24vCTD

I'm (finally) working on installing the gauges I got probably about 10 months ago...AutoMeter Ultralite pyro, boost, and electric fuel pressure in a pillar mount...and I've decided I want them on their own seperate dimmer control (the stock gauge cluster at full brightness does not bother me at night but since these gauges are going to be mounted higher I may want to tone them down depending on how much they throw a glare on the windows). They just have the standard incandescent bulbs, so I was thinking of mounting a potentiometer in the pillar with them. Question for somebody who might have an idea--what resistance value should the potentiometer have?

Thanks in advance

nkennedy

When I got my Auto meter EGT gauge it came with 2 little rubber boots to go ocer the light bulb. One green & one red. I used the green one, and along with giving it a green tint, it makes it a little dimmer. Check the boxes. You should have gotten something like that. If you do use a series resister/potentiometer, make sure it's rared for the WATTS you have in the line. Add all bulbs totals.

dzlfarmboy

I know I installed some autometers in a customers truck once and they were lite bright green, they would literally blind you how bright they were, and I can't remember what ohm resistor we put in them to make them manageable. I couldn't even believe how autometer could sell gauges that were that bright.

24vCTD

Thanks guys--
nkennedy, the boots were in the boxes but I'm looking to have something I can adjust on the fly vs. having to take the whole works apart if I ever change my mind about how it looks.

dzlfarmboy, did you use a fixed value resistor, or a potentiometer to make it adjustable? I'm looking to go the adjustable route; any ideas on where to start as far as resistance goes? As I stated previously these gauges are not going to be hooked to the stock instrument panel dimmer as I want independent control of the pillar-mounted gauges.

Thanks

DNRCustoms

Just wire the light power right to your factory power wire that runs to your dimmer switch. It is that easy. Just get a test light and find that wire. That is what I do. Run a lead wire off the hot/dimmer wire that goes to the factory switch. Put a plug male or female terminal on the end then tie all your power wires together for the guages into one wire and put the opposite terminal on there. Plug it in and your all set. Then just hook up a good ground. Hope this helps

CORed

I've been running my Isspros and one Westach lights off my stock dimmer wire for 4 years now and never had a problem. I dim occasionally and will dim more now that I have taken the load off for the main headlights with the LMC harness. Before, dimming would cause the switch area to heat up more than I was comfortable, so I didn't do it a lot.

CORed

wcbcruzer

Dang, I'll have to ask my uncle later on to be sure. But if I were doing this, I'd probably go with the 5k-ohm pot that Radio Shack carries (lowest one) and take it from there.

24vCTD

THANK YOU wcbcruzer,

I may pick one up later and give it a shot. In the meantime if you get a chance to talk to your uncle about it gimme a shout.

Much appreciated.

24vCTD

As I stated previously these gauges are not going to be hooked to the stock instrument panel dimmer as I want independent control of the pillar-mounted gauges.

Thanks

wcbcruzer

Ok, I just talked to my uncle. Since we're only dealing with 12 volts, you'd want to get the lowest resistance potentiometer (pot) you can find. 5k-ohms is too high but it all depends on the pot. I mean, if the range is from 0-5k ohms in one turn then you're lights will dim but it would be really sensitive.

Here's the calculations:

1) P=IV=V^2/R (watts = amps x volts = (Volts)^2/ohms)

To calculate new wattage of bulb with added resistance:

2) V-lamp = V-battery x (R-lamp/[R-lamp + R-pot])

and

3) I-lamp = V-battery/(R-lamp + R-pot)


So here's a theoretical example... Assuming the 194 bulb is 5 watts and your truck is running on 12 volts. Without any pot the bulb has a resistance of 12^2 (or 144) divided by 5, which equals 28.8 ohms; and a current draw of 5/12 = 0.42 amps.

If you add a pot and have it set on 100 ohms, then the new voltage (eq. 2) that the bulb sees is: 12 x (28.8/[28.8 + 100]) or 2.68 volts.

The new current draw of the bulb (eq. 3) is: 12/(28.8 + 100) = 0.093 amps.

Plugging into the first equation we get a new bulb wattage of 0.25 watts (0.093 x 2.68) with the pot set on 100 ohms.

As you can see, even with 100 ohms the wattage drop is quite significant, which means your bulbs will be really dim. If you have LEDs, they won't even light because they require a minimum working voltage.

Sorry for the long post but it might come in handy one day.

24vCTD

wcbcruzer-

Thank you so much for the info. I picked up a 5K pot (obviously before I saw this last post of yours) and tried it, and as you pretty much said if you even think about reaching for the **** the lights go out completely on the gauges... I found an online electronic parts supplier that has potentiometers starting at 100 ohms so I'll probably order from them. First I'm going to try and find out what the actual amperage of the bulb is (x3 gauges) so I can plug that into the formulas you posted to help me select the right pot. Once again, thanks for the excellent info.

biododge1

the red lamp cover mellows the light good-and red light is on the far side of the visible spectrum (for humans). green is the easyest to see, then 'white' light, then red(hardest to see)