Flasher Relay + Resistors

Flasher Relay + Resistors (Update)

Long story short: Does a resistor built into an LED signal's wiring conflict with a flasher relay designed for LED lights?

Long story:

I just got an integrated tail light, with the turn signals built in. All LED of course. I plan to use my existing LED turn signals at the back as well, just to maintain the high visibility. So in the end I will have an integrated rear tail light/turn signals, two rear LED turn signals, and two front-facing LED turn signals.

I have a flasher relay (not yet installed, but will be) that should give everything a normal flash rate. I plan to install it with my two front and rear LED turn signals.

However, the integrated tail light with turn signals comes with a resistor installed to complete the same task that the flasher relay would be doing. Considering that the flasher relay is already managing the proper flash-rate, would the resistor, simulating the load of an incandescent bulb, cause the relay to think something is wrong, or make the flash-rate even slower?

I'm assuming the answer is no, and that everything will be fine, but I figured I would ask. I can always cut out the resistor and solder the wires again effectively removing the resistor from the equation. Especially since it causes unnecessary heat anyway.

Thanks

You are correct good sir. It does not matter. I have the integrated Motodynamic tail light with the relay and the resistor and it works fine. I'm not very good with electrical, so someone can science this up a bit for me, but I believe the led relay works off of a timing system, amount of time goes by, turns on, another amount of time, turns off, so it doesn't matter the resistor, hell it'll work with stock signals.

Thank you! That's the exact tail light I got. I've just heard about people having very dim lights, or no lights, or a slew of other problems. I wanted to make sure having a resistor + flasher relay wouldn't be the source of any of those issues.

Thank you!

Short answer: No.

Long answer: The "LED" flashers don't care about resistance at all. Old style flashers use resistance to determine how long a capacitor takes to discharge/charge. This capacitor powers the inductor (electromagnet) on a relay. During the charge cycle, blinker is off, during discharge cycle, blinker is on.

LED flashers use discrete components to create an astable latch/vibrator type set-up that doesn't require any specific resistance or voltage to stay at a certain rate.


On a side note, the power usage from the LEDs comes from the resistors. If you cut them out anyway, that will cut the LED power draw down to almost nil. I would cut it out just for that reason. But if your bike isn't overloaded and that sort of thing doesn't matter to you, it shouldn't matter. I have 2 incandescent bulbs, 2 LED flashers, and an integrated setup in my bike. I haven't got around to changing out the front flashers to LED.

DownrangeFuture said:

Short answer: No.

~snip~

On a side note, the power usage from the LEDs comes from the resistors. If you cut them out anyway, that will cut the LED power draw down to almost nil. I would cut it out just for that reason.

Click to expand...

That's another thing I was considering as well. I'll definitely end up cutting it out then. It's huge, anyway, and I'd rather have a nice 'slim' setup.

Thanks for the detailed description. That makes sense.

I have this on another post somewhere, it sounds like exactly what you're trying to do with your signals.

[ame=http://www.youtube.com/watch?v=ExhylUrbgyY&list=UU_8KU8nZuJ94oedvnXPrPBg&index=2&feature=plcp]Yamaha LED demonstration - YouTube[/ame]

No problem. Guess what I do in the military? Bet you can't.






Electronics tech? Good guess.

Correct me if I'm mistaken....

but I believe the resistor in the LED is to allow for a fairly universal use of a 12v source. The LED should say the suggested rating on it...but in your case I'm assuming the resister is dropping a specific amount of voltage - thus leaving you with the remaining desired voltage to source the LED.

If you wanted to place more LED's in series, you would need to use different resistors as you would need to maintain the specific voltage/current required to illuminate the LED - but spread across more LED's. If you were using a different source other than 12v, you would need a different resistor.

The resistor simply controls the amount of current to the LED and drops any undesired voltage to prevent an excess current of being drawn into the LED. The only reason to cut your resistor is if you were changing to a source with a different rating, or adding more LED's in series.

The flasher relay is a completely separate component.

My best answer for ya.

Well, yes. Most high quality LED signals have a voltage regulator. With only the voltage regulator and the LED's it will only draw what it needs to run the LED's which is tiny. 30 LED's might only draw 30-40mA. Compared to the 30W bulbs that's a huge change.

The resistor most people talk about in automotive LED arrangements is to simulate the resistance of an incandescent bulb so the stock flasher will work properly. LED's, being diodes, when they're on they are basically a short. So very, very little current draw and no resistance. The flasher unit, the part the OP was asking about, works as described.

So yes, you're right. You're just thinking about the wrong part of the circuit. And it's not a resistor, it's a regulator usually. Remember that resistors can change current or voltage depending on how they're placed in the circuit.

And in the Motodynamic LED light he has, the regulator is on the circuit board inside the housing. The resistor that simulates the incandescent bulb (by drawing as much current as an incandescent) is a HUGE block hanging out the back. That part can be safely cut off. Don't pull the 8 pin IC chip off the circuit board.

Resistors that are used with LEDs that are in series are used for voltage drop. This limits the current through the LED junction. you could use a Voltage Regulator but it is not always needed.

Resistors that are added in parallel on a directional circuit are used to simulate the load of an incandescent lamp so that the mechanical relay operate at the proper flash rate when you switch to LEDs. The mechanical relay does not use R/C components to regulate flash rates. The mechanical flasher uses a bi-metal element that heats, and bends as it's temperature rises. When the bi metal bends it opens the circuit and then cools. As it cools it remakes it electrical contact and the process begins again. You can see as we change the amount of current that loads through the bi-metal the flash rate can change.

An electronic flasher is different and has and IC controlled flash rate that is not dependent on load. So the electronic controlled flasher can operate a solid state or mechanical junction at a preset rate, within it's load parameters at a constant rate.

In addition, depending on whether it is an electronic or mechanical flasher unit, the resting voltage in the flasher off mode may be floating or +12 vdc. Some may vary according to design...

DownrangeFuture said:

And in the Motodynamic LED light he has, the regulator is on the circuit board inside the housing. The resistor that simulates the incandescent bulb (by drawing as much current as an incandescent) is a HUGE block hanging out the back. That part can be safely cut off. Don't pull the 8 pin IC chip off the circuit board.

Click to expand...

Lol, yep, that's what I was referring to. The huge block I wanted to get rid of.

Thanks for all the great responses everyone, I've learned way more than I intended to about this but it will serve me well in preparing my lights the way I want them. You guys are awesome!

@Soupflakes - yep, that's exactly what I will be doing.

So I'm about to do some work on the bike and figured now is a good time to sort out this tail light. I've taken apart the block that is the resistors, and now have a question again.

As per this thread suggestions, I want to remove the resistors entirely (I am using a flasher relay instead). What wires should I be cutting, and which should I be soldering back together afterwards?

I can do my own splicing, cutting, crimping, soldering, etc. However, I have very very minimal experience with this sort of thing so I'm no good with wiring diagrams and whatnot. Here's a few pictures of what I have:

Instructions (shows the wiring layout/diagram)


Included "Sub-harness". White blocks being the resistors I want to remove. Is this as simple as cutting the black wires on either side and then splicing them together without the resistors?


And the rest:










Please let me know what I need to to do effectively erase the resistors from the setup. I've dabbled in turn signals and the tail light before so I recognize where each wire goes, but I don't know if getting rid of the resistors is as simple as cutting them out and soldering the black wires back together on either side, removing the white block from the equation, so to speak.

Thanks!

Criminal Pilot said:

So I'm about to do some work on the bike and figured now is a good time to sort out this tail light. I've taken apart the block that is the resistors, and now have a question again.

As per this thread suggestions, I want to remove the resistors entirely (I am using a flasher relay instead). What wires should I be cutting, and which should I be soldering back together afterwards?

I can do my own splicing, cutting, crimping, soldering, etc. However, I have very very minimal experience with this sort of thing so I'm no good with wiring diagrams and whatnot. Here's a few pictures of what I have:

Instructions (shows the wiring layout/diagram)

Included "Sub-harness". White blocks being the resistors I want to remove. Is this as simple as cutting the black wires on either side and then splicing them together without the resistors?

And the rest:

Please let me know what I need to to do effectively erase the resistors from the setup. I've dabbled in turn signals and the tail light before so I recognize where each wire goes, but I don't know if getting rid of the resistors is as simple as cutting them out and soldering the black wires back together on either side, removing the white block from the equation, so to speak.

Thanks!

Click to expand...

From the pictures you've shown the resistors look like they are voltage drop resistors because it looks as if they are wired in series and not in parallel.

If they are series leave them in. Do not remove them!

If the resistors are connected, for example, between the yellow wire and the black wire, then it will be parallel and you can just clip the resistor lead near it's connections and reinsulate the wires.

Edit: I've looked closer and they look as if they are parallel so just remove them..

I am also using a digital flasher but I decided to be lazy and leave the big block.
If you decide to go that route you can attach the block to the other thinger as seen in the attached pic.

Motogiro said:

From the pictures you've shown the resistors look like they are voltage drop resistors because it looks as if they are wired in series and not in parallel.

If they are series leave them in. Do not remove them!

If the resistors are connected, for example, between the yellow wire and the black wire, then it will be parallel and you can just clip the resistor lead near it's connections and reinsulate the wires.

Edit: I've looked closer and they look as if they are parallel so just remove them..

Click to expand...

Yay! :thumbup: I hope I can get rid of them. I don't need the heat or the extra draw.

Heres another picture just so you (and therefore, I) can be certain:











If I were to snip off the resistor and re-insulate the connection, doesn't that leave the black wire connected to nothing? All the other signals I've been working on have had one coloured (green or brown) and one black.

Thanks for the help. New front turn signals are installed. Relay is installed (and works a charm!) and her rear end is now exposed waiting for the tail-light transplant. Can't wait to get it all sorted.

@eh89 - thanks for the tip. If I end up deciding to keep it, I will put it exactly there, looks like a comfortable fit.

Criminal Pilot said:

Yay! :thumbup: I hope I can get rid of them. I don't need the heat or the extra draw.

Heres another picture just so you (and therefore, I) can be certain:

If I were to snip off the resistor and re-insulate the connection, doesn't that leave the black wire connected to nothing? All the other signals I've been working on have had one coloured (green or brown) and one black.

Click to expand...

Yup. Snip em!

Motogiro said:

Yup. Snip em!

Click to expand...

Huzzah! Thank you sir. Pictures will be here when installed. Thank you very much for your prompt responses and attention! :thumbup:

Criminal Pilot said:

Huzzah! Thank you sir. Pictures will be here when installed. Thank you very much for your prompt responses and attention! :thumbup:

Click to expand...

And in case anyone needs the info, I believe the the resistors are 7.5 ohm 5% tolerance at a 25 watt rating.

Like if you had to come up with a matching load for you OEM flasher...

Criminal Pilot said:

@eh89 - thanks for the tip. If I end up deciding to keep it, I will put it exactly there, looks like a comfortable fit.

Click to expand...

Thank you, I considered getting rid of the harness but took the easy route and just hid the wiring instead. Good to know that it can easily be modified if I feel like getting rid of it in the future.... so thanks for the helpful thread :thumbup:

I got the smoke motodynamic tail light and only installed it a few days ago, yet I already greatly prefer it over my previous integrated tail light.