making a one filament light "two stage"

[Kejtar]

I need a light to be bright and then brighter. I have 2 12V lines to drive it. Problem is that the light is unfortunately one filament. So, I thought about it and came up with the following:

D would be a Rated 6 amps. 50PIV (For replacement and power supplies )
R would be a 50 ohm 10W 5% Wirewound Resistor
L is the light which is rated at 14V .33 A

I figure that the 50 ohm resistor would drop the brightness by a hair over 50%. The diodes would prevent backeeding.

Here's the diagram (http://home.socal.rr.com/remi/pics/lights.jpg)

Would this work? Or did I overlook something?

[Lou]

Looks OK.

The only thing is that I don't think the bulbs have a linear current/output relationship. You may end up having to use a (much) lower value R to get 50% output.

[old_man]

Why not just swap out the socket. Most are available in both bases. Then you just wire up the two different filaments to the two sources you have. It keeps it clean.

If you can't find a socket easily, send me a pix. I have all the catalogs.

[Kejtar]

Why not just swap out the socket. Most are available in both bases. Then you just wire up the two different filaments to the two sources you have. It keeps it clean.

If you can't find a socket easily, send me a pix. I have all the catalogs.

Thanks, but I've been pretty much to every auto parts store as well as every automotive light manufacturer and what I need they don't make (I need 2.5" round that mounts in a grommet that fits a 2.75" hole). SOckets are not replaceable as those are sealed modules :(

[MoFo]

You will lose 0.6 Volts across the diode junctions. You don't need the diode in the brake switch line. All you need is the diode in the lights line, to keep the brake pedal from lighting all the marker lights. This will make the brake position brighter.

As I said, you will lose .6 V through the diode junction with the lights, so you should drop the R to compensate for the loss.

[Kejtar]

You will lose 0.6 Volts across the diode junctions. You don't need the diode in the brake switch line. All you need is the diode in the lights line, to keep the brake pedal from lighting all the marker lights. This will make the brake position brighter.

As I said, you will lose .6 V through the diode junction with the lights, so you should drop the R to compensate for the loss.

That makes sense, but I ended up putting a diode in both since I didn't see this message till now (might make new harness at a later time). In any case I ended up using a 20ohm resistor and it seems to be working perfect. Now I just have to wait for the housings to dry and wire up the big connector to connect the tail light box harness and the vehicle and it'll be all finished :D

[Kejtar]

You will lose 0.6 Volts across the diode junctions. You don't need the diode in the brake switch line. All you need is the diode in the lights line, to keep the brake pedal from lighting all the marker lights. This will make the brake position brighter.

I looked at the FSM schematics and the diode is needed on lines. If the diode on the brake line is ommited, when the marker lights come on, the third (hatch mounted) brake light would come on.

[MoFo]

You right! My bad.