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Yet another op-amp problem

  1. Jan 19, 2007 #1
    1. The problem statement, all variables and given/known data
    Determine the values of R1 and R2 to aquire a voltage gain (Vo/Vi) ranging from 2 to 10.


    2. Relevant equations
    [tex] A_Vt = A_V1 * A_V2 [/tex]

    3. The attempt at a solution

    I completely bombed this test question...even though my professor gave me a "hint". he said the first op-amp is a unity gain op amp...which I TOTALLY do not understand why, being that the virtual ground makes Vo1 = 0V, making Vo1/Vi = 0 which doesn't seem to have much unity...

    being Vo/Vi = 0 of the first op amp...there is no way i can multiply any gain from the second op amp to give an overall gain of 2...and that's when i decided my professor smokes crack.

    How the heck do I solve this?
  2. jcsd
  3. Jan 19, 2007 #2
    The gain of the first op-amp is neither 0 or unity. Since it has a potentiometer in the feedback loop, it's gain will vary between 0 and a certain value. What configuration is being used for the two op-amps? Do you know how to calculate the gain of this configuration?
  4. Jan 19, 2007 #3


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    First, you must consider that people who are often addressed as "professor" may know a thing or two about their subject. Having said that, I don't see how an opamp circuit with a variable resistor can be "unity gain".

    Does it? I thought you'd understood this bit before, but it may be that you've confused yourself. The feedback resistor is variable, and not = 0 ohms.

    All that the virtual ground establishes is that V1(-) = 0 (the voltage at the inverting input is zero). You must then apply KCL to find V1(out), using the "zero input current" rule.

    Edit: Crossposted with anto.
  5. Jan 19, 2007 #4


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    I think the circuit is drawn incorrectly. As antonantal points out, shorting out the pot kills the gain to 0, so there is no way to configure the second stage to give you a net gain of anything.
  6. Jan 19, 2007 #5
    well i thought it was gain of the first times gain of the second...

    gain of first: inverting op-amp = (-Rf/R1), but since Rf can be 0.....that brings me back to my original catch-22.

    gain of the second: inverting op-amp = (-R2/R1)

    Is this wrong?
  7. Jan 19, 2007 #6
    I guess I should of mentioned my strategy for doing these problems.

    For this problem, I get an equation for the overall gain when the variable resistor is at 0-ohms (shorted out) and set it equal to 2

    THEN, I get another equation for when the variable resistor is fully involved (10k) and set that equal to 10

    now I have two equations with two unknowns, and bingo I can get the answers.

    unfortunantely the first equation didn't quite work out. is there another way to do this?
    Last edited: Jan 19, 2007
  8. Jan 19, 2007 #7
    It's correct for the drawing you posted. But as berkeman said it seems that the drawing is incorrect. If you would have, at the first op-amp, R1 connected to the ground instead of Vin and Vin applied between the + input and the ground than you could solve the circuit for the demanded range of gain. But the output will be 180 degrees out of phase.
    Last edited: Jan 19, 2007
  9. Jan 19, 2007 #8
    Well it's possible that what my eyes saw didn't match up with what my brain thought it was while taking the test. and I did redraw this op-amp circuit from memory.

    So are these my options?

    1) I didn't get an answer because I didn't look at the question right = I'm an idiot.


    2) my professer gave me a bum problem without realizing it & gave me bum hints = my tax dollars go into his crack smoking habits
  10. Jan 19, 2007 #9


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    No, you are right. I misunderstood what you were saying earlier - my bad. Yes, if you set the trimpot to 0 ohms, the net gain of the circuit is 0.

    So, if you drew the circuit correctly, you have a valid point, but as you and others have indicated, you may be incorrectly recalling the circuit topology.
  11. Jan 19, 2007 #10
    It's hard to think that your professor gave you a wrong problem since it's pretty basic. But maybe he did. And it doesn't mean that you are an idiot if you didn't solve it right. You just need more practice. :smile:
  12. Jan 19, 2007 #11
    Well I will update this when I get my test back. Thanks guys (and gals if any of you are gals)

    I almost hope it was me that messed up instead of him...because I don't want to be taught this stuff from someone who doesn't understand it himself
    Last edited: Jan 19, 2007
  13. Jan 22, 2007 #12
    Well when I got my test back today, I was hoping to hear a "i omitted this problem because it was drawn incorrectly" speech or something, and to my surprise, he didn't mention anything about it and marked my paper wrong.

    here it is scanned straight from the test:


    I asked him about it when class ended, and he said the gain of the first op-amp was -1 so Vout of the first op-amp = -Vin (when the potentiometer is shorted). I still don't understand this being that there is a virtual ground connected straight to the output. He said the correct answers were R1 = 2k and R2 = 4K. Some of my classmates were also confused and wondered if he was following the ideal op-amp rules. Is my professor still visiting his crack dealer or can someone explain this?
    Last edited: Jan 22, 2007
  14. Jan 22, 2007 #13


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    Hmmm. Okay, let's set the pot to zero, that means the input current Vin/R1 gets sinked by the output of the opamp, so that its - input is held at ground. And since the - input is held at ground when the pot is shorted, that means that the voltage Vin has no effect on the output voltage (hence zero voltage gain in the first stage). So once the pot is shorted, you cannot get gain through the pair of stages, since 2 = 0 is nonsense.

    Ask him again to show you how the gain of the first stage is -1 instead of zero with the pot shorted out. Ask him what the gain is when the pot is set equal to R1 :rolleyes:

    Maybe build the dang circuit on a breadboard and show the gain is zero -- should get you some of those 10 points back.
    Last edited: Jan 22, 2007
  15. Jan 22, 2007 #14
    So it seems that after all your professor needs more practice :smile:
  16. Jan 22, 2007 #15
    lol, luckily my professor was doing a lab today...so i went and built it! and sure enough I got nearly 0V for Vout when 10k was shorted. I showed him and he goes "i guess that's when happens when i make up a problem at midnight". So now my whole class gets that problem omitted.

    lmao! don't mess with me when I have PF backin me up!
  17. Jan 22, 2007 #16


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    :rofl: :rofl: :rofl: :approve:
  18. Nov 25, 2007 #17
    Your professor is righ...
    You can use R1=2K and R2=4K as your professor said.
    It's a kind of tricky question. The key is the gain=2 does not start when Rvariable=0ohm but it start when Rvariable=R1. Rvariable can vary from 0 to 10Kohms but it does not state in this problem that Gain=2 must be start with Rvariable=0,right?
    So if you use R1=2K and R2=4K, at
    Rvariable=2K, the o/p of 1st op-amp= -Vin and the o/p of 2nd op-amp= 2Vin
    Rvariable=3K, the o/p of 1st op-amp= -1.5Vin and the o/p of 2nd op-amp= 3Vin
    Rvariable=4K, the o/p of 1st op-amp= -2Vin and the o/p of 2nd op-amp= 4Vin
  19. Nov 25, 2007 #18
    hmm...it's been awhile since that test...yeah your way is right, but if that's the case then you could set the starting value of Rvariable anywhere (but zero) as long as it solves out correctly. and if THAT was the case the question SHOULD have explained that the range needed to be set. on top of that, the professor's explaination later assumed the range was 0 to 10k and I got him to give me my points back and jokingly tell me that he doesn't need to make tests while watching football :p
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