ragindave
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| posted on 8/5/08 at 07:14 PM |
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LED indicators need slowing down!
I have LED Indicators that flash to fast is the simple and cheap solution to put a resistor in line?
And if so what resistor would be suitable.
Thanks Dave.
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BenB
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| posted on 8/5/08 at 07:18 PM |
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Simple answer- Yes. A big resistor which would use up lots of power, make the alternator work harder and rob you of BHP 
Better option is to get an electronic flasher relay off Ebay that will cope with the lower current drawn by LED indicators.
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Paul TigerB6
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| posted on 8/5/08 at 07:22 PM |
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As Ben says, change the relay to one that operates on a time base rather than loading base (which is why they are flashing so fast with the
LED's in there)
Something like
this on ebay
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iank
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| posted on 8/5/08 at 07:56 PM |
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No the resistor is put in parallel not in series (in-line). Don't bother and get a flasher unit.
A good idea is to use search for this kind of thing - the last thread and answers is only 5 or 6 down the list.
--
Never argue with an idiot. They drag you down to their level, then beat you with experience.
Anonymous
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smart51
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| posted on 8/5/08 at 08:50 PM |
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buy the relay and do it properly. You need something like a 20W resistor to put the load up enough for a conventional relay to work. That's
more than my soldering iron.
I bought a relay from VWP. It replaces the standard ford item pin for pin and just slots in. What could be easier?
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MkIndy7
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| posted on 8/5/08 at 09:51 PM |
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Or if your really tight or need a soloution quickly... hide another couple of bulbs somewhere ( 1 on each side)
Under the dash, engine bay or back of the car... just make sure there tied up well and have plenty of air around them, maybe even keep them in an old
side repeater casing.
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ragindave
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| posted on 8/5/08 at 10:13 PM |
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Thanks for the advice I will get the relay I have got to look after my BHP as I dont have a lot....
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dhutch
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| posted on 9/5/08 at 08:58 AM |
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A standard bulb is a round 21watt, led around 1. Hence you needing a 20watt resistor.
V=IR, P=IV. V=12 (voltage), P=20 (target power), I=8 (required current), making R=1.5, if my calcs are right.
- So you would need 1.5 ohm resistor, capable of taking 20watts. Or alteast 10 or so if taking into account the duty cycal, as the indicator will be
flashing on/off and not be running long term anyway.
However as said, but the time you've bought them, you might as well get a suitable flasher unit instead in my book. Ditto sticking two filalment
bulbs in. Although the latter would be a great shortterm/sva fix.
The argument about saved power however doesnt really hold water. 20watts at 50% duty cycal averages to around 0.01horse or just under an amp.
- So even if you have all the flashers on continuously, you talking about under a tenth of a hp. And if you do, your proberbly parked anyway!!
Daniel
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smart51
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| posted on 10/5/08 at 10:07 AM |
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you save 20W per bulb and they're on 50% of the time, but you have one front and one rear. 20W. Of course, that will be 40W if you have your
hazards on.
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