Op Amp Output Impedance

Discussion in 'Homework Help' started by eng kryptonite, Jan 25, 2010.

  1. eng kryptonite

    Thread Starter New Member

    Jan 25, 2010
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    In the circuit attached:

    R2 = R3 = R4 = R5 = 18.2 kOhm, R6 = 5.2kOhm. Whats the value of R1 that will make the output impedance thats seen by ZL infinite? Answer is in Ohm.
    I don't understand what he means by infinite output impedance. I still don't understand how to get the value of ZL to be infinity by setting up some value for R1.

    No given for Vin.

    Thanks a lot!
     
  2. JDT

    Well-Known Member

    Feb 12, 2009
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    A circuit with an infinite output impedance is a perfect current source. It would supply (or sink) a constant current regardless of what is connected to it or at what voltage the output load was at.

    A perfect current source cannot be made, of course. There are various circuits using op-amps that can produce a fairly high (almost constant current) output impedance.

    A classic circuit that does this is attached.

    I'm not sure if or how your circuit can do the same thing.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    What the question is really asking you to do is to find the Thevenin equivalent resistance at the load point - i.e. the Thevenin resistance seen looking back into the circuit where ZL is connected and with ZL removed.

    Remember Rth=Voc/Isc

    You can write this as a simple equation with the unknown R1 as the independent variable. Once you obtain this relationship you will be able to find the R1 value at which the Thevenin resistance is infinite.

    This concept only works for ideal operational amplifiers - which don't exist in practice.
     
  4. Audioguru

    New Member

    Dec 20, 2007
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    The original circuit uses positive feedback to try to have infinite output impedance.
    It is impractical because it will try to oscillate.
     
  5. steveb

    Senior Member

    Jul 3, 2008
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    Expanding a little further, you can find the formula for output impedance by shorting the input voltage source (Vin), and imagine that you are driving the output node (node with ZL connected to R6, but disconnect ZL to calculate) with a current source (Io), which then generates an output voltage (Vo). Calculate the transfer function Zo=Vo/Io. You'll see that the value of R2 and Vin do not matter, and that the equations are not too bad if you write them out. You'll have linear equations to solve to get an expression for output impedance (Zo). The equations will show that there is a positive feedback term (due to the R1 feedback) that can drive the denominator of the equation to zero, which equates to infinite output impedance.

    As mentioned, driving the impedance all the way to infinity is not practical because tolerances can bring you into the unstable region. The clue is that the output impedance can become negative if R1 becomes just a little too small. To be practical you would need to make R1 a little larger, and settle for a moderately high output impedance. I havent' fully analyzed this circuit, but my gut says that this is not a great current source. But, you can still solve the problem as asked.
     
    Last edited: Jan 25, 2010
  6. Ron H

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    Apr 14, 2005
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    Not necessarily. The Howland current pump uses positive feedback, and it will not oscillate, if used properly.
    [​IMG]
    Go to this Bob Pease article for the source of this schematic.
    The output impedance can be infinite (or can even be negative, depending on feedback ratios), but the net impedance at the load (including the load) needs to be positive to prevent oscillation.
    I see no reason why the OP's circuit will be unstable, as long as the net impedance at the load is positive.
     
  7. steveb

    Senior Member

    Jul 3, 2008
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    This is a very good point. When we calculate output impedance, we assume the load is infinite and look back into the circuit. Hence, the stability condition is implied to be only valid with infinite load. A proper stability condition would include the load.

    Assuming, finite load impedance, this circuit should be theoretically stable with R1 set to the infinite output impedance condition. However, Audioguru might still be essentially correct. If tolerances are considered, this may not be a practical circuit.
     
  8. eng kryptonite

    Thread Starter New Member

    Jan 25, 2010
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    Thank you very much guys. This was very helpful.

    I know its an impractical circuit but I'm required to do it as a homework :/ Now I've done the analysis (which I'm not sure if it is right as I'm not very good at complicated op amp circuits) and got the following:

    Vo = (Vin R1 R3 R5) / [R2(R1R4 - R3R5 + R4R6)

    To make the denominator goes to 0 You'll have to find a value for R1 which will do that since we had the rest given. I got 11kOhms for R1 and that will make the denominator goes to 0 which will make Vo infinite then?

    What I did .. I substituted ZL with noting and put Vo instead to calculate for that .. still not sure if thats correct.
     
  9. Ron H

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    Apr 14, 2005
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    Here is my approach: If you substitute a voltage source (call it Vout) for ZL, then for infinite output impedance, I(Vout)=0. This means that I(R1)=-I(R6). Since the summing node of A1 is at zero volts (virtual ground), the value of R2 is irrelevant (assume for now that Vin=0). You can omit it when solving for R1. R2 just determines the transconductance.
    Just solve for R1. I have solved and simulated it, and the answer is not 11k.:D
     
  10. steveb

    Senior Member

    Jul 3, 2008
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    I agree with Ron, but I'll give an answer from a different point of view.

    First of all, your approach is not really correct. You calculated the gain, under the assumption of infinite load impedance. You got that answer right which is a good sign, but the gain does not help you here. What you really need to calculate is the output impedance directly. My first post in this thread explains how to do this. It turns out that (coincidentally) your calculation should have given the correct value for R1, which is not 11K. You just miscalculated. However, your approach is not really correct and I think you are just lucky that you can get the right answer with the wrong approach.

    You need to learn the formal way to calculate output impedance as I described. If my explanation is not clear, I can post the equations and a schematic to illustrate better.
     
  11. eng kryptonite

    Thread Starter New Member

    Jan 25, 2010
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    Thank you Ron. :)

    Steve, it will be very helpful if you could explain how to calculate the output impedance the formal way because I never got the chance to be able to follow my teacher in op amps :( (I don't wanna say she was not clear enough when she explains :D) .. I would really appreciate it. I can analyze basic op amps (taught myself those .. non-inverting .. etc) but when it gets a bit complicated I have hard times figuring them out.

    Your help is very much appreciated! :)
     
  12. steveb

    Senior Member

    Jul 3, 2008
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    In my view, output impedance calculations are tricky. They are conceptually confusing, and the math (although simple) plays tricks on you at first. It's hard to explain why, but you will see it when you do examples. Anyway, I've attached a brief description on how to proceed and show your example. No doubt I've not included every concept, but this will get you started.

    Note that generally, you need to have (or you must establish) a linearized circuit model. You will then short out any ideal voltage sources and open up any ideal current sources in the circuit. This is the standard approach. The attached PDF describes the details from there.

    Feel free to ask questions if this is not clear. Good luck and note that you need to practice many examples before this approach becomes second nature.
     
  13. eng kryptonite

    Thread Starter New Member

    Jan 25, 2010
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    Thank you very much!!!!!!! :D:D:D:D:D
    Ok now I get what were you trying to explain. I think I was close somehow but needed a little correction. This was very helpful and I appreciate taking the time to explain this to me. :)
     
  14. eng kryptonite

    Thread Starter New Member

    Jan 25, 2010
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    Ok, so if you don't mind explaining a little further about op amps analysis (Sorry for asking too much questions, I have to wrap up quickly for class). I have attached a circuit containing Diode. Now I don't really know exactly how to analyze an op amp circuit containing a diode. But I went through a couple of powerpoint slides to figure out the answer but still no big idea about it.

    When, V1 = 3.2V, V2 = -0.3V, calculate Vo .. All resistors have the same value.

    My analysis came to:

    Vo = R3R6(V1R2 + V2R1) / (R1R2R5)
    Vo = 2.9V

    If anyone would help me with how to analyze this kind of circuits this will be wonderful.

    Thanks a lot :)
     
  15. steveb

    Senior Member

    Jul 3, 2008
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    There are a number of ways to handle OPAMP circuits with diodes. Generally, OPAMP analysis requires a DC analysis which makes doing a small signal AC analysis (as is often done with transistor circuits) less useful, typically. One method is to treat the diode as a 0.6V DC voltage source when diode current is flowing forward, and treat it as an open circuit when the voltage at the diode terminals is not forward biased. With this method, you have to carefully break up the solution into pieces based on when/if the diode switches from forward to reverse biased, or from reverse biased to forward biased. Sometimes this is easy, and sometimes it's a pain. When coils are involved, it's more difficult. This method is not perfect, and there are other methods too.

    I sometimes prefer to use the exponential Shockley diode law Id=Io*(exp(Vd/nVt)-1), but this requires nonlinear circuit analysis. The nice thing about this approach is that you don't need to break the solution into pieces because the Shockley equation works for both forward and reverse biased conditions.

    EDIT: A further clarification is needed here. With any approach you use to model the OPAMP circuits with diodes (or any other nonlinear element), at some point you may need to convert to an equivalent linearized circuit, if you want to do certain AC calculations like small signal gain, input impedance and output impedance. There are various linearization methods, but one of the first ones you typically use is to solve for an operating point (DC Q-point) and then create a linear small signal equivalent circuit which is valid for small perturbations around that operating point.
     
    Last edited: Jan 26, 2010
  16. t_n_k

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    Mar 6, 2009
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    Since I had suggested the Thevenin approach I have attached my solution which gives the same result as that found by steveb.
     
  17. Ron H

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    That yields the same result for R1 as I got with my method, which only solves for R1, and does not give an equation for Zth.
     
  18. t_n_k

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    Yes Ron your approach is much more succinct - particularly as the question was only asking for the "magic" value of R1 rather than a lengthy derivation of Zth or Zout.

    It's the annoying tendency in me to "over design" rather than look for the elegant solution.:)
     
  19. Ron H

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    Apr 14, 2005
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    Well, knowing how to derive Zth is important, and could yield valuable insight when combined with the load impedance. I just zeroed in on R1 because that was all that was asked for, and it was a lot easier.:p

    BTW, I disagree that this circuit is impractical - although it might not be optimum from a parts count standpoint.
    I have seen the general circuit configuration before, but I'm danged if I can remember what it's called.
     
  20. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Everyone agrees that to make the output impedance infinite, we need to
    set R1*R4 + R4*R6 - R3*R5 equal to zero.

    This gives an expression for R1:

    R1 = (R3*R5 - R4*R6)/R4

    but, what happens if R3*R5 = R4*R6?

    If R1 is zero, then is the output node shorted to the minus input of A1? And we still have Zout = ∞? Probably not. It's fighting infinities.
     
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