Op-Amp Tee to Pi transform

Discussion in 'Homework Help' started by J_Rod, Feb 15, 2015.

  1. J_Rod

    Thread Starter Member

    Nov 4, 2014
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    op amp based circuit.png
    Good morning...
    I was wondering if I can apply the above transformation on this op-amp based circuit, with a T resistor junction turned into a pi resistor network. I have the equations to do so since all three resistances are known, but is it okay to do that? My plan is to then use the value Rf or Rc in the red transformation to find the output voltage in:
    Vo = Vout = -Rf/Rs *Vin

    R1 = 60k ohm, R2 = 30k ohm, R3 = 40k ohm
    Rc = (R1 R2 + R2 R3 + R1 R3)/R3 = (60k*30k + 30k*40k + 60k*40k)/40k = 135k ohm
    Rf = Rc = 135k ohm
    Vo = -135k ohm/8k ohm *0.240V = -4.05V
    This makes sense since -15V < -4.05V < 15V, but is it acceptable to use the transformation?
    Thanks for your time folks.
     
  2. crutschow

    Expert

    Mar 14, 2008
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    That conversion does appear to work to give the correct gain.

    I would solve for the current into the summing junction due to Vo and use that to calculate the equivalent Rf.
    (Solve for the voltage at the junction of R1, R2, and R3 due to Vo. Use that voltage to then solve for the current through R1 into the summing junction (at 0V). From that current you can solve for the equivalent Rf).
     
    Last edited: Feb 15, 2015
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  3. J_Rod

    Thread Starter Member

    Nov 4, 2014
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    Ah that's good to know this will work ;) I don't follow what you mean - I think to analyze op-amps the prof said to use KCL or nodal analysis, which is what you mean. I just couldn't figure that out because of one too many unknowns... Can the 8k ohm and 60k ohm resistors be combined in series, since there is no current effectively entering the op-amp? Then I could use node analysis at the top of the tee, and a KVL walk to get the second equation to eliminate va.
     
  4. crutschow

    Expert

    Mar 14, 2008
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    Remember that the summing junction (- input) of this circuit is a 0V so the input resistor is not seen by the feedback resistor.
    There is no current going into the op amp itself but the node is at 0V due to the negative feedback, which forces the input current to be balanced by an equal and opposite current through the feedback resistance.
    So you can calculate the feedback resistance independent of the input resistance.

    So here's my approach:

    Calculate the parallel resistance of R1 and R3.

    From that calculate the voltage at the resistor's junction due to Vo (the other end of R1 being effectively at 0V as noted above.)

    Now calculate the current through R1 (Ir1) to the summing junction (again at 0V) due to this calculated voltage.

    Then calculate the effective Rf resistance as Vo / Ir1.

    That's not a strict nodal analysis but it gives the correct answer.

    You can do a more formal voltage or current nodal analysis if you like.
    Just remember that the left end of R1 is effectively at 0V.
     
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  5. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    I think the transformation works because it is after all a resistive transformation that, well, works.

    But im not sure why you would want to do this since other types of analyses are so simple anyway, since you elect to spec the op amp as having zero input offset.
    Since you elect to do that, then the current flowing in the 8k resistor is the same as that flowing in the 60k resistor, so the voltage drop across the 60k resistor must be equal to:
    60*0.240/8=1.8 volts

    and to keep the inverting terminal at zero volts this must be negative so we have at the center node -1.8v. Note that already we have the voltage at the central node with very little work, and that should be the initial goal because once we find that we can easily find other things using Ohm's Law.

    Next, the 40k resistor draws current:
    i40=-1.8/40000=-45ua

    and since the 60k resistor passes -30ua the total is -45ua-30ua=-75ua, so the output must supply -75ua through the 30k resistor, and that means the 30k resistor has a drop of:
    (-75ua)*30000=-2.25v

    and since the left end of the 30k is already at -1.8v that means we have at the right side of the 30k resistor:
    -1.8v-2.25v=-4.05v

    So we got the output voltage with just a few simple steps, and using the zero input current approximation to the op amp (ie current in the input also flows in the feedback).

    One catch with the transformation though is that if the resistances on both ends do not come out to be the same value, then you might have to be careful which value you choose for which end of the network because only one end resistor actually does anything when there is some input offset. The right side resistor does nothing, but the left side is critical to getting the transformation to be EXACT as the original down to the millivolts. For this particular network it is probably the larger of the two that goes on the left, and the right doesnt matter at all as long as it doesnt draw too much current from the op amp output (note the op amp output is taken to be an ideal source so the load resistor does nothing). In a model with some output impedance then the resistor would matter, but probably only down in the microvolts.
     
    Last edited: Feb 16, 2015
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  6. WBahn

    Moderator

    Mar 31, 2012
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    Yes, the transform is valid because, as indicated by Mr. Al, it is a transform that transforms between two equivalent resistor configurations. In this case it is even power-equivalent (overall), unlike most transforms.

    But there is some fine print that you may or may not have dealt with properly. Your approach is based on all of the current going through Rf. In this case that is true because Rb is at 0V on both sides (hard ground on one and virtual ground on the other) and therefore has no current flowing through it.

    Once you jump through the gymnastics to do the transform, the solution is then fairly obvious and simple. For that reason it also has an elegance to it. The real question is whether the time and effort needed to first recognize that the transform is even an option, that it will make things simple, and then to carry it out is a net gain versus using a more direct approach as proposed by crutchow. Hard to say. I almost certainly would have not done the transform, but mostly because I just don't look for or spot those options because I don't have the transform equations memorized and don't like looking things up (or deriving them from scratch too often).

    I would have done it by noting that the current from the signal source is 30mA and has to flow through the input resistor and R1, making the voltage at the junction between R1 and R2 equal to -1.8V. With that, I know that the current in R3 is 45mA upward and that this combines with the current from the source giving 75mA which then flows through R2 resulting in an addition 2.25V drop giving a total voltage at the output of -4.05V. In this case these are all simple enough to do without using a calculator or even writing anything down, but that's more the luck of the draw than anything else.
     
    Last edited: Feb 17, 2015
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  7. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    I am not sure why you would repeat the same approach i had outlined in my previous post, except the currents are in micro amps not milli amps.
    The currents should be in microamps. Do you see any other difference?
    This view of the op amp seems simpler to me for this kind of problem.
     
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  8. WBahn

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    Mar 31, 2012
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    Yes, it should be in uA. Thanks for the catch.

    I was walking through the approach I would take primarily in contrast to either the TS's approach or crutchow's. Frankly I didn't read yours closely enough to recognize that it was the same way I would do it, otherwise I would have just mentioned that.
     
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  9. MrAl

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    Jun 17, 2014
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    Hi,

    Well it is nice to see that we both would have done it the same way though, so that's a good point.
    I was also wondering if the OP had another reason for doing this, perhaps to look at the resistor network transformation itself but it was written more like it was about solving the op amp circuit so i decided to fly with that.
     
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  10. WBahn

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    I suspect that it was to solve the op amp circuit and that the approaches we would have followed just didn't suggest themselves to him. It's good that they looked in their toolbox and tried to see if any of the tools available there, such as the tee-pi transform, there could be employed.
     
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  11. crutschow

    Expert

    Mar 14, 2008
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    Your calculator needs new batteries. :rolleyes:
     
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  12. atferrari

    AAC Fanatic!

    Jan 6, 2004
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    I need to sleep more. I hope had deleted the post before anyone had seen it but no...
     
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  13. MrAl

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    Jun 17, 2014
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    Hi,

    Yeah it is funny sometimes if you make a typo and post it, when you go to edit it someone has already 'quoted' your post so the typo appears in the quote. Kind of nutty :)
     
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  14. atferrari

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    After going again to bed I started to revise mentally the orders of magnitude and said: "it is wrong". Too late; somebody had already quoted me!
     
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  15. J_Rod

    Thread Starter Member

    Nov 4, 2014
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    My reason for using this transform was to find the equivalent Rf, so that I could apply the gain formula Vout/Vin = -Rf/Rs. The question was to find Vo in the circuit. You're right that I couldn't recognize the straightforward analysis, so just saw something familiar in the tee-to-pi transform, which we had learned in class around that time also. At any rate, I'm so glad to have all these responses because I have learned more about the OP amp now.
     
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  16. atferrari

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    Jan 6, 2004
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    Resourceful man you. saludo con sombrero.gif
     
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  17. WBahn

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    The only concern that I have is that you might be looking too much at how to fit things into formulas instead of being able to analyze a circuit based on basic principles. On the flip side, you are not blindly throwing formulas at things, which is all too common, so I'm confident the needed analysis skills won't be too long in coming.
     
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  18. atferrari

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    A delicate balance, WBahn, isn't it? In my activity I see quite frequently Chief Officers that hardly could say why they are actually doing regarding weights distribution on board (no matter how intuitive it could seems but not always is).

    Now, back into Electronics, I can see that lot of people, besides the formal reasoning, tends to use personal "models" where the concept is put in action using surprising analogies.

    In my case I made my life simpler using the seesaw idea to see how a summing amplifier works.
     
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  19. MrAl

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    Jun 17, 2014
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    Hi again,

    Well i do loudly applaud you for finding another way to do this in part because i always advise readers to do a problem two different ways, and if both techniques calculate to the same result, the chances are better that you are correct than if you just do it one way. So do it both ways and then compare your results and i think you will be very happy :)
     
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