DC regulators can't drive car headlight DC bulb... why?

[picbits]

Maybe the original poster should measure the voltage going into the regulator when he connects the bulb and see how much it droops by ?

A poorly heatsinked / high input to output voltage will also cause issues as the regulator will thermally limit pretty promptly.

 

[MrAl]

Hello again,


Here is the simulation of a bulb being driven by a regulator ic chip similar to the LM317 but with higher output current ability.

For the simulation of the regulator without any lower capacitance the regulator current limit was shut off so that we could see the full peak during startup.
As shown, 10uf does almost nothing, while 100uf starts to have an effect, and lastly 500uf seems to do the trick.

The only thing that puzzles me is why didnt the natural built in current limit of the LM317 (or similar) kick in and limit the current in the first place? I would have thought that the current would have naturally ramped up.
One answer is that the bulb heating curve does not allow the bulb to heat up enough with the current limited current level, and so the resistance can never increase even though the LM317 (or similar) does not have fold back current limiting. One thing about those car headlamps is that they have a pretty big filament (high mass more or less). But that would imply that the regulator could never put out enough current to heat the bulb, ever, even with the lower capacitor connected as my drawing. In this case the only thing that would work is to have that extra energy of a storage cap so that during turn on it can get it's extra start up energy and get hot.

I think this demands some careful tests however. I would like to see a curve of the bulb current vs voltage for whoever wants to perform this test. I'd do it myself but i dont have a bulb that big around without taking one out of the car, which i dont want to do

The actual bulb models that work pretty well include feedback from the filament temperature and are driven by the energy input and the losses by filament conduction and radiation. But i think a voltage vs current test would be good enough for our purposes.

All the bulbs i have tested in the past (not that many through really) have a more smooth curve where the filament starts to heat up quite rapidly. So in other words the current is monotonically increasing with voltage and this never changes. But with this bulb it could be different, so a test is really in order.

So an actual test of the bulb voltage vs current would tell us exactly what is going on here and then we would know exactly what to do. To perform thist test might require a variable power supply that goes from say 0 to 15v at say 10 to 15 amps. As the voltage is raised, the current is logged. That will tell us the full story (well, almost).

 

[audioguru]

The regulator limits the startup current to 1.5A minimum. Then if the filament is 1.2 ohms the voltage to it is only 1.8V so the filament does not get hot enough for its resistance to increase enough for the regulator to apply full power to it.

 

[Qaisar Azeemi]

What voltage do you have to start with? We know you want 12 volts out, but we don't know what voltage the input to the regulator is.

i am using 2 lead acid battries in series. so a max of 27v down to a minimum of 20v may be there at the input of the regulator or transistor.

 

[picbits]

Ok - so you're dropping 27 volts to 12v and powering a 21 watt lamp.

Thats 1.75 amps and a 14 volt drop so you'll be dissipating around 24 watts of energy - unless you have a large heatsink on your regulator, it's going to get very hot very quickly.

 

[audioguru]

i am using 2 lead acid battries in series. so a max of 27v down to a minimum of 20v may be there at the input of the regulator or transistor.

The datasheets for the LM317, LM338 and LM350 regulators show that they reduce their maximum output current when the voltage from input to output exceeds a fairly low voltage. For the LM317 if the input is 27V and the output is shorted with a cold filament the current is limited to typically 1A but it might be as low as 375mA which will not light the bulb.

 

[MrAl]

He said he tried an LM338 right.

Qaisar:

I used two 7812's in parallel and also used lm338, 5A regulator. since load is very high 1.2ohm that cause an initial current of 10 to 15 amps and shuts the regulators off.

Why dont you do a test on the bulb. Run it at 1v, write down the current it draws. Run it at 2v, write down the current it draws. Run it at 3v, write down the current it draws. Do this up to 12v, writing down the current for each voltage level.
This way we will know at a glance what is going on. We can only guess unless we have the bulb data.

Also, what kind of heat sink are you using on the LM338?

 

[Qaisar Azeemi]

Ok - so you're dropping 27 volts to 12v and powering a 21 watt lamp.

Thats 1.75 amps and a 14 volt drop so you'll be dissipating around 24 watts of energy - unless you have a large heatsink on your regulator, it's going to get very hot very quickly.

The datasheets for the LM317, LM338 and LM350 regulators show that they reduce their maximum output current when the voltage from input to output exceeds a fairly low voltage. For the LM317 if the input is 27V and the output is shorted with a cold filament the current is limited to typically 1A but it might be as low as 375mA which will not light the bulb.

Thanks to all valuable meembers who replied. i think the small cold resistance of a bulb requires a very high initial current of I=12v/1.2ohms = 10A which is toooooo much current to shut the auto protector inside the lm317 or lm338 down to off. as i tested when i connected the bulb to regulators they shut off. i test the same thing using 2SD1047 power transistor thats of 100w using the following configuration :
View attachment 68240

It lit the bulb with slight varying voltages about 8.5V to 8.6V across the capacitor (the bulb is connected across the capacitor), and the transistor gets extremely heated up.

now i order to complete the voltages i perform another experiment and connect and inductor of 470uH with the output as shown below :

View attachment 68241

it gave a fantastic result.... guess what?????






Just blown out the Transistor after a mint ... the output was varying voltages about 8.2V to 8.65V across the capacitor and then voltages started increase more and more and transistor started burning smell and smoke and then........

so now thinking of Buck regulator.......

another question rises here is why a 100w tansisor can't wighstand with only 21w bulb???? the power dessipated in the transistor is P= (27V-8.6V)*1.5A = 27.6W in the transistor which is too less than 100W..... then why it blown out??????? i used a normal small finned heatsink...

View attachment 68242

 

[Qaisar Azeemi]

He said he tried an LM338 right.

Qaisar:



Why dont you do a test on the bulb. Run it at 1v, write down the current it draws. Run it at 2v, write down the current it draws. Run it at 3v, write down the current it draws. Do this up to 12v, writing down the current for each voltage level.
This way we will know at a glance what is going on. We can only guess unless we have the bulb data.

Also, what kind of heat sink are you using on the LM338?

its a good idea MrAl; i must test is tomarow. i wrote about My experiment with LM338T in post # 28.

 

[picbits]

another question rises here is why a 100w tansisor can't wighstand with only 21w bulb???? the power dessipated in the transistor is P= (27V-8.6V)*1.5A = 27.6W in the transistor which is too less than 100W..... then why it blown out??????? i used a normal small finned heatsink...

View attachment 68242

What is the heat rise per watt of that heatsink ? If it's something like 12 degrees / watt then putting 27.6 watts of heat is going to raise the temperature to 324 degrees C

 

[MrAl]

its a good idea MrAl; i must test is tomarow. i wrote about My experiment with LM338T in post # 28.

Hi again Qaisar,


I look forward to seeing your data on the bulb tomorrow some time. I think this will be very interesting.

As far as the iddy bitty heat sink you have there (he he) you do need a bigger one. A very very rough rule of thumb (or should i say rule of big toe) is one square inch of heat sink surface area per watt (room temperature ambient). This isnt 100 percent accurate but provides a quick means to show what it takes to keep a device cool, and even that means a 60 degree C rise over ambient at best.
So if you add up the surface area of all the fins on that little guy you will find it is much much too small. The transistor will burn out.

And what's up with that 50v supply you show in your drawings? Is that really 50v or is that just for the sake of the drawing?
Two lead acid batteries in series is only 24 volts, which may be ok but then we have the efficiency issue to talk about, which means wasting lots of energy. Maybe this is why you started to think about the buck regulator, which is a good idea to eventually replace your linear used for the experiments.

So when we see your bulb data we'll have a much clearer picture what is best to do here. But please be careful and take the readings as careful as possible to ensure at least some level of accuracy.

Also, you really have to be super careful and definitely wear eye protection when doing these experiments. Things blow up and they are usually very very hot by that time. If even one small piece hits the eye it turns a nice interesting experiment into a life changing disaster.

 

[audioguru]

When you measure a very low resistance (the cold light bulb), subtract the resistance of the leads of your multimeter that might be 0.5 ohms. So measure the resistance of the leads first.

 

[kinarfi]

LM2576 3 amp buck step down - almost as simple as LM338

 

[KeepItSimpleStupid]

Your best bet for a headlight is PWM. I do remember when I purchased a surplus Harrison Labs (Agilent) power supply, I tested it with a headlight and the power supply blew up. Overvoltage, rather than short. It used a bunch of 2N3055 transistors, but SOMEONE replaced a few with a lower rated transistor, so they popped.

www.Jameco.com has a nice Velleman PWM motor controller that will work.

A very unusual power supply. Claims "Selectable" 0-32 V, 0-10 A. A short drops the output to zero. "Selectable" comes from you have to change the tap on the transformer to avoid excess dissipation. It was basically used as a fixed rack mounted linear supply. I don't think it had an adjustment for current limiting, but it's very nicely short circuit protected. I've had it for nearly 25 years.

 

[Mr RB]

My old Weir linear supplies will give me up to 5 amps and go into current limit as soon as I hook up a high wattage 12v bulb. With a large capacitor in parallel with the output they will happily run the bulb - it's just overcoming the current limit with the cold filament.

The same happens with large motors with a high start current but minimal running current - without a large cap on the PSU or "jump starting" them with a 12v SLA they won't run.

To me that PSU sounds like "current shutdown" or "protection" feature rather than proper "current limiting". My 5A bench supply has current limiting at 5.5A, and will provide 5.5A into a short circuit all day long. No caps ont he output are needed. It's quite happy running 12v 50W halogens, and starts them fine with a 5.5A current limit they just take a second or two to get to full brightness.

...One answer is that the bulb heating curve does not allow the bulb to heat up enough with the current limited current level, and so the resistance can never increase even though the LM317 (or similar) does not have fold back current limiting. ...

I think the simulator might be failing to accurately model the thermal/time characteristics of the filament and bulb.

... One thing about those car headlamps is that they have a pretty big filament (high mass more or less). But that would imply that the regulator could never put out enough current to heat the bulb, ever, ...

This is unlikely. We are talking about a thermal runaway situation. The filament ismounted on thin metal rods in a low heat conduction low pressure gas. It's got to be reasonably good thermal insulation, and has a really tiny mass. Any decent constant current into the filament will heat it, and resistance will rise, and it will eventually go into thermal runaway and get white hot and the final temp is dictated by the supply voltage.

The only way the lamp would not start with a fixed current is if the power into the cold filament is less than the "critical power" (the heat the filament can dissipate), which must be quite small as the filament mass is very low and its thermal insulation is quite high.

I think that "critical current" needed to start the lamp and get it into thermal runaway would be far lower than it's normal rated running current. It would be fun to see what the critical currents are for certain lamps.

 

[Bob Scott]

How about bypassing the regulator in and out connections with another identical bulb. It will give the regulator a bit of assistance. Both bulbs wired in series will "half light" and then the linear regulator will automatically redirect power within it's current capability to your main bulb?

That is, if your unreg power does not exceed 24V.

Note: Your automotive headlight designed for 13.8 to14.2 volts will underpowered at 12V.

 

[MrAl]

Hello again MrRB and Bob,

MrRB:
Well i agree for the most part, that's what puzzles me too about this. It appears that it must be something else causing the problem but i figured with the bulb curve we would know the full story about the bulb and that would help greatly.
With the bulb at room temperature and assuming it has been that way for some time (not too unreasonable) it is already at equilibrium with it's surroundings, and so the only place to go is up (the resistance that is) because most materials when heated show an increase in resistance. But the question is, is the smaller power level (at the lower voltage level) enough to get the bulb heated enough such that the current falls enough such that the voltage rises and so more energy enters the bulb, and does so with some current limit in place?
Say we have a 24 watt bulb, With 2 amps and 12 volts we have 24 watts and the filament glowing brightly. But what do we have at 2 amps and 1 volt? Is that enough to get the bulb heated enough to allow the voltage to rise, or is that another equilibrium point?
So with a current limit in place we have limited energy entering the bulb. With no current limit we dont have that situation.
It's also a little difficult to judge the thermal characteristics and keep in mind that the bulb, when hot, can easily dissipate 24 watts of heat. If the bulb was completely covered in insulation (which might make for another test) only then would we have the situation you are talking about. And remember that if there were no thermal losses the bulb would heat up to an infinitely high temperature (assuming it could survive).
So make no mistake, the thermal losses are truly part of the model and can not be ignored. But of course the thermal characteristics of the regulator also have to be brought into question here.

Bob:
That's a very interesting idea and could very well work out in real life, but you might want to ask yourself if you really want a SECOND bulb hanging around just to be able to turn on the first

Notes:
I think we can simplify this problem by reducing it to one of trying to drive the bulb with a constant current where the current limit is somewhat higher than the bulb's normal operating current. So for a 1.8 amp bulb we might ask if we can light the bulb up with a power supply with a current limit of 2 amps for example. 2 amps is definitely higher than it normally needs, but will it heat up enough to allow the voltage to rise as high as (say) 12 volts?
1.8 amps at 12 volts is 6.67 ohms and almost 22 watts, so the cold resistance could be 1 ohm or less. If it was 1 ohm, then 2 amps would produce 2 volts which would produce 4 watts of heating which is 1/5 of the normal operating power. It does seem hard to believe that the filament would not heat up enough with 1/5 of the normal power level, so perhaps here it is the external drive circuitry that has some other limitation other than 'normal' current limit (which is based on current alone).
Another thing to think about is that while this bulb is heating up, the voltage across the BULB could be only 2 volts but look at what the REGULATOR has across it now... with a 24v power supply the regulator is putting out 2 amps with a 22 volt drop, that's a lot of power to dissipate, and that regulator heating probably limits the current even more.

So a simple test would be to just measure the current after the bulb is turned on and is running for a few seconds or so. If the current is not as high as 2 amps (as per the example above) but is only 500ma, we would have our proof that the regulator is limiting more than expected and so we have to improve the circuit in some way.

Also important, we have to remember that with a 24 volt power supply and 12v across the bulb even when it is up to full power and glowing brightly the regulator is going to be dissipating the same power as the bulb. So the regulator might not be able to handle the bulb EVEN when the bulb is actually already hot!! So it might shut down anyway

 

[crutschow]

............................................
It's also a little difficult to judge the thermal characteristics and keep in mind that the bulb, when hot, can easily dissipate 24 watts of heat. If the bulb was completely covered in insulation (which might make for another test) only then would we have the situation you are talking about. And remember that if there were no thermal losses the bulb would heat up to an infinitely high temperature (assuming it could survive).
So make no mistake, the thermal losses are truly part of the model and can not be ignored.

.............................

Thermal conduction losses are relatively small, it's the black body radiation from the filament that dissipates most of the power. And that doesn't become very significant until the filament starts to glow a dull red.

As I noted previously, I was able to easily start a taillight bulb which takes about a half amp at 12V and start it with the current limit at about 0.4A.

Look at it this way. Suppose you slowly apply voltage to the bulb while keeping the current at no greater than its rated value. I find it hard to believe that the bulb would never turn on. My experiment with the taillight bulb showed that it would start up and glow even if I never exceeded half the rated current.

 

[picbits]

I have over a thousand car bulbs in stock at the moment but zero time or I'd have a play and plot the results

 

[Qaisar Azeemi]

What is the heat rise per watt of that heatsink ? If it's something like 12 degrees / watt then putting 27.6 watts of heat is going to raise the temperature to 324 degrees C

i don't know about this. from where i come to know about heat sinks, there details, heat rise per watt and other calculations???? kindly tell me.