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DC Power Supply (5V 10A) - Questions

Discussion in 'Power Electronics' started by Dr.Ehsan, Nov 18, 2011.

  1. Dr.Ehsan

    Dr.Ehsan

    6
    0
    Nov 18, 2011
    Hello,
    I'm planning to design a circuit for a DC Motor with these parts:
    1* 7805
    2 * 2N2955 (paralleling two transistors to increase current)
    plus some capacitor and resistors.
    (the circuit image is attached)
    Here is my question:
    If I use a 12volts-6Ampere transformer ( 12V * 6A = 72W), does it provide enough power to have 5V-10A as output?
    theoretically: 72W/5V = 14.4Ampere


    Thanks,
     

    Attached Files:

  2. Resqueline

    Resqueline

    2,848
    1
    Jul 31, 2009
    That transformer will give you only 15V & 4.24A DC (= 64W) since you lose a couple of volts in the rectifier.
    So you can use that transformer and achieve those spec's only if you build a switch-mode (buck/step-down) regulator having an efficiency of 79% or better.
    Using that linear regulator you'll need a transformer giving 7V & 14A (minimum), and a heatsink good for 30W (minimum). The rectifier will dissipate 20W btw..
     
  3. Dr.Ehsan

    Dr.Ehsan

    6
    0
    Nov 18, 2011
    Thanks, as far as I knew, designing a switching (SMPS) circuit is a little hard and needs many parts.
    Is there any simple switching circuit to bu used?
     
  4. jackorocko

    jackorocko

    1,284
    1
    Apr 4, 2010

    If you look around you might be able to find a switchmode IC circuit that would only require a few external parts.
     
  5. duke37

    duke37

    5,201
    713
    Jan 9, 2011
    Your circuit is wrong, you have three resistors in parallel. you need only one.
    You need a low value resistor in the emitter of each transistor to ensure current sharing, dropping say 0.5V at maxixmum current.
     
  6. Dr.Ehsan

    Dr.Ehsan

    6
    0
    Nov 18, 2011
    There's a simple switching circuit (including LM2576), which has 5V - 3A output.
    (the circuit image is attached)
    Datasheet: http://www.national.com/ds/LM/LM2576.pdf
    Question: Is it possible to increase its current by power transistors?
     

    Attached Files:

  7. Resqueline

    Resqueline

    2,848
    1
    Jul 31, 2009
    There's also the 5A LM2678 (& LM22678) but I'm not so sure it's just a matter of parallelling & sync'ing them or that there's an easy way to add a transistor.
    It might be better to use a low-power switcher chip that is made to drive an external MOSFET. Then there's (almost) no limit to the current that can be had out.
    Or, if you can accept buying a ready-made 10A switchregulator module I found one example here.
     
  8. Dr.Ehsan

    Dr.Ehsan

    6
    0
    Nov 18, 2011

    Attached Files:

    Last edited: Nov 19, 2011
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,172
    2,689
    Jan 21, 2010
    I'm a little concerned that the current peaks through the 1N4148 will be quite high.

    I see a rated max of 450mA for repeditive peak current. I'd be far happier following the advice of selecting some other schottky diode in its place.

    And you're pronbably better of not quoting the entire article. A link to it is fine.
     
  10. Dr.Ehsan

    Dr.Ehsan

    6
    0
    Nov 18, 2011
    I suggest to use 1N5822 or BYV28 instead of 1N4148.
    I don't have BYV29, so I'll use BYV32 which has Average Forward Current of 20A.
    Also, L2 is 100uH and should be similar to the attached image.
    Question: may I use 1Amper inductor in this circuit?! OR I have to find a high-amper inductor?
     

    Attached Files:

    Last edited: Nov 19, 2011
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,172
    2,689
    Jan 21, 2010
    Your inductor carries the full current, so if your output current is > 1A... (and you have to consider that peak current will exceed the average current)

    This pdf may help you.
     
  12. Dr.Ehsan

    Dr.Ehsan

    6
    0
    Nov 18, 2011
    Thanks steve, but my output current is at least 8A
    What happens if I use 1amper inductor?
     
  13. Resqueline

    Resqueline

    2,848
    1
    Jul 31, 2009
    Overloading components will (as usual) make several things happen:
    1st; the inductor core will saturate, which means the current will "spike" (rise uncontrollably fast) very early in each conduction cycle.
    2nd; this will most likely destroy the MOSFET, and as a result you'll get the full 15V out on the 5V.
    3rd; if (as in by magic) the MOSFET should survive, the inductor wire will burn to a crisp.
     
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