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Question on Non-Inverting Amplifier

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider the circuit that attaches in the attachment. Calculate the Vout.


    2. Relevant equations
    Non-inverting om amp:

    Gain = Vout/Vin
    = 1 + (Rf+Rs)

    Golden Rules:
    i- = i+ =0
    V- = V+

    3. The attempt at a solution
    Gain = Vout/Vin
    = 1 + (Rf+Rs)
    = 1 + (10/20)
    = 6

    Vx = (5/4+5) Vin
    = 5/9 Vin

    Vin = 72mV

    KCL:

    (Vx - Vin /12K) + (Vx/9K) + (Vx- Vout / 84k) = 0
    Vx(1/12k + 1/9k + 1/84k) = (Vin/12k) + (Vout/84k)
    [(5/9)(72*10^-3)(1/12k + 1/9k + 1/84k)] - [(72*10^-3)/12k] = (Vout/84k)
    Vout = 0.189V
    = 189.33 mV

    I am interested to know how to find the value of Vx.
    Besides, I would like to know is my final answer correct or not.
    If not, please kindly point out my mistake.

    Your help is very much appreciated
    Thank you
     

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    Last edited: Apr 30, 2013
  2. jcsd
  3. May 1, 2013 #2

    NascentOxygen

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    Hi Judy9911! [Broken]

    Where did this formula come from? Where are the 10 and 20 data values from?
     
    Last edited by a moderator: May 6, 2017
  4. May 1, 2013 #3

    NascentOxygen

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    After adding an essential set of parentheses, can you explain the derivation of this equation.
     
  5. May 1, 2013 #4
    For the Gain = Vout/Vin
    = 1 + (Rf+Rs)
    = 1 + (100/20)
    = 6

    This is my analysis from the circuit. Sorry that I make mistakes for the equation, should be Rf = 100K ohms , while Rs = 20k ohms

    Vx = (5/4+5) Vin
    = 5/9 Vin

    The bias current of the positive input of the op amp is 0, so the 4k ohms and 5k ohms resistors act as a voltage divider.
     
  6. May 1, 2013 #5

    NascentOxygen

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    Okay. It is customary to reserve ❝Vin❞ to designate the input to the circuit, and not use ❝Vin❞ for some intermediate point.

    Can you take a fresh look at this? Which is greater, Vx or Vin?

    It would be better to designate the signal at the non-inverting input as something like v(+) rather than Vin.
     
  7. May 1, 2013 #6

    rude man

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    You should start by giving symbols (R1, R2 etc.) to all your resistors, then write KCL or KVL & solve the simultaneous equations by Cramer's rule or whatever.
     
  8. May 1, 2013 #7
    It's not in the exact form of the non-inverting amplifier, so you cannot use that equation in this problem.

    Best bet is to use nodal analysis and plug the matrix in matlab or something.
     
  9. May 2, 2013 #8
    Redo this question

    I try to redo this question by using the nodal analysis :

    I label my diagram in detail which is attached to this post.

    The following is my step to solve for Vout.

    Vx= 72mV
    At Node'A':

    Ix + Iy - Iz = 0
    (Vx-Va/12k) + (Vout - Va /84k) - (Va - Vb/4k) = 0 [Equation 1]

    At Node 'B':
    { Golden rule: i+ = i- = 0}
    Iz - Is = i+ = 0
    Iz = Is
    (Va - Vb /4k) = (Vb / 5k)
    (Va/4k) = (4Vb+5Vb)/5k
    5Va = 9Vb [Equation 2]

    At Node 'C'
    It - Iu = 0
    It = Iu
    (Vout - Vc)/100k = Vc/20k
    Vout/100k = (5Vc + Vc)/20k
    Vout = 6Vc
    Vc = 1/6 Vout [Equation 3]

    { Golden rule: V+ = V-}
    Therefore, Vb = Vc
    Sub [Equation 3],
    Vb = Vc
    = (1/6) Vout [Equation 4]

    Sub [Equation 4] into [Equation 2]

    5Va = 9(1/6) Vout
    Va = 3/10 Vout [Equation 5]

    Sub [Equation 5] & [Equation 4] into [Equation 1]
    (Vx/12k) - (Va/12k) + (Vout/84k) - (Va /84k) - (Va/4k) + (Vb/4k) = 0
    (72m/12k) - ((3/10 Vout)/12k) + (Vout/84k) - ((3/10 Vout)/84k) - ((3/10 Vout)/4k) + (((1/6) Vout)/4k) = 0
    5.04 = 42Vout
    Vout = 0.12V
     

    Attached Files:

  10. May 2, 2013 #9

    NascentOxygen

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    Looks right. You conclude with Vout/Vin = 1.667 and that's what I calculated. :smile:
     
  11. May 2, 2013 #10
    Thanks for giving me the guideline.
     
  12. May 2, 2013 #11

    rude man

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    I got the same.
     
  13. May 2, 2013 #12
    The solution to these sorts of problems can be solved in a very systematic way with linear algebra:

    attachment.php?attachmentid=58429&stc=1&d=1367528349.png

    I happened to notice that this circuit exhibits a peculiar behavior. Try analyzing the circuit with an opamp gain of 7 instead of the ideal gain of ∞.

    It is peculiar enough that I'm going to start a new thread in the Electrical Engineering forum.
     

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  14. May 2, 2013 #13

    rude man

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    Op amp gain of 7: output = 0.1020V.
    What is so peculiar about this? Just change 1 equation, and add 1 more.
     
  15. May 2, 2013 #14
    I get .0582 volts. Anybody else?

    But. actually I meant to explore something else:

    Sorry. I spent so much time playing with a small change that I forgot to mention it. I should have said to reverse the v+ and v- terminals (or let the opamp gain be -7 rather than +7).

    In the original circuit, exchange the + and - inputs of the opamp but otherwise let the gain still be ∞. If an analysis is performed using the "Golden rules" nothing changes because the golden rule is still just v+ = v-; no distinction is made between the v+ and v- terminals in the "Golden rule". The gain A of the opamp isn't even considered.

    But, try this. Obtain a general expression for Vout/Vin in terms of the resistor values and A for the case where the + and - terminals are as shown in the OP's schematic, and also obtain the expression for Vout/Vin with the + and - terminals reversed.

    Now take the limit as A→∞ for both cases and note that Vout/Vin is 5/3 in both cases. You get the same transfer function regardless of the orientation of the + and - inputs to the opamp.
     
  16. May 2, 2013 #15

    rude man

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    The 'golden rule' does not apply to just any op amp circuit. It only applies if the amplifier output is not saturated and if the gain is infinite (if gain = 7 the + and - inputs are different, for another example). By reversing the + and - inputs to the op amp in this circuit I'd bet the amp goes into saturation and V+ will be very different from V-. But I'm too lazy to do the computations, and I haven't used my PSPICE in years.
     
  17. May 2, 2013 #16
    The "Golden rule" apparently applies to the OP's circuit, doesn't it?

    Look at the OP's solution in post #8. Which equations would be different if the + and - inputs of the opamp were reversed?

    The OP made the assumptions that are usually invoked to justify the use of the "Golden rules", namely that A = ∞ (this one was unstated) and i+ = i- =0.

    How is the OP or NascentOxygen or rude man or anyone else to know if the opamp output is saturated as a result of solving these equations? What further mathematical tests (this excludes building the circuit or using a simulator) could determine if the opamp output is saturated?
     
  18. May 3, 2013 #17

    rude man

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    That's a very good question. And the procedure is as follows:

    1. assume the amp is unsaturated and let V+ = V-. Solve the equations. Then, if the output is within the amp's bounds (e.g. +/-10V for a +/-15V-powered amp), the assumption was warranted and the result is OK.

    3. If the output computes to saturation, V+ ≠ V- and V+ and V- now become separate voltages. The output is then defined (fixed) as + or - 10V and the equations solved for all the independent nodes. This procedure works for any condition where the output is saturated. Of course, this circuit would typically be useless.

    Textbook questions seldom try to trick the student into giving him/her a saturated amplifier, so most often the assumption of V+ = V- is warranted, and the output computes to a non-saturated value.

    BTW the same can be said for exceeding the common-mode voltage limits, which is why when the system of equations are solved, all nodes should be computed, not just the output. Some op amps have very limited common-mode voltage limits, others can handle almost to the + and - power supply rails. (The 'common-mode voltage' is the actual voltage at each of the two op amp inputs referred to ground, not to each other. So you can have V+ = V- = 12V which looks good at first blush (inputs equal) but the 10V common-mode voltage would probably exceed the max limit & the amp would most probably saturate.)

    Try it with this circuit (switch the op amp inputs) & see!
     
  19. May 3, 2013 #18

    rude man

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    I should add that the above assumes "fully-compensated" op amps, so any dc gain is OK, even very low gains. Also, it assumes resistors only. Op amp stability is an advanced concept. Analog IC suppliers like Analog Devices Inc. have excellent free publications if interested.
     
  20. May 3, 2013 #19

    rude man

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    Yes, but not to yours if you reverse the op amp inputs.
    I did not check the OP's equations, but the answer to that question should be obvious if you did. They would most definitely be different and probably compute to a saturated output.
     
  21. May 3, 2013 #20
    OK. Solving the OP's problem equations with the + and - inputs as originally specified, the output is .12 volts; not saturated.

    Now reverse the + and - inputs and solve again (using the golden rule method, not a simulator, as I specified when I asked the question). The output is once again .12 volts. This procedure says that the opamp doesn't saturate with either orientation of the + and - inputs. Is this true?
     
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