By using LM317T regulating IC i want to get an regulated 9volt DC output

Discussion in 'The Projects Forum' started by Sivalie kohkoh Samba, Nov 9, 2017.

  1. Sivalie kohkoh Samba

    Thread Starter New Member

    Oct 13, 2017
    4
    0
    By using regulating IC LM317 I had wanted to get an output of 9volts Dc by inputting 19.54DC as shown by the configuration below

    upload_2017-11-9_4-1-45.png

    This configuration gave an output of 7.54 volts DC. But yesterday I changed the electrolytic capacitor which was 10µF 25v to 3300µF 35v and I get 8.61 volts DC output, except that the LED died few seconds during the process. I want to know what component to change; the component value or the component it self to get an output of 9volts. And also if my circuit diagram is correct for this job. Thank you
     
  2. Bordodynov

    Well-Known Member

    May 20, 2015
    1,819
    548
    You need current and not voltage. On the Internet and in this forum you will find current generators on the LM317. The main thing is to know what current is needed for the normal operation of the LED. I know a couple of cases when the LEDs are connected directly to the voltage source. In one case, a $ 30 LED burned out.
     
  3. AlbertHall

    AAC Fanatic!

    Jun 4, 2014
    6,207
    1,445
    I think the circuit is designed to supply 9V to some other circuit and the LED is just to indicate that it is on.
    If this is the case then the problem is that you do not have a resistor in series with the LED to limit the current, about 470Ω would do do the job.
    Without this resistor the LED will draw a large current and drag down the output voltage as you saw. Also this will kill the LED as you also saw/
     
    ebeowulf17 likes this.
  4. AlbertHall

    AAC Fanatic!

    Jun 4, 2014
    6,207
    1,445
    I calculate the expected output voltage as 8.3V so you may want to increase the 1.3K to 1.4K.
     
  5. crutschow

    Expert

    Mar 14, 2008
    19,548
    5,449
    The closest R2 1% resistor value to use for 9V output is 1.37kΩ.
     
  6. crutschow

    Expert

    Mar 14, 2008
    19,548
    5,449
    Why?
    10μF is quite adequate for most applications.
     
  7. alfonsoM

    Member

    Nov 8, 2017
    41
    8
    In addition to what that was advised.
    Changing 10uF to 3300uF increases the voltage because you have very high ripple on the 19V input. Leave C2 = 10uF and put the 3300uF across C1, it will give you stable output.
     
  8. Sivalie kohkoh Samba

    Thread Starter New Member

    Oct 13, 2017
    4
    0
    using the 10uF gave an output of 7.54v and i needed an out of 9v

    ok, i will do just that thank you

    ok, thank you all for your help, i will get on to you again:)
     
    Last edited by a moderator: Nov 10, 2017
  9. crutschow

    Expert

    Mar 14, 2008
    19,548
    5,449
    The value of that capacitor is unrelated to the output voltage.
     
  10. AlbertHall

    AAC Fanatic!

    Jun 4, 2014
    6,207
    1,445
    That is true unless the input has a lot of ripple on it. If the value of the capacitor changes the output voltage significantly then it seems there is a lot of ripple.
     
    Sivalie kohkoh Samba likes this.
  11. alfonsoM

    Member

    Nov 8, 2017
    41
    8
    The symptoms of his problems indicate that he has no smoothing cap after the bridge rectifier.
     
    Sivalie kohkoh Samba likes this.
  12. Standisher

    Member

    Jan 16, 2015
    119
    16
    The figure of 19.45 VDC quoted leads me to believe that the OP is deriving input voltage from a laptop power supply. So, probably quite a lot of ripple.
     
    Sivalie kohkoh Samba likes this.
  13. Sivalie kohkoh Samba

    Thread Starter New Member

    Oct 13, 2017
    4
    0
    Hello guys, i have finally got a result which i would like to share.
    upload_2017-11-15_14-27-24.png
    instead of using the 220ohm resistor, i used 200ohm, but i could only lay hand on 100ohm resistor so selected two resistor of 100ohm each and connected them in series and i also change he C1 (0.1uF) to 3300uF 35v capacitor and C2 remain the 10uF and i connect the LED series with 470ohm resistor. So my output result was 9.47v DC.
    Thank you all for your help
     
  14. Sivalie kohkoh Samba

    Thread Starter New Member

    Oct 13, 2017
    4
    0
    I may like to post the whole project about this LM317T but i dont know how to upload it.
     
  15. Audioguru

    Expert

    Dec 20, 2007
    10,397
    1,140
    Calculations from the 555 datasheet:
    1) 1.25V/200 ohms= 6.25mA.
    2) 6.25mA x 1300 ohms= 8.125V.
    3) 8.125V + 1.25V= 9.375V plus 0.065V for ADJ pin current= 9.44V but it could be a little higher or lower.

    Your schematic still wrongly shows an LED shorting the output of the supply. The LED must have a resistor in series with it.
     
Loading...