zim Posted May 8, 2012 Share Posted May 8, 2012 Morning, I've got 3 stack gauges in my truck, but at night they're too bright. Apparently they have LED's as the backlight. So is it possible for me to dim them somehow ? A quick google says that putting a resistor in series is no use ? Cheers Gordon Quote Link to comment Share on other sites More sharing options...
Glue Posted May 8, 2012 Share Posted May 8, 2012 I don't know the gauges you're refering to, but is it possible to fit a piece of thick ish (0.08mm / 3 thou) plastic round them. If so it may be wirth investigating a "neutral Density" filter produced for photographers and theatre use. Google lee filters, product numbers 209/210/211 may be of interest. Quote Link to comment Share on other sites More sharing options...
CwazyWabbit Posted May 8, 2012 Share Posted May 8, 2012 Normal LED's brightness is affected by changing the voltage across them so adding a resistor would work. Of course this is assuming that there isn't some funky dc-dc converter or other electronics in between. EDIT Just happened to have a few resistors and a bog standard red LED next to me and I can confirm it is definitely dimmer with two resistors. I don't have any 12v white leds to play with so don't know if these work differently. Quote Link to comment Share on other sites More sharing options...
zim Posted May 8, 2012 Author Share Posted May 8, 2012 Oh. I'll be honest and say i didn't try a resistor it's just what google told me. I'll have a play when i'm back Maybe an adjustable pot to be posh G Quote Link to comment Share on other sites More sharing options...
CwazyWabbit Posted May 8, 2012 Share Posted May 8, 2012 Lol to be fair I don't expect most people to just happen to have a few resistors and an LED sat beside them Quote Link to comment Share on other sites More sharing options...
zim Posted May 8, 2012 Author Share Posted May 8, 2012 So what sort of resistance range variable resistor should i look for ? Maybe i'll have to play with some resistors first. Quote Link to comment Share on other sites More sharing options...
Bowie69 Posted May 8, 2012 Share Posted May 8, 2012 Get a 1K potentiometer, set the brightness, then measure and replace with the fixed value. You will probably find the dim to off very very sensitive..... Quote Link to comment Share on other sites More sharing options...
zim Posted May 8, 2012 Author Share Posted May 8, 2012 Thanks Quote Link to comment Share on other sites More sharing options...
muddy Posted May 8, 2012 Share Posted May 8, 2012 I took my gauge to bits and glued abit of black bin bag to the back of the dial face. Low tech approach Will. Quote Link to comment Share on other sites More sharing options...
Roverbo Posted May 9, 2012 Share Posted May 9, 2012 A quick google says that putting a resistor in series is no use ? Cheers Gordon LEDs are dimmable, but what differs them from ordinary bulbs, is that the colour temperature of the light won´t change (getting warmer when dimmed). No problem in your car, but you can´t get a cozy feeling in your kitchen, living room etc. with a dimmer. Quote Link to comment Share on other sites More sharing options...
Ed Poore Posted May 13, 2012 Share Posted May 13, 2012 Normal LED's brightness is affected by changing the voltage across them so adding a resistor would work. Of course this is assuming that there isn't some funky dc-dc converter or other electronics in between. EDIT Just happened to have a few resistors and a bog standard red LED next to me and I can confirm it is definitely dimmer with two resistors. I don't have any 12v white leds to play with so don't know if these work differently. Not quite correct. They will be affected by adding a resistor in series with them but because it's changing the flowing through the pn junction not the voltage. LEDs (like diodes because they are effectively the same thing) will always drop a constant voltage across them which relates to the wavelength of light that is output (see Wikipedia for a table) but typically ranges from ~1V to 3V. So you need to provide at least that voltage in order to "activate" the LED and provide enough of a potential difference for it to start emitting light. For a bog standard 5mm LED the forward current required is ~30mA. You should always limit the current somehow, usually with a resistor for simple stuff picking an appropriate sized resistor therefore controls how much current flows through the resistor and subsequently the diode. For examples sake say that for 100% brightness you require 40mA with an LED that has a diode drop of 2V. Therefore from a nominal 12V supply you have Vled = 12 - Vdrop = 10V and for 40mA you want I = V / R ==> 250R. If you want to dim the led (current is not directly proportional to perceived light output if I remember correctly) then say put 20mA through it so a 500R resistor would halve the current. Hope that helps Quote Link to comment Share on other sites More sharing options...
CwazyWabbit Posted May 13, 2012 Share Posted May 13, 2012 Funnily enough when I first wrote the reply I did in fact say current but then had a brain fart moment and went back and edited the post to say voltage. Anyway cheers for adding the calculations as it should give Zim some idea where to start with values Quote Link to comment Share on other sites More sharing options...
zim Posted May 14, 2012 Author Share Posted May 14, 2012 Ed - wow.... reading that post after a long night shift is like double dutch to me when i'm home i'll try different resistors to see what's good Quote Link to comment Share on other sites More sharing options...
Ed Poore Posted May 14, 2012 Share Posted May 14, 2012 Ed - wow.... reading that post after a long night shift is like double dutch to me when i'm home i'll try different resistors to see what's good Ok to put it another way. These funky little creatures called electrons give off light when they get woken up. In order to wake them up you need to give them a good kick (potential difference aka voltage). The more often you can kick the more electrons wake up and therefore the more light you get. Now sometimes you find you have to kick them through lots of mud (this is a Land Rover after all) therefore there is some resistance against your boot which means that your kicks per second fall off but you can still kick with the same amount of force. This can be written down by the following equations Brightness ~= Kicks / Second ~= Force of kick / Mud thickness. (Or I = V / R). Quote Link to comment Share on other sites More sharing options...
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