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What makes current flow in a transmission line?

  1. Nov 5, 2009 #1
    In a transmission line, what makes the current flow?

    I understand that the line is excited by an AC voltage at one end. This sets up a changing electric field between the conductors and therefore a changing magnetic field. However, there is no voltage along the line, so what makes the current flow along the 2 conductors?
     
  2. jcsd
  3. Nov 5, 2009 #2
    You ask a very important question, and one that I pondered for some time before realizing why there is a current at all.

    Consider a long transmission line of a specific impedance Z (e.g., RG-58 which is 50 ohms), meaning a specific inductance L and capacitance C per unit length, so that Z = sqrt(L/C). The velocity of propagation is v = 1/sqrt(LC). If you do not understand this, then review transmission line equations.

    Now, in the middle of this line, place a charge Q on the center conductor. Voltage is proportional to Q/C. Does the charge stay there forever, or does it flow away? If it flows away, does it go in on direction or both? Why? How fast? Eventually you will realize that the charge redistributes away at velocity v in such a way that there is no initial current at the point where the charge was initially placed. Why no current? Because there are equal and opposite currents in both directions.

    If you are an EE, then think of the transmission line as a series of inductances L shunted by capacitances C, and place the charge on one of the C's (the equivalent circuit model for a transmission line), and watch the charges move. Use time-domain differential equations; V = L dI/dt, and I = C dV/dt.

    Bob S
     
    Last edited: Nov 5, 2009
  4. Nov 5, 2009 #3
    There is indeed a voltage across the 2 lines & a current. Neither one exists w/o the other. Neither one "causes" the other. They are inclusive.

    The voltage across the 2 wires at any point on the line, and the current at that same point, are related through Ohm's law, where Z0 is the characteristic impedance of the line, so that V = I*Z0.

    What makes the current flow, as well as the voltage (propagate), is the source of power connected to the input. If this source is e/m radiation, such as an antenna would receive, then this incident power is why the current and voltage are present. Since the t-line has a finite Z0, the only way either I or V is zero, is when both are 0, which only happens when the incident source power is 0.

    Does this make sense?

    Claude
     
  5. Nov 6, 2009 #4
    Using the telegrapher equations and the distributed element model can explain the CHANGE in the phase and amplitude of V and I as it travels down a line, but it doesn't tell me how the wave propagates. A lot of people resort to saying things like 'the voltage charges up the capacitor and then it discharges through the inductor', but this is not a valid explanation of how the wave propagates, just how it is affected by the line parameters.

    I've tried hard to answer this question and I think the 'kink' model used to explain radiation is the most valid answer. The change in the electric field between the 2 conductors at the source caused by an alternating voltage creates a kink and this propagates down the line just as EM radiation propagates into space. The current is caused by the charge building up at subsequent points in the line as the electric field travels down the line. This also has a magnetic field associated with it, just like plane wave radiation. HOWEVER, the telegrapher equations only explain how the amplitude of the voltage and current at any point on the line is affected by self inductance caused by the current flow and the capacitance caused by the charge build up, so in itself, it is not a full explanation of how a wave travels down a transmission line, just how it is affected.
     
  6. Nov 6, 2009 #5

    vk6kro

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    Suppose you had a pair of wires parallel to each other and about a kilometer long.

    Now you put a battery across one end, between the wires. What would you measure at the other end of the wires?

    Nothing.

    Because it takes time for the voltage to get there. But it has no choice about getting there and it will get there eventually.
    This is how waves propagate down a transmission line.

    If you change the polarity or the level of the voltage, this will have to travel to the other end as well, because the wires are conductors.
     
  7. Nov 6, 2009 #6
    The point is there is no potential difference between one end of the conductor and the other. The potential difference is between the conductors, and even if the 2 conductors were not connected at the load, a wave would still travel down the line.

    Thnk about a HV transmission system. 2 substations are connected by a transmission line and both are at a potential of 400,000 volts. Power still flows from one to the other. Power may even flow from substation A to substation B if A has a lower voltage than B!

    I'm asking why this happens. I think it's because of a kink in the electric field caused by accelerating electrons at the start of the line. It's just like plane wave radiation, but guided.
     
  8. Nov 6, 2009 #7
    With wave propagation, the electric & magnetic fields play an equally important role. I don't think it can be explained in terms of just E, or just H, but rather both are necessary.

    As far as potential difference from one end to the other goes, there is a difference. If a battery and switch (initially open) are placed at one end and the switch is closed, note the following. After the switch closes, the voltage and the current (I=V/Z0), propagate towards the other end. Thus, the 2 ends of the same conductor are at differing potentials. The end nearest the battery has V & I, whereas the other end is at 0 & 0 until the wave gets there.

    A good e/m fields book will clarify this greatly. This question is well understood. Heaviside in the 1870's solved it.

    Claude
     
    Last edited: Nov 6, 2009
  9. Nov 6, 2009 #8

    vk6kro

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    It has nothing to do with any "kink".

    If you connect a power source to any part of a conductor, the whole conductor will eventually acquire this potential.
    The key word is "eventually". It takes time for this transfer of potential to take place because the effect can only move at the speed of light.

    In my example, power does not reach the other end of the two pieces of wire for about 3.3 microseconds. This is the propagation along a transmission line.

    If you wait around for 3.3 uSec, suddenly power from the battery is available at the far end of the wires. It got there at almost the speed of light.
    Because it is a voltage it does not need a voltage difference to move. If it was a current, it would.
     
  10. Nov 6, 2009 #9


    A voltage and a current need each other to move. The impedance Z0 of the t-line is neither 0 nor infinite. Thus neither I nor V can exist independently. I & V need each other.

    Visualize a group of charges propagating in one direction. The group is moving, constituting a current, but the charges are separated, constituting a voltage. The ratio V/I is equal to Z0. The voltage needs the current in order to propagate. If at the input side of the t-line, we have charges separated, we have voltage. In order for V to propagate, the charge groups must propagate as well. But this propagation of charge is exactly what current is. V cannot propagate unless charges move. But moving charges is current.

    I & V need each other. Both are equally important. Neither comes first.

    Claude
     
  11. Nov 6, 2009 #10
    But my question is WHY does voltage propagate down the line. I know transmission line theory quite well, I know it takes time to propagate and I know there is a distributed L and C in the transmission line model. What i'm stuck on it why the voltage, i.e., the charge between the 2 conductors, moves along the transmission line. I think there is a kink in the electric field lines, just as there is in plane wave radiation. This is explained in the EM textbooks that go beyond the telegrapher equations and attempt to explain it properly.
     
  12. Nov 6, 2009 #11
    That's a tough question. "Why" is always difficult to understand. But my futile attempt is that energy is transferred like that of a set of dominoes. Have you seen those gadgets where 5 steel balls are suspended with strings? By lifting one ball outward and upward, then releasing it, the ball hits the other 4 and the one at the other end moves away from the group. It returns, striking the "lattice" and so on. Energy and momentum are transferred between the steel balls, 2 at a time.

    Why does current "propagate" from one end to the other? It appears that energy and momentum transfer is taking place at the atomic level.

    This is a fascinating question. If you wish to really delve deeper, I'd suggest studying peer-reviewed textbooks on QM, & solid state physics including theory of solids/metals. This should give some insight.

    Have I helped?

    Claude
     
  13. Nov 6, 2009 #12

    vk6kro

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    If you had just one wire, connected to one terminal of a battery, forming a huge circle and coming back almost to a battery, the voltage from the terminal connected to the loop would still travel around the loop and become available at the gap.

    This is because there cannot be a voltage difference across a conductor without a current flowing. So, all parts of a conductor have to be at the same potential, even if it takes a few microseconds for this to happen.

    This effect is not related to charging capacitors and has nothing to do with current or with transmission lines. It is something that always happens with conductors.
     
  14. Nov 6, 2009 #13
    Good.
    If we delve deeply into the magnetic fields H of the current, and the electric fields E of the voltage between the wires, the power flow is actually a TEM (transverse electric magnetic) wave between the wires given by the Poynting vector

    P = ∫E x H dA

    where A is a surface between the wires. The currents in the wires support the fields, just like the currents in the walls of a waveguide support the E and H fields inside. The power is in the fields inside, not in the waveguide walls.

    A good example of a transmission line is the 300-ohm TV lead-in wire from the antenna. sqrt(L/C) = Z = 300 ohms.

    Bob S
     
    Last edited: Nov 6, 2009
  15. Nov 6, 2009 #14

    f95toli

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    It is perhaps worth remembering that "voltage" and "current" mean slightly different thing when dealing with transmission lines than in low frequency circuits. In a way it is more physically "correct" to just use the E and the H field to describe what is happening, it is just a propagating TEM wave: in a way we are just "confining" EM radiation to a tube, most of the energy is actually travling in the dielecric BETWEEN the conductors.

    Remember that there are many types of lines that can be used to send signals, there is no fundamental difference between e g. coaxial lines, waveguides and dielectric waveguides such as optical fibers. The only difference between them is the "configuration" of the E and the H fields. You can even get far infrared radiation (around 1 THz or so) to propagate along a twisted pair.
     
  16. Nov 6, 2009 #15

    vk6kro

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    The question was: why does the signal propagate along the transmission lines. The fundamental reason it propagates is the property of conductors to acquire the same voltage.

    What happens after that in the way of charging the capacitance between the lines etc is incidental to the question.
     
  17. Nov 7, 2009 #16
    I don't think that explanation is particularly useful. It's a lot more complicated than that. The voltage is between the forward conductor and the return conductors. To a good approximation, no voltage along the conductor is necessary. The signal propagates as a wave and maxwell's equations that govern the wave's movement. This is true, as I understand, for all conductors, even at low frequency and DC. Look up the lumped element approximation and study how reflections in a transmission line act to give the illusion that there is an electric field along the conductive path.
     
  18. Nov 7, 2009 #17
    Good. In the limit, consider a transmission line using superconducting wire. So the only voltage drop along the conductor is, in the EE notation, V = L dI/dt, where L is the series inductance. It is useful here to consider what the series inductance really means. Comparing the EE and physics formulas for inductive energy storage:

    E = (1/2) L I2 = (1/2) ∫B H dV =(1/2 μ0) ∫B2 dV,

    Where B and H represent the magnetic field energy between the conductors, and dV is the volume. So the signal propagation in a transmission line is actually the propagation of electric energy (the electric field) and the magnetic energy (the magnetic field) between the conductors. The flow of power is actually represented by the Poynting vector cross product of these two fields between the two conductors:

    S = E x H

    For this reason, the transmission line problem is best treated using Maxwell’s equations.

    Bob S
     
  19. Nov 10, 2009 #18
    Very well stated. Power flow is indeed the cross product of E & H integrated over the area, known as the Poynting vector. It "points" in the direction of power propagation. Without I, there would be no power. All V w/ no I requires that Z0 be infinite, requiring infinite spacing between conductors. Likewise, all I w/ no V would need Z0 = 0, an impossibility.

    E, H, I, V, B, & D, as well as P, the power, can only exist mutually under time-varying conditions. That is Maxwell's equations in a nutshell. When incident power impinges on an antenna, dipole for example, which is a modified t-line, I & V, as well as E & H, are mutually present. I, V, E, & H proagate at transmission speed less than c (light speed). They arrive at any point in unison. The current arrives at the same time as the voltage. Likewise for E & H.

    Claude
     
    Last edited: Nov 10, 2009
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