Negative Impedance Converter and its Sensor Application

Discussion in 'Homework Help' started by mjakov, Oct 16, 2014.

  1. mjakov

    Thread Starter New Member

    Feb 13, 2014

    How would you find the v-i relationship of the ideal op-amp circuit below for the interval of linear operation ?


    Let  V_{out} be the voltage between  R_3 and  R_2 . Then:
     V_{out} = v - R_3 i
    Because of  V^+ = V^- and KCL:
     V_{out} - (i + i_0) R_2 = v
     i_0 + i = \frac{v}{R_1}
    Solving this system of equations we get:
     v = - \frac{R_3}{R_2} R_1 i

    However, according to the book the correct solution should be:
     v = - \frac{R_2}{R_3} R_1 i

    Which of the two solutions is the correct one?
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    If Vin = 1V and R2 = 10K ; R3 = 1K; R1 = 1K.
    Vp = Vn = 1V so
    IR1 = 1V/1K = 1mA, this current will flow through R2. So voltage at op-amp output is Vout = Vin/R1 *(R1 + R2) = 11V
    From this we can find Iin = (Vin - Vout)/R3 = (Vin - Vin/R1 *(R1 + R2))/R3 = -10mA
    And after we solve this Iin = (Vin - Vin/R1*(R1 + R2))/R3 for Vin we have
    Vin = - R3/R2 * R1*Iin
    Vin = - 1K/10K * 1K*-10mA = -0.1*-10V = 1V
  3. The Electrician

    AAC Fanatic!

    Oct 9, 2007
  4. WBahn


    Mar 31, 2012
    Try solving from as different a perspective as you can and see if you get the same answer. For instance, because the two op-amp terminals are the same, you know that whatever voltage your drop across R2 you have to get back across R3. So that means that (I)R3 = - (I+Io)R2. See if you can use that to back your way into a solution.
  5. mjakov

    Thread Starter New Member

    Feb 13, 2014
    Thank you all for the helpful replies!

    The following circuit is supposed to contain the application of the negative impedance converter for a sensor. Element  v_s is the sensor voltage source,  r is the internal resistance of the sensor and  R is the load resistance. The goal is to make the current  i independent of the load resistance. It should be proportional to the voltage source only.


    So, here are the equations:
     v_A = v_C
    Using the node voltage method with nodes A and C:
     \frac{v_s - v_A}{r} - \frac{v_A}{R} - \frac{v_A - v_B}{R_3} = 0
     - \frac{v_C}{R_1} + \frac{v_B - v_C}{R_2} = 0
    Solving this we get:
     v_A = \frac{-v_s}{r \frac{R_2}{R_1 R_3} - 1 - \frac{r}{R}}
    If we choose  R_1, R_2, R_3 so that
     \frac{R_2}{R_1 R_3} = \frac{1}{R} , then:
     v_A = v_s

    So this would mean that i does not depend on the internal resistance r of the sensor. But, it does not seem that i is independent of the load R, because
     i = \frac{v_A}{R} = \frac{v_s}{R}  .
    Would you agree with this conclusion and is there some other way to make i independent of R ?

    Could you recommend some good information source as a reference or tutorial on these kinds of circuits? My textbook on the analysis of electric circuits sometimes omits details such as the names of the circuits like the negative impedance converter above. Could you please give some directions as to what part of electrical engineering does this subject matter belong to?
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
    Last edited: Oct 17, 2014