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Help, I don't understand the turn signal circuit

8K views 16 replies 5 participants last post by  Mydlyfkryzis 
#1 ·
I have a 72 CL450 with a single turn signal indicator light in the tachometer. I am rewireing my bike and can't get the turn signal circuit working properly.

So flashing power is supplied to the turn signal switch which then switches it to either the left or right turn signal circuits. However the two sides are connected by the common indicator light. When I have the bulb in both sides flash when I switch it to either side. When I take the indicator bulb out it works fine. Studying the wiring diagram, it looks like if the left side is switched to, the power goes to the indicator light and then goes to ground through the right side circuit. I just don't understand how it is supposed to work. What gets me is that I know it worked before I rewired it.

Anybody know? Thanks.
 
#2 ·
Are you installing LED bulbs? Or another low-power type, other than the 1156? When one side is selected, the cold bulbs on the other side provide the ground for the indicator. If they can light up on the small amount of current the indicator passes, you need to increase the load. LEDs require load resistors on both sides, for this reason, and so they don't flash really fast (bulb out indication).
 
#3 ·
Yes, they are LEDs. The indicator bulb is not, or else it wouldn't work. My knowledge of electrics is a bit fuzzy, been a few years since my college electrical engineering courses, but that doesn't seem to make sense. I'm probably wrong but it seems like all of the bulbs in the series, not just the first one, would light up.

I did get an electronic flasher so they would at least flash.
 
#4 ·
The D in LED stands for DIODE, which only allows current to pass ONE direction. Because of the way the indicator is wired, getting its ground from the light that is off, it will work ONE way, but not another. Because one way the current flows correctly, forward biasing that LED and making it glow. The other way it hits the brick wall that is what diodes do best, filter AC voltage to DC. You can rework the ground for that bulb for that LED to function or just run the standard bulb in it.
 
#5 ·
Other thing that comes to mind is that even a small incandescent bulb will draw a lot more power that an LED.
I dont know how that would affect this circuit.
Could you run 2 small LED's into the same position as the repeater in the instrument and connect them into each flasher circuit. That way one would illuminate whichever side was used.
 
#6 ·
Franz, the indicator bulb is not an LED. As you pointed out, the circuit would not work if it were. I fully understand why my current setup does not work. What I don't understand is how the original setup with the incandescent bulbs ever worked.

I spoke with one of the sparkys here at work and the best we could figure out is this. When you switch to left, the power goes through each left bulb in parallel (12v across each left bulb) but the power goes across the indicator and right bulbs in series (12v across the indicator and a right bulb). So the indicator is small enough to light up while the big turn bulb would not.

What I am going to do is run both left and right circuits to the same side of the indicator bulb, but put diodes in the path so the power doesn't feed back through the other side. Then run the other wire from the indicator to ground. I'll try to draw up a diagram.
 
#9 ·
I drew out the stock turn signal circuit (after the flasher) in a bit simpler manner. I also drew out my proposed circuit. I know my proposed one will work, but again, I just don't understand how the stock one works. Why do all the turn signals not come on at the same time.

See attached.
 

Attachments

#11 ·
Ok for what its worth, and as said before, I'm not an electrical engineer, however I think I understand it.
The power is drawn through the flasher causing the B/M strip to cycle.
This is then sent off to the appropriate pair of indicator bulbs (42 watts app). The repeater bulb also recieves this power too but only draws 3.4 watts, which is too small amount of current to illuminate the other pair of 21 watt bulbs, and therefor returns to the green wire and the negative side via the opposite bulb circuit.
[attachment=0:1jhxno5r]img270.jpg[/attachment:1jhxno5r]
 

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#12 ·
Thanks Nigel. That is pretty much what one of the sparkys said here. And since I now have LED turn bulbs, that little bit of power is enough to light them.

I will throw some diodes in and hope for the best. Thanks guys. Hopefully this helps someone in the future.
 
#13 ·
Flugtechnik said:
I drew out the stock turn signal circuit (after the flasher) in a bit simpler manner. I also drew out my proposed circuit. I know my proposed one will work, but again, I just don't understand how the stock one works. Why do all the turn signals not come on at the same time.

See attached.
With that circuit, you can replace the indicator with a LED, and get a brighter one, at that.
The stock circuit works because the stock turn lamps take about 24W each to light up, while the indicator is 7W or less. Current flows through the lamps on the off side, but so little that it doesn't even warm the filaments. If one bulb is out, and the lens is off the working bulb, you may just discern the filament glow a bit in a darkened room. You would never see that under normal conditions.
 
#15 ·
I know this is an old thread, but I have something to add to it. My NH750 is wired through the indicator the same way. What I did is get 2 small LED's (used resistors to drop the 12V) that replace the 2 diodes shown in the schematic. The 2 LED's were then inserted behing the yellow indicator lens. So for left or right, one of the LED's glows, but dehind the lens, it really isn't evident that there are 2 LED's back there. So I used 2 leds, instead of 2 diodes and an LED.

But otherwise, the circuit works great.
 
#17 ·
I am using a solid state flasher relay...KISAN makes it. Gives me 4 way flashers, wigwags, and auto cancellation.

That silly interconnected indicator light drove me nuts for a while.
 
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