LM338 Power Supply Current Regulator

Discussion in 'The Projects Forum' started by Umer_Farooq, Mar 14, 2010.

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  1. Umer_Farooq

    Thread Starter Member

    Mar 6, 2010
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    1. I am using Lm338 Power Suply I add Current Regulator. In the Lm338 Data Sheet Current regulator use befor voltage regulator.In the data sheet for 5A current regulator 0.24Ω 2W resistor connected while P=IxV and I=5A and V=1.25 then P=1.25*5= 6.25w resistor. How 2w resistor conecting in the data sheet Please help me

    2. Please send me a diagram whose help me add a current regulator 0-5A
     
  2. SgtWookie

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    Jul 17, 2007
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    This schematic is on the 12th page of National Semiconductor's datasheet for the LM138/LM338:

    [​IMG]

    You have calculated the power used in the 0.24 Ohm resistor correctly, but you should double the actual power dissipation to use for the resistor rating.

    Also, keep in mind that 1.25v is nominal (average) for Vref. However, it may be as low as 1.19v, or as high as 1.29v. You must plan for that, too.

    If Vref is as high as 1.29v, then you will have 1.29v/0.24 = 5.385A output; and power dissipation of 6.93375 Watts, for a resistor rating requirement of 13.8675 Watts.

    You will need a large heat sink for your regulator that is either water cooled, or cooled by forced air - a LOT of air.

    [eta]
    Measuring your LM338's Vref:
    1) Connect a resistor of between 25 and 100 Ohms between the OUT and REF terminals.
    2) Connect the IN terminal to a voltage supply positive terminal that measures from 4v to 10v.
    3) Connect the ADJ terminal to the voltage supply return terminal.
    4) Measure Vref from the ADJ terminal to the OUT terminal.

    Use the reading you measured in step 4 to determine the proper resistance and power rating for R1.
     
    Last edited: Mar 14, 2010
  3. Umer_Farooq

    Thread Starter Member

    Mar 6, 2010
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    Thanks SgtWookie
    But my problem how 2W Resistor Work
    If i am using 2W resistor than this resistor blown



    (Measuring your LM338's Vref:
    1) Connect a resistor of between 25 and 100 Ohms between the OUT and REF terminals.
    2) Connect the IN terminal to a voltage supply positive terminal that measures from 4v to 10v.
    3) Connect the ADJ terminal to the voltage supply return terminal.
    4) Measure Vref from the ADJ terminal to the OUT terminal.)

    Please help me to send a diagram
     
  4. SgtWookie

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    Jul 17, 2007
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    Here is a link to the datasheet:
    http://www.national.com/ds/LM/LM138.pdf
    Look at page 12, schematic on top right.
    Instead of a potentiometer, use a fixed resistor between 25 and 100 Ohms.
    [eta]
    Like this:

    [​IMG]

    V1 is a voltage source that must be at least 3.5v.
     
    Last edited: Mar 17, 2010
  5. Umer_Farooq

    Thread Starter Member

    Mar 6, 2010
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    Very Very Thanks SgtWookie
    Look at page 12, (5A Current Regulator) The Resister R1 0.24Ω and 2W.
    My Problem 2W Resistor will Work or Blown


    2. Please Send Look at page 12, (Adjustable Current Regulator) Diagram Schematic Discription and send R1 0.24Ω what Watt & R2 150Ω What Watt and R3 120Ω What Watt


    3. If i digitally select output current 0-5A then Circuit Diagram



    4. If LM338 use as voltage Regulator then R1 Must be 120Ω. If i use other value than what change in Circuit


    Thanks Very Very Thanks
     
  6. SgtWookie

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    Jul 17, 2007
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    If your Vref is 1.3v, then:
    1.3v x 5A = 6.5 Watts. Double that for reliability to 13 Watts.
    A 10W resistor might work, but it will get pretty hot.

    R2 and R3 can be low wattage, as there will only be about 10.4mA current flowing through them.


    You will need a digital potentiometer of 150 Ohms resistance to replace R2. You probably cannot get a digital pot that will "float". I do not have a good solution for you on this question.

    R1 is 120 Ohms to guarantee minimum current for proper regulation.
    If you increase R1, you will have to make certain that there is at least 10mA load on the output for proper regulation.

    If there is less than 10mA load on the output, the voltage output will rise.

    R1 should not be higher than about 1k Ohms.

    The formulas for calculating values for R1 and R2 are in the datasheet.
     
  7. Umer_Farooq

    Thread Starter Member

    Mar 6, 2010
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    Very Very Thanks SgtWookie

    Look at page 12, (Adjustable Current Regulator) If i not use Lm317 Cct then Change in circuit R2 connect direct to ground


    2. -ve supply why use if i connect to ground then what will change


    3. lm317 ic donot use heat sink because current 10ma
     
  8. Umer_Farooq

    Thread Starter Member

    Mar 6, 2010
    104
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    how Vref Change whie in data sheet vref not changes it equal to 1.25V
     
  9. SgtWookie

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    It will not work like that. The LM317 is there to constantly sink about 10.4mA from R2. If you remove the LM317 then it will no longer work.

    Why don't you try it and see what happens?

    Correct.
     
  10. SgtWookie

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    Jul 17, 2007
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    Look at the specification for Vref. It averages 1.25v from regulator to regulator. However, it might be as low as 1.2v or as high as 1.3v.

    If you want to make certain of what your individual regulator's Vref is, then you have to measure it under load.
     
  11. Umer_Farooq

    Thread Starter Member

    Mar 6, 2010
    104
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    1. How 10ma Current while R1=0.24Ω & V = 1.2V than I=5A


    2. R2 = 150Ω and V = ? Because R2 across Vout and gnd. My idea the voltage across R2 equal to 1.25V(VRef). I am Correct or Not.


    3. If i this cct use than out put current formula.


    4. Across R2 I = ? and V = ? than i calculated P


    5. The -Ve supply provide other transformer
     
  12. Umer_Farooq

    Thread Starter Member

    Mar 6, 2010
    104
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    Thanks a lot Sgt Wookie
     
  13. SgtWookie

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    Jul 17, 2007
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    Referring to this schematic:
    [​IMG]

    The LM117/LM317 has a 120 Ohm resistor between ADJ and VOUT. This causes that regulator to sink (nominally) 10.417mA current from R2.
    In order to be able to sink that current, the ADJ terminal of the LM117/LM317 needs to be connected to a voltage that is at least 4.56v less than the lower side of R2, since the LM117 drops at least 3v across itself when sinking current. If R2 has 10.417mA flowing through it, there will also be a 1.56v drop across R2. So, in order to be able to provide current regulation down to 0v, the ADJ terminal of the LM117 must be connected to a supply that is at least 4.56v less than the voltage at the junction of R1/R2.

    No.
    R2 will have a nominal 10.417mA current flowing through it due to that current being sunk by the LM117/LM317. Multiply that current times the resistance, and you will have your voltage drop.

    I do not understand what you are trying to say.

    This has been explained above.

    You will need a negative voltage source from somewhere, of at least -5v and able to sink at least -10.5mA current.
     
  14. Umer_Farooq

    Thread Starter Member

    Mar 6, 2010
    104
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    My mean that if i am using (Adjustable Current Regulator) then output current formula because vary R2 the vary the output current

    2. Attatch Image if i this cct use for current regulate than this cct is usefull or not. Current control in steps.
    Thanks a lot SgtWookie
     
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  15. SgtWookie

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    Jul 17, 2007
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    If you use the circuit that you attached in your last post, then the minimum output current will not be predictable, as you have no provision for sinking at least 10mA current to provide guaranteed regulation.

    In the previous circuit from post #13, R1 limits maximum current to approximately 5A.
    R2 then sets the output current.

    There is approximately 10.4mA flowing through R2.
     
  16. SgtWookie

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    Jul 17, 2007
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    2.5 Ω will only allow a nominal 0.5A current.

    Our OP needs to find out what is the Vref of their regulator, as I have already instructed them.

    Then R1 can be calculated from Vref.
     
  17. Umer_Farooq

    Thread Starter Member

    Mar 6, 2010
    104
    3
    Thanks a lot Sgtwookie
    Very very thanks a lot sgtwookie
    you are a knowledgeable man
    many problem solved this forum


    then thanks Alberto My 2W problem Solved
     
  18. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    God bless SgtWookie :p
     
  19. SgtWookie

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    Jul 17, 2007
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    Umer,

    Just to make certain that you understand what is going on in this circuit:

    [​IMG]

    The regulator attempts to keep Vref constant (the voltage difference between Vout and ADJ) by sourcing current from Vout.

    As I wrote before, Vref can be different for each regulator, as they are not all exactly the same. It is very typical for Vref to be in the range of 1.25v to 1.27v. However, if you wish for your regulator to be accurate, you must measure your individual regulator's Vref as I have already written.

    Then for R1, calculate:
    R1=Vref/5A
    The datasheet has R1=1.2/current, but that is only approximate. It does cover all Vref ranges from 1.2v to 1.3v, but the higher your actual Vref is, the less accurate the datasheet formula becomes.

    Power resistors will change their resistance somewhat when they are heated. If R1 is carrying 5A current, then power dissipation will be around 6.2 Watts, as has been previously mentioned. The standard "rule of thumb" is to use a resistor rated for at least twice the expected power dissipation for reliability. I suggest that you would be best served to use a resistor rated for 20W, or a pair (or more) of resistors in series and/or parallel which can dissipate that much power. Their combined resistance must equal Vref/5A.

    If you have plenty of power dissipation capacity, the resistors will be much more stable; they will not change in resistance as much under load.

    The 2nd part of the circuit is the LM117/LM317 regulator. This regulator sinks a constant current from the bottom of R2, which keeps the voltage drop across R2 constant. An LM117/LM317 must have a minimum 10mA current flow from it's output to provide guaranteed regulation. This is why they used a 120 Ohm resistor in the example schematic, as 1.2v/120 Ohms = 10mA. Since Vref is 1.2v to 1.3v, a value of 120 Ohms covers all possible values of Vref.

    If you wish for your circuit to have more predictable values, you should measure the Vref of your LM117/LM317 in the same way you measured the LM338 Vref. With the LM117/LM317, I suggest using a 100 Ohm resistor for the test; connecting it from Vout to ADJ, grounding ADJ, connecting 4.5v to 8v to Vin, and measure from ADJ to Vout to obtain Vref.

    Whatever voltage you measure for the LM117/LM317's Vref, multiply it by 100 and use that value for resistor R3; eg: 1.27v x 100 = 127 Ohms.
     
  20. Umer_Farooq

    Thread Starter Member

    Mar 6, 2010
    104
    3
    Thanks a lot SgtWookie
    You are a good knowledgeable man
     
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