the electrical system between 6V and 7V, you would get between 13.3mA and 18.8mA through 3 LEDs in series.