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How to drop 18.5V 6A DC to 15V 6A DC

Discussion in 'General Electronics Discussion' started by shumifan50, May 26, 2017.

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  1. shumifan50

    shumifan50

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    Jan 16, 2014
    I have found several circuits using a LM317 with pass transistor to up the current capability, but they all seem a bit like an overkill for my purpose.
    I have a Turnighy Accucel 6 LiPo battery charger. It requires 12-18V 5A input to it.
    I have a 18.5V 6A laptop PSU
    I would like to drop the voltage to suit the charger.
    Note: The charger cuts out on over-voltage, actually even when close to 18V, so ideally I am aiming for 15V. I will mostly use for charging 3S LiPos, so 15V means it will not need step up.
    I have a 3-15V brick PSU 4A, but ideally I would like to use the dedicated 18.5V for the LiPo charger.
     
  2. Tha fios agaibh

    Tha fios agaibh

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    Aug 11, 2014
    Perhaps parallel three or four lm7815 regulators to increase the output current.
     
  3. duke37

    duke37

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    Jan 9, 2011
    If you wish to drop 18.5V to 18V, then you only need to put a (10A) diode in series. A couple of diodes will give a little more latitude.
     
  4. davenn

    davenn Moderator

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    Sep 5, 2009
    will only give around 3 - 4 Amps ... and yes I know they state 1.5 A but in reality they seriously over heat even at 1A ( heatsunk)


    2 x LM338 5A reg's would be a much better and easier way to go much less circuitry to deal with

    he's going to 15V not 18V ;) else I would agree, it would be a simple answer


    here is the ideal choice and a great low price .....

    http://www.ebay.com/itm/6A-DC-DC-3-...969189?hash=item4b1468f1e5:g:N54AAOSwdGFY2SIA



    Dave
     
    Tha fios agaibh likes this.
  5. shumifan50

    shumifan50

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    I don't think you can parallel fixed regulators like the 7815 - the outputs will interfere with each other, which is why they show the pass transistors in the datasheets. I was wondering whether I could use a resistor/zener on the base of a TIP3055 to achieve the amps. That keeps the circuit really simple and I might be able to fit the parts in the original brick, with only an external heatsink.
     
  6. shumifan50

    shumifan50

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    They don't ship to South Africa and all 9 are sold out in any event. I have a bit of a problem here to get parts, so I try to stick to very common parts that I know I can get. RS has a branch here but it requires 2 trips to their branch; one to place the order and the next to collect once they got it from the UK. This also bumps the price significantly.
     
  7. davenn

    davenn Moderator

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    Sep 5, 2009

    well that was just one example .... surely you can find others .... or do you have that problem with all eBay sellers ( not shipping to SA ?)

    try amazon
     
  8. bushtech

    bushtech

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    Sep 13, 2016
    Have you tried mantech.co.za. They will deliver to a post box which RS wont do. And Riecktron also sell some interesting stuff
     
    davenn likes this.
  9. shumifan50

    shumifan50

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    I use Mantech and communika, but they carry limited ranges. Did not know about Riecktron.But I am still wondering whether a voltage divider made up of a 15V zener and resistor on the base of a well heatsinked TIP3055 might do the job. I am not sure what the value/ratingof the resistor must be. I think the 3055 amplifies the current 10 times, so if I put 1 amp on its base it should deliver 10A and I only need 6 amp. That would need a 3R 3w resistor (I think). Might even get away with 6R 1w. Opinions/advice on this much appreciated as the power brick has space to fit these components, but not much more.
     
  10. duke37

    duke37

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    If you wish to go with a voltage follower you could consider a LM317T instead of a zener. This would dissipate less power and could be easily set to the voltage required.
    Investigate whether 6A in one 2N3055 will be reliable.
     
    shumifan50 likes this.
  11. shumifan50

    shumifan50

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    Thanks Duke. I could use the circuit from the datasheet for the LM317, which would give all the over voltage, thermal protection etc. The problem is that that would require quite a big circuit that would not fit into the existing PSU box. A mate of mine suggested using two 1A diodes on the base of a TIP3055. That will give sufficient voltage drop while the TIP will provide sufficient amps for the Accucel 6 charger (5A) if heatsinked properly and it will fit inside the existing box with external heatsink. Thanks to all that responded.
     
  12. duke37

    duke37

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    Jan 9, 2011
    The LM317T is the same size as a TIP3055 and very little different to two diodes. A couple of very small resistors would be needed. It could be made very small, Manhatten style.
     
  13. shumifan50

    shumifan50

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    Jan 16, 2014
    @Duk The lm317 can only handle 1.5A at a push with heavy heatsinks. It would still require a pass transistor to handle 5A.
     
  14. duke37

    duke37

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    Jan 9, 2011
    I suggested using the LM317T as a substitute for the big resistor and big zener. It will only take the base current of the output emitter follower. You will need about three volts more than the output, Two for the 317 and one for the 3055.
    If you wish to use a pass transistor, then a PNP transistor would be needed but would give more stable output voltage which you may not need.
     
    shumifan50 likes this.
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