Gain of op-amp amplifier with potentiometer

Discussion in 'Homework Help' started by ham3388, Jun 10, 2016.

  1. ham3388

    Thread Starter Member

    Jul 3, 2012
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    Hi dears
    Below is a worked example with solution from the university material plus an assignment asking for solution. I did solved it but I am not sure about my solution. Could you please go through it and give some kind of help and support to understand it or give some means to solve it if it is incorrectly been solved.
     
  2. bertus

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    Apr 5, 2008
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    Hello,

    @shteii01 , you forgot to take the 100 K resistor (R3 in the schematic) between the output of the opamp and the 1 K resistor to ground into account.

    Bertus
     
  3. DGElder

    Member

    Apr 3, 2016
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    Ham, I think you may be confused about how a potentiometer works- Rp is not R3: Look up potentiometers in Wiki.

    It is the same circuit as the example with variations only in R3 and R4 and you are given the gain formula in the example. All you have to do is figure out R3 and R4 which depend on the pot wiper position, "x", and then plug the numbers back into the gain formula. For example: Part b) R4 = x*10K + 1K = 8k and R3 = (1-x)*10K = 3K.
     
    Last edited: Jun 11, 2016
  4. shteii01

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    Feb 19, 2010
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    I modeled Part A. Part A has the pot set to 0 Ohm.
     
  5. DGElder

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    Apr 3, 2016
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    shteii01, the model is not correct for any of the parts. Look up potentiometers in Wiki to see how they work.
     
  6. shteii01

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    Feb 19, 2010
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    Draw the circuits for each part of the problem and post them.
     
  7. ham3388

    Thread Starter Member

    Jul 3, 2012
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    Thanks friends. , I will go through the Wikipedia then I will get back to you guys. :)
     
  8. shteii01

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    Feb 19, 2010
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    Ok. I see what I did wrong earlier. The pot resistance does not just go away. It moves as the junction moves. When x=0, the whole resistance of the pot is "above" the junction, or, as bertus said, the 10 kOhm resistor is between output of op amp and the junction of the 100 kOhm and 1 kOhm resistors.


    x zero.jpg
     
    Last edited: Jun 11, 2016
  9. ham3388

    Thread Starter Member

    Jul 3, 2012
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    I went through the Wikipedia and other resources and recalculated it based on those resources plus guidance from Mr. DGELer and Mr SHTEII01.
    Could you please look on the below attachment and give your comments.
     
  10. The Electrician

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    Oct 9, 2007
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    I think you have the x=0 and x=1 cases reversed. x is the fraction of the potentiometer measured from the bottom.
     
  11. dannyf

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    Sep 13, 2015
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    the inverting end is at round. So current flows from V1 to the inverting end and then R2 and then to Vx. That current is I1 = V1 / R1.

    So Vx = - I1 * R2 = - V1 / R1 * R2.

    The current flows from ground to Vx through R4: I4 = -Vx / R4.

    I1 + I4 flow through Rp to Vout.

    Vout = Vx - (I1 + I4) * Rp = Vx - (V1 / R1 - Vx / R4) * Rp = (-V1 / R1 * R2) - (V1 / R1 + V1 / R1 * R2 / R4) * Rp = (V1 / R1) * (-R2 - Rp - R2 * Rp / R4).

    In this case, it is 11.1v.

    It can be re-expressed to be more intuitive.
     
  12. DGElder

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    Apr 3, 2016
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    Ham, only in part b) are the R3 and R4 values correct. If x=0 that means the wiper is at the bottom of the resistor (zero distance from the bottom), so the entire 10K is above the wiper. When X=1 the wiper is at the top so that the entire 10K is below the wiper. My formula for R3 and R4 applies for all wiper positions. Post #8 correctly represents the pots contribution for x=0.
     
    Last edited: Jun 11, 2016
  13. Jony130

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  14. ham3388

    Thread Starter Member

    Jul 3, 2012
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    Thanks friends. ..I think I got it
    Cheers
     
  15. dannyf

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    Sep 13, 2015
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    another approach would be to think of it from Vout's perspective: Rp and R4 are in parallel. So the feedback resistance is R2 + Rp//R4.
     
  16. The Electrician

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    Are you saying that the feedback resistance is R2 + Rp||R4, therefore the gain is -(R2 + Rp||R4)/R1? Is this for the case x = 0? It isn't the same as the correct expression you gave in post #11: (-R2 - Rp - R2 * Rp / R4)/R1
     
  17. DGElder

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    Apr 3, 2016
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    The text's gain derivation was a bit difficult to follow and cookbookish, but I got the same answer, -1020, deriving from basic ideal op amp principles - shown in attached image.

    For the potentiometer version I got:
    x=0, G = -111
    x=0.7, G = -14
    x= 1.0, G = -10
     
    Last edited: Jun 12, 2016
  18. ham3388

    Thread Starter Member

    Jul 3, 2012
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    For the potentiometer version I also got

    x=0, G = -111
    x=0.7, G = -14
    x= 1.0, G = -10
     
  19. dannyf

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    Sep 13, 2015
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    Think from Vout's perspective. Vout + Rp (serial) can be viewed as a current source of Vout/Rp + Rp (parallel).

    Rp in parallel + R4 in parallel -> Rp//R4 in parallel, with a current source of Vout/Rp -> a voltage source of Vout * (Rp//R4) / Rp + Rp//R4 in serial.

    The rest is similar to a typical inverting amplifier.

    This approach is used to form large feedback resistors.

    A good approach for a voltage amplifier. A terrible approach for a transimpedance amplifier.
     
  20. dannyf

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    Sep 13, 2015
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    if you want to work out a generic form with the wiper, just work the wiper position into Rp and R4.
     
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