Installed a set of Autometer gauges and tied the illumination wires to the factory dash lighting circuit. They are way too bright. Need to dim down the stock lights to near off just to get the new ones to dim.
Anyone know where I could get a stand alone dimmer switch?
OR, what size resister I could put in?
I think the Autometer gauges are LED lighted.
Thanks, Rob

 

Get yourself a variable resistor that you can adjust the brightness. I am pretty sure radio shack has them. Just wire it into the lighting circuit for your new gauges and you are all set.

John

 

The five new gauges combined draw a total of about 1.2 amps @ 12VDC, or around 1.5 watts.
All the Radio Shack pots are rated .5 to .75 watts and from 5 to 100k ohms.
There is one "rheostat" rated for 3 watts 25 ohms. Might that work? Is 25 ohms good for this application?
What's the difference of a potentiometer VS a rheostat?
Thanks, Rob

 

Replace the bulbs with ones with less candlepower. They had dash bulbs with bayonet bases back in the 50's & 60's.
<!--StartFragment -->A rheostat, looks a bit like a volume control potentiometer but generally wire-wound to handle the higher currents.

 

I get 1.2 amps x 12 volts =14.4 watts. You are correct that it would take a very large resistor to handle that. If they are indeed LED's, then they do not get very hot, and you could possibly paint over them to reduce the light and give them a cool colored glow. If they are regular lamps, I would change the bulbs light subford recommended.

 

That seems like a lot of amps for LED's to pull (or there's a ton of LED's). Most pull around 10 mA at the very most, so you'd have to have 120 LED's drawing 10mA each to total 1.2A (assuming the draw full rated current, which usually they don't). With that many LED's I wonder if you could wire some out of the circuit to reduce the brightness.

 

I didn't notice that, but I think you are right. They must be regular type bulbs.

 

Well I didn't do the math so good on the watts (thankyou Franklin), so now I'm not so sure of my other numbers. I've installed 5 new Autometer gauges, 4 Cobalt series (which I believe to be blue LEDs), and a tach that is regular incandecent. Befor install, I tested one of the Cobalt gauges illumination wires and thought I read .21 amps off of a 12V battery. All of the gauges illumination wires are tied together to a single conductor which is tapped to the factory dash light circuit.
I will pull the radio and check the total combined amp draw at full bright (12V) and see what I get.
I will still need something to dim them down reguardless, as is they are way too bright at night. The 4 Cobalt gauges are sealed units with no access to the bulbs.
Not sure why var resistors (pots/rheostats) are rated in watts when current should be the limiting factor in my mind?

 

Pots and Rheostats are mearured in watts becuase that is the real power it which things get done. Most don't consider the voltage because they are universally used. 25W rheostat would be ideal, try a local electric store or another electronic store than radioshack as they don't have much anymore. You could wire the gauges in series dropping the voltage across each bulb. That will effectively drop the brightness by half.

 

The gauges illumination has only one wire each and must share the ground with the function wires. They are new, fairly expensive, sealed units, so I realy don't want to tear into them to change the wires or bulbs.
Seems to me that the right resistor (variable or not) would do the trick.
Just checked the total combined load of the new gauges illumination circuits and got 317 ma @ 11.8 vdc.(that's 0.317 amps, right?) The 11.8 V is the factory instrument illumination voltage (with and without the new gauge lighting load) with the factory dimmer at full bright, engine running, and vehicle system voltage of 13.8. At full dim, the factory illumination circuit voltage is 3.6, and the new gauge light load is 29ma (.029 amps ?). At full dim the factory illumination is not visable, and though not nearly so bright, the new gauges are clearly illuminated.
If I can get my math right this time, I would need a resistor capable of handling about 3.75 watts, but at what ohms?
A 4 or 5 watt rheostat would prob be best, and perhaps I should just take this load from a regular lighting or ignition swiched circuit.
Does anybody know how to calculate the resistance I should use, based on the stats above?

I still don't get why these are rated in watts, and not amps?
A 1 watt pot/rheostat is rated for:
1 amp @ 1 volt,
0.1 amp @ 10 volts,
0.01 amp @ 100 volts.
?????????????????????

I need a device that will dim a load of about 1/3 of an amp, and that's just not alot of power.
Would someone tell me what I'm missing here?
Thanks, Rob

 

12V/.333A=36Ohms .333A*12v=4 watts So a 5 watt rheostat would work in your situation. Or if you put a 30 ohm 5 watt resistor in series with each bulb, only problem there is 5 watt resistor gets HOT, you will have to locate it somewhere with good airflow to help keep it cool. Not only that but a 5 watt resistor is probably going to cost more than a 5 watt rheostat. If you use a rheostat you will then be able to adjust the two sets of gauges independantly.

 

So, ohms = V/A eg: "12V/.333A=36 ohms" ? Is that only for a purely resistive load? Isn't the resistance in a circuit constant? Why, at 3.6VDC do I get 0.029 amps? OR, 3.6V/.029=124 ohms? Could the LEDs have impeadence that would account for the difference?
Even at full factory dim (3.6V), the new gauges are brighter than I might want them at times, so I'm not sure that 30 ohm or even 100 ohm will get me to near zero illumination on the new gauges. The R=E/I doesn't seem to work here.
Would hate to get a 1k ohm rheostat and only use 1/10 of the rotation.
I not only want to solve my light problem, but I'd love to learn what's going on here.
Thanks for helping, Rob

 

One other option would be to put a resistor on each individual gauge illumination wire. That way it would only have to handle the current of one gauge instead of the whole group. Although if you go with a rheostat you'd have to feed all of the gauges through that. You could figure out the current going to one gauge and then calculate the size resistor you need to drop from 12V to say something like 9V. You could even test that brightness with a 9V battery.

 

I get
Volts----------------Current (Amps)---------OHMS--------------Watts
11.8-----------------0.317------------------37.8---------------3.8
3.6------------------0.029------------------124.1--------------0.1044


Here is calcualtor for you:

http://www.opamplabs.com/eirp.htm

 

Thanks SUBFORD, Thats what I get. I just don't get how the ohms change.

The wattage should go down with the voltage (W=VxA). But voltage shouldn't change the resistance (should it?).