So let's see what your actually know

congo_90

ok dramatic title over with.. (sorry)
Simple question for those who know more on electricity than me.
I've a planned (in motion) modification of my car (not a sham, but am styling the car not sham wise).

The mod includes putting 5 LEDs (wide angle, not yet bought) into the door to match a certain theme. Problem is from my knowledge, an led runs on 3v and the battery is 12.
The idea is they come on when the door is opened. Possibly somehow interlinked to the current door trigger for the stock light?
anyways. . Is there a wat to get away with this? bearing in mind
it is essentially 2 doors open but usually one which must light the lED's on demand or perhaps the standard cabin light individually?

5x leds @ 3v = 15v (max exceeded) that is my main problem. I've Led's already in the car (totalling 6 so far) which are on a direct line to the battery but have a kill switch on them. The car starts no problem with these illuminated and the radio active which in theory is more than the car should handle? as your can see it's a dilemna. Naturally at engine stat up i'll typically switch off all non-neccassary electrics unless testing parts under stress.

Either way. Bottom line, can I get away with it again? we're talking 10 mroe LED's on a max (with engine running (14v) source whic so far holds up no problem.

If i'm not clear enough please point it out and thanks for all advice/ replies in advance.

Tribunius

Well for starters your mixing up current (amps) and voltage.

Automobile electrical systems are 12 volt systems. So as long as the leds are rated for that then they will be fine.

The current drawn by an led is very low it will be milliamps. I would be surprised if all the leds in your car total up to more that an amp or two. Which should be no problem.

Also rigging them to come on when the door opens using the door light switch should be straight forward aswell.

What leds are you planning on using?

Also what car is it (specifics year, trim level etc)?

Any other modifications done to it? What are the leds already installed?

congo_90

You sure? typically the forward voltage is 2.5-3v depending on type. I couldn't find the exact one's online but these are identical.
how would you recommend opening into the door circuit? bearing in mind there is 2 switches I was thinking of breakin off at the cabin light itself and using the negative (earth) as the switch somehow. Unless there's a much simpler method?

congo_90

Tribunius wrote: »

Also what car is it (specifics year, trim level etc)?

Any other modifications done to it? What are the leds already installed?

To add to this.
hyundai accent gls (more known as LC) 2000 reg. there was a 40amp direct current to power an amp but this has been disconnected due to a fault devolping in the amp itself. other than that just 6 LED's running fine no matter what for now.

Tribunius

Yes positive. If you don't believe me look at your battery. It will have 12v on it somewhere.

Hooking into the light is the way I'd go about it too. Can't get simpler.

Tribunius

Ok from what I can find you have a 75amp alternator and a 500amp/hr battery.

This will be more than enough for what you want but if you start adding a lot of other electronics these will not be enough.

congo_90

Tribunius wrote: »

Ok from what I can find you have a 75amp alternator and a 500amp/hr battery.

This will be more than enough for what you want but if you start adding a lot of other electronics these will not be enough.

Based on standard electronics in a car running (headunit, heat lights etc)
on top of these stats:

Forward voltage is 2.5V
Forward current is 30mA

Battery voltage with engine running is approx 13.5V

Bearing in mind i'm already running 4 of same power. (which are usually only powerer on with engine on).

I won't have to modify my alt or batt?

Akaik there is a constant voltage running to the door switch of 12v. Can I just "tap" into the wire at some point? basically cut it and drop the circuit in either side that controls the switch?

backboiler

With an LED you need either a current-limiting resistor or a constant-current supply. Some LEDs are sold with a resistor already moulded into the package so you don't need to provide your own but they would then be only usable with a specific power supply voltage (or greater).

Since this LED you're talking about seems to have Vf (forward voltage) and If (forward current) rating provided, it's likely that this is just a plain LED without any resistor so you will have to provide one of your own.

LEDs are non-linear devices meaning that a given change in voltage across its terminals does not necessarily mean the same change in current through it at different voltage starting points. This means that reading 2.5 off the LED document and dividing it into 12 (actually when the engine is running the value will be more like 14 V) won't work reliably. Tiny changes in the alternator output can - and likely will - overload your (expensive) LEDs and kill them.

With a 12-14 V supply, I'd say that you should use 4 of your LEDs in series, that is one connected to the other with the positive lead of one leading to the negative of the next. Yes, I know you want more than 4: that doesn't matter for now, I'll come back to it. As you realised the total Vf cannot be more than the lowest available supply (say about 11 V for a car battery) Now the science bit:

• R = (Vs - (n * Vf)) / If
• P = If * If * R

where,

• R is the resistor value you need in ohms.
• Vs is the power supply voltage in volts, lets say 14.
• n is the number of LEDs in series, we said 4 up above.
• Vf is the LED forward voltage in volts, you said it's 2.5.
• If is the forward current of the LED in amperes, you said 30 mA, which is 0.03 A.
• P is the power wasted by the resistor in watts. We need to know this to decide on the physical size of the resistor you need after we figure out its resistance value.

Right, plug in the values:
R = (14 - (4 * 2.5)) / 0.03
so R = (14 - 10) / 0.03,
so R = 4 / 0.03,
so R = 133 ohms.

Resistors are only available in "standard values" (see http://www.logwell.com/tech/components/resistor_values.html) and 130 ohms is the closest available to the value you need in the most common (E24) range.

P = If * If * R,
so P = 0.03 * 0.03 * 130,
so P = 0.117 W
The next standard power rating up (always go up) is the very common 0.25 W type.

So you've figured out that for every 4 LEDs you need a 130 ohm 1/4 watt resistor in line with them. This can go either at the positive or negative end of the chain of LEDs. It doesn't matter but it's probably best to go for the positive end for protection in case one of the LED leads touches the body of the car (assuming it's a negative ground car).

Say you want 8 or 12 LEDs, you just make a second identical chain of LEDs and a resistor and connect it in parallel.

congo_90

Hi backboiler,
Not only thanks for the great feedback but for also doing all the math and explaining your answer! thanks for that! saves me a lot of time!
From what I know. LED's in series can be a bit hit and miss. So is it going to be just one resistor at the start of each chain or am I doing it as you would individually? ie; one resistor inbetween each LED. say we go for 10. 5 Per side, and run it off the same switch (no problem there as plenty of current available). what does worrry me is the forwad voltage is 2.5 x 10 = 25V.

I'm sure i'm making some school boy error here in my calculations but is that figure accurate of would wiring in series reduce current and voltage demand due to the actual resistance of each LED?

You can buy strips of LED's online so I wonder how they get away with it. As for me, My idea is a complete custom install. Much like the LED's already in the car.

backboiler

congo_90 wrote: »

From what I know. LED's in series can be a bit hit and miss.

Don't know where you got that from. Direct parallel connections are the dodgy ones where LEDs are involved. The only way the series connection would give trouble is if the LEDs had the in-built resistor I mentioned at the top of my earlier post. In this case you must use the parallel connection. As I indicated these type will usually be advertised either as having the resistor or be specified for use at a particular voltage, e.g. something like http://www.oznium.com/led-flex-strips.

congo_90 wrote: »

So is it going to be just one resistor at the start of each chain or am I doing it as you would individually? ie; one resistor inbetween each LED.

You can do each one individually if you want but that's a bit wasteful since the resistor on each one would be burning away 0.345 W with the car running. Means you need a bigger and harder to find 0.5 W resistor and you now need ten of them.

congo_90 wrote: »

say we go for 10. 5 Per side, and run it off the same switch (no problem there as plenty of current available). what does worrry me is the forwad voltage is 2.5 x 10 = 25V.

I'm sure i'm making some school boy error here in my calculations but is that figure accurate of would wiring in series reduce current and voltage demand due to the actual resistance of each LED?

Kind of. The current you want is fixed at 30 mA. To get this to the right value you use the resistor, which as I said literally burns away (turns into heat) the excess electricity that the LED isn't able to handle.
The idea of putting more LEDs in series it to reduce the amount wasted in the resistor. The limit on how many LEDs you can put in series is decided by two things: the forward voltage of the LED and the available power supply. You have a (worst case) supply of about 11.5 V (car off, battery in poor shape) and a forward voltage of 2.5 V. That means that you can put a maximum of 4 (2.5 * 4 = 10 but 2.5 * 5 = 12.5: too high). Don't go thinking you'll get a fairly bright glow because it's "close enough". I know it's complicated but have a quick look at "Figure 5: I–V characteristics of a P-N junction diode." on http://en.wikipedia.org/wiki/Diode. The bit you're interested in is the green "Forward" section, in particular at "Vd" (we're calling it "Vf" in this thread - same thing) where the blue curve goes straight up in the air. Within limits you can think of the height of the blue line as how bright the LED is shining so you can see if the forward voltage (the side-to-side axis) is even slightly below the required value the brightness will disappear down to nothing very quickly. As well you can see if it's too high by a small bit the blue line goes way off high and your LED goes on fire.
That's the end of the theory.
For your purpose I'd say you could do with making 4 chains in total: two chains of 3 LEDs and a resistor and two chains of 2 LEDs and a resistor. Put your 2+3 in parallel and you have 5 per side. Here's a diagram.

You'll have to work out the figures for R1 and R2 for yourself using the stuff I posted yesterday, all that's changed is the value of "n" being 2 or 3.

congo_90 wrote: »

You can buy strips of LED's online so I wonder how they get away with it. As for me, My idea is a complete custom install. Much like the LED's already in the car.

The correct resistor on these is most likely built into the strip. They're sold in such a way so when you connect more than one strip module they are in parallel, which is the right way to connect this type.

backboiler

Oh yeah, meant to mention about your worries on power consumption. If you go with my suggested 2+3 way above you have 4 chains each passing 30 mA at 14 V max.
The power is therefore 4 * 0.03 * 14, which is 1.68 W.
To put that in context, your dip lights will probably be 55 W (or 80 W if you have 100/80's). Your audio setup could be drinking hundreds of watts at peak.

congo_90

HI backboiler.
Wow you really went outta your way there mate seriously! My physics teacher never ever explained it in such detail. even as just diodes and the principles of forward voltage.

I really like that idea of 2+3. It's truely genius. I'm not too worried about them being at their peak brightness. Just as long as they'll light up to a point of being visible.

So if my understand is what it is. Then this diagram (sorry it's a bit sloppy)
Should answer all. After all it'll be operating off one switch so I'll naturally be running some cable from the battery again.

backboiler

Yeah, you seem to have the idea. I wouldn't bother with all the fuses though (if that's what they are). Total current is about 120 mA so just a single 250 mA fuse near the switch will do the job. More important than a fuse is a solid construction and insulating the connections between components and wires from the bodywork of the car. Use heatshrink tube if you can get you hands on some. Remember to thread the wire or component lead through the heatshrink before soldering it.

When you're assembling things for the first time I'd recommend you use a 1000 ohm or so resistor instead of or as well as (in series) your calculated values just as a precaution against wiring the LEDs the wrong way. If they light then they're the right way round. Some of them are very sensitive to reverse voltages especially them blue ones that you're probably using given the Vf of 2.5 V (you can get a good idea of the colour from the Vf).

Post up a snap if the finished job - not the wiring or anything, just the lights. I'm curious.