corradoboy

Calling all clever sparky's,

I've just replaced the silver fusion bulbs in my 03 WRX detango'd headlights with LED's, as I was unhappy with them being almost un-noticeable on bright sunny days. The LED's are a bit better, but the tiny amount of power they use seems to be making the relay behave as if a bulb has blown and flashes really quickly. Can I cure this by adding a resistor into the circuit And if so, what size/type would be best ? It's a long time since I did O'Level physics, and I can't remember the equation for ohms, volts, amps and watts Obviously it's a 12V circuit but I don't know the ampage (I seem to recall car batteries may be 80 amps ???), so what resistance would 2x 5 watt bulbs give ?

Any help greatly appreciated

ru'

V=IR

P=IV

So, P=(V*V)/R

For one bulb, 5=(12*12)/R

Therefore R=144/5

R=28.8 Ohm (with at least 5 Watt powert handling)

If you want on resistor to replace both of these bulbs (i.e. a 10 Watt load) you would need one 14.4 Ohm resistor (capable of handling at least 10 watts).

corradoboy

ru'

Thanks for the reply. So, one 14.4ohm/10w resistor for each side should do it ? Would I wire that in line (serial ?) to the positive side, or across + to - (parallel ?) somewhere in the circuit ? Or would 2x 28.8ohm/5w wired across (parallel) each LED + to - or (serial) before/after each LED be better ?

ru'

Each replaced bulb would need a resistor across it (in parallel), I'd get higher power resistors than 5W; they're cheap anyway.


Get the nearest value, but always go to a higher ohm value rather than a lower one.

I'm assuming you've replaced two bulbs on each side? (front and back?) - you could just go for one resistor per side (14.4 Ohm or above, 10W or higher) in that case, but then the wiring would have to take more current where the rsistor was. This probably won't be a problem, however if you have one resistor per bulb it mimicks the old arrangement better.

corradoboy

So it's best to just add a resistor in place of each missing bulb, wired parallel across + to -. The bit I'm not getting is when you say if I use one resistor it needs to be 14.4ohm/10w, but if I use 2 per side (for each bulb) then 28.8ohm/5w. Surely this is building up too much resistance as I'll end up adding 57.6ohms to the circuit ? Or am I just dumb

I think I'll add one for each bulb, so can you definately tell me which resistor will best suit ?

Thanks again.

ru'

If you have two resistors (same value) in parallel, the total resistance is halved (basically an equal amount of current flows through each, so twice the current compared to one resistor). 10 ohm in parallel with 10 ohm gives 5 ohm.


If you have two resistors (same value) in series then the total resistance is doubled. 10 ohm is series with 10 ohm gives 20 ohm.

So, if you're putting on resistor across each bulb location then you want 28.8 ohm 5W ones.

Maplin seem to have 10 Watt wire-wound ones, but the sizes are 47 or 68 ohms. If you can, have two 68 ohms connected in parallel (to give 34 ohms) and then connect this in parallel with each 'bulb'.

http://www.maplin.co.uk/search.aspx?...23m7&worldid=3

Let us know how it goes.

Ru'

corradoboy

If I'm doing each bulb at 5w, then could I not use maybe the 7W/22R (L22R)

I suppose they're so cheap I could buy several of each to try.

ru'

22R (ohm) would give about 6.5 watts, which is higher than the original bulbs. As you say, maybe try a few values but I'd suggest always taking the next highest value compared to that calculated rather than one below.

corradoboy

So what would the 7W47R work out at Just trying to find the easiest solution, but if 2x 10W68R's is best then it will have to be. I don't think 6.5W will blow the car up though, as I've seen people put 21W bulbs in indicators without problems

I really appreciate your help on this, those equations you posted may as well be in Welsh

ru'

47R would be around 3 watts. I have no idea what is the minimum load needed to make them flash correctly; 47 ohm may be enough for a decent rate (could be slightly faster than normal, but should be loads slower than without the resistors).

corradoboy

OOOPS !!!

Just been thinking (dangerous I know) and I suddenly remembered that the indicator bulbs are in fact 21W

Would you be so kind as to run this new information through your formulae to ascertain the most suitable resistor.

Sorry !

ru'

You'd need a 7 ohm-ish one for that.


And it'd have to take 21 watts.

Closest thing would be three of the maplin 7-watt 22 ohm ones in parallel (that'd be about 7.3 ohms) - they'd share the load so if each could handle 7 watts, three of them will take about 21 watts.

If you can be bothered, try one of them on its own. If the flash rate is too fast then add another in parallel. Only if this is too fast add the third.

corradoboy

Right, that's what I'll do on monday. And thanks again for your help

corradoboy

Popped into Maplins today, and described my intentions before making a purchase. The knowlegable guy was all snooty and sneery about it and in any case, they only had 2 of the required resistors in stock. They did however suggest I speak to the guys in the motorbike shop next door, as they are often "@rsing around" with LED lights. Low and behold, they knew exactly what I was doing and stocked a ready made 21W resistor just for the job Front ones now done, just gotta wait for the replacements in the correct colour to do the backs.

Thanks again for all your help ru'

ru'

Cool! Does the 21W resistor have anything written on it? Be good to see that my calcs were up to scratch.


How much did they sting you for it?