1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What kind of op-amp is that?

  1. Dec 2, 2011 #1

    Femme_physics

    User Avatar
    Gold Member

    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Dec 2, 2011 #2
    I have never seen anything like this !!! Are you certain you have the correct diagram.
    There are applications where Vout is connected back to the - input (this gives a gain of x1) but this circuit seems to me to show some inconsistencies.
    I cannot see how Va can be equal to Vb.
    I look forward to see the response of others
     
  4. Dec 2, 2011 #3

    Femme_physics

    User Avatar
    Gold Member

    Oh, that reminds me! I forgot to ask my teacher about what you confirmed was correct! I have a class with him sunday, I will definitely remember.

    Yes, I'm looking forward to other replies as well-- even though you seem to have great knowledge at the field yourself.
     
  5. Dec 2, 2011 #4
    The first thing I did was re-draw your circuit.... I kept R1, R3 and R4 exactly as they are but I then drew R2 just above R3 connecting to the - input.
    The connection from Vout to point b remains as it is.
    This just did not 'look right'....
    Thank you for your comment:blushing:
    I love op amps.... they look very complicated but the principles are straight forward and I like sorting them out.
     
  6. Dec 2, 2011 #5

    gneill

    User Avatar

    Staff: Mentor

    It's a tricky divide by 4 circuit. That is, Vout = Vin/4. To analyze, recognize that the feedback from Vout to b will make a = b = Vout. Then forget about the op-amp and concentrate on the resistor network, putting in a controlled source "Vout" to stand in for the the op-amp:

    attachment.php?attachmentid=41442&stc=1&d=1322857270.jpg

    All resistors are 5KΩ.
     

    Attached Files:

  7. Dec 2, 2011 #6
    Thanks gneill I tried something like that to make Vout = Va =Vb and got tied in Knots !!!!
    It is not really a normal question about op amps.... difficult to see this as a practical application.
     
  8. Dec 2, 2011 #7

    gneill

    User Avatar

    Staff: Mentor

    Yeah, you might say that it's more of an academic exercise :smile:

    As for its practicality, I suppose one would have to check to see how accurate the division operation is for the given component tolerances as compared to other configurations.
     
    Last edited: Dec 2, 2011
  9. Dec 2, 2011 #8

    Femme_physics

    User Avatar
    Gold Member

    I forgot to add that the supply voltages are +-10

    Is there a name for this op-amp, gneill?

    REally? That's it? That's the answer?
     
  10. Dec 2, 2011 #9
    I can think of a name for it (if it doesn't already have one):devil::devil::devil:
     
  11. Dec 2, 2011 #10

    gneill

    User Avatar

    Staff: Mentor

    I don't know if there's any specific name, but it certainly falls under the general heading of non-inverting amplifier. I'll admit that the positive feedback loop in addition to the negative feedback loop makes it a tad unusual. And yes, the amplification factor is 1/4. You should see if you can derive it from the hints given.
     
  12. Dec 2, 2011 #11

    Femme_physics

    User Avatar
    Gold Member

    Can you elaborate more please? I can't see the logic of just dividing by 4.....is it purely based on KVL and KCL?
     
  13. Dec 2, 2011 #12

    gneill

    User Avatar

    Staff: Mentor

    Yes, KVL, KCL, the usual analysis tools. If you can see how the diagram I gave arises (here it is again):

    attachment.php?attachmentid=41442&stc=1&d=1322857270.jpg

    Then you can start from there. Suppose that all resistors are R. Apply nodal analysis at the node labelled Vx. You'll get an equation involving Vin, Vout, and Vx. Make Vx vanish by noting that i3 = Vx/(2R), and that Vout = i3*R. Re-arrange what's left to find Vout in terms of Vin.
     
  14. Dec 2, 2011 #13

    gneill

    User Avatar

    Staff: Mentor

    Just in case it's not obvious where the "reduced" circuit came from, here it is again but with the op-amp shown in place. Note that all I really did was to make the op-amp into a controlled source.

    attachment.php?attachmentid=41445&stc=1&d=1322869245.jpg

    If you think about it, the op-amp is in a voltage follower configuration with the output fed back directly to its "-" input. So it really is a controlled source: it's output is 1x the voltage across the resistor at its + input.
     

    Attached Files:

    • Fig1.jpg
      Fig1.jpg
      File size:
      11.8 KB
      Views:
      211
  15. Dec 3, 2011 #14

    Femme_physics

    User Avatar
    Gold Member

    Last edited by a moderator: May 5, 2017
  16. Dec 3, 2011 #15

    I like Serena

    User Avatar
    Homework Helper

    Yes, it is like that.

    But your Sum Va is not quite right.
    You're forgetting the voltage of the controlled power supply, which is the same as Vout.
    I'm afraid gneill left that out of his (otherwise neat and insightful) drawing.
     
  17. Dec 3, 2011 #16

    gneill

    User Avatar

    Staff: Mentor

    :confused: It's the diamond shape at the bottom left, labelled with the carefully concealed clue "Vout".
     
  18. Dec 3, 2011 #17

    I like Serena

    User Avatar
    Homework Helper

    Sorry!
    Yes it is there.

    @Fp: so you should include Vout in your equation.
     
  19. Dec 3, 2011 #18

    gneill

    User Avatar

    Staff: Mentor

    @FP: also note that your path V(b) should include the voltage drop across R1 due to I1.
     
  20. Dec 3, 2011 #19

    Femme_physics

    User Avatar
    Gold Member

    :eek:
    OK, rather surprising! but ok..
    :shy:
    -------------------------------
    So I3R4 = Vout?

    Ok, think I got it.

    http://img72.imageshack.us/img72/6636/pitaron.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  21. Dec 3, 2011 #20

    I like Serena

    User Avatar
    Homework Helper

    How so?


    Yes.


    Looks good!
    Don't forget to use Vout=I3R4.
     
    Last edited by a moderator: May 5, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What kind of op-amp is that?
  1. Op amp trouble! (Replies: 10)

  2. Thermistor with op amp (Replies: 9)

  3. Basic Op-Amp Problem. (Replies: 1)

  4. Op-amp Problem (Replies: 4)

Loading...