Can You Calculate The Input Impedance?

Discussion in 'Math' started by MrAl, Dec 3, 2018.

  1. MrAl

    Thread Starter AAC Fanatic!

    Jun 17, 2014
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    Hello there,

    (See attachment)
    Here is a little challenge. The idea is to simply calculate the input impedance looking into Z1.
    This is with all the impedances (Z1, Z2, etc.) made into a pure resistance with value equal to the subscript.
    So Z1=1 Ohm, Z2=2 Ohms, Z3=3 Ohms, Z4=4 Ohms, Z5=5 Ohms.

    Hint:
    Zin (which is actually Rin with all resistors) is either equal to 8/15 Ohms or 15/8 Ohms. Or am i wrong?
    Warning: This might be more difficult than it looks like at first glance.
     
  2. Tesla23

    Senior Member

    May 10, 2009
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    Assuming that the op-amps stay linear (so V+ - V- = 0 for both):
    denoting the input voltage as V, then the voltage at the middle node must be V as must the voltage across Z5. So the current through Z5 is V/Z5, which is the current through Z4, so the current through Z3 must be -Z4*V/(Z3*Z5), this is the current through Z2, so the current through Z1 is Z2*Z4*V/(Z1*Z3*Z5). This is also the input current, so divide it into V to get the given input impedance.
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    At first glance it seems pretty simple -- but looks certainly can be deceiving. So let's give it a go.

    Assuming the opamps are operating in their linear region, we know that the voltage at each input of each amp is Vin. So that's the voltage at {Vin, P23, P45} (I'm assuming people can figure out my nomenclature for the nodes -- if not, P23 is the node that connects Z2 and Z3.

    The voltage across Z5 is therefore Vin and the current in it is

    I5 = Vin/Z5.

    This current must flow in Z4, from which we can find the voltage at P34 (which I'll call V34).

    V34 = V45 + I5·Z4
    V34 = Vin + (Vin/Z5)·Z4
    V34 = Vin·( 1 + (Z4/Z5))

    This results in a current (flowing to the left) through Z3 of

    I3 = (V34 - V23)/Z3
    I3 = (Vin·( 1 + (Z4/Z5)) - Vin)/Z3
    I3 = Vin·[Z4/(Z3Z5)]

    Since this same current has to flow into Z2, this results in a voltage at P12 of

    V12 = V23 - I3·Z2
    V12 = Vin - Vin·[Z4/(Z3Z5)]·Z2
    V12 = Vin[1 - (Z2Z4)/(Z3Z5)]

    The current flowing (to the right) in Z1 is therefore

    Iin = (Vin - V12)/Z1
    Iin = {Vin - Vin[1 - (Z2Z4)/(Z3Z5)]}/Z1
    Iin = Vin·[(Z2Z4)/Z1Z3Z5)]

    The input impedance is therefore

    Zin = Vin/In = (Z1·Z3·Z5)/(Z2·Z4)

    If the resistance is equal to the subscript (in ohms), the Zin is

    Zin = Vin/In = (1 Ω · 3 Ω · 5 Ω)/(2 Ω · 4 Ω) = (15/8) Ω

    An answer of (8/15) Ω is not possible, since it would be (8/15) siemens. Track your units instead of just tacking them onto the end!
     
  4. MrAl

    Thread Starter AAC Fanatic!

    Jun 17, 2014
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    Hi,

    Ok, yes, great guys :)

    Nice to see two different approaches too.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    It's the same approach, I was just more explicit on the expressions and descriptions while Tesla23 left out intermediate steps leaving it up to the reader to fill in the gaps. Each approach has pros and cons. Leaving out steps forces the reader to think more deeply to fill in those gaps, which aids in learning and understanding more so than the spoonfeeding approach I took. But it also takes more time and many readers won't make the effort to fill in the gaps or will fill them in incorrectly because they didn't think about things deeply enough. For instance, it would be easy for someone to conclude from Tesla23's description that all of the resistors have the same current flowing in them, even though nothing along those lines was actually claimed. But since a common mistake people make when working with opamps is to assume that NONE of the inputs have nonnegligible current flow, many people would be primed to make that inference and not go any further.
     
  6. MrAl

    Thread Starter AAC Fanatic!

    Jun 17, 2014
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    Hi,

    Ok great. I just threw nodal at it and then calculated the difference voltage across R1 and then the current, then the input resistance.
    Yeah some of the voltage came out the same, such as vR5=Vin.
     
  7. LvW

    Well-Known Member

    Jun 13, 2013
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    Here is my approach for calculating the input impedance.
    When we remember how the circuit was developped (see the attached jpg file) it is very simple to find the input impedance.
    The "first" GIC was a simple series connection of two different NIC blocks (one input "short circuit stable" and the other one "open circuit stable" )

    * The input resistance of the first NIC (the second NIC replacedby Z6) is ZA=-Z1*Z6/Z2
    * The input resistance of the second NIC is ZB=-Z3*Z5/Z4

    When we now connect the second NIC instead of Z6 to the first NIC we must replace Z6 by the input resistance ZB.
    The resulting input resistance now is positiv:
    ZA=Z1*Z3*Z5/Z2*Z4

    This is already the desired expression - and it will remain unchanged if we introduce the modification of the circuit as proposed by A. Antoniou in 1969. Assuming ideal opamps, the diff. voltage between the opamp input terminals will be zero - and the potential of the non-inv.inputs of both opamps will be the same. Hence, we are allowed to interchange both non-inv. nodes.
    If we now redraw the whole circuit, we have the classical GIC (in Antoniou topology) as shown in the task description.

    Comment: The two-NIC circuit as shown in the attached jpg file works as a GIC - however, the Antoniou-structure has better properties with respect to real world conditions: The non-ideal opamp properties (finite and frequency-dependen open-loop gain) cancel each other up to a certain degree. This is the result of "cross-coupling" between both output nodes.
    This effect works best if we make the choice Z2=R2 and Z3=R3 with R2=R3.
     
    Last edited: Dec 5, 2018 at 6:00 AM
  8. MrAl

    Thread Starter AAC Fanatic!

    Jun 17, 2014
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    Hi,

    I didnt go through the math but that does look like an interesting way to do it too.
     
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