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Differentiating Op-Amp basic exercise

  1. Nov 15, 2011 #1

    Femme_physics

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    1. The problem statement, all variables and given/known data

    Given here is an Operational Amplifier, its feeding Voltages are +-10.

    A) Calculate Vin to get a Vout of -2V, point out the direction of the current in this case (From a to b or b to a)

    http://img72.imageshack.us/img72/2550/needheed.jpg [Broken]


    3. The attempt at a solution

    http://img844.imageshack.us/img844/4493/blury.jpg [Broken]

    Is this the right formula?


    The Vx's are at point a and c, that I know.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 15, 2011 #2

    I like Serena

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    How did you get this formula?
    And what is V1?

    Do you have a list of relevant equations?
     
    Last edited: Nov 15, 2011
  4. Nov 15, 2011 #3
    This is an amplifier with negative feedback. The main things you need to know about such an amplifier are:
    1) The V at the - and the V at the + inputs can be assumed to be equal (V at a = V at c = +4V)
    2) the input resistance is infinite, this means that no current flows into or out of the - and + input terminals.
    the thing to do now is calculate some currents.
    Current will flow from V in through R1 towards the - input, so Iin = V/R1, V is the voltage ACROSS R1
    Iin = (Vin- 4)/R1....... check that you are OK with this !!!
    This current does NOT flow into the -input, so it must go up through R2.
    This current passing through R2 will produce a PD = I x R2
    So a voltage of R2 x (Vin-4)/R1.....( 2Vin -8) ...... put in the values of R1 and R2) will be developed across R2 (this is the familiar form of the equation for the gain of an amplifier of this sort.... I hope you recognise it!!)
    I have the current flowing from left to right through R1 and R2 so this means that the voltage at the right hand end of R2 must be the voltage at the - terminal MINUS the voltage we have calculated across R2 (are you still with it?)
    So V at the right hand 'pointed end' of the amplifier must be 4 -(2Vin - 8)
    so V = 12 -2Vin .
    Because of the 4V battery connected to the end of the amplifier Vout = V at end of amplifier -4
    So 12 -2Vin -4 = Vout
    So 8 - 2Vin = Vout
    If Vout =-2 then 8-2Vin = -2..... gives Vin =+5V
    I hope you can follow this !!!!
    I will make my next post assuming that Vin = 5 volts and follow it through.

    There is a lot to follow here !!!! I hope you get something from it.
    I will make another post explaining what happens assuming that Vin = +5V
     
    Last edited: Nov 15, 2011
  5. Nov 15, 2011 #4
    Now here is what happens if we know that Vin = 5V.
    (You have to accept that the V at the - input = The V at the + input) This is ESSENTIAL
    V across R1 = 5-4 = 1 Volt
    Therefore current through R1 = V/R = 1/5000 = 0.0002A
    No current flows into the - terminal, therefore this current MUST flow through R2 (10,000)
    So the voltage developed across R2 = I x R2 .... = 0.0002 x 10,000 = 2V
    The voltage at the Right hand end of R2 is therefore 2 volts LOWER than at the left hand end
    Therefore the voltage at the RIGHT end of R2 is (4-2) = +2V (remember the volts at the - input =4V).
    There is a 4V battery connected with its + terminal to the Right end end of R2 so the voltage at the -terminal of the battery is -4 volts lower than at the + end.
    The + end is at +2V so the -end (Vout) must be -2V
    Hope this helps
     
  6. Nov 16, 2011 #5

    Femme_physics

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    Yes, teacher gave it to us....

    V1 is what in the drawing, I guess?

    http://img809.imageshack.us/img809/425/diffformulas.jpg [Broken]

    Because Vin goes into the minus, right?

    Yes, we called those Vx in class.

    I was trying to google "how op-amps" work after this statement. I didn't find one that refers to this line. But I'll try to ask my classmates/teacher.


    It's -4 and not +4 because this is occuring on the minus leg?

    I'll comment and ask more questions later, i'm trying to do few at a time...
     
    Last edited by a moderator: May 5, 2017
  7. Nov 16, 2011 #6
    Op amps have such a high gain (100,00+) that you can assume that any difference in volts at a or c is negligible.
    That is why it is OK to say Va = Vc = your Vx (in this question because of the 4 volt battery connected to the + input the voltage at the - input must be +4 volts
    An ideal voltage amplifier should have infinite resistance so that no current is drawn from the source. (in a similar way an ideal voltmeter should have infinite resistance)
    Op amps are designed to have an extremely high input resistance (Millions of ohms) so it is safe to assume that no (negligible) current flows into the input.
    The V at the - input has to be the same as the V at the +input which is +4V.
    The - at the - input tells you what the amplifier does to a Voltage at this input.
    If V at the - input increase in the + direction then the effect is to make the OUTPUT voltage decrease (go in the negative direction.
    The - input is called the 'inverting' input.
    Think of a see saw with the pivot very close to one end. Push the end down and the other end goes up.
    Hope this helps
     
  8. Nov 16, 2011 #7

    I like Serena

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    Well, your formula does not make sense to me.

    And I also do not understand why your teacher called it a differentiating amplifier.
    It isn't.
    It's also not a differential amplifier, although your new formula is for that.



    Here's a picture of a differential amplifier.
    200px-Op-Amp_Differential_Amplifier.svg.png
    The corresponding formula (symbols from the picture) is [itex]V_{out} = (V_2 - V_1) {R_f \over R_1}[/itex]



    What you have looks more like an inverting amplifer.
    Here's the picture:
    200px-Op-Amp_Inverting_Amplifier.svg.png
    The corresponding formula (symbols from the picture): [itex]V_{out} = -{R_f \over R_{in}}V_{in}[/itex].




    With the extra batteries included, your formula should be (I'm not going to explain it now):
    [tex]V_{out} = V_1 - (V_{in} - V_1) {R_2 \over R_1} - V_2[/tex]
    However, this does not look like your formula.
     
  9. Nov 16, 2011 #8
    I agree with 'I like Serena' final equation taking V1 to be the battery connected to the + input and V2 to be the battery connected at the output.
    In your case V1 and V2 are equal to 4V so the equation reduces to
    Vout = 4 - (Vin -4)R2/R1 -4
    Vout = (4 - Vin)R2/R1
    Substituting Vout =-2 and R2 =10k and R1 = 5k gives Vin = +5volts
     
  10. Nov 16, 2011 #9
    A completely irrelevant point!.... I have never met such a circuit in practice.
    It looks like an exercise to test your ability to analyse op amp circuits
     
  11. Nov 17, 2011 #10

    Femme_physics

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    Thanks technician, ILS!

    There is a second clause asking me what is Vin if Vout = 8V

    So I think I got it right now :smile:


    http://img412.imageshack.us/img412/1879/voutvin.jpg [Broken]

    Sorry, I may have made a mistake mishearing/miscopying...
     
    Last edited by a moderator: May 5, 2017
  12. Nov 17, 2011 #11

    I like Serena

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    Looks good! ;)

    But do you know how to get the formula?
     
  13. Nov 17, 2011 #12

    Femme_physics

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    Thanks, master! :smile:
    Well, yea, I ask Klaas :wink: ..........


    You mean the development of the formula? No, not really. It's a bit blurry to me.
     
  14. Nov 17, 2011 #13
    You have made a tiny mistake in the equation
    +8 = +4 - (Vin - 4) x 2 -4
    +8 = +4 -2Vin +8 -4
    +8 = -2Vin +8
    0 = -2Vin
    so Vin = 0
    Can you see where you missed a - x a -
    You have got it !!! well done
     
  15. Nov 17, 2011 #14

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  16. Nov 17, 2011 #15
    well done. Your original post was Op-amp basic exercise !!!! This was NOT a BASIC exercise !!!!!!
     
  17. Nov 17, 2011 #16

    Femme_physics

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    Oh, didn't know...though it seems pretty basic to me. What would you call it?
     
  18. Nov 17, 2011 #17
    It is unusual and there is one more thing you may need to worry about!!
    In the last calculation with Vout = 8V this means that the voltage at the output of the amplifier would be +12V (8V + that 4V battery)
    If you go back to your original post you will see that the power supply is +10 to -10.
    The voltage at the output of the amplifier cannot be greater and cannot be less than +10 and -10 so there is an inconsistency here !!! I wonder if, whoever set the question, was aware of this ???
    It is a detail that is easily overlooked.
    I would love to know what your teacher/tutor makes of this
    Try to let me know
    By the way, the most basic set up to look at is when there are no extra batteries added
    The equation then becomes Vout = -Vin x R2/R1
     
  19. Nov 17, 2011 #18

    Femme_physics

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    Oh! I was not aware of this, technician. I will ask my teacher tomorrow morning as I have a class with him.

    Thanks. I will let you know, I promise.

    Duly noted :smile:
     
  20. Nov 17, 2011 #19

    I like Serena

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    That's all right.
    yes-master-lord-pain.jpg


    He must be really smart then!
    Who is he?


    Oh well, perhaps some day soon the fog will rise.
     
  21. Nov 17, 2011 #20

    Femme_physics

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    Nice pic!

    He is Klaas, The Klaasmaster of all Classes :wink:

    I bet!
     
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