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My turn signals have 23 watt bulbs in the front and approx. 10 watt LEDs in the rear. I need a resistor to slow my blinker speed down. What size resistor do I need? And does it run in sequence to both front and rear after the flasher/controller? I do not have an uglies book with me, and Im not too sure I would get it right anyhow. Could someone give me a hand with this?
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Ok...I got my book and have come up with a 23 ohm resistor. Is this correct? Anybody...anybody? -
If the rears are LED's, the resistor should be inline with the rears. I don't know what size resistor you need, but since resistors are cheap, grab five, ten, fifteen, twenty, twenty five and thirty ohm resistors and try them inline until you get the results you want. Return the others...
Cheers
=-= The CyberPoetComment
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Hmmm I don't mind asking stupid questions.....
Do you need one on each wire? (2/side)...Comment
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You should be able to just put it in the feed (non-ground) wire ahead of the LED and have it work, if you have the right resistor. One per LED blinker.Originally posted by Dennis Wilson
Cheers
=-= The CyberPoetComment
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If the fronts dont see equal resistance...wont they blink faster? They will all need equal resistance..right?Comment
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And yes Cyber...only one lead...I think. OMG Im so confused!!!!Comment
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Let's try to deconfuse this some (remembering that I am not an electronics board jockey, but have a reasonable understanding of electricity)...
First, you need to understand a capacitor is effectively a form of battery -- it stores power by loading up flat plates of metal with electrons, to be used on demand. It can store a certain amount and that's it (kind of like peas on a flat plate -- you can put so many on before you can't add any more peas).
The flasher relay basically trips back and forth (on/off) based on the load state of a capacitor (switch off on empty; switch on when the capacitor is full), if I understand correctly (at least old-school flasher models from 30 years ago did; these days it could be done electronically instead of electrically -- or with smoke & mirrors for all I care, since the operating principle is the same for our purposes). The speed of the drain on the capacitor is normally dictated by the resistance of the filaments in the incandenscent blinkers, which will only let so many electrons slip though at a time (which controls the drain-rate on the flash relay capacitor). Basically, the bulbs' filaments act like a funnel to control the rate the peas can fall off the plate, so they all fall off in exactly the time we want the bulb to actually flash on. So far, so good?
The problem comes up when the LED comes into play, because the LED's have a much lower power draw and a much lower resistance level unless there is a resistor pack (think of it as an artificial pea funnel) built into the LED bulb (some LED bulbs have this integrated, usually in the bulb base if it's there).
Since the front blinker is still using a traditional incandescent bulb, it doesn't need to have it's portion of the circuit modified. That part remains the same...
The only part that needs to have any change is the LED blinker portion (i.e. - the loop that includes rear blinkers in your case). So we need to take the line going to the rear blinker and increase it's total resistance to offset the difference between the regular incandescent bulb that's normally there and the LED that's there now. You increase this by placing the resistor inline in the powerline (+) to the LED Blinker bulb. I suppose you could just as easily wire the resistor inline between the ground lead on the blinkers and the whereever the wire grounds, but it's traditional (for reasons unknown to me) to put it in the upstream (+) side of the current flow.
In theory, you should also be able to figure out the resistance difference between a traditional bulb and the LED-equivilent, and come up with the resistor rating needed, but I'm not sure if these values are the same under the traditional 13.2 volt load as they are when measured on a table without a load.
SPECIAL NOTES:
The OEM front blinkers are wired in reverse on Katana's (and many other bikes). If you replace them with aftermarket incandescents and get a fast-blink or leak-over (slower blink than usual, or just a constant glow), this is the reason. Reverse the polarity of the wires at the front blinkers.
Hope that helps.
Cheers
=-= The CyberPoetComment
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cyber - a standard bulb has very little resistance.
a standard LED is going to draw only around 20mA.
if you look at the resister kits - the resister used to get
the led to flash & at the correct speeds - is a high watt resister because
they are basiclly shunting current to ground. ( they will get HOT )
the resister needed would go in PARALLEL to the LED.
tim
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Originally posted by The CyberPoet
WRONG - it draws less current because it has more resistance.
timComment
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Or it has higher resistance because it draws less current. I understand... Duh.Originally posted by trinc
Thanks for straightening me out on thatOriginally posted by The CyberPoet
Cheers
=-= The CyberPoetComment
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I was told by one of my riding buddies that works for a dealership just go get a car flasher and put in it and it will work fine. Is this true or not? By the way I went ahead and bought a new set of blinkers that are not LED. I understand ohms and resistance, but did not want the head ache on trying to make sure I put the correct one in. I am afraid if I do not do it right then there goes my electrical system all together.Comment
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cyber - that was a bad way for me to say it.
an LED has a current rating - it can only handle so much current
( to a small degree you can regulate the brightness by how much current
you put through it - but there is a minimum for it just to turn on )
anyway......
the amount of current through the LED will be control not by the resistance
of the LED itseft but a current limiting resistor. (which is most likely built into the led bulb assembly ) the resistance of the diode + the current limiting resistor will by far be larger than the couple of ohms a regular
bulb.
to calculate the resistor ( current limiting ) you would take the power supply (12V) - 1.2v ( voltage drop across the LED ) = 10.8v/ 20mA
= 540 ohms resistor .
(with the effective resistance of the LED in the ON state of 60 ohms
(1.2v /20mA)
this is for a single LED - i don't know if the multi led's are made in series
or parallel or a combination of the two- i'll have think which would make more sence once this beer wears off
timComment
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