new to electronics in general

[exhumed07]

Hey everyone. new here of course. and well new to about everything electronic when it comes to building things and having them work lol. I have tinkered around here and there but mostly just ripping things apart like my last 4 computers. but now I thought I would start doing things constructivly. Well I think I may have found the perfect starter project for myself but still need some advice. I have a 55 gallon and 20 gallon fish tank that I would like to add LED moon lights to. I've already ordered my LEDs and they have resistors attached to them already. they are supossed to work on any 12vdc power supply. now here comes my problem. I have a 12vdc power supply but when it is plugged in it's registering a solid 16v. As well as 1.2A, As far as I know I should not exceed 40Ma if I'm correct with LEDs. so my main question is what size resistor should I put onto my positive lead to drop the voltage as well as amps to avoid burning out my LEDs but still retaine their brightness? I was told 10k by kipkay from make magazine but wanted some other opinions. thanks for any info.

 

[pendragon]

Hi nice to see someone who will have go with the 'soldering iron..

First I would ask, the voltage referred to, 16v,as against the required 12v.. Is this measured 'on load' (connected) or just the floating (un connected) voltage.. The l.e.d 's, you say; already have the appropriate limiting resistors in series ,so I would not expect any great problem. However if you feel a wee bit hesitant,then put another limiting resistor (50 to 200 ohms,in series with the common feedline .. All in all,the 'offload voltage' will always be higher than the 'loaded' voltage ..

 

[ElectronWorks]

10k seems a bit high. Try it. If they do not light, your value is probably too high.

Your LEDs sound like nothing more than a standard LED (with a voltage drop of 2V) in series with a resistor. Check the spec of your LED. They probably need about 20mA (for standard LEDs). High Brightness LEDs need a lot more current (upto 700mA).

For 20mA, use Ohm's Law:

you need to drop 4V (16v-12v) across a resistor with 20mA flowing through it. My calculator tells me this is 200 Ohms.

Just massage the numbers above for different currents and voltages.

 

[exhumed07]

ok first off for pendragon. it's measuring at 15.6 unplugged and 16.3 plugged in. and the amps never change either way. but the reason y I was thinking that I would have to drop the voltage is because the LEDs are only rated to be on a 12v power supply. I don't want to end up burning up my LEDs over something as simple as not putting on one resister. not sure if this will help but these are what I ordered: goldengadgets.com/LED-Light-Bulbs-DIY-LED-with-Resistors/c12_89/p361/L-LED3%3A-3mm-LED-Light-Bulb-with-Wire-Resistor-Attached/product_info.html in blue. they say they will work with 12v but nothing about amps.

Now to get to u Electron lol. now I understand what u are saying. I work with subs and amps as a hobby so I know ohms, the only thing is I don't know anything about resistors. so would a 200 ohm resistor just be that or is there a code to go along with that? like 10k? is that 10,000 ohms? I don't understand the terminology.

 

[pendragon]

exhumed07

First of all regarding your statement, 'you know ohms but not resistors.'. Well you really need to learn or have a copy of the 'The Resistor Colour Code' In this query you refer to 200 ohms ,which in the colour code equates to Red,Black,Brown. Now, I am not sure just how many L.E.Ds you are going to use, but typically a single L.E.D of say Green in colour has a forward voltage of 2.2 volts at 20ma..,But a Blue L.E.D, I must admit I am not sure of the forward voltage drop.. Briefly, the number of L.E.Ds you will use will be (again I assume) in series. BUT, you already have a series resistor on each L.E.D !! which implies they would be used in parallel ... Final query, how many L.E.Ds in total , and what are the colours(Banded) on each L.E.D. At a guess, you may see Red,Blue,Brown (250 ohms).. You say your'e aware of ohms law , then you will know that R = Vs - Vf / If. Vs = supply voltage Vf = Forward voltage and Is , is Supply current, hence total forward voltage drop is equal to each L.E.D in series......... .To save a lot of trouble , get yourself a 12 volt d.c transformer !!!!!

 

[pendragon]

exhumed07

To simplify ; Join all your Red (positive) leads together and the same with the Black (Negative) ,Put a 200 ohm resistor in series with the Red lead , attach your 16 volts dc,and switch on.. As Electron Works, pointed out , this is a nominal value, assuming 20- 25ma of current, and should be O.K -- Note, l.e.ds in parallel as your's are, use more current than l.e.ds in series, tho' not enough to break the bank, you have 1.2 amps

 

[exhumed07]

ok so basicly I'll have 1 positive wire with the 200 ohm resister running the entire length that I will tie the LEDs into as needed and same with a negative and I should be ok intstead of running them all together negative to positive all in a single line. I just want to make things as crystal clear as possible as my LEDs should show up tomorrow. I know I got more then enough extra if things don't work but still I'd like to get this first crack as I don't have the best soldering iron in the world for electronics this small and they are kinda a pain. Great for house and large electrical wires though. just burns to hot and tends to burn the rosin. but no worries if this works I'll be getting a temp control one.

 

[pendragon]

exhumed07

If you have a bundle of these l.e.d s, as stated, just to prove that all will be O.K, just join, by any means, all the REDs together,and put a 200 ohm resistor in series with them to the Positive (16v+supply), and then join all the Black leads together ,(no resistor) and take to the Negative(16v-supply) . .You are now 'feeding' them all in parallel, whether you use one led, or several, they will light..! When you have proved that its O.K, then you can of course,use any length of wire you like, spreading the l.e.ds with a common wire between all REDS;and a common wire between all Blacks.. That is as simple as I can explain, Good luck,