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hot relays and fast signals


Nevets

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I have been having a problem lately with my headlights staying on after I turn off the bike. I have noticed that if I “flick” the Load Relief Relay (or the items in it’s immediate vicinity) with my finger, they go out. Assuming I have a relay going bad, I stopped at the dealer last night to pick up a replacement. When I pulled the relay, I noticed it was REALLY hot. eek.gif Not sure if the source of the heat was the relay, or something else in it’s vicinity, but the metal connectors were way too hot to touch. If I look really carefully, I may be seeing a very small amount of melted plastic that has oozed out. Obviously a bad thing. My question to all is: what the heck?!?!? Do the relays typically get hot from the engine or something else, or is it symptomatic of some type of electrical “issue”.

 

I have been having a weird problem with the turn signals, which may possibly be related. Occasionally the turn signals (both sides) will flash fast, as if I had a bad bulb or socket. I am very sure this is not the case, as they have been checked/cleaned/secured/soldered completely. It’s doubtful it’s the turn signal module, as this has been replaced. I do note that if the headlights or the heated grips are turned off, the signals will resume to flash at the proper rate. This leads me to believe that it may be some type of “voltage level” issue, where turning off the other items raises the voltage to the level where the signals can resume to operate properly. I was unable to see that this is the case when I attached a DVM to the accessory socket and went for a ride.

 

I am replacing the battery today (with an Odyssey PC680), so I suppose there is a chance that this could solve the issue. I have been having a problem with a rough idel, but I’m assuming that this is non-related. Any ideas (besides see if the new battery solves all of my problems…)???

 

Thanks!!!

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ShovelStrokeEd

Likely the contacts on the relay have corroded or eroded or sumpin'd. This can cause a higher than normal resistance and lead to the heat. Maybe ditto for the turn signal business. A low battery and a weak charging system could be partially to blame for this as well as lower than normal voltages will lead to higher than normal currents.

 

As to the idle thing, well, your on your own there. If the battery does improve things then I would have a quick peek at the alternator belt. It could be on its last legs.

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Ed - thanks for your ideas. Please help my feeble mind understand how lower voltages can lead to higher currents. This seems counter intuitive (I=E/R). Also, the belt was replaced (after breaking) about 3K miles ago, but I'll look at it just in case... Maybe I'll slice open the old replaced relay and take a look at it's innerds.

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Ed - thanks for your ideas. Please help my feeble mind understand how lower voltages can lead to higher currents. This seems counter intuitive (I=E/R). Also, the belt was replaced (after breaking) about 3K miles ago, but I'll look at it just in case... Maybe I'll slice open the old replaced relay and take a look at it's innerds.

 

I've seen that statement alot on the internet, but 2 years of formal all day training and plus 30 years of electrical troubleshooting and maintenance do to not support lower voltage means higher current. I=E/R is correct which does show that lower voltage equals lower current. One thing that can happen with lower voltage and relays is that the contacts will not close as tightly with lower voltage. This leads to higher resistance at the contacts and higher temperature. A layman may interpret this higher temperature as higher current but it is not.

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ShovelStrokeEd

If the belt was replaced, was it re-tightened after about 600 miles of riding? It should be cause it tends to stretch a bit and may now be slipping.

 

As to the Ohm's law thing, yes but now there is a resistance, the eroded contacts, where there should be none, or to be precise, very low. That resistance will cause a voltage drop across the contacts which will generate heat. I got the science wrong but the effect right.

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This seems counter intuitive (I=E/R
Ohms law.

 

But that's not the applicable formula in this case. The applicable one is W=I*E. Wattage equals current times voltage. To produce its designed amount of light a bulb will consume its designed amount of current. Say 55W. Regardless. Up to the capacity of what's supplying it (the wires, etc.) if the voltage goes down, the current must go up to still equal 55W.

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Stan Walker

To produce its designed amount of light

 

But there is the problem, it won't produce it's designated amount of light when the voltage drops, it get dimmer, comsumes less current at a lower voltage.

 

If this were not so, your flashlight wouldn't get dimmer as the battery was used up and the voltage goes down.

 

Stan

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There are 2 sources of heat in a relay.

 

First is the heat created by the resistance of the coil. This will not change with time (CAN not change with time!), and is generally in the range of 1-2 Watts for a power relay of this type.

 

The other source, as Ed has pointed out, is heat created by bad contacts. For a high current relay like this, a set of resistive contacts can generate a LOT of localized heat. Normally, the contacts generate no significant heat at all. Teh melted plastic you are referring to strongly indicated contact oxidation or other similar contamination.

 

You can either replace it, or disassemble it and clean the contacts with fine emery. Being a cheapskate, I generally do the latter. They are not supposed to come apart, but it isn't so hard to disassemble them. Removing the crud from the contacts will generally render the relay like new.

 

Bob.

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This seems counter intuitive (I=E/R
Ohms law.

 

But that's not the applicable formula in this case. The applicable one is W=I*E. Wattage equals current times voltage. To produce its designed amount of light a bulb will consume its designed amount of current. Say 55W. Regardless. Up to the capacity of what's supplying it (the wires, etc.) if the voltage goes down, the current must go up to still equal 55W.

 

Sorry, but that is entirely incorrect.

 

Reducing the voltage to a purely resistive load will obviously result in a reduced current as well. For example, suppose a load has a resistance of 12 Ohms. With 12 Volts supplied to it, Ohm's law states that the current will be 1 Amp.

 

Reduce the voltage to 6 Volts, and the current will be only 0.5 Amps.

 

What is different here, is that a tungsten lamp is a NON-LINEAR load. The as the voltage is reduced, and the current starts to drop as a result, the filament gets dimmer. This means that the filament is getting COOLER. Like most metals, the resistance of tungsten decreases as it gets cooler.

 

Thus, the decreased filament resistance caused by reduced current, partially (but not completely) compensates, and results in in the current increasing a bit more than if the filament was a pure (unchanging) resistance.

 

So, in short, reducing the voltage to a lamp initially reduces its current (that is simply Ohm's Law in action). The reduced filament current results in the cooling of the filament (less current = less heat). The cooler filament results in a DECREASE in filament resistance, which in turn partially offsets the reduced current caused by reducing the voltage. IN other words, whenthevoltage is reduced, a lamp will draw less current, BUT still a bit more than you would expect if the lamp was simply a resistor.

 

But it is ENTIRELY wrong to state that if the light is 55 Watts, then if the voltage goes down, the current magically increases to ensure the light always draws 55 Watts!! The light draws 55 Watts ONLY when the voltage is at the lamp's rated value. If you increase the voltage too much, the lamp will produce so much heat (heat = Watts) that it will burn out immediately. If you reduce the voltage, the lamp generates LESS heat (= less Watts), which is why it gets dimmer.

 

If what you said was true, then at Zero Volts, the lamp would have to draw an INFINITE current to produce its 55 Watts of heat! Obviously that doesn't happen!

 

Bob.

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I have/had same problem on my wife's bike, you need to pull the relay block/wire harness out of the fuse box and check for melted/corroded wire under it. I had both and since I do not want to replace the wire harness, I cut all the bad wire out, splice in new wire and female blade conectors and connected the new relay to that bypassing the relay block since it was melted and BMW does not sell that block. Don't know if it will cure the heat as I have not tried it out on a long run.

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But you miss this key part of what I was saying - "Up to the capacity of what's supplying it (the wires, etc.)" The ability of the wires to carry the needed current level is the limiting factor. If you had 000 gage wire running to say your headlight, the rule would still apply.

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Stan Walker

But you miss this key part of what I was saying

 

The key part of what you are saying would only be true if a bulb was a constant wattage device at all voltages, and that isn't the case with a light bulb. Take a 55W bulb at 12 volts, lower the voltage to 10 volts, and you now have a 38 watt bulb that is much dimmer. And yes Bob, I'm ignoring the effect that lowering the temperature has on the bulb.

 

Stan

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But you miss this key part of what I was saying

 

The key part of what you are saying would only be true if a bulb was a constant wattage device at all voltages, and that isn't the case with a light bulb. Take a 55W bulb at 12 volts, lower the voltage to 10 volts, and you now have a 38 watt bulb that is much dimmer. And yes Bob, I'm ignoring the effect that lowering the temperature has on the bulb.

 

For the purposes of illustration, you are bang on.

 

Bob.

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