LM317 voltage regulator circuit

Discussion in 'Homework Help' started by notoriusjt2, Dec 15, 2010.

  1. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    [​IMG]



    allright. I pulled up the LM317 spec sheet and found the following


    [​IMG]

    so....here is the work I have based on this equation

    [​IMG]

    I am wayyyyy off base here. what am I doing wrong?
     
  2. t_n_k

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    Mar 6, 2009
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    The adjustment pin current is typically a max of ~5uA for the LM317. You appear to have Iadj = 0.01A - which is way off.
     
  3. Wendy

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    Mar 24, 2008
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    In R2 of your schematic (your diagram and the application notes have reversed designations) you have 100Ω. Since the method the LM317 uses is to maintain 1.25V between the output pin and adjust you should be able to figure current. The current through R2 is the same as through R1. From there it is gravy.
     
  4. Jony130

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    Feb 17, 2009
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    Iadj is 50uA typical, and 100uA=0.0001A max.

    But this circuit can be treat as a voltage divider.
    Since R2 = 100Ω and voltage across R2 is equal 1.25V we have
    IR1 = 1.25V/100Ω = 12.5mA
    And the current that is flow through R1 is equal
    IR2 = Iadj + IR1 = 12.6mA
    And we want Vo = 15V----> VR1 = 15V - 1.25V = 13.75V

    R1 = 13.75V/12.6mA = 1.09126984KΩ

    R1=R2*\frac{Vout-1.25V }{1.25V+I_{ADJ}*R2}

    Or

    R1=R2*(\frac{Vout }{1.25V}-1)
     
    Last edited: Dec 16, 2010
  5. Kermit2

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    Feb 5, 2010
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    The position of R2 and R1 are different in your problem and in the example you posted. Go back a redo the math and notice carefully how the fraction made by the resistor values is arranged in the example and how the Resistors in the problem are the reverse.
     
  6. t_n_k

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    Mar 6, 2009
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    Agreed - I misread the data sheet. Looked at the adjustment pin current change row rather than the actual value. I can only put it down to deteriorating vision.

    Apologies for the misinformation.
     
  7. notoriusjt2

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    Feb 4, 2010
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    what is the variable "Uwy" referring to?
     
  8. Jony130

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    Well in my country "Uwy" means exactly the same as Vout in english.
    And sorry for confusion
     
  9. notoriusjt2

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    Feb 4, 2010
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    no worries... what country is that?
     
  10. Jony130

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  11. notoriusjt2

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    thats a great explanation.

    I = 0.0125A

    R = 13.75/0.0125 = 1100

    which is basically the same answer that was derived by Jony130

    however, this answer was listed as wrong

    what are we doing wrong here?
     
  12. notoriusjt2

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    Feb 4, 2010
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    what is the dropout voltage? could that have anything to do with it?
     
  13. Jony130

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    Maybe answer is wrong

    As for the dropout voltage some times called headroom voltage read this
    http://en.wikipedia.org/wiki/Dropout_voltage
     
  14. notoriusjt2

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    Feb 4, 2010
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    per wikipedia

    "In electronics, the dropout voltage of a voltage regulator is the smallest possible difference between the input voltage and output voltage to remain inside the regulator's intended operating range."

    so if I understand correctly the dropout voltage has no effect on what the output voltage will be.
     
  15. notoriusjt2

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    Feb 4, 2010
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    I am to the point now where I am mindlessly staring at this problem... ugh
     
  16. bertus

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  17. notoriusjt2

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    Feb 4, 2010
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    but my schematic looks different from the one that is in that PDF

    for example

    [​IMG]

    what values should I use for Rwp and Rwn?
     
  18. bertus

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    Apr 5, 2008
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    Hello,

    Rwn and Rpn are the resistances of the wires from the regulator to the load.
    These will be dependend on the used wire, but can be dismissed when using small loads, as the values are mosttimes very small.

    Bertus
     
  19. notoriusjt2

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    Feb 4, 2010
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    allright so in the equation they are subtracting (IL*Rwp), which in reality is the voltage drop of the wire. since this loss is negligible we will disregard it.

    [​IMG]

    R2 should be -88

    also incorrect
     
  20. Jony130

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    Check your math.

    15 = 1.25 * ( 100 + R2)/100
    1500 = 1.25*(100 + R2)
    1500 = 125 + 1.25R2
    R2 = 1375/1.25 = 1100
     
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