Negative input resistance - OPAMP circuit

Discussion in 'Homework Help' started by xxxyyyba, Oct 2, 2015.

  1. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    Hi.
    I study OPAMP circuit with negative input resistance. Here is schematic:

    1.png

    To find Rin, I connect test generator:

    2.png

    Rin would be Vtest/Itest.
    I get Rin = - R1 * R3 / R2 = -10 Ohm. (for ideal OPAMP)
    For real OPAMP (LT1097CN8), input impendance will be very close to ideal one I got:

    3.png

    Rin = 0.01V / (-1.001mA) = - 9.99 Ohm.

    My question is, how can I use this circuit? For example, If I have this circuit:

    4.png

    How should I connect OPAMP circuit to this circuit to cancel source internal resistance R_internal? Current in that case would be:
    I = V1 / (R_internal + R_load - Rin) = V1 / R_load.
     
    Last edited: Oct 2, 2015
  2. Jony130

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  3. DickCappels

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