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

[Qaisar Azeemi]

Hi;

I often use various voltage regulators (lm78xx and lm338 and lm2576 etc) to bias and power different electronic circuits; but when i use these regulators to drive a DC 21W car headlight bulb all of these fail to drive it. for 78xx family i use various combinatios i.e; two 7812 in parallel and to use it with TIP127 to pass high current but still it failed to lit the bulb. i also tried lm338 that had easily passed up to 3A current to charge my 12V lead acid battery but cant conduct 1.7A to light a small 21w bulb that hardly need 1.7A of current :???: i don't know the reason. can any one tell me the reason for that?

also tell me what technique i should use to lit a 21w dc or higher wattage lamps/loads? waiting for replays of experts with a thank in advance.

 

[ChrisP58]

One thing you need to know, is that the cold resistance of an incandescent lamp is significantly lower than it is when the filament is hot.

This means the the inrush, or startup, current can be 10 to 20 times the running current. Whatever electronics you use to control the lamps needs to be able to deliver that startup current.

 

[Qaisar Azeemi]

whenever i connect the bulb to any of the above arrangments it just drop the voltages down to 0.2v and then to 0...

yes you are right chris..... i just checked the cold resistance of the bulb.. it was not more than 1.2ohms.. so it is really very low and needs I=12v/1.2ohm =10A initial.... its toooooo high current .... i even used a 2200uf capacitor but it also failed to start it... should i increase the capacitor?? is there any other technique instead of using the capacitors?

 

[picbits]

As above - the current taken to get a cold incandescent filament going can be many times the running current.

I've had success with large capacitors (>10,000uf) on a 6 amp regulated supply to get a 55 watt bulb lit and just about running. Without the capacitor, the overcurrent of the PSU kicks in as soon as the bulb is connected.

 

[aljamri]

beside all what the above mentioned,

* 1.7A is more for some regulators as it's 1A or 1.5A and rarely 2A.

* you did not mentioned what is your circuit limits before the regulator, that may limit your output.

 

[Qaisar Azeemi]

As above - the current taken to get a cold incandescent filament going can be many times the running current.

I've had success with large capacitors (>10,000uf) on a 6 amp regulated supply to get a 55 watt bulb lit and just about running. Without the capacitor, the overcurrent of the PSU kicks in as soon as the bulb is connected.

here the only solution seems to connect the capacitors at the output of the regulator or transistor to store the charge and apply it to the bulb.. i try it too but the 1.2ohm resistance of the bulb don't let the capacitors to get charged.. now i am confused how i design the circuit????

 

[Qaisar Azeemi]

beside all what the above mentioned,

* 1.7A is more for some regulators as it's 1A or 1.5A and rarely 2A.

* you did not mentioned what is your circuit limits before the regulator, that may limit your output.

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.

 

[picbits]

here the only solution seems to connect the capacitors at the output of the regulator or transistor to store the charge and apply it to the bulb.. i try it too but the 1.2ohm resistance of the bulb don't let the capacitors to get charged.. now i am confused how i design the circuit????

The bulb must be left out of the circuit while the capacitors are charged. When they are charged, connect the bulb and they will give the PSU the extra current capacity for a split second so the bulb filament can heat up.

 

[MrAl]

Hello,

It's amazing how hard it is to light a simple bulb sometimes isnt it? This shows us just how much more complicated things are than they look at first sometimes.

Another idea would be to connect an inductor in series with the caps between bulb and caps. It may work with just the inductor too if it is sized right. Have to pay attention to kick back when the circuit is turned off too however as inductors can generate high voltage when disconnected from a circuit while current is still flowing through them. You could probably get away with an inductor rated for the nominal current flow.
Alternately, you could design a transistor delay circuit that turns the bulb on after the voltage on the caps has built up to near max.

Best: phase the dang thing out and use LED's instead

 

[Qaisar Azeemi]

Hello,

It's amazing how hard it is to light a simple bulb sometimes isnt it? This shows us just how much more complicated things are than they look at first sometimes.

Another idea would be to connect an inductor in series with the caps between bulb and caps. It may work with just the inductor too if it is sized right. Have to pay attention to kick back when the circuit is turned off too however as inductors can generate high voltage when disconnected from a circuit while current is still flowing through them. You could probably get away with an inductor rated for the nominal current flow.
Alternately, you could design a transistor delay circuit that turns the bulb on after the voltage on the caps has built up to near max.

Best: phase the dang thing out and use LED's instead

its a good idea. .. how to calculate the value of the inductor for that purpose?

i don't wana use LEDs. my task is to lit the bulb. i.e; to drive a high power load.

 

[ronv]

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.

 

[kinarfi]

Not sure how you're using the cap, is it paralleling the regulator or the load?
Here's what I would try, the cap get's the load heated up, the inductor keeps the current down while the everything charges and the load lights/runs

 

[audioguru]

Simply ramp up the voltage fed to the light bulb so its current also ramps up.

 

[MrAl]

its a good idea. .. how to calculate the value of the inductor for that purpose?

i don't wana use LEDs. my task is to lit the bulb. i.e; to drive a high power load.

Hello,


Well from:
v=L*di/dt

But since you are using a regulator already we could probably just control that instead.
Find a capacitor value that when placed in parallel with the lower resistor of the LM317 type regulator takes roughly 500ms to charge.
I'm thinking maybe around 2500uf or so.
With this connected from the ADJ pin to ground, it will take some time for the regulator voltage to rise, perhaps giving the bulb time to warm up before getting the full voltage potential.
If you have a 2500 to 5000uf cap laying around, try it. Maybe even 1000uf might work.

 

[MikeMl]

Simply ramp up the voltage fed to the light bulb so its current also ramps up.

audioguru has it right. Just put a "slow-start" ramp on your regulator. An opamp to generate a two to three second ramp and to buffer it, applied to the adjustment pin of the regulator should do it...


It would take a HUGE inductor to generate a 2 sec slow-start.

 

[Jaguarjoe]

PWM ramp up not feasible? It'd be more efficient.

 

[MrAl]

audioguru has it right. Just put a "slow-start" ramp on your regulator. An opamp to generate a two to three second ramp and to buffer it, applied to the adjustment pin of the regulator should do it...


It would take a HUGE inductor to generate a 2 sec slow-start.

You dont need anything complicated here such as an op amp. As i said back in post #14, all that is needed is a capacitor from the Adj pin to ground. It has to be large enough so it doesnt charge up too fast, like maybe 2200 to 3300uf or something like that.

A couple of protections diodes might be needed too to bypass the regulator when the power is turned off so the cap discharges into the power supply or load.

The idea is that when the regulator is first turned on, the cap has zero volts across it so the output can only be 1.2 volts. As the cap charges up, the output ramps up higher and higher until it reaches the value set by the two resistors.

 

[Mr RB]

I don't see why a regulator won't work? A "12v 21W" headlight is less than 2A, and a LM317 will provide 1.75A even into a short circuit )ie at startup).

If 2A is enough to make that filament glow white hot then the 1.75A from a LM317 at startup will quite quickly get that filament up to temp. All the fussing with start up resistances etc is a sidetrack, it's purely a current issue.

 

[crutschow]

I agree with Mr RB. I had a taillight bulb that I tried with my power supply. I set the current limit to slightly below its 12V current. When the bulb was connected, the current limited and the voltage dropped to a few volts. But the voltage rose as the bulb heated and after a few seconds the bulb was lit.

If you have a foldback type limit that could cause a problem with a bulb, but a standard constant-current type limit should light the bulb.

 

[picbits]

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.