another op-amp transfer function

Discussion in 'Homework Help' started by derill03420, Apr 15, 2012.

  1. derill03420

    Thread Starter New Member

    Oct 15, 2011
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    I need some help on how to approach this one, the way i see it i can combine C1 and R1 and then because it is ideal no current flows into - terminal so same current is through R1//C1 and R2, then if that is true i have one node that i do kcl on which would be at the output of the op-amp, is this thinking correct?

    if it is correct then i would proceed by saying i1 - i2 = 0 (from where i re-drew circuit) and remove i3 from my drawing since the current would be i1
     
  2. panic mode

    Senior Member

    Oct 10, 2011
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    assuming ideal opamp, there is no current flowing in or out inputs hence i2=i5=i3+i4
    also input voltage is same at both inputs (v1=v1+=v1-).
    so we can also write v1=i5*R2

    BUT while we asusme that OpAmp inputs have infinite impedance, this is not the case with output (we assume that output has low impedance) so you CANNOT write i1-i2=0 because those two currents are not related (there IS current flowing in or out of OUTPUT of the OpAmp).

    but keep working and post if you get into any trouble.
     
  3. derill03420

    Thread Starter New Member

    Oct 15, 2011
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    ok here is my preliminary equations, but where im stuck is do i really need 2 equations? could i say i1-i2-i3 = 0? or should i just do kcl at output node and express my equations as i1-i2 = 0?

    Also i was in a hurry and forgot ground symbol on bottom of inductor and 10k resistor
     
  4. derill03420

    Thread Starter New Member

    Oct 15, 2011
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    attachment for previous post forgot to hit upload lol
     
  5. panic mode

    Senior Member

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    you cannot write KCL at opamp output - because you don't know amount of current flowing in or out of opamp's output. ok you CAN write it but it is useless because we can't determine current coming out of opamp. the only thing you can use for that node is KVL. so PLEASE get rid of i1-i2=0 nonsense.
     
  6. panic mode

    Senior Member

    Oct 10, 2011
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    look at this circuit, i added I+, I- and Io.
    we can neglect I+ and I- and assume they are zero.
    But Io is NOT zero, in fact it is quite significant, specially when compared to to I2.

    if you want to write KCL for output node, then it is:

    Io=I1+I2

    I don't know about your teacher but most would deduct points just for writing I1+I2=0 because it is way way wrong.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    As panic mode has already stated, you CAN'T do KCL at the output node because you have no way of determining what the current flowing into (or out of) the amplifier output is (until after you solve the circuit and THEN you can use KCL to find it). Or, said another way, if you use the equation that results from applying KCL at the output of the amp, the the amp output current has to be included as one more unknown.

    Ask yourself the following questions:

    1) Other than drawing additional current from the opamp (which, being ideal, has infinite drive capability), what impact does R3 have on anything?

    2) What components can affect the voltage V1? Consider this in light of the answer to the previous question.

    3) Given an output voltage from the amplifier, which components determine the voltage at the inverting input of the opamp (which I will call V2).

    4) Are there ANY components in the answer to question #2 that are also in the answer to question #3?

    5) What do you know about V1 (written as a function of the input Vg) and V2 (written in terms of V0)?

    Finally, I see that you still refuse to track units. Having seen people literally die because they couldn't be bothered to track their units, I will no longer assist you in becoming yet another incompetent engineer that puts people's lives and property at risk because you feel either that (1) an engineer has no obligation or responsibility to use even the most basic methods of checking their work and catching their errors, or (2) that you believe that such obligations and responsibilites simply don't apply to you.

    I will start helping you again if and when you decide to start tracking your units.
     
  8. panic mode

    Senior Member

    Oct 10, 2011
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    Proper analysis would go something like this:

    i2=i3+i4=i5+i-

    since i- is assumed to be negligible, we can write APPROXIMATE (!) equation:

    i2=i3+i4=i5

    similarly we assume i+ is negligible and BECAUSE of that we can APPROXIMATE input voltage as:

    v1=vg*R4/(R4+sL1)

    since for ideal opamp in normal conditions input voltages are assumed to be same:

    i5*R2=v1

    for reasons mentioned above, in feedback loop we have APPROXIMATELY:

    (vo-v1)/Zf=v1/R2

    where Zf=(R1||C1)
    Zf=(R1/sC1)/(R1+1/sC1)=R1/(1+sR1C1)

    Now we have enough equations to find vo and transfer function T(s)=Vo(s)/Vg(s)
     
  9. mlog

    Member

    Feb 11, 2012
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    It seems to me that the transfer function can be found by first finding V1, which is formed by the voltage divider of R4 and L1. When you get V1, you can then find the current through R2. The current through R2 is also the current through the parallel combination of of R1 and C1. The output voltage is simply that current multiplied times R2 plus the parallel R1-C1. (Of course, you need to observe the negative sign at the output.)
     
  10. derill03420

    Thread Starter New Member

    Oct 15, 2011
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    [QUOTE

    Finally, I see that you still refuse to track units. Having seen people literally die because they couldn't be bothered to track their units, I will no longer assist you in becoming yet another incompetent engineer that puts people's lives and property at risk because you feel either that (1) an engineer has no obligation or responsibility to use even the most basic methods of checking their work and catching their errors, or (2) that you believe that such obligations and responsibilites simply don't apply to you.

    I will start helping you again if and when you decide to start tracking your units.[/QUOTE]

    In physics it was very important to track units because you are very correct that is how i knew i had correct answer, but our professor has never included units in circuit analysis and thats how im used to working so far with this stuff. Plus he has taught us that in s-domain analysis everything is expressed in ohms so i guess i don't see the point, but i guess im wrong
     
  11. derill03420

    Thread Starter New Member

    Oct 15, 2011
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    [QUOTE

    similarly we assume i+ is negligible and BECAUSE of that we can APPROXIMATE input voltage as:

    v1=vg*R4/(R4+sL1)

    [/QUOTE]

    isn't the voltage at v1 or + terminal a voltage divider? so it would be [L1/(L1+R4)]*Vg if im wrong please explain why?

    So if i just use relationship of i2 = i5 to compute Vo and Vg that would be correct?

    which would be (Vo-V1)/(C1//R1) - V1/R2 = 0 C1//R1 means parallel combo of C1 and R1, im sure you know that just making sure were on same page
     
  12. WBahn

    Moderator

    Mar 31, 2012
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    Many professors don't care about units and, likewise, many textbook authors don't, either. Then again, for all too many professors and textbook authors a wrong answer is merely an academic experience. In the real world, wrong answers have real consequences.

    It is simply wrong that in the s-domain everything is expressed in ohms. Are voltages expressed in ohms? Currents? Inductance? Capacitance? Frequency? Are the units of Cs, RCs, LC all expressed in ohms? Even if they were, you can't add something in ohms to something that is dimensionless or something that has units of ohms-squared.
    Finally, if they 'should' all be in ohms, what if you make a mistake and use (1/(Cs)) where you should have used (Cs)? Guess what, it messes up the units and if you track the units you will catch it. But if you just blindly go along and don't bother to check because 'everything is in ohms', then you will produce completely wrong answers and be completely oblivious to the fact.

    If nothing else, consider this: Your grade is primarily based on the points you lose while working problems. Think about all of the points you have lost because of algebra mistakes or typos. Now think of what your grade would be if you could get 75% or more of those points back. I have seen many students go from C level work to A level work once they committed themselves to ALWAYS tracking units and ALWAYS doing sanity checks and almost all of them fought it at first because, after all, no one else did it.
     
    Last edited: Apr 16, 2012
  13. panic mode

    Senior Member

    Oct 10, 2011
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    hi Darryl,

    you are right, that was my mistake:

    correct equation should be:

    V1(s)=Vg(s)*(s*L1)/(s*L1+R4)

    but this is less important (do as i say, not as i do :p).

    anyway my point was that when doing analysis, one should write things out, and if something can be neglected that's ok but explain why. as you have seen, at different pins of the opamp, we apply different rules. while some assumptions apply to inputs,opamp output is different story. if the input current was not negligible we could not use voltage divider approximation, we would write more complex equation and take that current into consideration (if we knew what that current is). fortunately, usually we don't need to worry about that.

    I agree 100% with WBahn about taking care of units. there are many examples of things gone wrong because someone didn't bother to do it properly.

    Btw. How is the transfer function coming along? Anywhere close...?
     
  14. derill03420

    Thread Starter New Member

    Oct 15, 2011
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    here is what i get
     
  15. WBahn

    Moderator

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    Are you sure it's X/Y in your last line?

    Since you jumped in right away with the component values (and no units tracking), it is very difficult to do sanity checks on your final result.

    But one question you could have asked is what the transfer function should look like at DC (s=0). At DC, the inductor will look like a short circuit forcing V1 to be 0V which, consequently, will result in no current in the 20kohm resistor and, from there, no current in the 25kohm resistor which will therefore force Vo to be 0V regardless of what Vg is. Therefore H(s=0) should be 0. But in your answer it blows up toward positive infinity. Noting that this is not right, you could have gone back to the next-to-last line and discovered that, there, it does show that Vo goes to 0V regardless of Vg. So you would have known that there was a problem in going from the next to last line and the last line and probably would have found it within seconds.

    Gee, think of the points you could have avoided losing if you had only decided that it was worth the 30 seconds it took (literally, that's what it took) do a check that didn't even require a single calculation.

    Let me ask you this: Would you put your life in the hands of a surgeon that didn't believe it was worth making simple checks to avoid making mistakes? For instance, it is routine practice to do things like write "Not this one" on the leg that isn't to be amputated or counting all of the instruments and supplies before sewing up someone's chest cavity. Would you agree to be operated on by a surgeon that didn't see any point in bothering with those pesky little checks? I'm guessing the answer is no. But before you say that a surgeon could really hurt or kill someone if they made such a mistake, let me remind you that surgeons typically kill people one at a time with their mistakes while engineers kill people in job lots.
     
  16. derill03420

    Thread Starter New Member

    Oct 15, 2011
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    ok some checks like at DC s=0 i didnt know to do but now i know, and on my next look over this ill post with symbols and units. i appreciate ur help and this is only a study guide its not a test, the test is wednesday. My teacher moves very quick and its rough when we get a study guide with op-amps as transfer problems when he didnt cover this in class. i REALLY APPRECIATE ur help.
     
  17. WBahn

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    Mar 31, 2012
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    Good, because I'm not going to stop harping on this stuff. For context, when I teach a course (for which this is applicable), I tell the students Day 1 that they WILL track their units and check their answers or they WILL fail the course. I realize that most of them have not previously been expected to do so and, therefore, I use a ramped approach. At first, I note the lack of units and such and take off between 5% and 10% of the problem value. As the semester increases, the point penalty grows. By the end of the course, I will typically take of 50% for not tracking units, 75% for not catching an error that tracking units would have caught, and 100% for making a mistake as a result of not tracking units. However, if someone that has established a pattern of consistently tracking their units makes a goof and leaves units off in one spot or fails to spot a subtle units mismatch, then I treat that the same as any other math error and it usually costs about 5%.

    I once had a group of students that, early in the semester, tried to get up a petition to have me removed as the instructor. The Department Head got wind of it a few weeks after it started (I was completely unaware) and asked to speak to the class (without me present) one day about mid-semester. In our conversation afterward, he said that the biggest complaint was my units policy but that the class had settled down because I had backed off on it. I actually laughed out loud because I had, in fact, already tripled my penalty points but the students simply weren't making units mistakes anymore. I wasn't the one that had changed -- they had, and they didn't even realize it! As it turned out, all three of the instigators of the petition stayed in touch with me for several years and each of them ended up being very grateful for the painful lesson (and it WAS painful for them and me, believe me!).

    I've had students come back years later, either from industry or from grad school, and thank me for being a Units Nazi. But that was only fair; I had made a point many years earlier to thank the professor in my past who had driven home the lesson for me (and I had always been pretty good about units ever since high school chemistry, but I just carried them through and used them to help with unit conversions and did not use them as a general means of error detection).
     
  18. panic mode

    Senior Member

    Oct 10, 2011
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    hi derill,

    i tried to derive same transfer function and got following:

    (vo-v1)/Zf=v1/R2
    vo-v1=Zf*v1/R2
    vo=v1(1+Zf/R2)

    note:
    v1=vg*sL1/(R4+sL1)
    v1=vg*s/(s+2000)

    note:
    Zf=R1||C1
    Zf=25000*(1/s*10*10^-9)/[25000+1/(s*10*10^-9)]
    Zf=25000/[1+25000*s*10*10^-9]
    Zf=25000/[1+s/4000]
    Zf=25000*4000/(s+4000)

    vo=[vg*sL1/(R4+sL1)]*(1+Zf/R2)
    vo/vg=(1+Zf/R2)*sL1/(R4+sL1)
    H(s)=vo/vg=s*(s+8000)/[(s+2000)(s+4000)]

    aside from 8000 vs 9000 in one of terms, this seem to be reciprocal of what you got. one or both of us did something wrong. anyway, it is super late and i'm dead tired, i'll check it again later on.
     
  19. WBahn

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    Mar 31, 2012
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    I only scanned your solution, but noticed that there didn't seem to be anyplace that I could recognize as the 20kohm resistor. Is it possible you used the 25kohm value of the other resistor in both places? I think that would account for the discrepancy.

    And his solution is incorrect only in being the reciprocal of the correct answer (assuming I did it right).

    The reason for his error can be seen by looking at what he defined X and Y in the next to last line to be and then how he used them in the transfer function on the last line.

    BTW, derill, had I been grading this (units and such aside), the fact that you identified what X and Y were and then showed how they were used in the final equation would have kept the point loss to a minimum since it was obvious (before I even looked at the details of the final expression) that you had made a mistake and that the mistake was a minor math manipulation blunder. So good job on that count.
     
  20. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I make the transfer function as ...

    H(s)=\frac{s(s+9000)}{(s+2000)(s+4000)}
     
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