Electrical Noob, need help with LEDs

Electrical Noob, need help with LED's

Ok, I'm a total noob when it comes to fabricating electrical circuits so maybe some of you can help me with this project.

I wanted to do a LED setup on my interior cuz that will sort of complete the interior for a while. What i wanted to do was to habe about 12 LED's of variable length connected to the battery with a switch in the middle. i know that the 12V from the battery will fry them instantly so i will need to get a resistors or more than one resistor. what i need your help with is if some one can possibly give me a setup of what kind of LED and what sort of resistor i should get as well as what GA of wire, and if possible a diagram. I would much appreciate your help with this.
thanx
Achint

The voltage drop across an LED is always going to be around 1.7 volts. What you want to do is control the current flow - LEDs are always rated in terms of current. A series resistor will control current flow.

If the car's electrical system is at 14.4 volts (engine running), the voltage across the series resistor will be about 12.7 volts. If you're using 10 mA LEDs, the required resistance would be 12.7V/10mA = 1.27 kOhms.

When the engine is not running (battery voltage 12.6), you'll get 10.9 volts accross the resistor, which will give you 8.5 mA - close enough to not make a discernable difference.

lol, thanx for the effor oxymorone, but i am a complete idiot when it come to this and that made no sense at all. here are a few links of what i was looking at

http://www.radioshack.com/category.asp?catalog%5Fname=CTLG&category%5Fname=CTLG%5F011%5F006%5F002%5F000&Page=1&find=LED(keyword)

those are LED ^^^^^


http://www.radioshack.com/category.asp?catalog_name=CTLG&category_name=CTLG_011_002_014_000&Page=1

Those are Resistors ^^^^^

now as an example can you explain what would i need in a combination for say 12 - 14 LED's. sorry if i'm being a bit difficult but i have no idea how this electical stuff works in terms of down sizing power and stuff.

The link has a bunch of different LEDs. Which one are you looking at?

lookin at these 2

Cat # 276-320 first row first column

Cat # 276-307 red one second row third form the left

OK. They're both rated at 20 mA max, but different voltages. Give me a few minutes...

lol, ok thanx, sorry for the late reply, i was gettin something to drink

OK. Here goes...



Wire each LED in series with its own resistor (R).

For Radio Shack cat. # 276-320, the LED voltage is 3.6 V and the maximum current is 20 mA. The electrical system voltage (alternator volatge) is 14.4 volts, so the voltage across the resistor will be 14.4 - 3.6 = 10.8 volts. To get a current of 20 mA (0.02 A) you need a resistor with a value of 10.8/0.02 = 540 ohms. This is the minimum resistance. Do not go below this value or you will fry the LED. To add a safety margin add 20% to this minimum value. The power dissipated in the resistor is going to be the voltage across the resistor squared divided by its resistance. In the case of 540 ohms, the power will be 10.8 squared divided by 540 = 0.216 Watts, so a 1/4 watt resistor is all you need.

For radio Shack cat# 276-307, the LED voltage is 1.7 volts and the max current is 20 mA. The voltage across the resistor will be 14.4 - 1.7 = 12.7 volts. For a current of 20 mA (0.02 A) a resistance of 12.7/0.02 = 635 ohms is required. Again, this is a minimum resistance. Go 20 % higher than that. With a 635 ohm resistance, the power dissipated in the resistor will be 12.7 squared divided by 635 ohms = 0.254 watts, which is just above 1/4 watt.

As for the wire you use, it really doesn't matter. The current flowing in these circuits will be really small. 22 gauge will be just fine.

By the way, Radio Shack is a rip off, and their selection of resistors really sucks. Try to find a local electronics shop to get your parts.

BTW the Trianlge type image is your LED. its actually also a diode, so its the same symbol.

You can solder the reistor to the end of ONE of your LED legs and connect them to ANY 12V source after that. Because you just made your 3V LED really need 12V with that resistor in place.

Huh?

thanx oxymorone, one last question, do i have to solder cuz i dont have a soldeing iron at home, i'll need to purchase one.

When you solder that resistor on the end of a leg from a LED it makes that LED need 12 to light up. You are dropping the other 9 volts across the resitor. Acording to your diagram the source is paralleled, which means it doesnt matter if you have one LED hooked up like that or 1000 of them, they will all have a 14.4V (12V for arguments sake) supply to each one of them.

AND THAT is why he does NOT need the same 12V source for all of them, he can tap some of them off hius cigarete plug, and some of the others of his headlight relay if he wanted. They will all light up the same.

You pretty much hace to solder. Any other connection isn't going to last too long. Find a good electronics shop and you can buy an iron with all the money you'll save on the LEDs and resistors.

do you by anychance know how much those cost, i'll try to look online as well

is this one good for occational use

http://www.familydefense.com/solderingiron.htm

and does that price seem normal i men 4 bux

Should be less than $10. Try to get at least a 25 watt iron.

Why not just hook them all in series? 14.4V divided by nine LEDs will deliver 1.6V to each led.

A LED will light brightly with as little as 1.2V, so you could actually use 12 LEDs in series and they'd light up fine.

But when the engine is off, you'd have 1.1V across each LED. They'd light but not as bright as normal.

hey john, is there any chance that i can talk to you over AIM tomorrow morning???

My father's having his entire knee replaced tomorrow, so I have to bring him to the hospital at 5 in the morning. I'm going to stick around for the surgery and until he's awake, so I probably won't be home until late.

I'll be on after I get home, if it's not too late then.

Sorry

that's fine, i just wanted to discuss some stuff about the issue of the LED's i'll talk to you tomorrow night then i guess, i'll be on about 9 ish cuz i get off school at 730 and then driving home and dinner. thanx

K, I'll look for you then. I'd go on now but I gotta get up at 4 and it's already 10 here.

Have a great nite, ttyt.

J

Might work, but probably not. LEDs are single junction semiconductors. If you drop the voltage across the junction only slightly the current flow reduces dramatically. The relationship between voltage and current is completely non-linear, unlike resistors. Using resistors is the simplest way to ensure that the voltage across the LED stays constant regardless of whether the engine is running or not. Also, if you wire them all in series, if you lose one, you lose them all.

That'll do.

Oxy I respect your knowlege, but it will work. The relationship between voltage and current is fixed in stone, IE: Ohm's Law. LEDs are "avalanch" semiconductors, once they're sufficiently biased, they're on, period. Using a bunch of LEDs and resistors in parallel will work also, but as you reduce the voltage to the input of the ladder, you will reduce voltage at the individual LEDs also. Six of one, half dozen of the other. Series eliminates at least nine parts requiring 18 more solder joints, a good thing for someone who's new at soldering.

LEDs, if not over-voltaged, will light today, and still be lit 100 years after you and I are and buried. So losing one and losing all is not a issue. If you're really worried about over-voltaging them, it's pretty easy to use a 12V reg (7512 I think, there's another part number but it's CMOS and doesn't appreciate noise spikes like car regulators switching) in the circuit.

I hand soldered over 500 LEDs together in a series / parallel config and made a American flag about 10 years ago. It still works fine today. (no I didn't buy them, I worked at Lockheed and engineering bought a couple thousand wrong parts, they threw them in the trash so I nabbed them, and yes, I was bored as hell that month).

Getting a 25W soldering iron would be good, because it could also be used for a lot heavier soldering jobs (general purpose). But if you're going to solder LED leads with it, use a heatsink. Or melt the LED. A alligator clip (eg roach clip) works fine.

I'm in the waiting room at Lahey Clinic in Mass so I won't be able to keep up with this thread. Can you believe they have computers in the waiting rooms for us to surf!?!

Um, Ohm's law applies to linear conductors not semiconductors. Before you get the bias voltage established the "resistance" is pretty much infinite. Once switched on the relationship between voltage and current is logarithmic.

The point of using series resistors in a ladder configuration is that changes to the input voltage will not result in significant changes in the LED voltage. If at 14.4 V the LEDs have 1.7 volts across them, dropping the input voltage to say 10 volts will reduce the voltage accross the LEDs to perhaps 1.69 volts. It's best to think of LEDs (or all semiconductors for that matter) as current devices, not voltage devices. If you concentrate on controlling the current, the voltage will work itself out.

A compromise that would work quite well would be to wire a bunch of LEDs in series with a single series resistor. With five 1.7 volt LEDs wired in series, a 295 ohm resistor will control the current to 20mA at 14.4 volts. When the engine is not running and the voltage drops to 12.6, the current would drop to about 14 mA.

Hope everything goes well at your end.

"Um, Ohm's law applies to linear conductors not semiconductors. Before you get the bias voltage established the "resistance" is pretty much infinite. Once switched on the relationship between voltage and current is logarithmic."

Very true. So as long as the supplied voltage remains high enough to keep the LEDs in conduction, they'll lite. If the battery gets low enough for them to drop off, probably a good thing they're not lit."

"A compromise that would work quite well would be to wire a bunch of LEDs in series with a single series resistor. With five 1.7 volt LEDs wired in series, a 295 ohm resistor will control the current to 20mA at 14.4 volts. When the engine is not running and the voltage drops to 12.6, the current would drop to about 14 mA."

Exactly! Except instead of using five LEDs and a dropping resistor, he can wire 12 in series and skip the resistor.

How is it we always end up on the same page? Great minds, huh!

My father went into surgery at 7:30 to get his knee replaced, takes three hours (guesstimate). The power keeps blinking here and crashing this computer, nurse says it's not blinking in the OR (probably a good thing). So I gotta type fast if I want to not type everything over and over.

Thanks for the wishes!

Usually yes, this time no, but I'm too tired to deal with it. Got some real design work to do tonight.

I trust everything went OK?

I hear you on the tired part. What a long day...surgery took almost four hours, then waiting for him to be awake enough to talk to was another three...a hour driving there and a hour and a half driving back. After getting all of 4 hours sleep last night. Ugh.

He did very well doc says. They have his leg in some Rube looking device that keeps bending it back and forth over and over. On morophine, so he actually smiled at me once.

Now he's got a month on his back to deal with.

Tnx for the concern.

Was thinking on the way home about you being right, needing a dropping resistor in series to control the current. Wondering why my flag setup works so well in a series / parallel circuit. Will think more tomorrow.

It works fine because you will get current as a result of your voltage and the resistance of each LED. they do still have it or you wouldnt get a voltage drop across them.

I have wired them that way before too.

I am also too lazy to actually figure out what size resister I need so I usually use a linear Pot and just turn it up till they look bright enough. this can of course get you into trouble if you go to high!

so let me get this straight, whay stinky and 03Toyman are saying is that
if i was to just take a wire and add say 12 LED's to it, it would still work and i wouldnt need resistors and all that stuff.