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

[picbits]

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.

Have a read through the design part of this page - https://en.wikipedia.org/wiki/Heatsink

Basically when a heatsink is sold, part of its specifications are its temperature rise per watt of energy put into it. A small heatsink may have a temperature rise of 20 degrees for every watt you ask it to get rid of while large heatsinks may have a 0.1 degree per watt. This is greatly simplifying things but it will give you the general idea.

 

[MrAl]

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.

Hello,

Yes i more or less believe that too. But in the quest to find out what is going on i consider all possibilities before coming to a conclusion.
I think we all agree here that it is more likely that the regulator is turning off, and a simple measurement is all that is needed to prove or disprove this fact. I have yet to see some carefully controlled tests performed. Raise the input voltage after the bulb lights and see if you can find a point where the bulb turns off again.

Qaisar:
Measure the current getting to the bulb after everything is turned on. If it is lower than the current limit of the regulator, then the regulator is shutting down due to overheating or too high of a voltage differential.
As another test, you could use an input that starts out at only say 2v, then gradually bring up the regulator to full power (with the bulb connected all the while). You may reach a point where the bulb lights, but then raising the input voltage more eventually makes the bulb turn off again because of the regulator chip limitations we've been talking about.

Notes:
The test of raising the input voltage to the regulator chip slowly to see that any bulb lights and then raising the input voltage some more to see if the bulb turns off again is similar to how some simulator programs handle difficulties in convergence during start up. We're simply doing the same thing but in real life.

 

[Mr RB]

Ok I just went out to the workshop and run some tests on a 12v 50W halogen which runs about 4A at 12.0v.

Constant current : lamp voltage : (time)
400mA 0.12v (remains stable)

1A 0.30v (0 mins)
1A 0.45v (1 mins)
1A 0.46v (2 mins)
1A 0.46v (3 mins) (filament still looks cold)

1.4A 0.7v? (0 mins)
1.4A 1.13v (1 mins)
1.4A 1.13v (2 mins) (filament a dull red)

Below 1A I could see the voltage creep up a bit after the constant current was applied, but quickly topped out as the filament was dissipating that power with ease.

At 1A it was close to the "critical current" but still a touch below it. The voltage rose over about a minute but still topped out at 1A 0.46v (0.46W) with the filament "cold" and the bulb body barely warm to the touch.

At a constant 1.4A it was definitely over the "critical" current, the voltage rose very quickly over maybe 10 or 12 seconds and the filament went a dull red but obviously "hot" in terms of a metal wire. Obviously well over the thermal runaway point needed to activate the filament.

So I would say the critical current for my 50W 12v halogen lamp is somewhere between 1A and 1.4A, and its normal running current is just over 4A.

I also want to point out the filament of that 50W 12v halogen is a few turns of a very thick looking wire, in comparison with something like a 12v 25W turn signal bulb that has a much longer filament of much thinner wire. It's likely most normal bulbs would start at a lower current (proportionally) when compared to that halogen that has such a thick filament?

 

[MrAl]

Hi MrRB,


Oh yes, very nice. So it appears that the bulb can 'start' with much less current than the normal operating current. This makes a good case for the regulator shutting down due to thermal issues.

Just to note, when we bang a bulb with the full operating voltage the current jumps way up, then comes back down and eventually reaches operating current. That assumes the full surge current is actually available from the power supply of course. But with less current available it should still start unless the regulator being used goes in to foldback current limiting. The LM317 and similar types dont use foldback current limiting so it should start.
So there has to be some other issue like the thermal overheating that is preventing the proper start up, or the fact that the voltage differential is just too high and the chip is cutting back (as per data sheet).

I'll see if i can find a car bulb around to test also, but i dont think i have one. Next time at the auto parts store

 

[Mr RB]

For sure, there's something funny going on with the OP's setup, some type of shutdown.

The OP said he was using two 12v lead acid batteries in series, which is unlikely to "shutdown", it could even be something silly like a mis-wired 338 regulator. Or maybe thermal shutdown as you said, but with most regulators that is obvious and they "smell hot" or you burn your fingers when poking around etc.

Re buying bulbs, those 12v 50W halogens can be bought from the supermarket quite cheap for home lighting use. They make a handy dummy load or PTC type device in the toolbox.

 

[Diver300]

A thermally limited regulator might well go into thermal shut down before the bulb was lit.

I haven't done the calculations but I could imagine that you could get a bistable condition where either the bulb or the regulator is hot.

 

[MrAl]

Hi,

[See attachment]

Well we have to remember that if the voltage differential across the regulator is too high it wont put out the full current either. That's why the last test i'd like to see is where we include the regulator as well as the bulb, and gradually raise the input voltage to the regulator (LM317 or similar with higher current ability). As we start to raise the input, we should see the bulb get more and more current and start to heat up, and with a limited differential voltage we probably see it light up to full brightness as the input gets to around 14 to 15 volts, then as we go above that 16, 17, 18, i think that eventually we will see the bulb turn off because the regulator starts to limit current just based on the voltage differential alone. And once the current in the bulb decreases, the resistance again decreases, and the regulator limits even more and more and so the voltage differential continually increases and this 'locks' the regulator into a low value current limit that could be a lot lower than it's normal current limit. This would keep the bulb off indefinitely until the circuit power was removed and then replaced again, but of course it would turn off again.

A related test would be to simply see what current it puts out with a large voltage differential. Tune the output for say 3v and use a 3 ohm load perhaps, and see what input voltage causes the current to go down in the output load.

So the first current limit is the value of the rated current limit and that is due to a pure overcurrent load, but the second current limit is due to a high voltage differential and this current limit will not reset automatically but will latch because the bulb resistance goes down and that causes an even higher voltage differential.

Looking at the diagram, we can see that if the bulb started normally it would probably run ok if it only drew about 1.8 amps nominal, but if the bulb didnt start first then it depends on that curve whether or not the bulb starts. It does look like it would start but it might be too close to call. [If the bulb didnt get hot at all the regulator would see V2, but if it started ok the regulator would see V1, so during startup the regulator diff voltage follows the curve from V2 to V1 progressing to the left].
At some point maybe i'll see if i can include this behavior into my model for the regulator and test it with my model for the bulb.

This is what i believe could be happening, barring a simple wiring error. I dont think the spice models model that aspect of the device either so we'd have to use a real life device to test this theory.

[LATER]

Ok, i threw together a quick behavioral model of the secondary current limit effect, modeling the curve as an offset quadratic for any differential voltage above 11v because that's what the curve looks like in the data sheet. It's an approximation but i thought it might tell us something about the behavior even if it wasnt dead on accurate. The 'normal' current limit was set to 2.6 amps so it would be above the 11v secondary limit point which was set at 2.2 amps.
What i found was that with input voltages (to the regulator) less than some max level, the bulb voltage would rise slowly at first and then a bit faster until it went up really fast (very steep slope) and then it would reach max (12v for a 12v bulb).
Raising the input voltage the bulb would turn on slower and slower, but still reaching a point where the voltage rises very quickly until it reaches 12v, so the bulb still started ok.
But finally reaching some set level of the input voltage (in the case of this actual model set it was at 28v) the bulb current would rise slowly as usual but never reaching the point where the voltage started to increase more quickly. In other words, the voltage would taper off to some set level and never rise above that, so the bulb never started completely.
Then testing it with no time constants (thermal time constants all set to 0 so it would take no time to start in a real setup) the bulb still would not start at 28v.
The reason for any lack of start condition was always due to the secondary current limit internal to the LM317 type regulator, which is based on the voltage differential across the regulator.
Note that other regulators might give different results though if they have a different curve than this one does.
The difference between 24v and 28v is significant, but even with that spread it's hard to say if the regulator was a little different or the bulb a little different the limiting action might occur at 24v instead of 28v. We'd have to test a real regulator/bulb combo.

What else this says is that the regulator doesnt need time to heat up in order to see this limiting effect. It's not the thermal limiting it's the voltage differential limiting. That's not to say that it doesnt eventually heat up anyway though and start to limit because of that too.

 

[Mr RB]

...
The reason for any lack of start condition was always due to the secondary current limit internal to the LM317 type regulator, which is based on the voltage differential across the regulator. ...

Very cool! I think you found it.

With the OP's original post he mentioned a LM338 capable of 3A, and a 1.8A bulb, so I'm still of the opinion that regulator *should* have started the bulb.

And the million dollar question; if the LM317 is cutting out and fails to start the bulb from that high input voltage, what is needed to fix it? Maybe a good sized electro cap from Vin to Vout? And/or a resistor from Vin to Vout (as we can make assumptions of the two voltages and the current)... Or something more sophisticated?

 

[MrAl]

Hi MrRB,

Yeah that really does look like the problem alright

And i agree that with his setup it should have started, or at least that would be my guess. I'd like to find out more by doing a few simple tests myself with some bulbs and a regulator like the LM317, but i've been looking at bulbs and i see it's not going to be easy to pick the right bulb for the test because i'd have to get one that works with the regulators i have (LM317) and im not entirely sure about their *actual* current limit (data sheet says 1.5 amps i think so i'd be going for a 1 amp bulb maybe? but it could be 2.2 amps?).
Add to that we expect one of the biggest storms ever to touch down in my home state so im not sure what is going to happen.

How to fix this...
Yeah, i dont see any way to fix it because sometimes the bulb will still shut down even if it got up to speed solely because of the overly large voltage differential. But yeah there is probably a sweet region where it works if you can start it. We'd have to be sure we could always guarantee that it will stay in the sweet region even after some filament loss and other problems that could come up. so it would probably take an active circuit of some kind.
Probably the best bet is to make sure that the input voltage never goes too high to cause a problem based on worst case.

Any other ideas?

 

[kinarfi]

Did you try the LM2576 simple switcher?
How did that work?

 

[Mr RB]

I use LM317 a bit, and from what I have seen the hard current limit is about 1.7 or 1.75 volts. It might be different with different brands though.

...How to fix this...
Yeah, i dont see any way to fix it because sometimes the bulb will still shut down even if it got up to speed solely because of the overly large voltage differential. But yeah there is probably a sweet region where it works if you can start it. We'd have to be sure we could always guarantee that it will stay in the sweet region even after some filament loss and other problems that could come up. so it would probably take an active circuit of some kind. ...

Thinking about it a bit more, I think a simple resistor in parallel with the LM317 would do it. Assuming 28v Vin and 12v desired Vout, and say a 12v 12W 1A bulb, that means when running at 12v out the regulator is dropping 16v. The resistor would pass about half the current, so would be say 16v 0.5A ie 33 ohms. The regulator would still regulate voltage fine and supply the other 0.5A, and keep the bulb voltage at a clean 12v.

At startup even if the regulator supplies no current at all, you get 28v through 33 ohms which is almost 1A from just the resistor. That's almost the filament full current so the bulb should "start" no problem, then when the bulb voltage rises from its resistance increase the regulator should fill in the rest of the load current.

So this is replacing some of the huge regulator dissipation with resistor dissipation... that may or may not be better. Personally I would not use a linear regulator to drive a large 12v bulb from 28v, when PWM or SMPS could be used instead.

 

[shokjok]

Have you considered a slow-start light circuit, one that gradually illuminates the light from darkness to full bright? Unless your circuit is a spotlight device, the slow-start idea would need a heat-sinked regulator and transistor capable of the high current you require. An LM317K and 2N3055 pair would be a good start.

 

[picbits]

As an obvious thought, why not just run the bulb from one of the 12v SLA batteries ? It seems a bit silly using two in series then a regulator to drop it back down to 12 volts.

By the time the bulb starts to dim anyway, you should disconnect the battery as you'll damage them if you discharge them too deeply.

 

[MrAl]

Hi again,

picbits:
Yes i was thinking that too because i realized that i never had this problem because i always try to find a bulb that fits the purpose rather than try to fit a low voltage bulb to a higher voltage source. I do a lot of LED light sources though, but they dont have this 10:1 surge problem either.

MrRB:
Yes that's a good idea i think. A resistor could supply some nominal current like 1amp as you said and for a 1.5 amp bulb that would mean the reg would only see 0.5 amp maybe 0.75 amp with some input variation. I think this would be doable.
The other thought is the power dissipation of the required resistance, but i would think that if we intended to use a 12v regulator with a 24v supply then we must have already realized that we are going to see some significant waste power heating somewhere
So yes, i think this would solve the problem and it would be nice to test too.
SO maybe pick a resistor value based on 1/2 the current requirement at nominal input, or even for a little higher current.
If the user gets worried about power waste then a buck is the only way to go, with slow start.
As i mentioned above, i usually pick a bulb that fits the application, but then again i dont use bulbs anymore unless it's just a replacement for something that has already been running for years.

 

[Qaisar Azeemi]

I am very thankful to all my good friends who replied and helped me. i at last get success and lit the bulb using switching regulator (LM2576) successfully. it lit the bulb with 28v input to it and fed 12v to bulb and it turned on normally as it is connected to a 12V battery. the problem was the over heat dissipation than the rated amount in LDO regulators and Transistors. the problem was only the heat sink in case of 2sd1047 transistors which is 100w power transistor and the voltage difference across it was of 16V (28V-12V) and current through it was 1.75A and hence power dissipation on it was P=16*1.75=28W which is very very much less than the rated 100w for the transistor ............. but still it blown out the transistor because of insufficient heatsink.
and LDO fail to lit because of large voltage difference as i used LM338t but it couldn't lit the bulb and couldn't pass the desired ampere of current as i applied 28v dc at its input and tried to drive the 21w load at the out put......................so with this large difference it was not able to conduct 1.75A of current .................. but WHY the same regulator charged my 12V PB-Acid batteries with and input of 35V at its input and out put was set to 14.9V thus with a difference of 35-14.9=20.1v and i conducted a current of 1.5A approximately and a power dissipated across it was P=20.1*1.5 = 30.15W ?????? it there any difference to pass current to drive a load and to charge a battery????? i can't understand this?

 

[MrAl]

Hi,

Statically the battery and the load are the same, in that they will both draw current and if you didnt know which load you had you would not be able to tell the difference by measuring the current. However, the dynamics are different. That means the time related behavior is different than when it is just sitting there drawing current, and in this case the linearity is different too.
The battery does not draw any current until after the output voltage of the regulator gets up to 12v or more, while the bulb starts to draw current even with just 0.1 volt output from the regulator. So the loads are different right from the start.
A resistor is different yet, because although that draws current even with just 0.1 volt output for example it draws much less current than it does when it reaches full voltage, unlike the bulb which can draw more current at the start and less current later, and the battery which draws no current at the start and then more current later.

 

[Qaisar Azeemi]

thanks to all. i want to know if a high amount of current for a fraction of second due to cold resistance can turn the LDO regulator LM338 off then why this cold resistance is not effecting the switching regulator LM2576??? it lit the bulb just if it is directly connected to a 12V battery. what is the reason??

 

[ChrisP58]

The current limit function between linear and switching regulators is quite different.

In a linear regulator, the limit is also linear, and reacts very fast to shut the output off. The voltage falls to zero so no effective energy is delivered to the lamp to warm it up

In a current mode switcher, the current is evaluated each pulse and is used to terminate the switching pulse. But the pulse turns back on a few microseconds later. A narrow pulse of current is delivered to the lamp even when the regulator is in current limit.

If you were to look at the switch node (the junction of the regulator output, the freewheel diode and the inductor) with a scope, you would see the duty cycle when first turned on would be narrow, but each pulse still delivers current into the lamp. This current heats up the filament so that the next pulse is a tiny bit wider than the previous pulse. The duty cycle increases rapidly until the voltage control loop kicks in to regulate the voltage.

 

[ronv]

The linear regulator has what is called safe operating area protection to keep the power disapation within limits that are safe for it. It is in watts so the greater the voltage across the regulator the lower the current it can supply. This would explan why it works okay with the battery and not the bulb. The switcher runs in on-off mode so it can supply it's rating without this restriction.

 

[dr pepper]

Picbits with ref to post #20, I think theres something amiss with your supply.
I also have a weir, 0-30v 2a or 0-60v 1a it is 2 identical isolated supplies which are either paralled or series'd.
Mine has a max o/p of 2a and it lights a 12v 21w vehicle indicator lamp immediatley no stress.
If yours is a 5a and it struggles something's wrong, when I got mine a couple of years back some of the cans had to be replaced to get a clean o/p, maybe yours has dried caps too.