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Op-amp with potentiometer

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Here's the network:
    CCI11102011_00000.jpg


    R1=100
    The gain ranges from -1V/V ---> -10V/V and the potentiometer R4=10.
    Is R4 and R3 in parallel with each other? I cant seem to get the right answer....

    Moreover, this is what was done:

    -(R2/R1)(1+(R4/R2)+(R4/R3))=-10

    Can someone please explain this. I know what the -10 means, I just do not really understand what was done here, whether it be a voltage divider, current divider or neither.

    Regards, D.
     
  2. jcsd
  3. Oct 11, 2011 #2
    I don't know, but I'm subscribing to this threat because of that reason.
     
  4. Oct 11, 2011 #3

    gneill

    User Avatar

    Staff: Mentor

    Neither and both :smile:

    An expression for the maximum gain was derived and set equal to the given value, -10.

    In order to find that expression for the gain, consider dividing R4 into two pieces, x*R4 and (1-x)*R4, where 0≤x≤1. A KCL node equation at the "wiper" of the potentiometer (which is now where the two new resistances meet) would be a good start.

    Once you've got the full expression for the gain, plug in x=0 for one extreme and x=1 for the other. It should be simple to pick out which of the two resulting expressions represents the maximum gain.
     
  5. Oct 11, 2011 #4
    Okay, I'll show you my work. I cannot seem to get it. Thanks in advance.

    I let the bottom half of the wiper symbolize xR4, and the top half (1-x)R4. The current passing through R2 and entering the node at the wiper is vi/100.

    I can then do a current divider at the node to determine how much of this current will enter the bottom half of the wiper.

    (vi/100)((1/((1-x)R4+R3))+(1/xR4))^(-1)(1/xR4)

    Now. multiplying this by xR4 regards the voltage drop for the bottom part of the wiper.

    vo= -vi(R2/100)-((vi*R4*x)/100)(((1-x)R4+R3)/(xR4+(1-x)R4+R3))

    This for sure does not get me the right answer.
     
  6. Oct 11, 2011 #5

    gneill

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    Staff: Mentor

    Can you justify using a current divider formula here? After all, one end of the divider is terminated at ground potential while the other is driven by Vo.

    Here's your circuit redrawn to indicate the parts you've described. I suggest that you write a KCL node equation for the node labeled V1. Note that V1 is also determined by the input current passing through R2.

    attachment.php?attachmentid=39899&stc=1&d=1318386683.gif
     

    Attached Files:

  7. Oct 11, 2011 #6
    Hmm, I managed to determine R2 but I am assuming it was via fluke.. Here's the equations I set up.

    i1=i2+i3, where i2 going to towards vo through xR4, and i3 goes towards ground potential.

    i1=i2+i3
    (vi/100)=((v1-vo)/xR4) + v1/((1-x)R4+R3)

    where v1 = i1*R2=(vi*R2)/100

    I'm pretty sure it is at this point I have already messed something up as when I solve for R3 I get a negative resistance..

    Thanks
     
  8. Oct 11, 2011 #7
    Nevermind, I totally did have it right, I just messed up a negative sign. v1=-(vi*R2)/100

    Thanks a lot.
     
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