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Help! - Printable Version +- Forums (https://www.studioscalemodelers.com/forums) +-- Forum: Index (https://www.studioscalemodelers.com/forums/forum-623.html) +--- Forum: General Modelling (https://www.studioscalemodelers.com/forums/forum-680.html) +--- Thread: Help! (/thread-355.html) |
Help! - Wombat - 12-30-2009 OK,i have even less idea about lighting,than i do about building,so you can tell i'm in trouble a bit here! I'm looking at getting some stuff for the shuttle and need advise.The light's themselves are pretty much sorted thank's to Doug,it's more the transformer that'll be best for running them i need help with. OK,there will be 2 X 12V 5W Halogens along with probably 8 X 12V 5mm LED's. I've been looking at transformers and i see there are different output's even though they are 12V.I'd like advise what is the best to use? I can get a 12V 1A(12W) output or a 12V 2A(24W) one.I figure the Halogens are 10W on there own but am unsure how much the LED's will be. Do i get the 1A (12W) or the 2A (24W)? Thanks Phil RE: Help! - erospawn - 12-30-2009 LED's draw about .25 - .30ma ..., some even less than that.. Usually a 1amp 12 volt power supply will work fine.. With those Halogen bulbs, the 2 amp power supply... Keep in mind Halogen bulbs also burn pretty hot too... RE: Help! - Wombat - 12-30-2009 (12-30-2009, 11:13 PM)erospawn Wrote: LED's draw about .25 - .30ma ..., some even less than that.. Usually a 1amp 12 volt power supply will work fine.. With those Halogen bulbs, the 2 amp power supply... Keep in mind Halogen bulbs also burn pretty hot too... Cheers,i appreciate the help and will now grab the 2 amp one. Yeah the heat from the halogens is a small concern,but the globes look like the original models globes pretty much and considering 99% of the time they wont be on they'll look perfect.For the other 1% it'll be short run's only and again the result is gonna look pretty damn good going on Doug's testing.Now,if i could get some aluminum engine bells made up that'd be even sweeter and help with the heat concern. Thanks again mate!!
RE: Help! - 427 Cobra - 12-31-2009 Is this battery powered,or a straight from the wall AC/DC supply? You'd be better off with LEDs,as they don't give off heat. As long as you don't have anything with a heat source near it - these will work fine. You'll need a converter that will work with these,but you'll also need resistors that won't overheat. I'll have to see what rating resistors will work with your lighting setup,as I'll be using a similar system. Superbright LEDs will work in the engine bells,and not produce heat. LEDs usually are rated at 26 m (Milliamps) to 35 m. You just need resistors that will allow you to use these will a 12 volt adapter without blowing them up. RE: Help! - erospawn - 12-31-2009 The resistors used are based on the amount of current needed to light the LED's and based on the voltage supply desired. It will take a large number of batteries to generate enough current (1 amp or more) and last a long time without going dead in this case. The resistors will give off heat .. however, at 12 volts.. with a 3.4v drop across a single LED with 25mA of current.. you would need to pick our a resistor of the appropriate Ohms .. so we know there is a voltage drop of 3.4 volts across the LED.. which means we will have a voltage drop of 8.6 volts across the resistor.. no matter what resistor rating we choose.. so, if we need .25mA current.. this means 8.6v / .025 = ~ 350 Ohms.. If you can find that value..It just needs to be with 20 - 30 points of the calculated value.. The wattage, ie size of the resistor to get can be calculated as 8.6v * .025 = .215 Watts.. So, you might think about getting 1/2 watt resistors if you are using a 12 volt power supply.. Again, the formula is Voltage - LED forward Voltage / LED current = Resistor size in Ohm's. Wattage is Voltage * current .. THe foward voltage drop required across the LED can vary between 1.4 volts and up.. just look at the specs on the LED package.. Also, the amount of current to light it up can vary too.. The best approximation to make.. is get the current with in, say 5 mA of MAX current and you should be ok.. you'll know it wasn't good if you let the smoke out I typically use 18,000 mcd @ 30 degree viewing angle.. about 3.4 volt forward voltage at 25mA.. ( About 5mm in size) You can shave the plastic down to a point for tight fitting spaces..
RE: Help! - Wombat - 12-31-2009 All of this just reinforces that i have not the slightest idea about lighting and resistors and anything else either pretty much. I thought as the LED will be 12V ones they be hook straight up to a 12V source and that's it.I haven't recieved them yet but i don't think it said anything about there values when i ordered them.They were just 5mm 12V LED's.Now i read i need to have resistors also which leads me to another question.Do i need a resistor on each LED or one resistor that'll handle all the LED's on it's own and wire it into the feed that goes to all the LED's instead.How i was planning it was to have the power in source go to the three point's.Two wire's running to the halogens and one wire that'll feed the 6-8 LED's.I'm really lost on it all,even with your explaining there.But that's why i posted here,i know some of you guy's do know a thing or two about all this stuff as you've shown. Chris,the halogen's i want purely for there look mate,mostly while they are not switched on,but also the effect when they are turned on as well.The original model had globes that look very similar in there appearence,so it's mostly just for the look while they are not on and as i mentioned that'll be 99% of the time.I've also gone away from trying to run this on batteries,the transformer is a betta option. RE: Help! - erospawn - 12-31-2009 I understand really... I have a degree in Electronics eng.. and designing circuits has been kind of a handy hobby to have in addition to the modeling skills... Now, you can wire a single LED to the anode (long lead) of an LED.. or you can use a single resistor per LED.. Its also possible, the 12 volt LED's you bought, already have the resistors included.. in which case, all you do is wire them up.. where did you order your LED's from ? Do you have a link to the site and the model number of the LED's? I can take a quick look at them... I usually order my LED"s from http://www.superbrightleds.com Here's a link to LED's and design considerations: Links http://led.linear1.org/1led.wiz http://ledz.com/?p=zz.led.resistor.calculator This is a basic tutorial on LED's http://www.theledlight.com/ledcircuits.html Here's a pic that should give you some idea of what Im talking about: RE: Help! - Wombat - 12-31-2009 Thanks for those link's especially the last one there and also the picture you posted. OK,i like the look of the one shown as series in that last link.So if i'm reading it correctly and i use 8 LED's in this fashion each will have 1.5V running through it.Heres where i'm lost again,the LED's are listed as 12V so will they run having just 1.5V going through them?I've purchased them just off of evil bay and heres a link to the one's i'm getting.It doesn't say anything about resistor's,they are just pre wired,but it does say something about there power being 0.2W. http://cgi.ebay.com.au/20X-5mm-White-LED-Light-Lamp-Set-Pre-Wired-12-Volt-25cm_W0QQitemZ190359955973QQcmdZViewItemQQptZAU_B_I_Electrical_Test_Equipment?hash=item2c52566a05 Another thing,as it'll just be one transformer powering not only the LED's,but the two halogen's also,wont they have to be taken into account when working it all out or don't they effect anything whats needed for the LED's? Here i was thinking i could just hook it all up and it'd work! RE: Help! - erospawn - 12-31-2009 (12-31-2009, 01:29 PM)Wombat Wrote: Thanks for those link's especially the last one there and also the picture you posted. If wired in parallel, you won't have to worry about voltage.. current will be determined by the resistor wired to each LED.. Series... the voltage drop is the total voltage across all LED's... 3 LED's = 10.2 volts. with 1.8 left over.. so, that will be the voltage over the resistor.. so, again.. voltage / I = 1.8v /.025 = 72 Ohms.. Putting 4 LED's in series ( 3.4 * 4 = 13.6) probably won't light brightly, if at all.. So, you could wire 3 in series.. and then add the extra one wired in parallel.. LED's are basically DIODES... which means almost 0 resistance.. Current is what makes things happen.. You can have 50,000 volts with .000001 amp and not get hurt at all.. ---- 3.4v ----- 3.4v ------ 3.4v ----- 1.8v ® ----- 12Volts | ---- 3.4v ----- 8.6v ® ------------------------------| PS..Based on that Link , you bought LED's with resistors already attached.. So, just wire them in Parallel... IN otherwords, each LED Anode is connectd to 12 volts..each LED Cathode is wired to ground. +12 --------| @ (-) --------| RE: Help! - JAWS - 12-31-2009 Oh man there has to be an easier way to understand this stuff. Im with Phil & it seriously twists my brain to figure out or more so decipher what you just described. Can someone do a step by step wiring photo sequence for baby's? I might understand it then. RE: Help! - Wombat - 12-31-2009 Alright,again thanks,i think i'm slowly getting my head around it all.I see now that running 3 LED's in series is the most from a 12V supply as theres not enough volts to run more than that.I also get i think,that the resistor for 3 LED's run that way needs to be for the 1.8V that is left over.Still not sure about the Ohms part but i'm getting there. If i'm lucky then those LED's that are coming will infact have a resistor already included in that wiring,although i can't see one in the picture.If they do have a resistor then my understanding from all this is each one will have a resistance for the 8.6V that is not needed from the 12V supply to run an LED and that's how they are selling them listed as 12V ones,otherwise they'd just be pre wired LED's? Infact these could be listed as 9V ones but the resistor if i'm reading it right would be different,thats all. So for 12 Volt's it's 8.6v and for 9V it'd be 5.6V even though the LED is the same one? I just hope they have them resistors or then my next problem will be trying to find them also.So much for a Tandys shop being like they use to down here.Years ago you could get anything like this stuff off the shelf.Not any more,they carry FA these day's in that sorta stuff.Still,theres always the internet,it just more postage like everything else these day's. RE: Help! - erospawn - 01-01-2010 Exactly! You are getting it.. the LED is a diode.. which means its a semi-conductor and will always have a forward voltage drop.. IE, the amount of voltage neccessary to make the electronic gate close in the diode and thusly complete the circuit. The resistor has whatever voltage is left over from the forward voltage drop acros the LED. The resistor in this circuit, since the LED has 0 resistance, acts as a current limiter. The resistor value chosen, sets the maximum amount of current allowed to flow through the LED. A super conductor would allow for an unlimited amount of current....But super conductors are only found in those giant "Colliders" because they are two expensive to make and even then they aren't perfect.. (Super conductor is a device that can pass current through witout any loss whatsoever) An LED is a "semi-conductor" that can only handle a certain amount of current. Ohms is a measure of resistance..Resistance can be used to guide the electrons of current where you want them to go. Thats also what the color code on the resistors means...but basically, we need to have 25 - 35 mA current. (miliamps) for the LED to light.. We don't measure the resistance across the LED because there should be zero resistance... thats what a semi-conductor is ... as close to a super conductor as you are going to get.. Watts is a measure of energy dissipation on the resistor..( A super conductor would have 0 watts of engery dissipation) You cant have a transfer of energy without some cost involved...in this case, the cost is heat disipation.. Now, if I have 2 resistors in series .. 12v --------- 220 Ohm ------- 320 Ohm ----- ground I would add up the total resistance .. to get the total current... now once I have that I can then determine the exact voltage across each resistor... In this case 220 + 320 = 540 Ohms 12 / 540 = .022 amps or 22 miliamps.. Now ( V = R * I) .. 220 * .022 = 4.8v Standard Formula is V (Voltage), I (Current), R (Resistor),W (Watts) V/I = R V/R = I I * R = V Picures coming up .. RE: Help! - erospawn - 01-01-2010 The first picture below shows a typical LED with a resistor attached to the Anode (The longer lead) of an LED.. The other end would go to the negative batter source.. Think of the LED ( or any other Diode) as a gateway that only allows current flow in one direction..(See the picture on the right of the second image) If you put anode to positve and cathode to negative, the gateway opens ... put in the other direction, the gateway is closed.. Also, remember the forward voltag drop? Well, if the gateway is postioned to be opened, it will only open if the voltge exceeds the forward voltage drop.. LED's 3.4v as an example.. For Diodes (.7 volts) ..so you can see how they would act like a switch.. [attachment=2012] The second image -- Top pic is an example of two LED"s wired in parallel.. ( Standard way of wiring).. As you can see, each anode is connected to a resistor that then connects to +12 volts.. The other end of the LED, the Anode, is connected to ground.. As you can also see, the Arrow point in the direction that the current is supposed to flow.. (Trivia: The arrows are actually pointing in the wrong direction because current actually flows from negative to positive, NOT postive to negative) So, wiring in Paralell is like having several different circuits connected to the same power supply.... The second picture shows 2 LED's wired in series. Not the most ideal way of wiring them... Here we have a single resistor connected to the power supply to the Anode of the first LED.. The cathode of the first LED then connects to the anode of the second LED and the cathode of the second LED then goes to the negative supply of the power supply.. It bascially two gateways one right after the other.. What happens if I postion the arrow of the 2nd LED so the arrow points in the other direction (cathode connected to positive)? Answer: The entire circuit will not work because the gateway will be closed.. so that curent cannot flow to ground.. In that same respect, if one or the LED fries.. the entire cicuit will not work. [attachment=2011] RE: Help! - Wombat - 01-01-2010 Gee mate,you really are putting in some effort trying to explain all this for me and others to understand. Just when i think i'm sorta getting a handle on some of it though i see i'm still pretty lost. I think whats best is i just make sure i buy LED's that have a resistor fitted already for the voltage i wanna use and go the parallel route when wiring them all up.It look's to be the easiest way for someone with my abilities to take all this in. But it's not from your lack of trying that i'm still not getting it all,your help has at least got me understanding some of the ways these things can be run and i do thank you for that! RE: Help! - erospawn - 01-01-2010 (01-01-2010, 12:58 PM)Wombat Wrote: Gee mate,you really are putting in some effort trying to explain all this for me and others to understand. You actually understand better than you realize. Parallel is the best to wire them up... And based on the link you sent earlier.. those LED's already have the resistors soldered on the LED's. So, all you really need to do is worry about wiring them up. RE: Help! - Wombat - 01-01-2010 I really hope you are right mate about them LED's,it'll make life pretty easy if it's so. Least if nothing else i do see how thing's can be run and i agree,the parallel way seem's the easiest for someone like myself. Thanks again Ron,those links and your words have helped me to work out how to go about lighting this thing and also it'll help in some future project's no doubt as well.
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