Hi, so here's my problem, I recently installed new indicators on my 1980 XS400G, everything is hooked up properly and works, except for one huge problem, the indicators wont flash. They just stay solid when ever I switch on either the left or right side. I have done some research and I think I have narrowed it down to relay not flipping because of the difference in wattage being pulled from the new bulbs in relation to the old ones. The bulbs that came in the new indicators are, 12v 23/8w and the original bulbs are 12v, 27w. What would be the most simple way to make this work? would I need to find a new flasher relay in order for this setup to work? if so does anyone know where I could find one? I dont exactly understand light bulb or electrical information all that well so if someone could help me out id appreciate it, thanks!
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New turn signals wont flash, just stay solid.
- Thread starter Ethan
- Start date
Ben_Wagner
XS400 Member
Assuming that the 27 watts is important to the function of the relay, I believe that the installation of a resistor in parallel with the blinkers would work. Someone else may be able to tell you whether or not the power drawn by the LED's is a factor, but I think the following would increase the wattage drawn through the relay, hopefully making it work.
Since the new lights are 12v and have two different "filaments", one that's 8w and the other that is 23w you will need a different resistor depending on what side you are using. Note that Watts = Volts * Amps
Use this formula to determine the resistance required:
voltage^2 / (watts requried - bulb watts) = resistor
So if you used the 23w side, you would need (12v^2) / (27w-23w) ~ 36 ohms of resistance.
If you use the 8w side, 144/(27-8)~8 ohms of resistance.
This formula works because loads in parallel (the LED and the resistor) will have the same voltage but will vary in current. As a result, the bulb will get it's requisite voltage and current, while the resistor will pull additional current and therefore increase the watts "seen" by the relay.
(1) V = I*R
(2) W = V*I
(W is watts V is volts, I is current (amps) and R is resistance)
If we use equation (2) and solve for I, we get W/V = I. Next we can plug I = W/V into equation (1). This gives us V = W/V * R. Solving for R we get V^2/W = R. We want the relay to "see" 27 watts, we need the relay to pull the rest of the watts that the LED wont be pulling so that's why we do (watts requried- bulb watts) and due to the LED and the resistor being in parallel, they both have 12 volts.
I'm not sure how much you know about circuits but when something is installed in parallel, it will be connected across the LED. This means that one side of the resistor would be connected to the ground of the LED and the other would be connected to the "hot" wire coming into the LED. In series, the resistor would be put in, replacing a section of the positive lead coming into the LED.
Sorry for the long post, I hope it helps.
TL;DR
Installing a resistor with resistance R in parallel with the LED should increase the power drawn from the relay.
Where:R = 144 / (27-LED wattage)
Since the new lights are 12v and have two different "filaments", one that's 8w and the other that is 23w you will need a different resistor depending on what side you are using. Note that Watts = Volts * Amps
Use this formula to determine the resistance required:
voltage^2 / (watts requried - bulb watts) = resistor
So if you used the 23w side, you would need (12v^2) / (27w-23w) ~ 36 ohms of resistance.
If you use the 8w side, 144/(27-8)~8 ohms of resistance.
This formula works because loads in parallel (the LED and the resistor) will have the same voltage but will vary in current. As a result, the bulb will get it's requisite voltage and current, while the resistor will pull additional current and therefore increase the watts "seen" by the relay.
(1) V = I*R
(2) W = V*I
(W is watts V is volts, I is current (amps) and R is resistance)
If we use equation (2) and solve for I, we get W/V = I. Next we can plug I = W/V into equation (1). This gives us V = W/V * R. Solving for R we get V^2/W = R. We want the relay to "see" 27 watts, we need the relay to pull the rest of the watts that the LED wont be pulling so that's why we do (watts requried- bulb watts) and due to the LED and the resistor being in parallel, they both have 12 volts.
I'm not sure how much you know about circuits but when something is installed in parallel, it will be connected across the LED. This means that one side of the resistor would be connected to the ground of the LED and the other would be connected to the "hot" wire coming into the LED. In series, the resistor would be put in, replacing a section of the positive lead coming into the LED.
Sorry for the long post, I hope it helps.
TL;DR
Installing a resistor with resistance R in parallel with the LED should increase the power drawn from the relay.
Where:R = 144 / (27-LED wattage)
MoonClipper
XS400 Enthusiast
Wanted to chime in with a question. Wouldn't it be more beneficial to instead swap out the relay for one that's designed for LEDs? It would be more straight forward, and would retain the energy cost of having LEDs in the first place. Wouldn't a resistor increase the energy cost back to what a standard bulb draws, negating the benefits it LEDs?
I could be wrong of course, I'm not super savvy when it comes to electricals, at least not to that degree.
Edit: I answered my own question, so I figured I would edit my post with what I found. Apparently with the way LEDs work, they actually need resistors. The resistors and the LEDs together would still consume significantly less power than a traditional bulb.
I could be wrong of course, I'm not super savvy when it comes to electricals, at least not to that degree.
Edit: I answered my own question, so I figured I would edit my post with what I found. Apparently with the way LEDs work, they actually need resistors. The resistors and the LEDs together would still consume significantly less power than a traditional bulb.
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Idk if this changes any of the information you gave me, but I am not running any LED's Im using all incandescent bulbs, the newer bulbs are just a lower wattage. Thank you so much for the info, means a lot knowing people are so dedicated to these forums! haha
WelderDave
XS400 Addict
The stock is mechanical, it has to draw enough current through the relay to lift the strip of metal from it heat warping basically. You need an led relay.
edit: sorry, missed the part you weren’t using leds. The function of the stock relay is the same though. You need a certain current to heat up the metal strip enough to lift. Then it cools and makes a connection again, repeat. If it’s staying solid, there’s either not enough current drawn or the relay is bad
edit: sorry, missed the part you weren’t using leds. The function of the stock relay is the same though. You need a certain current to heat up the metal strip enough to lift. Then it cools and makes a connection again, repeat. If it’s staying solid, there’s either not enough current drawn or the relay is bad
Do they flash with the bike running? Make sure you have 12.7 volts with the bike off and 14.5 with it running. Switching to a lower watt relay will help with the smaller watt bulbs. Auto parts store with have 2 prong ones has will work.
thanks for the help everyone, after ordering various led relays, I found that not a single one of them worked. I found my solution to my problem was to find a replacement bi-metallic relay that matched the wattage of the bulbs I was using. I swapped in a 23 Watt flasher relay and it works absolutely great.
Well done good you got it fixed instead of messing with resistors. resistors heat up to very warm temps if blinkers are left on too long.do you have a picture of the relay you bought. thanks
