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Current direction in dipole antenna

  1. Nov 29, 2011 #1
    Is the current direction shown correct ?
     

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  3. Nov 29, 2011 #2
    I am no expert, I just studied some basic antenna theory. I use Balanis and EM book by David K Cheng. This is my understanding. According to both books, it is very hard to find the exact current distribution, it is only an approximation.

    The approximation assume current is sine wave and the antenna center at z=0 and the antenna total length is 2h. So using sine wave:

    [tex] I_z=I_m \sin[\beta(h-|z|)] =\;\left \{ \begin{array}{cc}I_m \sin [\beta(h-z)] & z>0\\ I_m\sin [\beta(h+z)] & z<0 \end{array} \right\}[/tex]


    If h< [itex]\frac{\lambda}{4}\;[/itex], then max current at z=0. If h> [itex]\frac{\lambda}{4}\;[/itex], then max current is somewhere along the line. Current always zero at the end of the line.
     
  4. Nov 29, 2011 #3
    i am just concerned with the direction. I think current flows outwards from the positive terminal of the source. But in the figure it is opposite.
     
  5. Nov 29, 2011 #4
    could be talking about electron flow rather than conventional current....
     
  6. Nov 29, 2011 #5

    Averagesupernova

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    The antenna is being fed by AC so why are you concerned with direction? Think of it as electrons bunching up on one end and then running back to the opposite end over and over. I would say the diagram is correct. It is a snapshot in time though, not a steady state representation.
     
  7. Nov 29, 2011 #6
    But according to the polarity shown , is the current flow direction correct ?
    In the next cycle , the polarity of the source will change and the direction changes as well.
    I have seen this same diagram in so many tutorials.
     
  8. Nov 29, 2011 #7
    The formulas I provided give you all the information you needed. Just plug in the data you can find the amplitude at any point. This is phasor form and done according to what you specified that you assume +ve on z>0.

    As you can see, it really depend on the length of h and z. Just put in this two parameters and you'll get your answer. I have not work out the direction, it's for you to verify, just use the formulas.

    Yes the current flows out of the +ve side which in your case is the +ve voltage direction so Im is +ve.
     
    Last edited: Nov 29, 2011
  9. Nov 29, 2011 #8

    sophiecentaur

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    Well, it shouldn't be. There are enough problems here without bringing those little devils into it!

    I think the direction shown is wrong. If you replaced the antenna with a capacitor (appropriate for a short antenna) then the current would be flowing INTO the Capacitor from the positive terminal. and Back Out into the negative terminal.

    But you need to remember that the two are out of phase (quadrature), in any case, so the direction thing is not meaningful as one 'changes direction' whilst the other is going through its [edit: max or min] value. Basically, don't worry. Once you get as far as the SUMS, it will sort itself out.
     
    Last edited: Nov 29, 2011
  10. Nov 29, 2011 #9

    Averagesupernova

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    The diagram is correct. If the antenna is actually radiating, the current has to flow the same direction on BOTH sides of the feed line. Could the antenna be fed if the current was flowing in opposite directions? Think about the current in the feedline. You are correct in saying that in the next cycle everything changes around.
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    As I stated in a previous post, and yes I'm sure sophie is cussing me at this moment for bringing those little devils into this, think of it as electrons bunching up on one end and then running back to the opposite end over and over.
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    It is easier for me to envision it like this: Think of a skate board ramp. The skater is an electron. The skater runs back and forth from side to side with max velocity in the middle (lowest point) of the ramp and least velocity at each side (highest point) of the ramp. Never do max velocity and highest elevation both occur at the same point on the ramp. The skaters direction changes every time he hits the top of the ramp the same way the current polarity changes with each cycle. I know there are people on here that HATE analogies, but I think it may help you. Velocity and direction of the skater represents the magnitude and direction of the current. Don't confuse this with velocity of current. No analogy is perfect of course.
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    The analogy does work for feeding the antenna as well. Think about where the skater gets his extra push to continue oscillating. If he were to get his push in the middle, which is a low impedance point on a dipole, he has to be pushed with max velocity which in the anology represents max current (lowest impedance). If he gets it towards the top it is the opposite.
     
    Last edited: Nov 29, 2011
  11. Nov 29, 2011 #10
    Of cause the direction keep changing. But you can just set it with a reference direction like op defined +ve at one direction. That's how books doing it too.
     
  12. Nov 30, 2011 #11

    sophiecentaur

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    Just BE CAREFUL when talking about electron motion in cases like this. Do you realise how far they are likely to move from their notional 'rest position' if their maximum speed is 1mm/s and they keep changing direction tens of millions of times a second? You just can't consider them as sloshing up and down the wire. Your skater is just standing still and vibrating a minute amount. Stick with current and you can't go wrong; it's abstract enough to follow the rules of EM waves and wires. If you insist on a 'mechanical' model then how do explain that the current further out gets less? Where do these electrons 'go', looking along the wire, because the current doesn't follow Kirchoff's First Law?
    Electrons are not a 'free pass' to understanding how electricity works. The bottom line is that they introduce as much confusion later on as the confusion you think they have eliminated at the start. This is a great example of why I object so strongly to using them.
     
  13. Nov 30, 2011 #12
    great explanations.........
    I got it.......thank you all.
    One more question
    I read this somewhere for current and voltage distribution that voltage will be minimum at center where current is maximum. How can it be possible for the voltage to be minimum at center? May be i read wrong
     
  14. Nov 30, 2011 #13

    Averagesupernova

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    I don't really know what to say here sophie. I already admitted no analogy is perfect. You question: ...how do explain that the current further out gets less? Well I don't. I stated, this is the third time now I believe, that no analogy is perfect. Get over it already, and I'm taking the liberty of remembering your acceptance of analogies from previous threads. If there is an explanation in this analogy it is that the current is represented by the velocity of the skater. And I already said DON'T CONFUSE THE VELOCITY OF THE SKATER WITH THE VELOCITY OF THE CURRENT!
    -
    You are correct in saying that electrons are not a free pass to understanding. Anyone who believes that has never probably done much with electronics. It is a matter of how far down you want to go. As far as most residential elecricians are concerned the wiring they install just as well be carrying invisible marbles. But if you are in the business of designing semiconductors it is unlikely you will be thinking of invisible marbles hopping between PN junctions.
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    From what I take from the OP in this case is that they don't understand how the current can be the same polarity over the length of the antenna and the voltage not be. I thought I explained how that is possible pretty well.
    -
    Edit: Looks like the OP got it, hmmmm...
     
  15. Nov 30, 2011 #14

    Averagesupernova

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    It is correct that the current is max at the center and voltage is at minimum. You read correct.
     
  16. Nov 30, 2011 #15

    sophiecentaur

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    That's right. If you take the two halves of a dipole and bend them so they are parallel, you will have a quarter wave transmission line. This line will have an open circuit at its end, with a current minimum and a voltage maximum. It is terminated with an open circuit (high impedance) and the quarter wave line transforms it to a low impedance at the feed point. As you spread the wires apart, the same thing applies but the impedance of the line is different.
     
  17. Nov 30, 2011 #16

    sophiecentaur

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    Sorry. I am not really that stroppy about things but I have so many experiences of people picking up a dodgy analogy and running away with it. Then drawing all sorts of conclusions all by themselves. Whilst I know we all have private pictures in our heads to explain stuff, when you try to communicate them the result can be just what you didn't want. If you want to describe something sloshing up and down a dipole why not just call it charge? That would be accurate and quite meaningful, actually. After all, they are no more likely to 'see' electrons doing it than to 'see' a charge.
     
  18. Nov 30, 2011 #17
    It is not true, it all depend on the length of the dipole as shown in my equations. Books claimed it is too difficult to get the exact expression of the current, so they approximate using phasors with open termination at the end as show in my first post. The equation specified for the length h compare to quarter wave that if h> quarter wave, the max is somewhere in the line, not at the end due to sum of the forward and reverse voltage phasor.
     
  19. Nov 30, 2011 #18
    Have you put in h and z in the equation? You cannot say anything until you specify the length of the dipole. It differ with length. Go look up in the web or find a way to read in google books or other site for "Field and Wave Electromagnetics" by David K Cheng , you will find more detail explanation under "Linear antenna" around Page 615.
     
  20. Nov 30, 2011 #19

    sophiecentaur

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    But it's true for up to and a bit beyond a half wavelength dipole and particularly for the dipole shown in the original diagram. Yes, you can arrange for a high drive-point impedance if you make the dipole still longer or if you drive it at a point not at its middle. That just adds complication I think. The exact current distribution is very difficult and the 'end effect' is also there to upset the voltage distribution. I'll bet all that wasn't included in your first introduction to antenna theory! :wink:
     
  21. Nov 30, 2011 #20
    let l=λ/2
     
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