@missleman
If the load is missing (blown bulb), how is that not that an open circuit?
If the circuit is open, then there should be no current flowing through the resistive wire (wrapped around the bimetallic strip) in the flasher.
Somewhere between our different understandings is reality.
@JoeMario: As I discussed earlier, the front and back bulbs are wired in PARRALLEL.
For an āopen circuitā to occur, there has to be NO path to ground. This isnāt how turn signals work: the current still makes it to ground via the working bulb. (Same with headlights, too: burning out one dies not cause the other to go out.)
I wasnāt upset earlier; I was (attemting to be) humorous. Iām not upset now, but might suggest you google āseries vs parallel circuitā because I am about through trying to explain the difference.
@JoeMarioā¦you are right. A blown bulb is an open circuit but as @meanjoe75fan has already pointed outā¦the open circuit is in a parallel circuit. That means less of a load on the blinker which will result in a faster blink. You must be thinking of a series circuit (like cheap Christmas tree lights) because if one blows none of them function.
The OP needs to clarify whether or not the left front blinker with the replaced bulb is not illuminating or if itās illuminating and not flashing. Could make a difference.
The OP is having the same problem I had with the front turn signals on my 99 Buick Century. The problem is probably a bad socket that needs to be replaced. Go to a U-Pull-R lot, find another GM of the same type (They all used the same type of sockets) and get two, one for a spare. For whatever reason, it was a recurring problem for me.
A blown bulb is an open circuit but as meanjoe75fan has already pointed outā¦the open circuit
is in a parallel circuit. That means less of a load on the blinker which will result in a faster blink.
You must be thinking of a series circuit (like cheap Christmas tree lights) because if one blows
none of them function.
@missleman, @meanjoe75fan
I am indeed thinking of a parallel circuit.
Take a look at the attached simplified parallel flasher circuit. Suppose each bulb for the āleftā side has a resistance of R. That means, if both bulbs are working properly, then the resistance of those two is (R*R)/(R+R) = R/2.
Now if one of those bulbs blows out, the resistance of the circuit increases from R/2 to R, and the current through the flasher is cut in half - which should slow the blink rate.
When you say the blink rate increases with a blown bulb, are you thinking of a different setup?
(By the way, I was originally thinking about a series circuit when this thread began. Thanks for setting me straight on that.)
@JoeMarioā¦the resistance actually decreases if there is only one bulb out of two that is working. Thatās why the blinker operates faster. Nice drawing by the way. Thermal dynamics are involved here as the spring steel snaps back quicker if there is less of a load on it. The more the load (resistance) the slower the spring steel cools which slows down the flash.
@missilemanāNo, if one bulb is out, the resistance increases. The bulbs are in parallel. Suppose each bulb has a resistance of 4 ohms. The total resistance follows the formula
1/R(total) = 1/R(1) + 1/R(2). Therefore 1/R(total) = 1/4 + 1/4 = 1/2
So if 1/R(total) = 1/2, then R(total) = 2. If both bulbs are working, the total resistance is 2 If one bulb has a broken filament, which means that the circuit through that bulb is open, then R(total) = 4.
Think about it like this. Using a Click and Clack explanation, think of the electrons as little workers who leave the their home, the power source, but need a return path back to their home. If you have two bulbs in parallel, you have provided them with two paths, twice as many workers are employed and , using the formula I = E/R, where I is the current, E is the voltage, and R is the resistance, E is 12 volts. Then I = 12/R, so as R decreases, I increases.
The way it was explained to me in an automotive service school about a 1000 years ago in its simplest form was that the flasher relied on several bulbs in the circuit to be operative (one side) and the heat generated by the current draw of those bulbs would cycle the flasher. One bulb burns out and there is not enough current draw to pop the flasher.
@JoeMario-- I have a problem with the diagram. If the hazard switch is closed, I donāt see how power would be able to flow to all the bulbs. The indicator switch would be in series and the circuit would be open. Also, there are usually two dash indicator bulbs and I donāt see how power ever gets to the KLB terminal.
@triedaq - I agree with you about the diagram and the hazard switch. I grabbed that diagram because it was simplified with a clear view of the parallel lights. Iād ignore the other oddities with it.
For me my secondary concern is why and my primary concern is what. Replace bulbs,socket, wire, etc. and be happy, happy, happy if it works, even though ignorant of why.
Flasher speeds up when bulb burns out? This is indeed a strange thing. It seems to me when one bulb of two burns out, the resistance of the bulb circuit is higher, so the current to the bulb circuit would be less. Iām guessing here, but perhaps when the flasher turns off, the time it takes for it to reconnect depends on how hot it was at the time it turned off. The hotter it was, the longer it takes to cool off. When one bulb is burned out, thereās less current, so the flasher doesnāt get as hot, so it doesnāt take as long to cool off, increasing the flasher speed.
Iām pretty sure the flasher blink speed is much more a function of the current during the heat-up (when the bimetallic strip is bending) and not during the cool down (when there is no current flow and the bimetallic strip is straightening itself out).
Iām going to have to agree with @Bing here. Regardless of the theoryā¦flashers tend to click faster when one bulb is blown. I just did an experiment on my truck. I removed the front bulb from my turn signal circuit. The flasher clicked faster with the bulb removed. I did the same for the rear bulb with the same results. If both bulbs are removed from the same side then there is no click at all.
Isnāt there always current to the heating element in the flasher, whether the light bulb circuit is connected, or not? I donāt think the only heating source is the current through the bulbs. I wonder, is there a diagram of how a flasher is constructed somewhere on the net we could look at?
@GeorgeSanJoseā¦try this link. Itās animated.
I think the problem here is that we are assuming that the flasher is a thermal unit that acts like a self-resetting circuit breaker. Indeed, this is the way the flasher units used to work. However, I think todayās flashers are a solid state device. Rather than a bi-metallic spring, I would guess that there is a quartz crystal similar to the clock device in a computer that oscillates when a voltage is applied making and breaking the circuit. I had a circuit board that would illuminate light emitting diodes at 1 flash per second, 10 flashes per second, 100 flashes per second, 1000 flashes per second and 10,000 flashes per second. I used this circuit board in a computer hardware course that I taught when I discussed the internal clock in the computer. Of course, all the students could observe was 1 flash and 10 flashes per second. I used an oscilloscope to show the higher frequencies.
Mu guess is that modern flashers are solid state that have an oscillator for a particular flash rate under a specified load. When the load decreases, maybe another oscillator is with double the frequency is used. The clicking might be a relay that switches the lights on and off. If it is the case that modern flasher units are solid state rather than a thermal unit, this might explain why the rate increases when a bulb burns out on one side of the car. Iām going to dig into this further.
Triedaq is correct, this conversation is mixing old with new.
If this is an old Impala, 1985 or older it would use the 551 lamp (two lamp) or 552 flasher (three lamp). If one bulb is out the flasher wonāt cycle, the other bulbs will be on without flash. If the OEM flasher has been replaced with a universal āHeavy Dutyā flasher the lights may flash slower when a bulb is out.
If this is a front wheel drive Impala it will have a solid state flasher relay. Open one up and you will usually find a circuit board with timer controlling the flash. They are generally designed to flash twice as fast when a bulb is out.
@Triedaqā¦I have a question. If the click of the flasher is caused by the bi-metal spring on an old style flasherā¦where does the click come from on a solid state flasher? Solid state devices normally generate no noise at all. Just something to ponder.