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Op Amp circuit analysis

  1. Mar 15, 2007 #1
    1. [​IMG]
    Assume that the op amp in the circuit shown is ideal.

    a.) Find the output voltage when the variable resistor Rx is set to 60k

    Assuming Ideal Op amp. All currents leaving nodes are negative.

    V+ = V- = 0 and In = Ip = 0

    -ip - iRx + iR2, so iR2 = ip + iRx, as ip = 0

    iR2 = iRx

    -in + iR3 -iR1, iR3 = in + iR1, as in = o

    iR3 = iR1

    iR2 = Vs/R2 , iRx = Vs/Rx (is Rx in parallel or series?).

    iR3 = Vo/R3, iR1 = VR1/R1

    so far is this correct? Is Rx in series or in parallel with R2? Im pretty sure its in parallel(as the current has 2 different paths to go.) but im not sure. The V+ = V- = 0 and i+ = i- = 0 assumptions are confusing me at the moment.

    Some of the other unknowns can be reduced by voltage dividers. My main problem is determining the Nodal Voltage formulas on these circuits. I had a go anyway.

    (Vn-Vo)/R1 + Vo/R3 + (Vs-Vp)/R2 + Vp/Rx = 0

    im guessing this is probably wrong. If anyone could give me some pointers in the right direction I would be very grateful. Im going to review Nodal analysis at some point tomorrow and have another go at the problem.
    Last edited: Mar 15, 2007
  2. jcsd
  3. Mar 16, 2007 #2


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    V+ =V- but not zero, to be determined
    iR2 = iRx as you have shown so iR2 not equal Vs/R2
    iR3 not equal to Vo/R3
    i+ = i- =0 is ok
    Last edited: Mar 16, 2007
  4. Mar 16, 2007 #3
    Ok, I was really tired when I made this post but I think I get your hints. Is Vo divided over R3 and R1 as they are both in series as a voltage divider? I see what you mean about iR2 = iRx but not equal to Vs/R2, would it be equal to Vs/Rx? (actually, ignore that....its a current divider between Rx and R2).
  5. Mar 16, 2007 #4
    This is more complicated than I need to learn for my exam, but I just see it as...If I know how to complete more complex problems my understanding can only get better
  6. Mar 16, 2007 #5


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    first question: yes
    second question: don't think so... if iR2=iRx then R2 and Rx are in series... do a KVL loop if still unsure but it is simple to work out iR2 or iRx
  7. Mar 16, 2007 #6


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    in that case, you better clear up those ideas, for otherwise you may confuse yourself.... this is a standard op-amp problem...
  8. Mar 16, 2007 #7
    ok so Rx and R2 is another voltage divider, actually I have just spotted where most of my confusion lies, so this has just become so much clearer, thanks. I do still need to review nodal analysis.
  9. Mar 28, 2007 #8
    hmm tried doing the q...do u have the answer to this q...? I got 4.8V but i didnt use any current thingy, just all on current divider rule
  10. Mar 28, 2007 #9
    i have 12.6
  11. Mar 28, 2007 #10
    hey, the answer is 4.8V i got the answer in the end and my understanding of op-amps is now pretty solid. Thanks Physforums
  12. Mar 28, 2007 #11
    i've rounded off

    [(Rx+R2)/Rx]*.4 = V+ or [(60000+15000)/60000]*.4 = .32

    [R3/(R3+R1)]*V+ = .29 or [63000/(63000+4500)]*.32 = .29

    [(R3/R1)*V+]+.29 = 4.77 or [(63000/4500)*.32]+.29 = Vo

    close enough to the 4.799 on the sim, don't know what i was thinking :tongue2:
    Last edited: Mar 28, 2007
  13. Aug 14, 2011 #12
    Hey,the answer is 4.8V. Remember in case of OP AMPs u need to apply KCL at every node.
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