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Differential amplifier with zener diodes

  1. Dec 10, 2011 #1

    Femme_physics

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    I think I'm getting better at this ((thanks to gneill, ILS, ehild and technician :wink:


    I'm quite sure I got it right using the easy methods (formulas). I'm not too sure about my "check up" to my solution using KCL and KVL...I didn't solve the check-up all the way because I wanted to see if my formulas are correct first.

    http://img46.imageshack.us/img46/6050/helpwithit.jpg [Broken]


    (IGNORE the mini-circuit drawn in (B)!)
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Dec 10, 2011 #2
    I think you have V+ incorrect. The 10V from that Zener is connected across a potential divider...(5k and 10k)....I am sure you can sort that out !!
     
  4. Dec 10, 2011 #3

    Femme_physics

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    Wait......I do't get it....do you see a mistake?
     
  5. Dec 10, 2011 #4

    gneill

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    Voltage divider:

    attachment.php?attachmentid=41792&stc=1&d=1323535207.gif
     

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  6. Dec 10, 2011 #5
    Can you now see what V+ is?....Then what V- MUST be....then......
     
  7. Dec 10, 2011 #6
    The way they have drawn that potential divider is very sneaky. If it had been drawn directly across the zener diode I think you would have seen it:wink:
     
  8. Dec 10, 2011 #7

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  9. Dec 10, 2011 #8

    gneill

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    For (III), isn't the path indicated in blue the one you're describing?

    attachment.php?attachmentid=41802&stc=1&d=1323554191.gif

    If so, what happened to R4?
     

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    Last edited: Dec 10, 2011
  10. Dec 10, 2011 #9
    This is a strange one FP...you got 4V as Vout, I got 4.1V. There is a nasty coincidence in the numbers which means you got the right answer but by the wrong numbers.
    Here are my calculations...see what you think.
    1) Potential divider means V+ = 6.7V (I have rounded 6.666 to 6.7)
    2) this means V- = 6.7V
    3) This means the V across the 5k into the - input = 8-6.7 = 1.3V
    4) Current through that 5k = (1.3)/5 mA This also equals current through feedback 10k
    6) Voltage across feedback R = 10 x (1.3)/5 = 2.6V
    7) left hand end of feedback R ( the - input) voltage = 6.7V therefore Vout = 6.7 - 2.6 = 4.1V

    Can you see it ?
    Hope this is not judged to be too much help !!!
     
    Last edited: Dec 10, 2011
  11. Dec 10, 2011 #10

    gneill

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    Keep a few more decimal places in your intermediate calculations (particularly the V+ voltage). Round when the smoke clears :smile:
     
  12. Dec 10, 2011 #11
    If you want to set yourself a challenge change resistor to 5k so the potential divider is 5k and 5k in series. If you apply your formula to this you will not get the correct answer.
    You got an answer of 4V because the difference between 10V and 8V is 2V and the gain with 10k as feedback resistor and 5k as input resistor = 2. coincidence:mad:
     
  13. Dec 10, 2011 #12
    The difference in the 2 answers is not due to rounding off numbers !!!!!
     
  14. Dec 10, 2011 #13

    gneill

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    Sure it is. Regardless of whether one was obtained via correct method or not, one answer was 4.0V and the other 4.1V. The difference is 0.1V.

    With a few more decimal places kept your result would be 4.0V also, and the difference would be 0V. Of course, this still doesn't address the issue of whether or not an answer was obtained via an incorrect procedure :smile: Just sayin', that's all.
     
  15. Dec 10, 2011 #14
    Sorry.... we must disagree!!! we need a 'correct' method. If the 2 numbers are the same as a result of a coincidence there is something wrong.
    Change the potential divider to 5k and 5k in series (V+ = V- = 5V)
    The output voltage is not 2 x (10-8) (that is the formula that FP used)
     
  16. Dec 10, 2011 #15

    gneill

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    Oh no, we are in complete agreement there! I was just pointing out that the difference in the result, 4 vs 4.1, was due to rounding. The fact that the '4' was obtained via an incorrect method is another issue.
    Again I agree.
     
  17. Dec 10, 2011 #16
    I am genuinely pleased about that gneill:wink:. This is a horrible question with nice round numbers chosen which make a trap for a student. it is easier to get the correct answer by the 'wrong' method than it is to get the right answer, and any student will learn absolutely nothing.
    As you can probably see, FP did not find out the value of V+ (6.66V) and fell into a terrible trap.
    I am certain we have the same mind set.... I love physics and love teaching, I have made some mistakes in my contributions and what I like about this forum is the variety of approaches to problems.... I have learned a lot about Kirchoffs laws (which I have never been comfortable with !)
    Cheers
    Tech
     
  18. Dec 10, 2011 #17

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  19. Dec 10, 2011 #18
    FP:
    The person who made up this circuit happens to have picked values for resistances that mean you got the correct answer for the wrong reasons.... it is not your fault !!!
    You calculated the gain of the amplifier to be Rf/Rin = 10k/5k = 2.
    There is nothing wrong with that but what it means is that the voltage across Rf is 2 x the voltage across Rin...... it does not tell you directly what Vout is !!!!
    You took Vin to be the difference between 10V and 8V = 2V (the zener diodes) and a gain of x2 made you come up with 4V as the Vout.
    Can you see that V+ = 6.66V due to the 5k and 10k potential divider??
    Therefore V- = 6.66V
    Therefore V across Rin = 8-6.66V = 1.34V therefor V across Rf = 2 x 1.34 = 2.68V
    The voltage at the left hand end of Rf = 6.66V and therefore the voltage at the right hand end of Rf (Vout) = 6.66 - 2.68 = 3.98V ( I am going to say 4V !!!)
    This is the voltage you got but by a different way. And this is the 'trap' produced by the question maker (probably unintentional)
    I imagine this will give you nightmares (when you wake up:wink:)

    It would have been better for you if the resistance values in this question had been more 'random'

    If you are not totally knocked out by this try the calculation with the potential divider across the 10V zener as 5k and 5k
    I will do a solution for you and see if we agree. Sleep tight :redface:
    Cheers
    Tech
     
  20. Dec 10, 2011 #19

    gneill

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    attachment.php?attachmentid=41804&stc=1&d=1323556508.gif
     

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  21. Dec 10, 2011 #20
    I agree with
    10V - (R3 + R4)*I2 =0
    which means that I2 = 10V/(R3 + R4)
    The voltage at V+ of the op amp is the voltage across R4 = I2 x R4 = 10V x R4/(R3 + R4)
    V+ = 10 x 10/15 for the values in this circuit
    i.e V+ = 6.666.....=V-
     
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