LED tail light problem and my solution

[PeterR]

Actually jeffa clearly made his post in reference to HALOGEN bulbs, which have a totally different operating characteristic than incandescent bulbs which you are talking about. ?

But halogen lamps are still a filament lamp with all the implications of cold start.     Perhaps you were thinking of a metal halide discharge lamp which is a different issue altogether.

My recollection from a looong time ago is that the cold resistance of a halogen lamp is around 5%-10% of the operating resistance.   This means if fed from a low impedance 12V source the inrush current for a typical headlamp would exceed 100A.   In a practical vehicle the resistance of the feed wiring and source resistance of the battery limits it to about half that value.   Nevertheless, the stiffer the supply, the more the lamp is punished.

When halogen lamps first appeared on the market they were a frightful price so in the interest of extending their life I designed a relay which limited the cold current then after the filament approached operating temperature applied full voltage. With the prototype in my pocket I approached one of the worlds largest manufacturers of auto elec components expecting them to greet me with checkbook in one hand and pen in the other. Instead they explained that even though this is was a very effective piece of equipment, it had no commercial value at all because by drawing attention to the adverse side effects of relays in this application it would likely harm the sales of relays and aftermarket lights.

[HOTRODSRJ]

What happens to incadescent bulbs is this.

Due to the high temperature that a tungsten filament is operated at, some of the tungsten evaporates during use. Furthermore, since no light bulb is perfect, the filament does not evaporate evenly. Some spots will suffer greater evaporation and become thinner than the rest of the filament. These thin spots cause problems. Their electrical resistance is greater than that of average parts of the filament. Since the current is equal in all parts of the filament, more heat is generated where the filament is thinner. The thin parts also have less surface area to radiate heat away with. This "double whammy" causes the thin spots to have a higher temperature. Now that the thin spots are hotter, they evaporate more quickly. It becomes apparant that as soon as a part of the filament becomes significantly thinner than the rest of it, this situation compounds itself at increasing speed until a thin part of the filament either melts or becomes weak and breaks.

What goes on when you turn on a bulb and the light goes out?

The answer here is with those thin spots in the filament. Since they have less mass than the less-evaporated parts of the filament, they heat up more quickly. Part of the problem is the fact that tungsten, like most metals, has less resistance when it is cool and more resistance when it is hot. This explains the current surge that light bulbs draw when they are first turned on. When the thin spots have reached the temperature that they would be running at, the thicker, heavier parts of the filament have not yet reached their final temperature. This means that the filament's resistance is still a bit low and excessive current is still flowing. This causes the thinner parts of the filament to get even hotter while the rest of the filament is still warming up. This means that the thin spots, which run too hot anyway, get even hotter when the thicker parts of the filament have not yet fully warmed up. This is why weak, aging bulbs can't survive being turned on due to the inherent inrush of current.

What are the difference of halogen bulbs?

A halogen bulb is an ordinary incandescent bulb, with a few modifications. The fill gas includes traces of a halogen, often but not necessarily iodine. The purpose of this halogen is to return evaporated tungsten to the filament. As tungsten evaporates from the filament, it usually condenses on the inner surface of the bulb. The halogen is chemically reactive, and combines with this tungsten deposit on the glass to produce tungsten halides, which evaporate fairly easily. When the tungsten halide reaches the filament, the intense heat of the filament causes the halide to break down, releasing tungsten back to the filament. This process, known as the halogen cycle, extends the life of the filament somewhat. Problems with uneven filament evaporation and uneven deposition of tungsten onto the filament by the halogen cycle do occur, which limits the ability of the halogen cycle to prolong the life of the bulb. However, the halogen cycle keeps the inner surface of the bulb clean. This lets halogen bulbs stay close to full brightness as they age. In order for the halogen cycle to work, the bulb surface must be very hot, generally over 250 degrees Celsius (482 degrees Fahrenheit). The halogen may not adequately vaporize or fail to adequately react with condensed tungsten if the bulb is too cool. This means that the bulb must be small and made of either quartz or a high-strength, heat-resistant grade of glass known as "hard glass".
Since the bulb is small and usually fairly strong, the bulb can be filled with gas to a higher pressure than usual. This slows down the evaporation of the filament. In addition, the small size of the bulb sometimes makes it economical to use premium fill gases such as krypton or xenon instead of the cheaper argon. The higher pressure and better fill gases can extend the life of the bulb and/or permit a higher filament temperature that results in higher efficiency. Any use of premium fill gases also results in less heat being conducted from the filament by the fill gas, meaning more energy leaves the filament by radiation, meaning a slight improvement in efficiency.

My last shot here is. . . . I don't think we should even worry about longivity of bulbs to any extent. In either case, incadescent bulbs do age and get dimmer and dimmer losing their efficiency.  That means simply changing the bulb to a new one will brighten your path.  Why wait for failure?  Just get'er done!

[PeterR]

My last shot here is. . . . I don't think we should even worry about longivity of bulbs to any extent. In either case, incadescent bulbs do age and get dimmer and dimmer losing their efficiency.  That means simply changing the bulb to a new one will brighten your path.  Why wait for failure?  Just get'er done!

You are absolutely right, ----but for an incorrigible tightwad, who still remembers blowing his first set of halogens when they were over ten times their present price, old memories linger on.

[58Apache]

While I agree, it's probably not necessary to worry about a large shock when turning on a high demand device such as headlights,.....but if a person were to want to reduce that shock just to make things better when designing a new system, ....reduce the sudden shock on the lights AND the relay contacts (de-bounce?).... and wouldn't it also cushion the change in output the alternator/regulator tries to keep up with therefore reducing the shock to those circuits as well? An overall reduced stress on three different devices all at once by adding a filter into the design.

...and with a capacitor being a voltage device, and a choke, or coil, being a current device, an L or pi filter would work nicely to reduce the overall shock to the devices involved when connecting a high current demand device.

I am not familiar with industrial chokes, but that's what that choke would be to handle the change in current and still allow a large amount of current to pass without adding too much resistance into the line and becoming an unintended fusable link.

The capacitor, since it won't pass DC, will connect between the positive line and ground, with the coil/choke being "inline" in the positive line to oppose any changes in current.

I believe this is the direction the younger generation is going in when designing in high current draw amps for car stereo systems. Big time overkill in my opinion, and someone is doing one heck of a business selling the equipment, but some of it can be used for regular car wiring jobs and will look nice as well.

                                                      Steve

[PeterR]

I am not familiar with industrial chokes, but that's what that choke would be to handle the change in current and still allow a large amount of current to pass without adding too much resistance into the line and becoming an unintended fusable link.

Steve

Yes, a series choke could be used to control the starting current, but there are a few practical restrictions.

The inductance required for this function requires a thumping great iron cored choke, and as you mention the copper used in the windings must be of sufficient cross section to keep the resistance low or the whole purpose of running a low resistance feed will be negated.

An inductor of this size will store so much energy that at switch off it will throw a spark big enough to cause serious arcing problems.

There are simpler solutions which do not require reactive components.

[HOTRODSRJ]

While I agree, it's probably not necessary to worry about a large shock when turning on a high demand device such as headlights,.....but if a person were to want to reduce that shock just to make things better when designing a new system, ....reduce the sudden shock on the lights AND the relay contacts (de-bounce?).... and wouldn't it also cushion the change in output the alternator/regulator tries to keep up with therefore reducing the shock to those circuits as well?

There are such called "soft start" devices that can affect this very thing. They vary in design. In fact, I have seen in an aircraft application bulb/filament warmers that sustain a very low voltage to warm the filaments for full application. These somewhat remove the "shock" from the start current. But, if you understand the real physics of what is happening to the filament such as I have offered above.....weakening of the filament itself due to metal evaporation, then it's really a waste because the filament is in a disrepaired condition and eventually will break anyway due to vibration if nothing else.

Newer alternators are far better at regulating the spikes than yesteryear so this too has been addressed and actually having a battery on the system is a BIG capacitor that sucks up large voltage spikes. In-so-far as inductive loads, only capacitors such as the battery can keep the voltage up but the current inrush is another issue.  IMO chokes are not practical for this application and in fact will cause other issues as pointed out by Peter.  Some motors which cause inductive loads are now coming with motor controls that bring them online slowly and this certainly aids the current inrush issues.

On a similar note, you can save relay contacts by overrating the relay for continous/momentary current and use "bucking" diodes to dissipate contact opening surges caused by motors. I use them on all my cooling fans with big current draw.

Just way toooooooooooo much information.

[PeterR]

There are such called "soft start" devices that can affect this very thing. They vary in design. In fact, I have seen in an aircraft application bulb/filament warmers that sustain a very low voltage to warm the filaments for full application. These somewhat remove the "shock" from the start current.

We have all seen a form of these without realising it.   Turn indicator flasher capsules do not switch the turn lamps off completely during the flash cycle, but place a series resistor in line so the lamp current diminishes and the lamp dims, but the filament does not chill out between flashes.

[Bruce Dorsi]

Just way toooooooooooo much information.

H-E-L-L,  NO!  ....I'm learnin' stuff here!  

On a similar note, you can save relay contacts by overrating the relay for continous/momentary current and use "bucking" diodes to dissipate contact opening surges caused by motors. I use them on all my cooling fans with big current draw.

Can you elaborate on the placement and sizing of these diodes, please?

Are the diodes inside the relay, or are they wired externally?

I have seen (non-automotive) circuit boards where diodes and capacitors were used to keep the relay contacts from arcing when the contacts were opened.  ....Is this what you are trying to prevent?

I do not know when/if a capacitor is needed in conjunction with a diode.  ...Can someone explain, please?

Thanks again, guys! ....Some of you get very technical, very quickly, but I always manage to learn something from you!

[HOTRODSRJ]

Can you elaborate on the placement and sizing of these diodes, please?

Are the diodes inside the relay, or are they wired externally?

I have seen (non-automotive) circuit boards where diodes and capacitors were used to keep the relay contacts from arcing when the contacts were opened.  ....Is this what you are trying to prevent?

I do not know when/if a capacitor is needed in conjunction with a diode.  ...Can someone explain, please?

I give it a shot!

I call them "bucking" diodes and also are known as "freewheeling" diodes as well.  Capacitors are really not needed for this application

We all know that when a DC motor is turning or "windmilling" by itself (without being connected to a source) that it becomes literally a generator.  Some of you may have seen this phenom by hooking your LED/indicator lights across the motor and to ground to find out that the LED/indicator will actually "glow" or turn on to some extent when you are running down the road with the fan OFF? My 32 roadster was like this...the faster I went the brighter the glow of the LED fan indicator.

What happens when a motor is turned off by either the switch or let's say a thermal switch for a cooling fan, when the relay contacts open the fan tries to keep the energy constant because of the stored up mechanical energy in the form of blades/fan shaft turning converting this to of course electrical generation. This energy will finally dissipate when the motor shaft completely stops.

Because the motor is an inductive storage device, the energy that is stored in the motor has no place to go since it's removed from the circuit at that moment of relay contacts opening. The voltage increases (negatively) and a spark is created across the relay contacts, greatly reducing the contact service life (some cheap relays can actually weld themselves together).  If a diode is placed (reverse biased) across the motor, it will not conduct in normal operation but will conduct when the relay contacts are opened only, thereby giving the energy somewhere to dissipate and greatly supressing the spark. The current is said to "freewheel" through the diode and the motor after the supply current from the battery is interrupted.  This voltage spike can get unbelievably high, hundreds of volts for a 12V system.

I recommend a 1N5404 which is rated at 3A, 400PIV. You may use larger diodes if you want both from a current and/or reverse voltage specification. Radio Shack also carries 5amp/400PIV diodes too.

Note the diagram below. This is for a cooling fan with two relays, one for regular manual and thermostat switching and the other for the airconditioning compressor circuit allowing the fan to run when it runs. The "freewheeling diode (1N5404) is placed reverse biased across the motor input

[Bruce Dorsi]

Thank you, Steve!

Your explanation was clear, concise, and easily understood.

The schematic helped a lot also!

I've saved this info for future reference.

~~~~Bruce