Soft Start circuit for automotive headlamps

Discussion in 'General Electronics Chat' started by rsicard, Aug 6, 2012.

  1. rsicard

    Thread Starter New Member

    Feb 21, 2010
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    I am looking for suggestions for a DC soft start circuitry for 130 watt automotive headlamps. The best that I am able to envision is a Power FET driven by a PIC Micro DAC output such that the voltage/current ramp up in a more linear manner yet short amount of time so as not to stress the Power FET too much. Any thoughts or suggestions are welcome. Thanks.
     
  2. k7elp60

    Senior Member

    Nov 4, 2008
    478
    69
    Here is another way. The schematic needs some changes but should work. I calculate the hot resistance of the lamps is about 1Ω. If you put a 0.5 resistor in place of the MOV shown in the circuit and change the 1M resistor to about 2KΩ., it will give approximately 20millisecond soft start.
    I have used the attached circuit in RV light figtures for bed reading lights for over ten years. The bulbs last over 10 years with this circuit. You can the about a 25W resistor at 0.5Ω. It should work fine as the time it is in the circuit is very short. The only problem is that the polarity is critical.
    The N channel Mosfet will easily handle the full current.
     
  3. rsicard

    Thread Starter New Member

    Feb 21, 2010
    4
    0
    K7elp60:

    Thanks very much for the suggested circuit. You have been more than enough help!
     
  4. rsicard

    Thread Starter New Member

    Feb 21, 2010
    4
    0
    K7elp60:

    The polarity is a given. Understand the RC time constant of the resistor and capacitor. Is the MOV the one in parallel with the Power FET? What would be the spec and part number of the MOV? What is the reasoning in replacing the MOV with a 0.5 ohm 25 watt resistor just curious as in all these questions? I will mock up and adjust the circuit with a headlamp and try to get current and voltage readings from the circuit. Again, THANKS VERY MUCH!
     
  5. #12

    Expert

    Nov 30, 2010
    16,298
    6,810
    More reliable to reduce the capacitor size. 2 uf and 10k would be proportional to the suggested 2k resistor and 10 uf of capacitor. I would use:

    http://www.mouser.com/ProductDetail/Panasonic-Electronic-Components/ECQ-E1225JF3/?qs=sGAEpiMZZMv1cc3ydrPrF0%252bjlB8SXIRuF%252b2wzcOpP5Y%3d

    2.2 uf, 100 V, 97 cents
    and a 10k resistor to make it more resistant to electrical noise.


    130 W @ 12 V =10.8333 amps (standard running condition).
    1/2 ohm would allow 24 amps when the MOSFET is off. That would allow a start surge.
    1 ohm (cold) plus .5 ohms would allow 8 amps of start current instantly. Then the MOSFET takes over. At 4.5 volts Vgs, the MOSFET is at .028 ohms
    .028 ohms in parallel with .5 ohms will waste .287 volts at 3.11 watts, most of it in the resistor.
    Both parts will be safe from melting.

    The MOV is used to suppress voltage surges that happen in automobiles. It would have to be rated between 12 volts and 60 volts.

    http://www.mouser.com/ProductDetail...=sGAEpiMZZMv1TUPJeFpwbiME22DOKZL64BMBpb243zg=

    19 cents

    From the equasion: time = -RC (Ln Vo/dV)
    Time = 21.6 milliseconds to full on.
    You can change that if you want to.
    My blog #3: R = -time/C (Ln 4.5/12) to find a new R for whatever time you require.
     
    Last edited: Aug 7, 2012
  6. rsicard

    Thread Starter New Member

    Feb 21, 2010
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    0
    #12:
    THANKS MUCH for all your work on this situation! I can't believe how much I have forgotten about the math and calculations since my schooling in Electronics. It is obvious you have much experience in this type of circuitry and calculations. This forum is very valuable to someone such as myself. Thanks again!
     
  7. #12

    Expert

    Nov 30, 2010
    16,298
    6,810
    There's a button for that...just click on it.
     
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