how to h-parametered?

Discussion in 'Homework Help' started by stupid, Mar 16, 2010.

  1. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    hi,
    how to express h-paramters in R form?

    my working:
    h11 = V1/I1 when V2=0

    when V2 is shorted,
    R= R1//R3

    I1=V1/(R1R3/R1+R3)

    h11= R1R3/R1+R3

    is that correct?

    about h12=V1/V2 when I1=0 which mean V1 is open circuit,
    h12= R1/R2

    is that correct?

    thanks
    atupid
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Your expression for h12 is incorrect. You will be driving V2 with a voltage source, so R2 will be shorted by the zero output impedance of the voltage source. Then you just have a voltage divider to get the voltage at V1, which is open circuited.
     
  3. stupid

    Thread Starter Active Member

    Oct 18, 2009
    81
    0
    hi, The Electrician

    i dun quite get what you mean.
    but i would assume,

    V1= V2R1/R1+R3

    h12= V1/V2= R1/(R1+R3)

    is that a right assumption?

    thanks,
    stupid
    [​IMG]
    [​IMG][​IMG]
     
  4. The Electrician

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    Exactly correct.
     
  5. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    hi the electrician,

    if V2 zero impedance shorted out R2, wouldnt the R1 also shorted out in the attached diagram?

    thanks
    stupid
     
  6. The Electrician

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    Oct 9, 2007
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    In the diagram you have attached in this post, yes, R1 is shorted. But in your original diagram there is R3. What happened to R3? In your original diagram, V2 would only short out R1 if R3 = zero.
     
  7. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    i m trying to prove a point using the last diagram.
    i m curious:
    r there 2 types of voltage supply, namely one that wouldnt short out a resistor & another would?

    if there were 2 resistors connected in parallel as per attached,
    which resistor would be nuked? R1 or R2?

    thanks to the electrician
    stupid
     
  8. The Electrician

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    Oct 9, 2007
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    In circuit analysis, when you speak of a voltage source, without any further qualification, it is assumed that the output impedance of that source is zero; that's what is meant by "voltage" source.

    When you speak of a current source, without any further qualification, the output impedance is taken to be infinite.

    Of course, it's possible to have a signal source with an output impedance which is neither zero nor infinite. In that case, it's usual to specify what the output impedance is, and represent that source as an ideal voltage source in series with an impedance equal to the output impedance specified. The Thevenin equivalent of the signal source, in other words.

    In the circuit you've shown in post #7, if V2 is an ideal voltage source (zero output impedance), then it will short both R1 and R2. But, if there were another resistor such as R3 in your original circuit in post #1, then V2 would only short R2.
     
  9. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    thanks the electrician for the explanation.
    going back to the first post,
    to solve h11 when V2 is shorted,
    should a zero-impedance source be connected at V1 & so nuke R1?

    stupid.


     
  10. The Electrician

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    Oct 9, 2007
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    When you're calculating h11, V2 is zero. That is because there is a short placed across the V2 port. But you need to know the input current, I1, so R1 has to be there.

    If you connect a voltage source at V1 to make the measurement, you still have to know I1, and R1 contributes to I1.

    When you're calculating h12, you apply a voltage source at V2, and you use the voltage at port 2 to calculate h12; the current at that port doesn't enter into the calculation. Any resistor in parallel with a voltage source doesn't change the voltage of the source. If only the voltage at such a port matters, and not the current, then you can delete any resistor in parallel with the voltage source.

    But, if the current at a port matters, as it does when you're calculating h11, the any resistor in parallel with the source matters, because it takes some of the current.
     
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  11. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    hi to continue...

    h21=I2/I1 when V2=0,
    I2=V1/R3,
    I1=V1/R1
    so h21=R1/R3


    H22=I2/V2 when I1=0
    total R= [(R1R3)/R1+R3]+R2
    I2=V2/R
    H22= 1/R= (R1+R3)/(R1R3+R2(R1+R3))

    r they correct?

    thanks
    stupid
     
  12. The Electrician

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    I1=V1/R1 is not correct. If you short the V2 port, what happens to R3? Don't you have R1 in parallel with R3 as the load on the V1 port?

    If the V1 port is open circuited, then you have the (R1+R3) combination in parallel with R2 as the total resistance, as measured at the V2 port

    Try again.
     
  13. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    oh no..
    i simply glossed over without careful thought.

    here they go,

    I1=V1/(R1R3/R1+R3)
    h21=R1/(R1+R3)

    R=((R1+R3)R2)/(R1+R2+R3)
    h22= (V2/R)/V2
    =1/R
    =(R1+R2+R3)/((R1+R3)R2)

    hope i got them right.

    thanks the electrician
    stupid

     
  14. The Electrician

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    These are correct.
     
  15. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    the values of h12=h21.
    is there any implication?

    thanks
    stupid
     
  16. The Electrician

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    Oct 9, 2007
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    This is because the network is bilateral. For a transistor, h12≠h21 because it has a different transfer characteristic in one direction than in the other. But, when the network consists of just resistors, the network is bilateral:

    h12=h21
    y12=y21
    z12=z21
     
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