Discussion in 'General Electronics Chat' started by 1-3-2-4, Nov 25, 2013.

  1. 1-3-2-4

    Thread Starter Active Member

    Dec 26, 2008
    I have a 18V input and I need a 12V output, in the datasheet they list this

    VO= 1.25 V (1+R2/R1)+I ADJ R2

    I was wondering if a 100K pot on the adjustment pin was enough range to get to 12V

    So if I do something like 100 ohm on R1 and 100k on R2 I should have around 12V? 860 ohm is .86K right?
  2. wayneh


    Sep 9, 2010
    Use a spreadsheet and plot Vout against knob position (percentage). That's how I do it. Then, you can set your target voltage at ~50% of the pot.

    You might find these links about using potentiometers useful.

    tailoring potentiometers.pdf

    This and this, "The Secret Life of Pots", too.
    PackratKing likes this.
  3. Dodgydave

    AAC Fanatic!

    Jun 22, 2012
    I would use 100 ohms for R1, and a 1k pot for R2,that will give you upto 13.25v out.
    PackratKing and 1-3-2-4 like this.
  4. 1-3-2-4

    Thread Starter Active Member

    Dec 26, 2008
    Yeah I put a 100 ohm for R1 don't have any 1k pots but will 5K do? I can't see why not.
  5. PackratKing

    Well-Known Member

    Jul 13, 2008
    Google " Pot shunt " ... It is possible to reduce a 5K pot to some lesser value, at a slight-to-severe loss of linearity... depending on how far afield you go with the parallel resistor.. bear in mind that [ check on this ] you place the parallel resistor on the pots element -- Not the wiper... and draw your voltage off the wiper as you normally would...

    That based on the Ohms Law that states two 1000Ω resistors in parallel will net approx 500 Ω\

    Juggling, is an art that is worth learning :D
    Last edited: Nov 25, 2013