2 port network problem

Discussion in 'Homework Help' started by StasKO, Jan 10, 2013.

  1. StasKO

    Thread Starter Member

    Apr 28, 2012
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    Hello again,

    i need help understanding how to approach this problem.

    we need to find the output resistance Ro.
    my first approach was to use Z-parameters, and more specifically only the Z22 parameter (port 2 is the one where Ro is). so i set i1 (the current from the source Vs) to zero (effectively disconnecting Vs) and then solving for v2/i2. my answer in this method is incorrect.
    i tried to use thevenin but this got me nowhere as well.

    now when i finally gave up i looked at the full solution and what our teacher did is setting Vs=0 and then finding v2/i2.

    i cant understand why its ok to do this and how in the first place to know that this is the correct method (if i didnt look at the final answer i would not know that using z parameters was giving me wrong answer)?

    setting Vs=0 is Y parameters and therefore should be i2/v2 and not the opposite.. right? and if we wanted v2/i2 (Z parameter) we need to use the transform formulas (the ones with determinants)...?
    and why set Vs=0 but leaving the dependant source gV untouched..?

    im confused. please help
    thnx
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You don't want to set I1 to zero, thereby disconnecting Vs, because the Thevenin impedance of a voltage source is zero. Replace Vs by a short and analyze again.

    Show your work so we can see where your mistakes are (assuming you don't get the right answer after following my advice above).
     
  3. StasKO

    Thread Starter Member

    Apr 28, 2012
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    but shorting Vs is equivalent to calculating y parameter y22=i2/v2, no?

    EDIT: i would like to make my question clear - can Ro be solved with Z-parameters? if not then why?
     
    Last edited: Jan 11, 2013
  4. The Electrician

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    Oct 9, 2007
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    What's wrong with that? The output resistance with the input shorted is just 1/y22. Since Vs has a Thevenin impedance of zero ohms, the output resistance with the input shorted is what you want.

    What is the correct answer? I'd like to compare with what I got.

    If you short the input you should be able to use any parameter set you want. However, it you want to use Z parameters, please show what you've done so far. It's hard to help much more without seeing your work.

    You are seeming to treat this circuit as a two-port, but there are internal nodes. How did you deal with the internal nodes? Please show your work.

     
  5. StasKO

    Thread Starter Member

    Apr 28, 2012
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    solutions attached
    mine is the hand written one
     
  6. The Electrician

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    I get the same result that you do if I don't replace Vs with a short.

    I get the same result as the full solution if I do replace Vs with a short.

    Try doing your analysis without setting I1 to zero.

    Edit: It's not whether you use Z parameters or Y parameters that makes the difference. It's that you didn't replace Vs with a short. You should be able to use Z parameters with Vs shorted.
     
  7. StasKO

    Thread Starter Member

    Apr 28, 2012
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    yeah i tried this already and i did get the teacher's solution. but the point is that how can i tell which one is the correct approach (without seeing the answers)? why calculating Z22 gives a different answer? shouldnt it be always the same answer no matter the method?
     
  8. The Electrician

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    Calculating z22 doesn't give a different answer if you replace Vs with a short.

    You mention Z parameters in your posts, and your initial diagram seems to show a two-port with an input port and an output port. A two-port is fully characterized by 4 parameters, z11, z12, z21 and z22. Do you know how to calculate all 4 parameters?

    It's true that the definition of the z parameters of a two-port assume that the input is open circuited when measuring (or calculating) z22, and the output is open circuited when measuring z11. But you can still calculate z22 of the network with a finite source impedance if you move the source impedance into the internal part of the two-port.

    Suppose you replace Vs by an impedance, Zs, the source impedance of Vs. Can you calculate the z parameters of the network as if Zs were internal to the input port of the network?

    If you could do that, then you could derive an expression for Ro which would involve Zs. Then you could take the limit of that expression as Zs approaches zero (a short, in other words).

    Are you aware that this circuit includes a model of a 3 terminal device with transconductance, such as a FET, transistor, or vacuum tube? With the coupling between input port and output port such as you have here, the output resistance is affected by the source impedance driving the input. When a voltage source is driving a two-port, its source impedance is taken to be zero. If the input were driven by a current source, the source impedance would be taken to be ∞ (I1 would be zero in that case).

    But no matter what the source impedance, you can solve the circuit with either z parameters or y parameters; you just have to take the source impedance into account.

    Just because you choose z parameters to solve the network doesn't preclude having a driving impedance less than ∞. You just have to move that source impedance into the internal part of the two-port.

    Why even bother with the two-port concept anyway? Just solve the network with Vs replaced with a short and calculate V2/I2.
     
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  9. StasKO

    Thread Starter Member

    Apr 28, 2012
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    the thing is that i didnt study analogue circuits yet (only next semester) so all of this active circuits dont mean nothing to me.

    this question is 1 out of 8 questions about the subject of two port networks which is the last subject we learned and we were taught to systematically solve the circuits with the basic definitions of the z- y- h- g- parameters.

    so this is what i did. without seeing the answer i would have been sure that my solution was correct.

    i also talked to some of my study friends about the question and they also didnt understand why Vs needs to be shorted and not disconnected (open circuit) since they also thought that i1 should be set to zero.

    thnx for your help though. i really appreciate it!
     
  10. The Electrician

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    I suppose what you should know is that voltage sources are assumed to have zero impedance (a short) and current sources are assumed to have ∞ impedance (an open circuit). That's why in this case you shouldn't set I1 to zero.

    So, if you have a problem like this again, and it's driven by a voltage source, use y parameters. If it's driven by a current source, use z parameters. :)
     
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  11. StasKO

    Thread Starter Member

    Apr 28, 2012
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    thnx!!
    i'll use that advice!
     
  12. StasKO

    Thread Starter Member

    Apr 28, 2012
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    hi electrician, i have another similar problem which again im not so sure about my answer..

    i attached the problem and again its about finding Ro.
    hm=85, gm=1ms

    what i did is short out V1 (its the arrow where Is is).
    then i calculated v2/i2 (i2 is not marked but its the current going into the upper terminal from where v2 is)

    my solution is Ro=105.43 ohm

    can you please solve it so i can compare (i dont have solutions for this question :) )

    PS:
    ignore v2/v1 and io/is. i solved them correctly

    thnx!!
     
  13. WBahn

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    Mar 31, 2012
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    You need to learn to check your work yourself -- as a practicing technician/engineer you will seldom have someone to check your work. Assume your answer is correct and then determine what the currents and voltages will be for a couple of signals that you make up. Then see if those answers work out on the original circuit as well.
     
  14. StasKO

    Thread Starter Member

    Apr 28, 2012
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    ok i did that and i believe that i have the correct answer. thnx

    here's another problem which i solved and checked my solution and it is indeed the correct one. when i checked the full solution, the teacher immediately assumed i=0 when he calculated h12 and h22 (i1=0).
    now i dont know if he is really smart and somehow understood it from the circuit itself or he did made mathematical calculations and was lazy to put them in his solution.

    can it be assumed just from looking at the circuit that i=0? if so then how? because all i see (when i1=0) is 2 resistors in parallel and both of them in series with dependent current source. i "imagine" that the current i2 enters the parallel resistors and splits into current -i (in R1) and i2+i (in R2).

    is it possible to intuitively understand that i=0?

    thnx!
     
  15. WBahn

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    Mar 31, 2012
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    There are several i's in the circuit. Which one did he assume is zero? I don't think he is saying that that currrent is always zero. He is saying, let's force the situation in which this current is zero (probably the output current) by open circuiting the output, which means we have an unknown voltage there, but we know we have no current. Playing games like that force some terms to disappear letting you solve for the others more easily. But, usually, you have to come back and do another analysis where you don't impose that simplifying condition in order to finish the solution, but doing so leaves you with a simpler problem to solve because you have already gotten part of it.
     
  16. WBahn

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    Wait, are you saying that he assumed that the current in the top branch (the one that shorts the input to the output) is zero?
     
  17. StasKO

    Thread Starter Member

    Apr 28, 2012
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    he wrote in his solution that the 'i' with no subscript was equal zero when he forced i1 to be zero (the input current) since he was solving for h parameters h12 and h22 (the question itself was to find the circuit's h-parameters).
    so can it be immediately understood that 'i' is zero when i1 =0 ?
     
  18. StasKO

    Thread Starter Member

    Apr 28, 2012
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    yup.. when i1 is forced to be zero
     
  19. WBahn

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    Mar 31, 2012
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    Okay, I read your post more carefully and see that the condition he is starting from is that i1=0.

    Your observation that you are left with two resistors in parallel and with that combination in series with a dependent current source is correct. But the key is that if you apply a test voltage V2 to the output, the current i will be negative, meaning that the current in the dependent current source will be negative. It's a simple matter to show that, with this constraint, that the voltage at the junction between the two resistors has to be V2 and, hence, no current flows at all. But I don't quite see how that can be stated intuitively. I sort of see it, but can't quite put it in firm words.
     
  20. WBahn

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    Mar 31, 2012
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    Okay, consider this.

    Label the junction of the dependent current source and the two transistors Node A.

    Lets assume that i is nonzero and is positive. This means that the current through R1 is such as to make V2 less than Va (the voltage at Node A). This means that the current in R2 also has to be flowing from the current sourse toward the output terminal. But, the current source is forcing current to flow from Node A to ground. So we have all three branch currents at Node A leaving Node A and none coming in, violating KCL. The same is true if we assume i is nonzero and negative, except now we have all the current flowing toward Node A and none flowing out. So the only way KCL can be satisfied is for there to be no current flowing at all.
     
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