transmission lines

Discussion in 'General Electronics Chat' started by gorgondrak, Nov 17, 2014.

  1. gorgondrak

    Thread Starter Member

    Nov 17, 2014
    61
    1
    I'm having a hard time getting a foundation for transmission lines so that I can build to firm understanding and move into some of the math. Firstly, the only difference between a transmission line and a regular copper wire that I can tell is that a piece of copper wire doesnt have distributed capacitance along its length. Is the characteristic impedance of a copper wire its intrinsic inductance in series with its resistance? If it were, then how would that tell me the value of voltage and current associated with the incident wave when the voltage is applied when that inductance and resistance is overall from end to end. Basically, getting rid of the resistance and assuming a lossless copper wire, how would the intrinsic inductance of the wire overall be used to calculate a characteristic impedance for that wire. In addition if a copper wire has almost zero capacitance, wouldn't that move the characteristic impedance toward infinity using the formula sqr(L/C)? Additionally using the formula V\Zo to find the initial current through a conventional transmission line, by what effect will the resistance in the line have in limiting the total current generated by the propagation of the incident wave? For example, consider a transmission line with 50 ohms impedance and assuming that the intrinsic resistance in the wire is in fact not included in the calculated characteristic impedance. If the intrinsic resistance of the wire is very small say .001 ohms then with 100 volts the incident wave will be generating 2 amps of current while the resistance will be limiting the amplitude of that wave by .001 ohms along its path. How would I calculate the amplitude of the incident wave for current or voltage after having reached its destination, whether a terminating impedance or a short circuit without a load...
     
  2. alfacliff

    Well-Known Member

    Dec 13, 2013
    2,449
    428
    there is only one single wire transmission line I know of, and it works completely different from normal ones. transmission lines are two conductor or more, either shielded or open wire. the impedance is calculated with the distributed using the distributed inductance and capicatance. the only one wire line I know of id the "G line" which is completely different, being more of waveguide based.
     
  3. crutschow

    Expert

    Mar 14, 2008
    13,052
    3,244
    A single copper wire does have capacitance to ground even if it is very small. And because of the logarithmic relation between its characteristic impedance as determined by the wire diameter and its distance to ground, the wire characteristic impedance does increase but doesn't go much above a few thousand ohms, even for a wire far away from ground, as shown by this.

    The line resistance will cause a reduction in the voltage based upon the value of the resistance and the transmission line current.
     
    Sensacell likes this.
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    The characteristic impedance of a single wire, radius r, height h above ground, is

    {Z_0} = \frac{1}{{2\pi }}{\cosh ^{ - 1}}\left( {\frac{h}{r}} \right)\sqrt {\frac{\mu }{\varepsilon }}

    A good text on the subject is

    Transmission Lines for Communications

    by C W Davidson (Heriot-Watt University)
     
  5. gorgondrak

    Thread Starter Member

    Nov 17, 2014
    61
    1
    Thank you and is there an online source to where this formula came from? I think mostly right now I'm stumped on there now being 2 ohms laws for 1 wire transmission line or otherwise. 1 being the standard v/r and now v/zo. Is the incident wave just a voltage generating a current?
     
  6. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    Unfortunately the AAC Ebook does not treat this exact case, although it can be derived from the two parallel wire case which is treated, where the formula is similar.
    http://www.allaboutcircuits.com/vol_2/chpt_14/3.html

    There is only one Ohm's law and it still applies to transmission lines.
    But Ohm's law is a point equation. That is it applied on a point by point basis, and the parameters are unaffected by variations at other points along the line.
    These are called lumped parameters.

    On the other hand transmission lines and waveguides admit periodic solutions to their defining differential equations will take account of the effect of other (all) points in the transmission region.
    Traditionally the term 'transmission lines' refers to the transverse electromagnetic wave, (TEM) solution and waveguides to other modes (TE and TM).
     
  7. gorgondrak

    Thread Starter Member

    Nov 17, 2014
    61
    1
    Say for example I had a 50 ohm resistive load with a transmission line of 50 ohms characteristic impedance and 1 ohm of "regular resistance" in the line. If I attached a voltage source of 100 volts then the incident wave would be generating 2 amps correct? What would be the value of the current once the wave reaches the load?
     
  8. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
  9. gorgondrak

    Thread Starter Member

    Nov 17, 2014
    61
    1
    The characteristic impedance of a lossy transmission line is given by
    Z0= The square root of R+jωL/G+jωC
    How is this possible? As G approaches 0 wouldn't the resistance of the line approach infinity meaning nearly no current at all?
     
  10. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
     
  11. alfacliff

    Well-Known Member

    Dec 13, 2013
    2,449
    428
    why do you keep mixing up resistance ad impedance? resistance is simple resistance, impedance is complex reactance. a resistor ins eries with a transmission line is not the only current limiting part.
     
  12. gorgondrak

    Thread Starter Member

    Nov 17, 2014
    61
    1
    In the formula for Zo provided in the previous post, is this division performed as complex number division or just regular division. Also alfacliff, can you explain to me in more detail the mistake I'm making?
     
  13. MrAl

    Well-Known Member

    Jun 17, 2014
    2,440
    492
    Hello,

    A circuit element (even a wire) is not considered a transmission line until the wavelength of the frequency becomes comparable to the physical length of the component. That's because if the wavelength is long then it appears that every point within the element is at the same potential, but if the wavelength is short then the physical position (x) of the potential becomes important.

    For a simple example, consider a long resistor say 1 meter long. For a sine wave of 1v peak amplitude at 1kHz if the left end (x=0) is at 0v, then the right end (x=1 meter) would be around 21uv, a difference of only 21uv (or something very small like that). This means it looks like the whole resistor is at roughly the same potential so any external view will show it acts the same at both ends. At 75Mhz however, if the left end is at 0v then the right end will be at 1v, which is a huge difference then what we saw at 1kHz.
    In the first case we could look at the resistor as a lumped circuit element, but in the second case we have to consider the physical dimensions.

    Mathematically in the first case we only have to deal with functions of time like F(t), while in the second case we must use functions of both time and space like F(x,t) because there is now a difference in potential along 'x' as well as with time 't'.

    Most lumped circuit equations use functions like F(t) while equations for true transmission lines use functions like F(x,t). If we happen to be unlucky enough to not have good symmetry along the width also, we might even need functions like F(x,y,t), and for the full three dimensional view we need functions like F(x,y,z,t), which cover all space and time.

    Probably a good way to get into the higher dimensional responses is to start with the heat equation and go from there. The heat equation is a partial differential equation which is a single equation not a coupled set of partial differential equations like the transmission line equations. The heat equation is:
    Ut=k*Uxx

    (subject to boundary conditions and initial conditions)

    where
    Ut is the partial derivative of a function U with respect to time, and
    Uxx is the second partial derivative of a function U with respect to x.

    This is a simpler way to start than jumping right into the transmission line euations, which are coupled not single like this one is. The function we would try to find would be U(x,t) which is a function of space and time, and it would satisfy all the boundary conditions and the initial conditions.
     
    Last edited: Nov 24, 2014
    studiot likes this.
  14. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    Good explanation Mr Al.
     
  15. alfacliff

    Well-Known Member

    Dec 13, 2013
    2,449
    428
    matching impedance of a transmission line is important whether the line is one wavelength or shorter. a 50 ohm output connected to a different impedance line will have loss. sometimes, you can use a different impedance line to match source to load, as in transmission line transformers, using a length of line reelated to the wavelength to "transform " impedance to a different impedance load.
     
  16. gorgondrak

    Thread Starter Member

    Nov 17, 2014
    61
    1
    Mral I like your example, thanks. That is measured across an open break right?
     
  17. Papabravo

    Expert

    Feb 24, 2006
    10,163
    1,796
Loading...