1. Martino Chiro

    Thread Starter Member

    May 1, 2015
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    The specs of the LT3012 shown a 250mA max output current.
    There is a solution to get more current from this device ?
    It is possible to parallel more LT3012 ? Or, put some power transistor in the basic configuration
    shown in the data shhet that boost the 250mA limit ?

    Thank You
     
  2. crutschow

    Expert

    Mar 14, 2008
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    Why not just use a regulator that has a higher current capability?
    How much current do you need?
    What is the input and output voltage?
     
  3. Martino Chiro

    Thread Starter Member

    May 1, 2015
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    Thank You Crutschow.
    I need the high input voltage (80V) of the LT3012 (48V minimum in my case) and 1A of output current.
     
  4. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    An external transistor could be used, but would have to dissipate, for example, (80-5)*1 = 75W for an 80V input and 1A output at 5V :eek:.
     
  5. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    What output voltage do you need, and what input range do you have?
     
  6. Martino Chiro

    Thread Starter Member

    May 1, 2015
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    The input voltage is 48V (+/- 10%, and the output voltage required is 24V and 1A continuous.
     
  7. Alec_t

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    Sep 17, 2013
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    So power dissipation by the outboard transistor would be 24W +/- 10%. Still a lot to handle!
    A switch-mode converter would be a lot more efficient.
     
  8. mcgyvr

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    Oct 15, 2009
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  9. Martino Chiro

    Thread Starter Member

    May 1, 2015
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    But how it need to be connected to the LT3012 ?
     
  10. mcgyvr

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    Oct 15, 2009
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    How do you intend to handle the dissipation?
    Its a total waste trying to use that IC for your needs..
     
  11. Dodgydave

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    Jun 22, 2012
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    What about a Lm2596, max input 40V, you just need to drop from 48 to 40v, takes 3amp output.
     
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  12. Martino Chiro

    Thread Starter Member

    May 1, 2015
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    Interesting the use of he LM2596. But again i need to put a transistor to get a drop of at least 8 - 10V.
    It is not possible to parallel more LT3012 ?
     
  13. spinnaker

    AAC Fanatic!

    Oct 29, 2009
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    The lm2596 can only have 40 in as I recall but it would not surprise me if TI had a chip capable of 48v in.
     
  14. Alec_t

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    In theory you could, but in practice the ICs will not be exactly matched (even if from the same batch) so they won't share the current equally. One will be taking more than its fair share of the load. You will have to ensure that this share does not cause the IC to go into thermal shut-down. The total power dissipated will still be 24W, which is going to need very effective heat-sinking. As mcgyvr asked, how are you going to handle that? Unless you have a sackful of 3012's looking for a use it will be cheaper and more efficient by far to use a switched-mode supply.
     
  15. Dodgydave

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  16. benta

    Member

    Dec 7, 2015
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    The TI LM317HV would fit your needs, I think.

    Regards,

    Benta.
     
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  17. Martino Chiro

    Thread Starter Member

    May 1, 2015
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    Both the L4978 and the LM317HV are good solutions for me.
    When Vin - Vout = 24V the L4978 seems can deliver 2A of output current max.
    When Vin - Vout = 24V the LM317HV seems can deliver 1A of output current max
    from the Figure 17 "Current Limit" for T and K package type, in the TI data sheet.
    [​IMG]
     
  18. benta

    Member

    Dec 7, 2015
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    I don't know the L4978, but I've worked a lot with the LM317/317HV.
    Your main concern is thermal management = cooling, and this is true for both devices.
    You'll need to dissipate ~24 W to the surrounding air, and that means a relatively large heatsink.
    Thermal resistance junction-case is 0.9 K/W for the "T" version, you need to add insulation (mica or silicone) which is around 1 K/W (worst case), let's say 2 K/W from junction to heatsink. This means a temperature rise of 48 Celsius.
    Say your maximum surrounding temperature is 35 C (I'm guessing here).
    To keep the junction temperature below 125 C, your heatsink needs to have a thermal resistance of:

    125 - 35 - 48 = 42, meaning your heatsink must have a thermal resistance of no more than 42/24 = 1.75 K/W

    That is quite a big chunk of aluminium, please check the heatsink suppliers' websites for example products.

    A solution with a switching regulator or a DC/DC-converter module might be better.

    Best Regards,

    Benta.
     
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