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OP amp: Find Vout of this amplifier

  1. Nov 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Find [itex]V_{out}[/itex] of this amplifier

    50a1dfdfe4b0e22d17ef351d-richyw-1352788537348-screenshot20121112at9.48.59pm.png

    2. Relevant equations

    [tex]V_{out}=G_v(V_{in}+V_F)=G_v(V_{in}+B_vV_{out})[/tex]

    3. The attempt at a solution

    I have been stuck on this for a week. I really don't know where to start. Have read the chapter multiple times, tried to create it on a circuit program, and looked through just about every website for an example where the positive and negative inputs are connected. So incredibly stuck (on something that might be trivial...) I thought that an op amp had really high resistance between the positive and negative input, so I thought it might just be whatever the voltage drop is across those two resistors. Sorry but I just have no idea!
     
  2. jcsd
  3. Nov 13, 2012 #2

    CWatters

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    Imagine Vin = 1V (for example). Then V+ = 1V. To be in a linear mode V- ≈ V+ = 1V.

    What output voltage would give V- = 1V ? ...

    If you assume the current going into the -ve input is zero then the circuit reduces to two resistors in series, with Vin one end and Vo at the other. What value of Vo gives V- = 1V ?
     
    Last edited: Nov 13, 2012
  4. Nov 13, 2012 #3
    uh I'm not sure. aren't you just plugging in a number for [itex]v_{in}[/itex]? If I could figure it out in terms of that then I could figure it in terms of [itex]v_{in}[/itex]. If [itex]v_+\approx v_-[/itex] Then I think that everything to the left of the right hand 10k resistor would have to be 1V, and if there was 1V on each side of the left resistor, then no current should be moving across it?
     
  5. Nov 13, 2012 #4
    I just don't see why anything is happening at all, and why it's not just the same potential everywhere.
     
  6. Nov 13, 2012 #5

    CWatters

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    Perhaps that's a clue :-)
     
  7. Nov 13, 2012 #6

    CWatters

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    PS: What happens with the switch in the other position? What effect does the switch have on the amplifier? The gain is similar but differs in one important respect?
     
  8. Nov 13, 2012 #7
    I know the other way is just an inverting amplifier with gain=1 (-1 i guess?) . Not because I understand what's going on, but because it's basically the first example every textbook has. For all I know there could be a wizard inside that triangle shooting lightning bolts.

    I can see that if v+=v- then since v+ is grounded, v- must be 0. so then vin would have to lose all of its potential across that resistor. I don't get how this wouldn't lead to current flow...
     
  9. Nov 13, 2012 #8

    CWatters

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    Basically you had the answer. With the switch as shown the circuit is a non-inverting amplifier (eg Vout = Vin, gain = 1).

    You have...

    Vin....10K...V-...10k....Vout

    As I suggested pick a value for Vin = 1V (or any other voltage you choose). If Vin = V+ = 1V then V- must also be close to 1V for the thing to be in a linear mode (Edit: see next post for more on that). Looking that the resistor chain above the only way for V- to be 1V is for Vout to also be 1V. Same applies for other values of Vin.

    You should also check what happens if Vout was forced away from the predicted value. Lets say Vin = 1V and we think Vout will be 1V. What happens if Vout was somehow forced to be 1.1V. Then you have...

    1V....10K...V-....10K....1.1V

    So V- would be betwee 1.0 and 1.1V eg higher than V+ which is still 1V. If V- is higher than V+ the amplifier will drive the output voltage lower eg it tries to correct the problem with the output voltage and demonstrates -ve feedback.

    So basically the switch just changes the amplifier from non inverting to inverting. In both cases the |gain| is unity.

    A non inverting unity gain amplifier does have practical uses. For one thing the output impedance might be a lot lower then the source impedance.
     
    Last edited: Nov 13, 2012
  10. Nov 13, 2012 #9

    CWatters

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    Ah that means my previous reply might be confusing..

    Think of an opamp as an ideal amplifier. The output voltage of the opamp...

    Vout = A * (V+ - V-).

    In other words it amplifies the difference between the voltage on the V+ terminal and the voltage on the V- terminal. A is a large number called the Open Loop gain and can be as much as 1x106.

    Because the gain is very large V+ and V- have to be at a similar voltage or the output will be huge (eg limited by the supply rails). So if the amplifier is operating in a linear mode you can normally assume V+ and V- are at roughly the same voltage.

    When you analyse a circuit you can normally connect nodes at the same voltage together without effecting it's operation. In the case of the opamp circuit you couldn't actually connect them together in real life but you can to help analyse the problem.

    At this point its worth looking at the circuit for an inverting opamp set up. Lets call the input resistor Rs and the feed back resistor Rf.

    If V+ is at 0V then for it to be in a linear mode V- must allways be close to 0V as well. Lets assume it is at 0V.

    The current through Rs is therefore (Vin-0V)/Rs = Vin/Rs

    Likewise the current through Rf is (Vout-0V)/Rf = Vout/Rf

    but where do these currents go? The V- input has a very high input impedance so it doesn't go in there. Applying KCL at the V- node you find...

    Vin/Rs + Vout/Rf = 0

    Rearrange to give

    Vout/Vin = -Rf/Rs

    This is the closed loop gain.

    If Rf was 100K and Rs was 10K the gain would be 100K/10K = 10x
     

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    Last edited: Nov 13, 2012
  11. Nov 13, 2012 #10
    thanks so much for your help!
     
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