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Non-Inverting Opamp with Voltage Source resistance

  1. Nov 30, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-11-30_18-28-51.png
    2. Relevant equations


    3. The attempt at a solution
    (a)
    We can assume for ideal op-amps that there is:
    • infinite input impedance
    • low output impedance
    • the potential for infinite amplification
    (b) Non-inverting op amp

    (c) This is the part I have trouble with.

    There is three resistances. The only equation we have used before for non-inverting op amps is

    Vout/Vin = 1 +R2/R1

    Vin = 8V

    R2 = 80K

    I'm guessing the 6.8k resistor is part of R1 with 20k but I'm not sure how to add it....
     
    Last edited: Nov 30, 2016
  2. jcsd
  3. Nov 30, 2016 #2

    gneill

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    Staff: Mentor

    Given your first stated assumption about ideal op-amps, what effect do you think the 6.8 kΩ resistor is going to have on circuit behavior? (Think about how much current can flow through it and what potential drops can occur across it)

    There's another assumption that can be made about the behavior of ideal op-amps that is tied to your third listed assumption, that of infinite amplification. What assumption can be made about the potential difference between the input terminals of the op-amp (when the circuit provides negative feedback, of course)? Can you analyze the circuit to determine Vout for a given Vin?
     
  4. Nov 30, 2016 #3
    Honestly, I don't know. These assumptions were only ever mentioned in fleeting in lectures and never discussed. Being honest, I don't entirely know what effect these assumptions mean for the any given op amp circuit.

    Is it that the current that could potentially flow through the resistor just be is V/R, in this case 8/6.8 approximately 1.18 A ?

    Is that that V- = V+ ? So there would be no potential difference between the two....

    As in nodal analysis? I' not sure how to do that with the op amp in the circuit...
     
  5. Nov 30, 2016 #4

    gneill

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    Staff: Mentor

    Something for you to investigate then. They are important factors that greatly simplify analyzing op-amp circuits.
    Think about it. If the input resistance is infinite, how much current could flow through the 6.8 k resistor and into the input?
    Yes!
    Could use nodal analysis, yes.

    Strangely enough the op-amp being there actually simplifies things:
    1. Infinite input resistance, so no current drawn from the nodes they're connected to.
    2. Input terminals at the same potential, so if you know one voltage you know the other, too.

    Can you determine the potential at the +input terminal?
     
  6. Nov 30, 2016 #5
    QUOTE="gneill, post: 5632418, member: 293536"] Think about it. If the input resistance is infinite, how much current could flow through the 6.8 k resistor and into the input? [/QUOTE]

    Well, V = IZ, so if Z is infinite then I has to be 0 ?

    So, could I treat R1 just as 20 k and use it in the above equation?

    Ok, but isn't it from nodal analysis that the equation Vout/Vin = 1 +R2/R1 comes from?

    So, could I use that or do I have to derive a specific one for this circuit?
     
    Last edited: Nov 30, 2016
  7. Nov 30, 2016 #6

    gneill

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    Staff: Mentor

    Yes. So that input resistor has no effect.
    I suppose so. You haven't specified exactly where R1 and R2 are or shown why you've chosen the formula you did, but as it turns out, your choices are fine.
    Well, the equation certainly can be derived using nodal analysis, but there are other ways too. You can use the formula if you're sure it's applicable (it is in this case). But you should always be prepared to derive the formula for a given circuit. Sometimes you'll be given a circuit that isn't exactly like one of the "standard" layouts and you'll have to do the work. And, you could always suffer a memory lapse and forget which formula applies in a given situation. You should be prepared to derive it quickly.
     
  8. Nov 30, 2016 #7
    The choice for R1 and R2 was based of my lecture notes as was the equation choice.

    upload_2016-11-30_20-25-52.png

    (I realise now, that the notes suggest no current flows into the terminals of the amplifier, so by extension of that the resistance in the Vs is unimportant, like you pointed out)


    Ok. I'll certainly keep that in mind for the future.

    So in this case Vout = (1 + 80k/20k)(8) = 40 ?
     
  9. Nov 30, 2016 #8

    gneill

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    Staff: Mentor

    Yes indeed!
     
  10. Nov 30, 2016 #9
    Ok great. Thank you very much !
     
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