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Homework Help: Electronic Circuits-Compute voltage gain for OP-amplifiers

  1. Dec 30, 2012 #1
    1. The problem statement, all variables and given/known data

    We got the following circuit:


    u_in is input
    u_ut is output

    3. The attempt at a solution

    I assumed that the upper amp has voltage u_in on the positive terminal.
    And that the lower amp has voltage 0 V on the positive terminal.

    then i calculated u_ut but the ratio then becomes: u_in/u_ut = 0

    Because the current from the upper amplifiers output terminal to the output terminal u_ut is


    which then gives us

    u_ut=u_in -(i*2R) = u_in -u_in

    the right answer should be -1


    help please.
  2. jcsd
  3. Dec 30, 2012 #2


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    Staff: Mentor

    I would analyze it by saying the voltage on the lower input is vx, and that on the upper input terminal is uin+vx, with vx being the common mode voltage.

    Otherwise you are losing all the information associated with the lower pair of R resistances.
  4. Dec 30, 2012 #3


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    Staff: Mentor

    First step: determine, in terms of the input, the voltage at the (+) input of the third op-amp.
  5. Dec 31, 2012 #4
    So this is how i did it:

    On the positive input for the 3rd amplifier we have v_x/2.

    And i suppose that the voltage on the output terminal of the upper amplifier is v_x+u_in

    The current through the resistor then becomes:

    [tex]i=\frac{\frac{v_x}{2}-(u_{in} + v_x)}{R}[/tex]

    Then the expression for the output u_o is:

    [tex]u_o=\frac{v_x}{2} - R*i = {\frac{v_x}{2}-\frac{v_x-2u_{in}-2v_x}{2R}*R}[/tex]

    after simplifying we obtain:


    u_o/u_in = -1 (if v_x = 0)

    Have i done correct?
  6. Jan 1, 2013 #5


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    Staff: Mentor

    If we use your above expression for i, then the "-" here should be "+"
  7. Jan 2, 2013 #6
    It is for the resistor after the output of the upper-most amplifier. I thought the current goes out of this amplifier and ends up at the output. So then it should be "-"

    But i re-calculated this task with my method again, and got a quite different result. That turned out to be correct. I calculated without adding v_x.

    The same current: i=(u_in/R)

    u_out=0-Ri= -R(u_in/R)=-u_in

    The gain then becomes -1.

    I think this is sufficient.
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