Perfect LM338 Power supply

Discussion in 'General Electronics Chat' started by korawy, Mar 21, 2013.

  1. korawy

    Thread Starter New Member

    Mar 20, 2013
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    0
    hi all
    i have built a power supply using lm338T with NO HEAT SINK :( ...but then i tried it after i have finished building it, and it works perfectly with a motor 24vdc and 2.5-3 A.

    Unfortunately the day after that i tried it again but it constantly delivering 42v out, after checking each component i found that the lm338T Vout and Vin pins are shorted :mad: .

    i changed the lm338T and the trans 220 to (16-0-16) 3A with trans 220 to (12-0-12) 3A, But still the same..

    This is circuit diagram
    http://powersupply88.com/simple-vaia...a-power-supply
    But without using the lm317 section (Current limiter)

    Then i found that i was wrong in using LM33T without using heat sink (50C/W),
    and found also that i must use LM338K with a suitable heat sink (1C/W)
    Then i have decided to rebuild the following circuit:
    http://www.eleccircuit.com/adjustabl...a-using-lm338/

    Is there any advices before I start??

    Im using a Trans 220 to (12-0-12) 3A
    LM338K with heat sink
    I have made calculations according to SgtWookie :) post in another thread and found that outputting 15v to 30v with 3A will be OK (15v through connecting a fixed Resistor series with the variable resistor).

    Am I right??
    [​IMG]
     
  2. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    ABSOLUTELY IMPOSSSIBLE without heatsink. You need 2.5 - 3A of current, and the regulator needs at least 2.5V drop across it to function as a voltage regulator. That gives you >6W of power dissipation.

    You need to calculate the power dissipation:

    Vin - Vout X load current = power

    Then select heatsink that can dissipate that and keep the semi die temp below about 130C.

    Read this it explains how.
     
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    Last edited: Mar 21, 2013
  3. Dodgydave

    Distinguished Member

    Jun 22, 2012
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  4. korawy

    Thread Starter New Member

    Mar 20, 2013
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    0
    Well..
    Doing simple theoritcial calculations :
    Power=(Vin-Vout)*I, If we need 10V out, and Vin after rectification will be 35v So:

    Power=(35-10)*3=75 watt..

    And, θJ-A = (TJ (max) - TA (max)) / P

    Assuming that working temperature is 25C, and knowing that the Max Temp is 125C, Then..

    θJ-A=(125-25)/75=1.33C/W

    And we know that θS-A = θJ-A - θJ-C - θC-S
    where θS-A is the thermal resistance of the heat sink, This means that θJ-C + θC-S must be very small to deliver 10V (I dont know the practical value of the thermal resistance of the available heat sink.. This is the one i found

    http://ram-e-shop.com/oscmax/catalog/product_info.php?products_id=1337

    and this is another one

    http://ram-e-shop.com/oscmax/catalog/product_info.php?cPath=146&products_id=1338 )

    Thanks for the replies...
     
    Last edited: Mar 22, 2013
  5. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    Correct. In practical reality, dissipating 75W means you must divide the power between more than one device. Even the TO-3 package will typically have a theta J-C value of maybe 1C/W and another 0.5C/W for theta C-S to connect it to the heatsink. That means you get a juntion temp rise of over 100C going from the junction to heatsink, then you have to add the heatsink rise..

    75W of power dissipation is not feasible for any single IC device.
     
  6. korawy

    Thread Starter New Member

    Mar 20, 2013
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    0
  7. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    Not even close.

    with 35V in and 15V out @ 3A, you have 60W power. The theta going from junction to heatsink (TO-3) is about 1.5C/W so you are already 90C above ambient just GETTING to the heatsink. That only leaves maybe 25C rise for the heatsink, so the theta can be no more than 25/60 = about 0.4C/W

    That is either a very large heatsink or some kind of air cooled heatsink assembly.
     
  8. korawy

    Thread Starter New Member

    Mar 20, 2013
    8
    0
    Should i use a cooling fan??
     
  9. korawy

    Thread Starter New Member

    Mar 20, 2013
    8
    0
    Using Cooling fan requires 12V, This means Using LM7812
     
  10. korawy

    Thread Starter New Member

    Mar 20, 2013
    8
    0
    i have rebuilt the circuit using a heat sink and a cooling fan of 12v (Using LM7812), But Unfortunately the trans i have 16-0-16, so using this i must select the proper outputs of the trans.

    Using the two outputs of 16 and 16 will deliver 32v (actually 32/0.707=45.2v) this is not allowable.

    Using the two outputs of 16 and 0 will deliver 16v (actually 16/0.707=22.6v),
    this will output from LM338K MAX 20v, I used this connection and it worked perfectly .

    The heat sink i use is a piece of u-shaped aluminium, i bought it from an Aluminuim shop.

    How can i measure the current derived to a load(ex.motor)? I connected a Multimeter pin to the output of LM338 and the other Multimeter pin to the load and i grounded the motor, But nothing happened...

    when i connect the fan the voltage is dropped by 2 volts (≈19-2=17)
     
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