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Op-amp circuit help

  1. May 21, 2012 #1
    1. The problem statement, all variables and given/known data

    in the attachment

    2. Relevant equations



    3. The attempt at a solution
    for q.1) its a voltage follower non-inverting
    Vcc has to be greater than 5V for the zener diode work?
    formulas?

    q2) does anybody know how to use multisim ?
    if i copy the figure in multisim would i be able to work out a,b,c,d
     

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  2. jcsd
  3. May 21, 2012 #2

    NascentOxygen

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    Looking at Fig. 1, if you have studied op-amps you should be able to figure this out. Assuming an ideal op-amp, what are its important characteristics?
     
  4. May 21, 2012 #3
    positive feedback?
    i read that the input draws no current.. but output has a value
     
  5. May 21, 2012 #4

    NascentOxygen

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    There is feedback. What input terminal does the op-amp's feedback connect to?
    Good. That, together with the op-amp's differential gain being "very high" is all the op-amp detail that is needed to analyze the circuit.

    Since the inputs of the op-amp itself draw no current, how much current will be flowing through the "horizontal" 10kΩ resistor?
     
  6. May 21, 2012 #5
    its midnight so ill get back to you in a couple of hours.thanks.
     
  7. May 21, 2012 #6
    the input it connects to is( - ) so its negative feedback.
    i dont think theres current flowing in the horizontal 10 K. because of no voltage drop?
    how do we find Vcc
     
  8. May 22, 2012 #7

    NascentOxygen

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    Negative, yes. What percentage of the output voltage is fed back to the inverting input?
    What will be the voltage on the left hand side of that horizontal 10kΩ resistor?
    Usually we don't. Vcc is given, or we simply assume it is appropriate for the circuit to work properly. I don't think you are asked to find a suitable Vcc, are you?

    You are asked to describe how the circuit works---or to explain why it does what it does. :smile:
     
  9. May 22, 2012 #8
    it has to balance to reach a point of equilibrium 99.99% ?

    0 Volts?

    true.
     
  10. May 22, 2012 #9

    NascentOxygen

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    The correct answer is 100% of output is fed to the inverting input.

    Do you understand how the zener diode is being used here?
     
  11. May 22, 2012 #10
    not really i know what it does like, it wont work if it doesnt get 5V or something
     
  12. May 22, 2012 #11

    NascentOxygen

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    Well, let's assume it does get sufficient voltage to develop 5V here, otherwise if we consider the circuit is not being powered there is not going to be much to explain. :wink:

    Perhaps you are thinking Vcc could be any old time-varying waveform? I'd say it can't be. It is a respected convention in electronics to denote only a fixed DC voltage as Vcc. (If you want to put a figure on it, think something anywhere in the range 9V - 18V. The particular value should not be critical.)
     
  13. May 22, 2012 #12
    if we dont put a figure on it how are we going to find the output current for part b and is Rf important to note down is Rf the horizontal one on the left 10K?
     
  14. May 22, 2012 #13

    NascentOxygen

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    That would be important only if the output current were dependent on Vcc. :wink:
    Rf is your feedback resistance, I presume? Rf connects the feedback (in general) from output to inverting input. In your circuit, the resistance in this position (viz., from output to inverting input) is a piece of wire, and nominally 0 ohms.

    Instead of worrying about component values, at this stage you should be trying to figure out what the circuit does and how it works. These details typically don't require specific component values, usually.
     
  15. May 22, 2012 #14
    is part a done ?
     
  16. May 22, 2012 #15

    NascentOxygen

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    You have worked out the formula for IL?
     
  17. May 22, 2012 #16
    i=v/r

    we makeup v (9-18)
    and how about resistance we just use the 10K before ground?
     
  18. May 23, 2012 #17
    :confused:please reply
     
  19. May 23, 2012 #18

    NascentOxygen

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    You may be under a misunderstanding that the only voltage source in the circuit is Vcc. That would be quite wrong. Not shown in many schematic diagrams are the + and - DC supplies powering the op-amp itself.

    The Vcc shown serves to supply current to the zener diode. What voltage do you expect to find at the junction of the two 10k resistors and the zener diode?

    Bearing in mind what you said about the horizontal 10k resistor, deduce the voltage that you would expect to measure on the right hand side of that horizontal 10k resistor (i.e., as measured at the inverting input (-) of the op-amp).
     
  20. May 23, 2012 #19
    V1 and V2 my lecturer kind of gave us an idea that there both 15 V and Vcc is 10V now im more confused
     
  21. May 23, 2012 #20

    NascentOxygen

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    The voltages powering the op-amp do not dictate the output voltage of the op-amp, meaning you can largely ignore those voltages---which is why they are not even shown on some circuit diagrams. What is important is that the output voltage of the op-amp is controlled by the input voltages of the op-amp.
     
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