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Inverting amplifier input impedance (and shunt-shunt feedback)

Discussion in 'General Electronics Chat' started by Elerion, Dec 13, 2015.

  1. Elerion

    Elerion Member

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    Doing some calculations, I recently reached a contradiction. I'm sure there must be an error in some step.

    Input impedance of a basic opamp inverting amplifier is given by input resistor (Zin = Rin).
    Gain, by ratio of feedback by input resistor.

    I treated this circuit as a shunt-shunt negative feedback, and did some calculations, which gave a much lower input impedance.

    In shunt-shunt configuration, close-loop input impedance is given by Zin / (1 +AB).

    Based on this assumptions, I see a contradiction.
    Zin or Zin/(1+AB) ??
     

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  2. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    It's simple, don't make incorrect assumptions and then blame reality for been wrong :D

    However, I'm completely baffled by your entire post?, in an inverting opamp how could the input impedance possibly be less than the input resistor, as the signal has to pass directly through it so it can't be less.
     
  3. Elerion

    Elerion Member

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    I know I'm making a wrong assumption, but I don't see where.
    If I knew, I wouldn't be asking :)

    The sheet of paper I uploaded tries to apply a method for converting a feedback circuit into a "standard" circuit, so the close loop gain can be found as:
    A/(1+AB)
    and input / output impedances can be found by the relationships shown below.

    shunt-shunt.jpg

    rin =Vin / Iin
    r'in = open loop input impedance.

    cl_impedance.jpg


    The question is, why does this method apparently fail ?
    What did I do wrong?
     
  4. dave

    Dave New Member

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  5. AnalogKid

    AnalogKid Well-Known Member

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    R22 is not equal to Rg.

    In other inverting opamp circuits such as a gyrator and some active filters, the input impedance can be less than the input R.

    ak
     
  6. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    Where's the input resistor in that partial circuit? - pretty obvious where you went wrong.
     
  7. spec

    spec Well-Known Member Most Helpful Member

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    Elerion,

    The circuit you have shown is an inverting current to voltage converter (Vout= -Iin* R1). It is a classic function and is sometimes called a transconductance amplifier, or something like that. Input impedance = zero, output impedance = zero. RL does nothing.
     
    Last edited: Dec 14, 2015
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  8. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Vos and temperature effects are the biggest problem with I-V converters. There is a voltage drop which is about Vos and there is a current burden of ib of the amplifier.
    Usually, thee are methods to "zero check" and correct the amplifier. You can basically turn it into an amplifier and amplify Vos.
     
  9. Elerion

    Elerion Member

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    An exercise from an electronics book (attached) confused me.
    It is trying to illustrate and be an example of the kind of exercise I attached at the top of the thread, applied to a shunt-shunt negative feedback.

    The input resistor is 1k.
    The answer (at the bottom) states that feedback circuit input impedance (Rent) is 150 and not 1k. It happends to be almost Rin / (1+AB). Same goes for output impedance (92 ohm - look at not below).
    ( -870 V/V is, as you can guess, voltage gain).

    Note: opAmp input impedance is not supposed infinite, but 100k (Rid). Output, 1k. Just to practice with non-ideal opamp.

    EDIT: Ok, I see my mistake... (to late! just a few minutes before Nigels reply) R_ent is the impedance seen passed the input resistor. So a voltage inverting amplifier's input impedance, in this case, would be its input resistor + 150 ohm. I just didn't see the point of calculating impedance at opamp input,... but I do now. Thanks to all.

    Sorry Nigel, I wan't mean to confuse anyone. I spent a couple of hours amazed on something I KNEW it couldn't be, but didn't see WHY.
     

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    Last edited: Dec 15, 2015
  10. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    You're confusing yourself (and everyone here), by talking about the supposed impedance of the opamp, and NOT the impedance of the circuit.

    The input of the circuit is 1K (as near as makes no odds) - you don't need to dig any deeper to use opamps, the whole idea of opamps is infinite input impedance, and zero output impedance - as this makes the calculations trivial, and it's so close to reality that it's well within component tolerances, then it's all you need.

    So no 'contradiction', just 'real life' as opposed to 'made up' theories.
     
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  11. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    I'm more of an "intuitive OP amp person". Actually got through 2 years of EE and a year AAS degree without studying OP amps, nor simulations. No or little PC's back then.

    I had a couple of "intuitive" screw ups too. When something with a zero output Z is fed with something with a bias current and the bias current has no place to go, it's like I=2pA/0 ohms. Driving capacitive loads is troublesome too.

    In DC circuits you really have to worry about Vos. Guarding. Never quite understood "noise gain" Thermoelectric (seebeck) effects. Use or non-use of solder.

    In an I-V converter, my Offset compensation relied on the D/A outpointing 0V for digital 0 in. Well that didn't happen. The offset was 40 pA which we could easily live with.

    In AC circuits you have to worry about frequency, capacitive loads, parasitic capacitances. That got me too.

    With one person doing everything linearly and managing someone else doing mostly programming in a language we both didn;t know. I would have yelled for more help. The overall design was based on a snapshot of (money/technology) and the technology was rapidly changing. The era was just after the MacII series where the IBM had the 6 character filenames and the crappy memory model and IEEE488 monochrometers didi't exist. There was a push to re-use stuff that we had.

    I picked the "right platform-Mac/LabVIEW" at the time. The instrumentation (SMU) was turned down (cost), but later used in a 17 year later upgrade. The second iteration used the PC/Labview/SMU.

    In another project I did not want a slot based data acquisitions system, but was turned down. They filled like a 6 lot MAC to the max and then found only 3-slot Mac's available which I predicted. I wanted known interfaces and the acquisition stuff separate.

    No PLC experience. It's hard doing "research" grade "production" and it's hard working where the mentality is "paper logging" because the the operator knows what's going on.

    It's also hard using "state of the art" technology to what it isn't ready for. e.g. Computers in the early moon shots.

    Sure, I went off on a tangent. When I left they had no documentation on a safety system that i was forced to build "on the fly". All I had was notes. No initial drawings, nothing. When it worked, it was DONE, I had documented my normal job. All of the "not working" stuff worked very well and better than it ever had. The electron-gun evaporator and a 1940's X-ray diffraction set. Prior to that I rebuilt a vapor dryer. There were some used ebeam evaporators that should have been fixed, but my new boss was territorial. My old boss let me help "anybody" UNLESS the time would be excessive. He knew that a few minutes of my time would save lots of time overall.
     
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  12. Ratchit

    Ratchit Well-Known Member

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    This appears to be a good time to introduce a little known theorem that is not covered in any textbooks of which I am aware. It is the General Immittance Theorem (GIT). It basically says that if you know or can calculate the transfer function (TF) (V2/V1,V2/I1,I2/V1,I2/I1) of a linear network, you can find the input impedance and output impedance of the network by solving for Zin = -rin and Zout = -rout of the denominator of the TF, where rin is the impedance of the input voltage and rout is the impedance of the outut voltage generator. Lets see how the theorem works for this problem. The current existing in R1 is V1/(R1+rin), and the current present in R2 is V2/(R2+rout). All the current present in R1 will be equal to the current existing in R2, because no charge flows through the minus terminal of the op-amp. Therefore V1/(R1+rin) + V2/(R2+rout) = 0. Manipulating the equation we get the circuit TF = -(R2+rout)/(R1+rin) = V2/V2 . -rin of the denominator is Zin = R1, and since no rout term exists in the denominator, the rout = Zout = 0 . This theorem works for all representations such as the time domain, Laplace, Steinmetz notation, etc. As long as the circuit is linear, this can be a timesaver if you know the TF. So don't calculate Zin or Zout separately if the TF is available, GIT with it and use the General Immittance Theorem.

    Ratch
     
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  13. Elerion

    Elerion Member

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  14. Ratchit

    Ratchit Well-Known Member

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