Guest analouge Posted February 3, 2006 Report Share Posted February 3, 2006 Hey,Some things I don't seem to understand that you guys hopefully can help me with...I'm currently building a system that includes dimming a lightbulb. It's 12V DC. And I'm not really sure how many watts the bulb is going to be yet, but I've decided for a maximum around 20W.So here is the thing. I've been looking for a nice and easy way to dim the light. I could of course use a PIC for PWM since it's already a part of my system - but for other purposes. So since there is only one lightbulb I need to dim, I decided that it would be a lot easier for me (who is a lousy programmer) to skip the whole PWM stuff and just use a potmeter instead, at least I understand how to use them :).But I have NO idea how to choose a good potmeter that could do the job well without setting itself on fire :D.Some related questions...Correct me please if I'm wrong; A 20W lightbulb should draw pretty close to 2 amps right? So if I'm right, obviously I can't drive the lightbulb directly from a NPN transistor? It would smoke itself right out of this world, right? So I decided to use a board mounted relay that's driven from the transistor.Again I assume that I'm right, and 20W equals about 2 amps. Is it then so that it's enough with a 2A rated relay? Or does the bulb in reality consume more power when it is dimmed by the pot? Or would the pot just get very hot and burn out? I'm a man of theory, and I don't test much in real life before I've made a simulation of it (electronic-wise of course :)), so I've really no idea how this would work.So I kinda hoped some of you could help me choose a pot that is suitable for the job? I've read some datasheets and catalouges for pots, but I don't seem to understand it very well, coz most of them have a max watt rating for like 0,1W to 0,6W within certain temperatures. And I'm not sure how to interpret that... does it mean that I can't even think of using a 0,6W rated pot to dim a 20W lightbulb?Thanks for any answers :)Bye,Ingebret Quote Link to comment Share on other sites More sharing options...
ImAlive Posted February 3, 2006 Report Share Posted February 3, 2006 Wattage on Pots means max. loss power generated in the part itself. You get this by multiplying current through the pot with the voltage over the pot (this depends on the resistance of the bulb AND the pot) - good old P = U * I.So, if you drive 2 Amps through a Pot which eats away 1 V out of 12 (2W generated), and the pot is rated 0.6W, it would probably not stand the heat. Quote Link to comment Share on other sites More sharing options...
illogik Posted February 4, 2006 Report Share Posted February 4, 2006 i'm probably missing the point; but do you just need a fader for a light? no automation?you can just buy special faders for lights in electronics shops round 15 euro's Quote Link to comment Share on other sites More sharing options...
Guest analouge Posted February 4, 2006 Report Share Posted February 4, 2006 Hmm... not sure I'm quite following...Let me set up a scenario for you guys so that I can better explain what I don't understand. This is all going on in my simulation program, please read, and correct me if I'm wrong:I have a lightbulb of 10 watts and a 10k linear pot. (in fact I have no idea how much res the pot should have, all I know is that it should be linear).I ground the bulb, connects the other pole to the pot, ground the pot, and connects the third pin of the pot to + pole on a 12V DC source (battery) and ground the negative pole of the source. This works.Then - between the pot and the bulb I connect a multimeter in series. Lets call this multimeterAanother (multimeterB) goes in series between the pot and the bulb.I simulate this and get the following values:MultimeterAPot 0% resistance (light fully on): 834mAPot 100% resistance (light off): 1,2mAMultimeterBPot 0% resistance (light fully on): 832,8mAPot 100% resistance (light off): 812nAThis tells me that when the pot is at 0% resistance it only eats 1,2mA wich according to the famous law of Ohm should be about 0,15W. Correct me please! So assuming this is right everyone sees that the pot (rated 0,6w) has no problem when it is in 0% resistance position.So what happens when I turn the pot to 100% resistance (about 10k)? This is where I FEEL i get lost. And I might say, there is a big big chance that I'm lost long before this too :) I'm very new to this, and should probably go and buy some books on basic electronics. :Pwow... I really am lost... :oCan someone please help me sort this out? :)you can just buy special faders for lights in electronics shops round 15 euro'sHehe. The funny thing in here is that someone always comes with a easy solution :) And that is VERY VERY good - it should be that way in a diy forum. The great thing about this forum is that amateurs can learn about anything electronic-wise (and much more) because of all the skilled and nice people in here!But I'm the type of guy that don't want easy solutions... hehe sounds odd even to me when I think about it :P. But this is just as much for me learning it, not just building it. Of course I could just go to a part supplier and say: I need to dim a 12V 10W light bulb and I need a pot that do the job! And the guy would give me one that I could just run home and connect three wires to and everything was fine. But then I wouldn't know why it works, and how it works would I? :) And it's really neccesary for me to learn it, so that I don't have to run to a part supplier making him do the work for me, when I have the brains to learn it myself :PAnyway - thanks for the help guys :). Quote Link to comment Share on other sites More sharing options...
Guest analouge Posted February 5, 2006 Report Share Posted February 5, 2006 Hey guys!I've been thinking about this stuff for a while now, and after I found out how f...ing simple it is to build a pwm controller from a 555, some resistors, a pot, and some caps I decided to never think of doing it the "pot" way again :P.Out of the 555 I get a frequency of about 250hz if my simulation is right, with varying duty cycle. And that would do the job!Now I just gotta design a mosfet amplifier to do the hard work. That shouldn't be any problem. :) I got a bunch of really heavy duty mosfets lying around for some years now, and finally I get to try them out :) They can handle about 200A at peak (PWM-ed!!). Should be good enough for my 10-20W lightbulb right :P.Anyways, thanks for help, but as it turned out it was more time consuming finding a adequate pot to do the job in a poor way than it was to design a 555 PWM system that works perfectly well ;) I can use the same design for big motors speed control too... nice ;DSee you around... Quote Link to comment Share on other sites More sharing options...
SLP Posted March 4, 2006 Report Share Posted March 4, 2006 Hi, why don't you just usea [glow=red,2,300]LM 350 T[/glow]?It's a Voltage regulator that can be regulated (???)...you know what I mean.Imax=3AUmax=33V (just choose the right pot for controlling or use another resistor in line with the pot)Pmax=25Whope this can help youMatthias Quote Link to comment Share on other sites More sharing options...
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