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Adjustable Voltage Regulator

Discussion in 'General Electronics Discussion' started by koie06, Feb 18, 2014.

  1. koie06

    koie06

    13
    0
    Feb 18, 2014
    Hi everybody,

    i just start playing around with electronic part. my 1st project is to make a variable voltage regulator.

    Part:-
    -LM317T
    -220ohm & 570ohm resistor
    -10uf Capacitor
    -0.1uf Capacitor
    -5k ohm potentiometer

    my problem is why the voltage range between 1.2v to 5.4v and the potentiometer is too sensitive(1/4turn already reach Max Vout). am i doing something wrong?

    Vin = 7.4v
    min Vout = 1.21v
    max Vout = 5.47v

    how can i make the Max Vout around 6v and make the potentiometer accurate?

    Hope you guys can help me to solve this problem.

    Thanks.
     
    Last edited by a moderator: Feb 18, 2014
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    8,881
    1,751
    Nov 17, 2011
    Please show us a schematic of your circuit so we can see how the components are connected to each other.
     
  3. duke37

    duke37

    5,145
    693
    Jan 9, 2011
    Presumably the 220 ohm resistor is connected between the output and the sense terminal. These differ by 1.21V so the current will be 5.5mA.

    To increase the voltage output, a resistor has to be connected from the sense terminal to ground which passes 5.5mA so to get an exra 2V you will need 364 ohms.
    Obvously a 1k potentiometer is as big as you need.

    If you only have a 7.4V supply there may not be enough to get up to the voltage you want, depending on the current drawn
     
  4. koie06

    koie06

    13
    0
    Feb 18, 2014
    miss information.

    the correct one is:
    -LM317T (ADJ, Vout, Vin)
    -510ohm resistor (connect between ADJ & Vout)
    -0.1uf capasitor (connect between Vin & GND)
    -1uf capasitor (connect between Vout & GND)
    -5k Potentiometer (connect between ADJ & GND)

    i dont have my phone usb cable in hand. so, i hope it can help to describe the situation a bit.

    is there something i need to adjust? like the resistor, capasitor or potentiometer value?

    this schematic is same like mine...
    [​IMG]
     
    Last edited: Feb 18, 2014
  5. gorgon

    gorgon

    600
    22
    Jun 6, 2011
    From the datasheet the LM317 need a headroom of 1.5V and upwards depending on the current drawn. 5.4V @ 7.4V input is not too bad.
     
  6. koie06

    koie06

    13
    0
    Feb 18, 2014
    i see..so how about the potentiometer?..should i low the potentiometer resistance?
     
  7. duke37

    duke37

    5,145
    693
    Jan 9, 2011
    There is a difference from the diagram. The diagram shows R2 as 240 ohm, you have used 510 ohm. This may not be important but high value resistance here is reputed to lead to instability.

    To make the potentiometer less sensitive, use a lower value.
     
  8. Arouse1973

    Arouse1973 Adam

    5,162
    1,077
    Dec 18, 2013
    I think you have your formula wrong. You need to swap the R1 and R2 designators around if your going to use the standard formula shown. so for 6V out it's this. Vout= 1.25*(1+2000R/540R)+(50uA*2000R) = 5.98V. So it's your R1/R2 not R2/R1 and also Iadj*(your R1) not R2.

    Adam
     
  9. koie06

    koie06

    13
    0
    Feb 18, 2014
    if im not mistaken I already try to swap from 510ohm to 220ohm resistor but it seem no different in Vout and potentiometer sensitivity.

    last night I try to stack 3pcs 3.7v battery. so, total Vin is 11.1v. the result is the potentiometer can adjust more than 1/4turn. now it increase to 1/2turn. how can I fully utilize the potentiometer with 7.4v Vin.

    need suggestion on:
    -ohm for R1(potentiometer)
    -ohm for R2(resistor)
    -value for C1 & C2

    maybe u all can suggest other schematic that work well..

    Thanks
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,108
    2,654
    Jan 21, 2010
    The schematic is fine, you just need to understand the role of the variable resistor.

    This sets the output voltage according to the formula given above. However, the output voltage can never be less than 1.25v (with a resistance of zero) and your input voltage less about 2 V.

    In your case, let's say the input voltage is 11.1V. The maximum range of adjustment is 1.25V to 9.1V.

    So you want to determine the value of resistance required to get 9V and find a potentiometer with a total resistance close to this.

    If the voltage is given by 1.25*(1+ Rv/Rf) (Rv is the variable resistor and Rf is the fixed resistor)

    then if we want 9V, 9 = 1.25*(1 + Rv/240)
    => (9/1.25 - 1) * 240 = Rv
    => Rv = 1488 ohms

    If you could find a 1.5k pot, that would be ideal.

    Incidentally, 1488 ohms is about 1/4 of 5000 and that matches well with your observation that 1/4 of a turn provided all of the adjustment.

    A 1.5k pot might be hard to find, but a 1k pot will not be. With a 1k pot, reduce the value of the fixed resistor to 150 ohms and you'll be pretty close to using it's full movement.

    You might like to verify what resistance is needed with 150 ohms as the fixed resistor to get 9V output, or alternatively what a 1k value will give you.
     
  11. koie06

    koie06

    13
    0
    Feb 18, 2014
    Thanks Steve.

    let say i I want max Vout 5v and fully utilize the potentiometer.

    I will need to replace:
    Rv = 500ohm
    Rf = 160ohm

    am I right?.
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,108
    2,654
    Jan 21, 2010
    Yep, that would be very close (certainly close enough)
     
  13. koie06

    koie06

    13
    0
    Feb 18, 2014
    ok thanks.

    will update later.
     
  14. Arouse1973

    Arouse1973 Adam

    5,162
    1,077
    Dec 18, 2013
    No read my post again. The labels R1 and R2 are wrong to your formula.
     
  15. Arouse1973

    Arouse1973 Adam

    5,162
    1,077
    Dec 18, 2013
    Yes the circuit is fine but the formula is wrong to his component labels I think. Can you look again. It's a give away when Iadj is apparently flowing in his top resistor.

    Thanks
    Adam
     
  16. Arouse1973

    Arouse1973 Adam

    5,162
    1,077
    Dec 18, 2013
    I thought you wanted 6V not 9V. What voltage do you want out?

    Thanks
    Adam
     
  17. koie06

    koie06

    13
    0
    Feb 18, 2014
    yes...actually i want 6v for Vout...since i was running on 7.4v for Vin and i got arond 2v drop voltage with LM317T..i think its impossible to get consistance 6v unless i raise up Vin value...

    in the schematic R1 refer to Potentiometer? and R2 is fixed resistor?

    please enlighten me..im new in electronic and this is my 1st project..
     
  18. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,108
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    Jan 21, 2010
    The R1 vs R2 confusion is why I used Rf and Rv. However you label these resistors, they have to be the right way around in the formulas or you can't calculate anything useful.
     
  19. Arouse1973

    Arouse1973 Adam

    5,162
    1,077
    Dec 18, 2013
    I think I worked this out for you earlier. 2000R and 540R = 5.98V
    Adam
     
  20. koie06

    koie06

    13
    0
    Feb 18, 2014
    ok thanks...will try 2000R/540R also...
     
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