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Pspice inverting opamp input resistance

  1. Jan 3, 2012 #1
    I am working with Pspice student verison 9.1. Is there any simply way i can find out the input and output resistance in pspice?

    400px-Simple_design_daisy_inv_ampl_1000.jpg
    basiclly my opamp is same as this without Rg resistor

    Thank you
     
  2. jcsd
  3. Jan 3, 2012 #2
    I don't use pspice, but you can look at the circuit and immediately see the input impedance is Ri and the output impedance is zero.
     
  4. Jan 4, 2012 #3
    sorry.. i mean the Rin in opamp
     
  5. Jan 4, 2012 #4
    someone said it can find in output file...could anyone tell me how can i do it?
     
    Last edited: Jan 4, 2012
  6. Jan 4, 2012 #5
    What you can do is open up the library with the OrCAD Model Editor (a separate program) and find the part. I found something like this.

    Code (Text):
    *-----------------------------------------------------------------------------
    * connections:   non-inverting input
    *                |  inverting input
    *                |  |  positive power supply
    *                |  |  |  negative power supply
    *                |  |  |  |  output
    *                |  |  |  |  |
    .subckt uA741    1 2 3 4 5
    *
      c1   11 12 8.661E-12
      c2    6  7 30.00E-12
      dc    5 53 dx
      de   54  5 dx
      dlp  90 91 dx
      dln  92 90 dx
      dp    4  3 dx
      egnd 99  0 poly(2) (3,0) (4,0) 0 .5 .5
      fb    7 99 poly(5) vb vc ve vlp vln 0 10.61E6 -10E6 10E6 10E6 -10E6
      ga    6  0 11 12 188.5E-6
      gcm   0  6 10 99 5.961E-9
      iee  10  4 dc 15.16E-6
      hlim 90  0 vlim 1K
      q1   11  2 13 qx
      q2   12  1 14 qx
      r2    6  9 100.0E3
      rc1   3 11 5.305E3
      rc2   3 12 5.305E3
      re1  13 10 1.836E3
      re2  14 10 1.836E3
      ree  10 99 13.19E6
      ro1   8  5 50
      ro2   7 99 100
      rp    3  4 18.16E3
      vb    9  0 dc 0
      vc    3 53 dc 1
      ve   54  4 dc 1
      vlim  7  8 dc 0
      vlp  91  0 dc 40
      vln   0 92 dc 40
    .model dx D(Is=800.0E-18 Rs=1)
    .model qx NPN(Is=800.0E-18 Bf=93.75)
    .ends
    Using this model, it may be possible to draw out the subcircuit and solve for input resistance.

    Alternatively, you could attempt to run a simulation and divide input voltage by input current.
     
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