LED Indicators triple speed

Hi Guys,

I know its been asked a million times by many people including myself, and yes I have done a few searches to see if I could get the answer on my own.

what resistor do I need to get to slow the damn indicators down?

I recall someone suggesting some from Maplins that were like 20p each

thanks for any assistance in advance

Mark

[mattSV]

I got 10watt 10ohm from Maplins - can't recall how much but they were 'pence'.

Had them on for about 18 months with no issues.

That entirely depends on the voltage that the LEDs take as an input, and the amount of current they use. Find out & post back, then I'll be able tell you exactly.

Basically, this covers all the gubbins behind it.

Resistance = (SupplyVoltage - LEDVoltage)/LEDPowerRating

If the LEDs use 1200ma (1.2A) and use for example 3v, you'd be looking like this:
R=(V1-V2)/A...R=12-3/1.2 ... R = 7.5 Ohms.

Double check your values, and if in doubt, post here or call into your local Maplin store, they should be able to work out what resistor you need, and if they can't, they need the sack IMO!

Edit: MattSV, I know you mean well, but your advice is a little flawed. All LEDs take different voltages and use different amounts of current. If the resistor doesn't drop the voltage enough, you can kiss goodbye to your shiny new LED setup. If it drops too much, you get pulled over for not having working lights.

If the setup doesn't use LEDs, you'll be needing either a new relay (more advanced type) or higher power bulbs.

When the bike is running I hope it produces more than 12volts.

I have put LEDs on the front and rear of my bike. Both kits came with their own resistors. The rear ones went on and flashed at the correct speed, put the front ones on and with or without the resistors they flash at a same (to fast) rate.

Played with a few resistor values and it made no difference.

Going to try and find someone when does and proper LED compatible flasher relay for the later model suzukis.

Originally Posted by ivantate

When the bike is running I hope it produces more than 12volts.

Coming out of the alternator yes it does. The bike's electric system (except HT leads etc) is 12volts though - hence the use of a 12v battery. The Reg./Rect. helps to convert the (IMO) massive voltage and AC current to a more bulb-friendly DC one.

Don't believe me, try sticking an LED to a HT lead & see how long it lasts.... :P

[northwind]

Originally Posted by Baph

Edit: MattSV, I know you mean well, but your advice is a little flawed. All LEDs take different voltages and use different amounts of current. If the resistor doesn't drop the voltage enough, you can kiss goodbye to your shiny new LED setup. If it drops too much, you get pulled over for not having working lights.

Not really... You're not actually trying to affect the indicators, it's the relay you're trying to trick- the LED indicators resistance is lower than a conventional lamp, and so the relay thinks there's a short. You're not aiming for a precise resistance, you just need to get it into the rather wide range where the relay is fooled.

Originally Posted by northwind

Not really... You're not actually trying to affect the indicators, it's the relay you're trying to trick- the LED indicators resistance is lower than a conventional lamp, and so the relay thinks there's a short. You're not aiming for a precise resistance, you just need to get it into the rather wide range where the relay is fooled.

Add a resistor, drop the voltage, dim the bulb. I'd rather not use a resistor with a normal bulb (non-LED) setup to be honest, but each to their own.

I know that there will be a range of resistor values where the voltage will be enough to be visible, and resistance high enough to fool the relay, but I still wouldn't like to do it.

[northwind]

Originally Posted by Baph

Add a resistor, drop the voltage, dim the bulb. I'd rather not use a resistor with a normal bulb (non-LED) setup to be honest, but each to their own.
.

Um, yep, I wouldn't either. Unless it was a very low resistance . But this topic is about LED indicators

The wiki link you sent isn't really what's being done here- that's just a regular LED circuit, which will always contain a resistor in series, you're not messing with that. Those do need to be of the right resistance to avoid problems What you're doing here, though, is adding a resistor in parallel with the LED instead- the original resistor is untouched.



I actually got this a wee bit wrong up the page- the relay doesn't pick up resistance, it picks up the current draw- what you're doing is increasing the draw by increasing the resistance. The means is the same but the purpose isn't quite how I had it, my bad.

I know that this setup is a LED based one, hence my original post (tell me voltages, I'll tell you what resistor would do the trick :P ). The original bulbs work with a power usage of around 20W-25W with the standard relay, new circuit needs same power usage in total.

I didn't know that LED kits for bikes come with a built in resistor, but thinking about it, it makes sense otherwise customers would just get p****ed off buying resistors etc. In this case it'll need to obey the old Rt=(R1+R2+R3...Rn)/n principal.

Edit: Make that the Rt=1/((1/ R1)+(1/R2)+(1/R3)...+(1/Rn)) principal

[northwind]

Originally Posted by Baph

I didn't know that LED kits for bikes come with a built in resistor, but thinking about it, it makes sense otherwise customers would just get p****ed off buying resistors etc. In this case it'll need to obey the old Rt=(R1+R2+R3...Rn)/n principal.

Every LED circuit has a resistor in series connected to the cathode. That's what's in the wiki link you sent Every LED circuit for bike use will be 12V. The formula you supplied in your original post was all for calculating the correct resistance for that in-series resistor, not for the parellel one, which doesn't have anything to do with this, since it's already there and we're not going to mess with it- chances are it'll be sealed into the indicator.

Not going to argue with the last post though, since it's exactly the equation I was trying to remember and failing dismally