jossey
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| posted on 27/5/10 at 10:51 AM |
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Lighting Issue for TINTOP working out resistance.
For side lights issue i have 5w 12v side light bulbs.
i want to change them to LED bulbs but when i do it kicks off saying the bulb isnt working although it is.
i hear this is due to the change in resistance.
the led is 12v 20ma bulb.
what resistor will i need.
i think the resistance is 28.8
but im not sure if a 28 ohm resistor is going to sort it.
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matt_claydon
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| posted on 27/5/10 at 11:02 AM |
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P= V^2 / R
R= V^2 / P = 144/5 = 28.8 Ohms.
So you need a resistance of that or less to convince the ECU that the bulb is still there.
You will need a 5-10W rated resistor though, not your usual tiny little thing with coloured stripes. It should be wired in parallel with the LED.
[Edited on 27/5/10 by matt_claydon]
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MikeRJ
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| posted on 27/5/10 at 11:07 AM |
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R = V^2 / P
= (12*12) / 5 = 28.8 Ohms
Obviously this resistor will be dissipating 5 Watts, so it's going to get hot, so you need to choose a resistor with a suitable rating (10Watt
at least) to ensure reliability.
EDIT: Just as Mat has said, I shouldn't have gone and got a coffee whilst writing that!
[Edited on 27/5/10 by MikeRJ]
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dhutch
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| posted on 27/5/10 at 11:10 AM |
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Voltage = Current * Resistance (V=IR)
Power = Current * Voltage (P= I V)
Hence from the above P=V^2 / R (or P = I^2 * R)
Re arrange that, R = V^2 / P
Hence for a power of 5 watts, at 12v, the resistance is 144/5 which is as said, 28.8 ohms.
The LED is 20miliamp. P=IV = 0.02*12 = 0.24watt. (20th of the 5w blue and probably enough to ignore!)
So so yeah, just thinking aloud. As you say, you need a 28/30 ohm resistor capable of taking 5watts or more (ie, not a tiny little surface mount one!)
and to stick that across with the LED. Remembering that the resistor will be dissipating the same heat as the bulb used to and will hence get quite
warm even at 50% duty cycle (flashing).
You may also find that in actual fact you dont need to take the same as the original bulb, just a fair bit more than the LED. A 50ohm resistor might
do it nicely....?
Daniel
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dhutch
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| posted on 27/5/10 at 11:10 AM |
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quote:
EDIT: Just as Mat has said, I shouldn't have gone and got a coffee whilst writing that!
Yeah and me!
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adithorp
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| posted on 27/5/10 at 11:13 AM |
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If you have to put a resistor in to bring the load back up to what it was with a filament bulb... why bother?
adrian
"A witty saying proves nothing" Voltaire
http://jpsc.org.uk/forum/
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t16turbotone
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| posted on 27/5/10 at 11:18 AM |
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......or you could leave the origonal bulb in circuit (covered up somehow!) and just connect your led lamp in parrallel to it
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jossey
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| posted on 27/5/10 at 11:33 AM |
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thanks everyone.
you wise people :O)
i was thinking of leaving the bulb in line and the led but obviously it gets hot to covering it is a pain.
i want to replace the led to get it more white than wee colour.
anyway ill find a resistor to try and see how i get on.
thank you all for your help.
cheers :O)
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nitram38
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| posted on 27/5/10 at 12:03 PM |
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Long term its easier to change the flasher unit to one suitable for leds.
Then you have the benefit of using the led in the first place.
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dhutch
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| posted on 27/5/10 at 12:06 PM |
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quote:
Originally posted by nitram38
Long term its easier to change the flasher unit to one suitable for leds.
Then you have the benefit of using the led in the first place.
This isnt a problem with the flash speed on a kit car though.
- This is the onboard computer of a tintop thinking the filament bulb is blown.
- I presume that the flash is already electronically controlled and hence flashes at the correct speed.
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britishtrident
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| posted on 27/5/10 at 06:44 PM |
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Can bus a real pain the backside ------ it can even tell the difference between a 21 watt bulb and a 10 watt bulb.
Two of these is series but wired in parallel with LED is as close as you will get
ebat item 380088716724
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