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Designing a voltage divider

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data

    Using a 100-kOhm potentiometer and suitable external resistances, design a voltage divider circuit such that varying the wiper from end to end varies the gain over the range:
    a) 0 to 0.75 V/V
    b) 0.2 V/V to 1 V/V
    c) 0.1 V/V to 0.9 V/V

    2. Relevant equations

    Not Sure to be honest

    3. The attempt at a solution

    I really do not know where to start on this problem at all. I don't think it is as complicated as it seems to be but my professor is absolutely horrid at teaching, and I really have no idea to go about solving this. What equations are relevant here?
  2. jcsd
  3. Sep 18, 2011 #2


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  4. Sep 18, 2011 #3
    The book had this diagram drawn for the answer in the book along with the answers to parts a-c.

    [PLAIN]http://img16.imageshack.us/img16/5962/voltagedivider.png [Broken]

    a) R1 = 33.3 k Ohms, R2 = 0
    b) R1 = 0, R2 = 25 k Ohms
    c) R1 = R2 = 12.5 k Ohms

    And I looked at the voltage divider information on that site you linked but I'm still as lost as ever. For one it doesn't look like the picture above. I'm a little lost on how to find R1 and R2 in those equations when I have 2 unknowns and only one equation as well. And what does the 100 K Ohms fit into in those equations? I don't know perhaps I'm missing something here.
    Last edited by a moderator: May 5, 2017
  5. Sep 18, 2011 #4
    Vo/Vi = 1/[1+R1/R2] and I plugged in 3/4 V/V for Vo/Vi and when I solved for the ratio, I got R2/R1 = 3. The answer to part a is R1 = 33.3 k Ohms, but I get R2 = 33.3 k Ohms, since R2 was in the numerator when I solved for the ratio. So basically I just take the 100 K Ohm Potentiometer and divide by the ratio and that gives me the value for the resistor in the numerator? I think I'm on the right track but if someone could point out a mistake or something, it'd make this go easier as I'm just winging it at the moment.
  6. Sep 18, 2011 #5


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    The 100K resistance is given by the problem. You are perhaps missing the understanding of how a potentiometer works.

    Let's look at question (a) first. Imagine the slider being at the top position, i.e, between R1 and the 100K resistor. Can you find V0 in terms of Vi? Now imagine the slider being all the way to the other end so that R1 and the 100K are on one side of it and R2 on the other. Again can you find V0 in terms of Vi? Note that you have two conditions that should give two equations from which you can find R1 and R2.

    Similar reasoning applies to questions (b) and (c).
  7. Sep 18, 2011 #6


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    You need to find Vo/Vi at an "intermediate" position of the slider, i.e. when the resistance is R1+r on one side and (100K-r)+R2 on the other. Then evaluate Vo/Vi when r = 0 (slider all the way up) and r=100K (slider all the way down) and set each result equal to what you want it to be. This will clarify to you what you are doing.
  8. Sep 18, 2011 #7
    I'm still a bit confused here, but I came across this formula. Vo = [kR/R]Vi or Vo = k*Vi. But then according to this, k is that ratio Vo/Vi. I don't even know anymore this book doesn't have k in any of the main equations, that's one reason I'm confused. I see that the position between the wiper and the bottom lead is kR, where k is Vo/Vi and R is the total resistance of that potentiometer. But my problem is understanding where this is useful at in the equation Vo/Vi = 1/[1+R1/R2]. Perhaps I'm understanding you wrong, if I am I apologize, I'm just a little confused at the lack of information to go on in my notes and in the book. Perhaps if someone could write one equation for part a, I might be able to write the other one for part a and go from there.
    Last edited: Sep 18, 2011
  9. Sep 19, 2011 #8
    [PLAIN]http://img705.imageshack.us/img705/5962/voltagedivider.png [Broken]

    Does this drawing make the problem clearer?
    Last edited by a moderator: May 5, 2017
  10. Sep 19, 2011 #9


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    This does not match your original drawing. I believe the drawing below is the one you should work from.

    Attached Files:

    Last edited: Sep 19, 2011
  11. Sep 19, 2011 #10
    too late now I guess, but the majority of the class had no idea how to work this problem. It truly is worded oddly, plus the fact our teacher did no examples on it at all.
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