Superposition on output of amplifier?

Discussion in 'Homework Help' started by jp1390, Sep 19, 2011.

  1. jp1390

    Thread Starter Member

    Aug 22, 2011
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    Hey all, this is just a super quick question regarding negative feedback and superposition.

    The book was going over how to find the gain in terms of finite gain and feedback factor beta for some simple amplifier circuits but I am a bit confused in one of the steps they used for the inverting amplifier.

    [​IMG]

    Solving for the non-inverting amplifier gain is easy because the the feedback voltage V- is a simple voltage division of the output, but it gets a bit trickier for the inverting amplifier.

    What confuses me is that when they solve for the voltage V-, they use superposition:

    V_{-} = V_{in}(\frac{R_{f}}{R_{in} + R_{f}}) + V_{out}(\frac{R_{in}}{R_{in} + R_{f}})

    They do this by setting first Vin = 0 and solving for V- in terms of Vout, but then they set Vout = 0 and solve V- in terms of Vin...

    I don't understand how they can set Vout = 0 if it is not an independent source. Can someone explain?

    Thanks

    Btw I am using the Microelectronic Circuits textbook by Jaeger and Blalock.
     
    Last edited: Sep 20, 2011
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    You probably mean independent rather than dependent - students are often incorrectly told that superposition can't be used in analyzing circuits with dependent sources. One needs to be careful in the application.

    This notion that superposition can't be used with dependent sources has been addressed by WM Leach Jr. Google that name with the search terms "superposition with dependent sources".
     
    Last edited: Sep 20, 2011
  3. jp1390

    Thread Starter Member

    Aug 22, 2011
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    Sorry, yeah I meant independent. Well, I know that for what I was taught is that in using superposition, dependent sources remain on but independent sources are turned off, depending on which source is not being looked at.

    In this example, Vo = A(0 - V-) = -AV- which is a dependent source...
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    One should keep an open mind.

    The attached paper demonstrates how superposition may be applied to op amp circuits.
     
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  5. steveb

    Senior Member

    Jul 3, 2008
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    :p And keep a open eye to watch out for bonehead editors, and backstabbing reviewers.

    Wow, did you see that trick the reviewer played? He claimed that a resistor can be considered as a dependent voltage source, OR as a dependent current source. Then he sets up a circuit that forces constraints that allows only one voltage AND one current in the resistor R1. Hmmm, don't voltage sources allow arbitrary currents, and don't current sources allow arbitrary voltages?

    Note that V1=(I1+I2)/g and the current through R1 is (I1+I2)/(R1 g). Both values depend on parameters that are pre-specified. So, the resistor R1 acts simultaneously as a voltage source and as a current source, which doesn't give any freedom in the external loading.

    Consider a case where I1=I2 , not equal to zero, and R3=R2. This circuit is symmetrical, so V1 must be zero, hence the current in R1 is zero. This also means that the dependent current source has no current, and the currents in R2 and R3 are equal to zero. So, we have two current sources I1 and I2 pumping current into the circuit, but that current doesn't flow into any of the allowed current paths.
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    If one is desperate to defend a long-held belief in the face of strong (incontrovertible?) evidence to the contrary, then sometimes deception (or is that blind stupidity) is the last resort. At least Marshall Leach wasn't subjected to a similar fate to that of Galileo or (worse still) Socrates. I suppose having a perfectly good paper (like Leach's) rejected is bit like being "sent to Coventry".

    NB: I must thank our excellent fellow forum member "the Electrician" for the original "heads up" on this unpublished paper.

    I return to edit on a personal note & reflection. In his preface Marshall Leach mentions studying circuit theory using Ronald E. Scott's (then unpublished) notes. I still have a copy of Scott's (eventually) published book entitled "Linear Circuits". I was introduced to the book as a college student by a wonderful English gentleman who taught circuit theory and gave his often wayward students some good advice about living wisely. Scott's book has been a valuable & cherished resource for me over 45 years. It's now somewhat worn looking, but hopefully it will still be on my bookshelf when I die. As Leach points out, Scott never included controlled sources in his work and hence doesn't give any clues as to his take on the use of superposition in that situation. I wonder whether he thought it best not to go down that road lest his work be rejected by the publishers.
     
    Last edited: Sep 20, 2011
  7. jp1390

    Thread Starter Member

    Aug 22, 2011
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    Okay, I'm sorry to bring this thread back to life but I am still confused as to how superposition can be applied to this circuit. Disregarding the pdf that was presented earlier, can anyone give an explanation on how the output of the op-amp can be treated as an independent source?

    Thanks!
     
  8. steveb

    Senior Member

    Jul 3, 2008
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    Why would you ask us to disregard the PDF to answer your question? The PDF gives the answer, and without that PDF, we here would be on very shaky ground to make a claim that is generally not supported by the rest of the literature. You saw how even the author could not convince two "respected" electrical engineering professors of his method.

    Section II of the paper gives the proof. To convince yourself, you need to understand the proof, and for any given circuit you need to make sure that your conditions are meeting the starting assumptions of the proof. This procedure might not be straightforward for your circuit depending on what equivalent circuit you use for the OPAMP, but I'm not sure because I haven't tried yet. Or, you can prove your particular case by noting that you get the correct answer using the method. Note the final quote as follows.

    "The above proof does not imply the controlling variables of a dependent source are deactivated when applying superposition to the source. Only the output of the source is set to zero. This procedure makes it is possible to write circuit equations by considering only one source at a time or to one side of a source at a time. This considerably simplifies the use of superposition with dependent sources compared to the way that it is presented in most circuits texts."

    I think this answers your question since you don't actually treat the output as an independent source, but you do treat it as a dependent source. It's just that you appear to be treating it as an independent source in the application of the method. But, the proof of the method does not treat the source as an independent source.

    A word of caution about OPAMP circuits is in order also. OPAMP circuits have all sorts of nonlinearities associated with them. Aside from the power supply limits that saturate the output voltage, there are also slew rate limits and distortion components in the output. However, we usually keep that knowledge apart in the OPAMP solution, and draw an equivalent linear circuit that we believe represents the circuit under certain constraints. For example, we make sure the output voltage is not saturated and we make sure the change in voltage on the output is not exceeding the slew rate limit. I know this caution applies to many nonlinear circuits, but somehow students are less cautious with OPAMPs and the slew rate limit can really sneak up on you if you are not paying close attention.
     
    Last edited: Oct 16, 2011
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