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Output voltage amplitude

  1. Oct 29, 2012 #1
    The circuit in the figure is driven by a 9 mV amplitude signal generator with
    an output impedance of 2 kΩ. What is the output voltage amplitude from the amplifier?



    lab3.jpg


    So what I did basically is use the formula Vout = -R2/R1 x Vin
    So.. I know Vin is 0.9 V (9mv) and R2 = 12 R1 = 1 that gives me 12 x 0.9 = -10.8V what do you thin guys?
     
  2. jcsd
  3. Oct 29, 2012 #2

    gneill

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    Staff: Mentor

    9mV is not 0.9V. a millivolt is a THOUSANDTH of a volt.

    You should perform a circuit analysis on the given circuit; the formula you're using for gain does not apply to this particular configuration.
     
  4. Oct 29, 2012 #3
    ok so which formula do I have to use? cause I was looking at my teachers notes and this the only thing I find similar
     
  5. Oct 29, 2012 #4
    Don't use a formula, figure it out -- it will improve your understanding greatly.

    You've got an ideal op amp there. They have huge (ideally infinite gain). If that's the case, what can you say about the difference in voltage between the - and + terminals of the op amp? This is what they call the virtual ground approximation.
     
  6. Oct 29, 2012 #5
    MmMMmM this is really hard for me I struggling a lot but I think is 9mV it doesnt have any resistor affecting before the Vout but Im not sure at all :/ there is no way using a formula or something like that?
     
  7. Oct 29, 2012 #6

    gneill

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    Staff: Mentor

    You won't be able to memorize all the formulas that will suit every possible configuration. You need to be able to derive them "on the spot" by analyzing the circuits.

    Ideal op-amps allow you to use some very handy simplifications to help in the analysis. For example, an ideal op-amp has essentially infinite input resistance at its input terminals (no current drawn by the op-amp terminals), and if there's a feedback path from the output to the - input, the op-amp will try to adjust the output voltage to drive the potential difference between its two inputs to zero.
     
  8. Oct 29, 2012 #7
    is open notebook test lol is why the teacher wants to show us the work and I still not understanding what is the answer ? is not -10.8 V?
     
  9. Oct 29, 2012 #8

    gneill

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    Staff: Mentor

    No, it's not -10.8V.
     
  10. Oct 29, 2012 #9
    So Im still stuck I dont know how to do this what formula can I apply? can you please explain me
     
  11. Oct 29, 2012 #10
    The circuit in the figure is driven by a 9 mV amplitude signal generator with
    an output impedance of 2 kΩ. What is the output voltage amplitude from the amplifier?


    Provided the opamp has infinite input impedance, you can disregard the output impedance variable. The rest should be a classical textbook example. However, if the opamp had its output impedance equal to 2 kΩ, the result would certainly be different.
     
  12. Oct 29, 2012 #11

    gneill

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    Staff: Mentor

    KVL. KCL, Ohm's law... the usual circuit analysis methods. Take advantage of the op-amp characteristics. Start by labeling the circuit with known quantities, then calculate currents and potentials...
     
  13. Oct 30, 2012 #12
    The op-amp places some constraints on the circuit. An ideal op amp has infinite gain, no current flowing into its inputs and zero output impedance.

    If we label the node at the negative terminal of the opamp Va, then the output of the opamp is Vout = A (Vin -Va) --> Va = Vin - Vout/A . where A is the amplifier gain. If this opamp is ideal, its gain is infinite so what can you say about Va? This is called the virtual short circuit and from now on you will be able to take this shortcut when you see an ideal opamp. Note that it's called virtual because a real short circuit would allow current to pass between the two points.
     
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