Intake lighting idea?

[Schudini]

Thanks for the backup cutlsp,

 

However, you can also do damage if applying overvoltage.

 

I don't want to get technical, but I will if y'all make me.

 

What I would do to avoid the hassle is go to (favorite autoparts store) and get one tiny neon stick (made for automotive apps, so they can handle 11-14V or whatever), wire it up with a hidden switch, and enjoy.

 

It's the old K.I.S.S. principle 8)

 

On second thought, what would look really cool (and not add extra heat) would be to get some of that 'neon' glow string and outline the plexi of the intake box, so that only the plexi glows and you can't see the actual wire. That would really trick it out.

[glock19]

It depend's on how you wire the LED's. For example If you wire 6- 2V LEDs in series with a 12v source, you don't need a resistor.

Glock

 

Ha, that is funny. Don't do it. If you try to send that much current through the LED's in series, you will burn out the junction of the first LED almost instantly. The resistor is the current limiter. Also, the LED's will get progressivly dimmer along the series.

 

In retrospect, just use the cold cathode idea. It will put out more 'flood' light than a bunch of LED's and will look better. LED's are more directional, and even mounting an array will give you more of a disco ball effect than an even flooding of light.

 

OK please get technical with us. What value resistor would you recommend with 6 2v led in a series circuit with 12 v supply

[EurosportZ34]

fuckin wanna be ricer...

 

you read my mind! LOL

 

 

i personally wouldn't do it sl3196, but that's just my 2cents

[Schudini]

I wouldn't wire LED's in series. Period.

 

Here's what I would do...

 

1. Select LED's (high mcd, wide voltage range, etc.)

2. Find MAX microAmps that each one will take

3. Reduce this # by 15-20%

4. Multiply the above number by the number of LED's to wire in PARALLEL!!

5. Find voltage drop across single LED. should be listed on box (we will use 2V for example)

6. Assume 12V supply

7. Using Ohm's Law, calculate the total resistance needed to give you the microamps from step 4. (remember V=IR, so R=V/I)

8. Calculate ratio of resistances needed to give proper voltage (2V) to LEDs. This is known as a voltage divider. (Will give formula tomorrow, if requested)

9. Make sure resistance values add up to give # from step 7, but are in the ratio from step 8. (Can be written to solve two equations for two unknowns, but is less fun)

10. Wire the junk up. The first resistor between source and PARALLEL LEDs, and the second resistor after LEDs to ground.

 

There, does that satisfy?

[sl3196]

I'm thinking I may just put an underhood light in since it doesn't even have one.

[glock19]

So basically you are going to drop 10V across a 500 ohm 1/4w resistor for each LED you plan on using( assuming 2V and 20 ma). Yes it will work. However you will have alot of extra parts and power dissipation for no reason.

 

The reason that 6 in series with no resistor will work is as follows:

1. Given 2V drop and 20 ma, the LED has a resistance of 100 ohms (ohms law).

2. If you put 6 of these in series you will get 600 ohms (simple resistor addition).

3. If you apply a 600 ohms resistance to a 12 v source you draw 20 ma of current (12/600= 20 ma).

4. Kirchoffs current law states that all current entering a node must equal the current leaving the node. Therefore the 20 ma goes thru the 6 LED's and there is no large current applied to the first LED in the chain.

 

BTW. The voltage divider rule is Vout = Vin*R2/(R1+R2).

 

Glock

[Dannymik]

I'm thinking I may just put an underhood light in since it doesn't even have one.

 

yeah, you would think of having one of those for starters!