OpAmp circuit - Help needed

Discussion in 'Homework Help' started by FiReStArtEr21, Mar 8, 2009.

  1. FiReStArtEr21

    Thread Starter New Member

    Mar 8, 2009
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    Hey, this exercise is from my electronics exam (which i failed). My approach in finding a solution to these kinds of problems is to strategically pick out the KCL equations at the input nodes of each opamp. Since we can find the proper values of the voltages at the inputs, its all very straightforward. The thing is, i dont know how i should get the voltage at the inputs of opamp A2 (normally, i'd use superposition, but since we're working with controlled voltage sources...im at a dead end). My result is:

    Va2+=[R2/(R2+R5)] * (K5Vx + K1Vy) (voltage divider)
    V2 = [R11/(R11 + R14)] * (K4V4 + Va2+) (idem)

    Another thing i found hard to determine was the voltage Vx. Since we dont have current flowing to the + input of the opamp, theres no voltage drop and Vx = K3V1...but what about the Vi source? should i use superposition to find Vx?

    Sorry if these questions are stupid in some way...but i need to be sure of what im doing ;)

    Help would be very much appreciated...:)
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    At the top of the page, you have given numerical values for the various controlled sources, except for K2 which is given as -R^(-1). Do you have a numerical value for R?

    Are all the opamps assumed to be ideal (infinite gain, no input current, etc.)?

    What are you supposed to solve for? All the numbered nodes?

    And, yes, it looks to me like Vx = K3 V1

    Was this a take home, or were you expected to solve it on a regular exam?
     
  3. FiReStArtEr21

    Thread Starter New Member

    Mar 8, 2009
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    Hey, the opamp is ideal, the resistors have all the same value R and the arguments for the controlled sources are listed on the top of the page. The objective is to get the voltage V0 at the central point of the circuit (this value should be a function of R and/or Vi).
    This exercise is from an exam (we should actually be able to solve this and 2 more exercises similar to this one in 2 hours) and im studying for the next one, so i need to understand this.

    Thanks for the help;)
     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    OK.

    The solution would seem to involve several equations. What method do you use to solve a set of simultaneous equations? Do you use a high-powered calculator, or do you use Matlab or something similar?

    Also, are you allowed to use any method you like to solve, or do you have to use some particular method?
     
    Last edited: Mar 8, 2009
  5. FiReStArtEr21

    Thread Starter New Member

    Mar 8, 2009
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    We solve this exercises on paper, by inspection of the circuit. What we do is to get the equations for each sub-circuit that surround each opamp. after we determine the output voltage of each opamp we apply KCL on the central node. Since the resistors have the same value, you can easily reach the final solution.

    To sum it all up, we use KCL as the main technique to solve this. What's hard about this is that it tests all your knowledge on analysing a complex circuit node by node...and theres where i need help. How can i approach a problem as complex as this one so i can easily solve it?

    Thanks;)
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I think the solution is as follows:

    1. Start at amp A3 with the +ve terminal at earth

    2. Vy = 0 hence V3 driven (by ideal follower amp) to 0V. V3 = 0

    3. At amp A1 +ve terminal, source K1Vy = 0 from above

    4. The current source K2Vi drives downwards current 1/R*Vi through R6

    5. Drop across R6 is then +Vi and V1 = Vi

    6. Source K3V1 = 2Vi. So Vx = 2Vi (no drop across R16 with no bias into amp A4)

    7. Hence V4 = Vx - Vi = 2Vi - Vi = Vi

    8. Source K4V4 = Vi

    9. Source K5Vx = 5/2 X 2Vi = 5Vi

    10. Hence at Amp A3 V3 = 5Vi - Vi = 4Vi

    11. So we have V1 = Vi, V2 = 4Vi, V3 = 0 and V4 = Vi

    12. From this you will then be able to find Vo = 3/2 Vi (I think)

    Your examiner is a tough customer!
     
  7. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    There's a typo; item 10 should be: Hence at Amp A2, V2 = 5Vi - Vi = 4Vi

    I was thinking Sadist!

    Do you know what the definition of a Sadist is?

    Someone who's nice to a Masochist.
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Mercifully, I haven't met either of those people ...... maybe you could swap classes Firestarter21
     
  9. FiReStArtEr21

    Thread Starter New Member

    Mar 8, 2009
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    Tell me about it...thanks for the help...the results are a little bit different from mine, and i've spotted where i was wrong ;) I'll be using this forum for help to study for the exam, so I'll be posting some more tricky exercises for u guys to help me...and just so u can see how mean my professors are :S

    Cheers
     
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