Pics of my ride

[T4 Primera]

Originally posted by kenchan
im thinking about running the wires from the fusebox (driver side, not engine bay) since dont wanna mess with running stuff through the firewall. in this case, from which fuse should I tap it to? and how much amps do I need for this neon?

thanxa bunch.

Pick a fuse that works in the key position you want the neon to work in. Say something that works with the key in the ACC position like the cigarette lighter etc. Tap into the busbar side (supply) of it with an in-line fuse right there so you protect all of the circuit. If you're lucky there might even be a blank space on that busbar to use with a normal fuse.

To work out your fuse size you need to know the power requirement of the neon and how many you are going to use. You should wire the neons in parallel and make sure you use a supply wire thick enough to ensure you don't get a significant voltage drop over the length of it. Now for the math (well, you asked the question)

Amps = Power or Watts/Voltage
e.g if your 12V neon is rated at 1W then
the current (amps) it will draw = 1W/12V =1/12 of an amp = 83mA.

Next you work out the resistance of the neon:
Resistance (ohms) = Voltage/Current (amps) therefore the resistance of one neon
Resistance (ohms) = 12V/0.083A = 144 ohms

When you wire them in parallel (say 5 neons) then the resistance for the complete circuit (RC) is like this:
1/RC = 1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5
or alternatively if all neons are the same rating
1/RC = Number of neons X (1/resistance of 1 neon)

So using the example values above, for 5 neons rated at 1W and wired in parallel, the current for the complete circuit is like this
1/RC = 5 X (1/144ohms)
= 5/144ohms
= 0.03472222222222 ohms, therefore,
RC = 1/0.03472222222222 ohms
= 28.8 ohms Note the resistance of the total circuit is less than for just 1 neon because, when you wire them in parallel, the current now has 5 paths to flow through instead of 1.

So the resistance of the total circuit is 28.8 Ohms. We can now work out the current for the complete circuit using
Current (Amps) = Voltage/Resistance
= 12V/28.8A
= 0.4166666666667 amps

Now we'll double check that the easy way like this:
For 5 neons in parallel, the total circuit current (A total):
(A total) = 5 X 0.083A
= 0.415 amps which is near enough.

so a 1/2 amp fuse would do to protect the circuit from overcurrent if something should short out. Remember, smaller is more protection not bigger but too small and you'll keep blowing fuses. You'll also have to ensure that the circuit feeding the busbar you tap into has enough capacity to run all circuits fed by that busbar. If all your accesories stop working and the fuses are ok then it will have blown a fuse or fusible link that feeds that busbar.

The figures I'm using are just arbitrary to step through the theory as I have no idea what wattage the neons are. Any questions? If you have trouble following the above just tell me what you want to do and I'll have a look at it. Funny the stuff you remember, eh...... now if only I could remember my name again......