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What options are out there to fixing the fast turn signal blink experienced from using LED type bulbs?

I know there are load equalizers you can buy, but need to tap into wires.
Same with JoeKD's fix of tying resistors to the wiring.

Does anyone make something that is plug and play and does not require cutting or tapping into wiring?
 

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i thought a fast blink was caused by one of the bulbs being out? maybe LED is different
 

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When you switch to the LED lights, you'll get a fast blink because the whole system runs off of resistance. When a light burns out, there is more juice flowing, thus the rest of the lights blink faster. LEDs, don't use that much power, so there is less resistance. That's why they blink faster. I think the only way to solve that problem is to install those load equilizers. Im not aware of any other solution.
 

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You've got to add load back to the system. Like Creepingdeath said, the flasher works off of current flowing in the system. The more current, the faster the flash. Thus the less resistance, the faster the flash. LEDs have only about a .7v drop accross them meaning they are little more than a short. You have to add more resistance to the circuit to bring the load back to what it used to be with the standard filament bulbs...
 

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Guys, might I offer a little electricity 101?

>When a light burns out, there is more juice flowing ...

When a light burns out, the filament is open an NO juice flows.

>Thus the less resistance, the faster the flash.

Right. Resistance is load. Lower resistance is more load.

>LEDs have only about a .7v drop across them meaning they are little more than a short.

Typical LED's can't handle 12 vollts directly, there's a 100 ohm resistor in series to limit current, or they burn out in seconds. An LED and resistor offers much higher resistance than a tungsten lamp filament.

Solution? Locate the flasher on the fuse panel and replace it with an electronic type. They use an R/C (resistor/capacitor) circuit to time the pulses and so timing is independant of load. They're available at any automotive supply for a few buks.

Wire2 (licensed electrician since '72)
 

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wire2 said:
Solution? Locate the flasher on the fuse panel and replace it with an electronic type. They use an R/C (resistor/capacitor) circuit to time the pulses and so timing is independant of load. They're available at any automotive supply for a few buks
I was going to do this and ran into two roadblocks on my 2000 Intrepid

1st - just try to find the flasher, it isn't inside the fuse panel, it's buried somewhere inside the dash

2nd - no one sells a variable load or electric flasher that is made for the intrepid, they will only sell you universal one that they cannot guarantee to work

been there, done that, still settled for using resistors in-line for my lights
 

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wire2 said:
Guys, might I offer a little electricity 101?

>When a light burns out, there is more juice flowing ...

When a light burns out, the filament is open an NO juice flows.

>Thus the less resistance, the faster the flash.

Right. Resistance is load. Lower resistance is more load.

>LEDs have only about a .7v drop across them meaning they are little more than a short.

Typical LED's can't handle 12 vollts directly, there's a 100 ohm resistor in series to limit current, or they burn out in seconds. An LED and resistor offers much higher resistance than a tungsten lamp filament.

Solution? Locate the flasher on the fuse panel and replace it with an electronic type. They use an R/C (resistor/capacitor) circuit to time the pulses and so timing is independant of load. They're available at any automotive supply for a few buks.

Wire2 (licensed electrician since '72)
Let's try some real electronics 101 here along with basic wiring. When resistance decreases load decreases because resistance is load. Current flow is NOT load, load is a resistance to current flow.

There are 12v LEDs out there, they aren't uncommon. The problem is that you must limit the current flow through them as they are little more than a short cicuit (i.e. they have little to no resistance). Even if you have a 12v LED and connect it to a 12v source it will blow unless the current is limited. An LED and resitor actually offer much less resistance than a lamp fillament. A lamp filament only glows because of the amount of head dissapated because of its resistance to flow. This why filament bulbs are so inefficient compared to other types.

Now, I do have it backwards as I always seem to do with flashers, more resistance to a flasher will cause it to flash faster, not less resistance. All the bulbs in a car are wired in parallel, not series so that one dead bulb won't stop the rest from working. When you pull a resistor out of a parallel circuit the total resistance of the circuit actually increases. Thus the flasher will start to flash faster. The load equalizing resistors are there to make the LED bulbs appear the same resistance as a fillament bulb.

dbaudiopro (BSEE '99)
 

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Ooooops!

You're right, Joe, I should have researched a bit more before I posted.
The flasher is plugged into a socket 4" left and 2" up from the brake switch mounted on the brake pedal arm.
But it's a specialized 5 wire "combo" flasher that can't readily be replaced by a generic electronic one. It has separate feeds for turn and hazard as it does both.
 

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>When resistance decreases load decreases because resistance is load.

Not so. As resistance decreases, (ohmic value goes towards 0, or dead short) load increases.

Ohm's law states; voltage = current divided by resistance. In other words, increasing resistance must =less current for the equasion to hold true.

>Current flow is NOT load, load is a resistance to current flow.

Current =voltage divided by resistance, load is voltage times current. They're all related but not the same thing.

Keep 'em coming, db, I love a good debate! Heading to work now tho. Have to keep these LX's coming together.
 

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It must be a difference in terminology between the trades, because in electrical engineering, a load is anything that causes a voltage drop. The larger the voltage drop, the bigger the load. Therefore, in our profession a load is equal to a resistance. Now, voltage times current is power. Power is what's delivered to a load.

According to dictionary.com their "electricity" definition is this:

a. The power output of a generator or power plant.
b. A device or the resistance of a device to which power is delivered.

Appearantly, the electrician's definition is the first one, and the second one is the electrical engineers definition (the proper one :p). Just depends on your profession I guess and what they teach you. So in our own way, we're both right...
 

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db, here's your misconception;

>An LED and resistor actually offer much less resistance than a lamp fillament.

Just the opposite. Put an ohm meter on a lamp. Probably 2 or 3 ohms (low). Now an LED and resistor in series. 100 ohms? (high).
I believe you're thinking in terms of conductance, the inverse of resistance.
 

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OK, there you did get me. A 55W bulb in a 14v system calculates to about 3.5ohms resistance, where as the LED is about 100ohms. It all goes back to my constant reversal of the flasher circuit. It ends up reversing everything in my head... :crazy:
 

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Hoo boy, I'm gonna be late for work, but;

>The larger the voltage drop, the bigger the load.

A load connected to a voltage supply will alway drop all the voltage (12v, 115v or whatever) It's a series circuit. Current and therefore power (in watts) are limited by resistance. If resistance is lowered to zero, current will increase to infinity.

>Therefore, in our profession a load is equal to a resistance.

Power(load)=current x voltage, while resistance = voltage divided by current, so you're calling Mr. Ohm a liar?

>Now, voltage times current is power.

Agreed.

>Power is what's delivered to a load.

Not quite. Power is drawn by and dissipated by a load.
 

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I'm not calling anyone a liar. I'm saying that in all the books I studied in college, load was equal to the resistance of the circuit. The higher the resistance the more load on the circuit. Just like in definition "b" from dictionary.com. I have one of the books right in front of me and there is a clear description of "load" in terms of resistance. It has nothing to do with power.

Right in your last statement, you contradicted yourself and said exactly what I've been saying:

Power is drawn by and dissipated by a load.
Therefore, a "load" does not equal "power." A load equals the resistance of the circuit placed on the power generation source... ;)
 

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INTREPER said:
WTF?:crazy:
What happened to my thread about LED's?
Sorry. :(
It's all related if you read between the lines though... :D
 

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Quite honestly, I think we're just running in circles around the fact that terminology is used differently in different trades. Simple as that...
 

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JoeKD said:
bottom line, as far as I know there is no company making LED's with built in load resistors.
Well, there is resistance in there, but it's there to limit the current, not to equalize the load on the light circuit... ;)
 
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