The last post on the 15 meter antenna brought up the radiation angle and the height of the ionosphere. I thought I would switch gears here and discuss some radio wave propagation. To make effective use of any antenna or to choose the right antenna it helps to know something about propagation or how a radio wave gets from here to there.
Most of our high frequency radio communication depends on the ionosphere bending the radio signal so it will return to earth. The ionosphere is an area of the upper atmosphere where the air is much less dense than here on the surface of the earth. It is also gets the full force of the radiation from the sun. It is the radiation from the sun that causes the air molecules to form ions. Positive and negative charges are created by the radiation from the sun. The strength of the radiation from the sun varies considerably. To put it simply, the radiation varies somewhat predictably with the number of sunspots following an eleven-year cycle. In other words there is usually a peak in the number of sunspots every eleven years. This peak in sun activity coincides with extremely good radio conditions on the higher frequency amateur bands such as 20, 15 and 10 meters.
In addition to the effect of the sun spot cycle, there is a seasonal and daily variation. The daily variation is mostly due to the fact that as the earth rotates, areas of the atmosphere or ionosphere are in view of the sun or being shielded by the sun. We find that during the day the ionosphere is “stronger” than at night. This is really a two edged sword. The higher daytime ionization helps return the higher frequencies such as 14, 21 and 28 MHz back to earth. It has a detrimental effect, however, on the lower frequencies (longer wavelengths). During the day a dense lower ionosphere layer forms. It is called the “D” layer. This strong daytime or “D” layer of the ionosphere absorbs the longer wavelengths. For this reason there is practically no sky wave propagation on 160 or 80 meters during the day. The 40 meter band is right in the middle. There is not much absorption on 40 meter signals so there is some sky wave propagation on 40 during the day. During the evening the “D” layer disappears and there is good sky wave propagation on the 160 and 80 meter bands and signals even get stronger on 40 meters.
Very high sun activity causes so much ionization that the 20, 15 and 10 meter bands can stay open for sky wave propagation 24 hours a day. Except for these periods of high sunspot activity, these bands are mostly considered “daytime” bands.
In general the higher the frequency, the more ionization necessary for sky wave propagation. This means that in the evening 10 meters will close first, then 15 meters will close and the last high band to close will be 20 meters. In the morning the reverse is true. 20 meters opens first, then 15 and then 10. The times all depend to how much radiation there is from the sun at any given time.
Saturday, March 6, 2010
Tuesday, March 2, 2010
The 15 Meter Dipole and elevation angles
The higher bands (15 and 10 meters) are now starting to open for DX.
If you do not have an antenna for these bands yet, I suggest building a simple dipole fed with coax. For lengths of up to 100 feet, RG 8X is not a bad choice. The loss for 100 feet of RG8 X should be less than 2 dB at 15 meters. Less loss will require larger and heavier coax. I feel it is most important to get the dipole outdoors if possible. It does not have to be real high as long as it is away from any nearby objects, especially metalic objects. On 15 meters I consider 20 to 25 feet high. That height puts the antenna a halfwave above the ground. Any antenna that is a halfwave above the ground is frequently a better general purpose antenna than one twice as high. I will go into more detail about antenna patterns at a later date. For now let's just say that I think the magic number for the height of a horizontal antenna (not a vertical antenna) is one half wavelength!
That's 16 feet for 10meters, 22 feet for 15 meters and 33 feet for 20 meters. Antennas that are in the clear, away from any nearby objects work very well at these heights. One problem may be 16 feet on 10 meters. It may be hard to get a 10 meter antenna in the clear at only 16 feet. In that case the best solution is to double the height to 32 feet. The same goes for the other bands. Doubling the height to one full wavelength will give added performance at long distances but will produce a reduction in performance at an intermediate distance. I am most interested in DX on 15 meters, so I chose to put my 15 meter dipole at about 44 feet. That height is acheivable because I have some nice tall trees I can use for support. I chose to do the same for my 10 meter dipole. If I ever put up a tower again I expect to only go about 35 feet. I have had my 20 meter beam at 70 feet and if I put it back up I think it will do almost as well at a halfwave as at one full wave. I also will not have the null at a vertical angle of 30 degrees.
All horizontal antennas will exhibit a decrease in performance or null at an elevation angle of 30 degrees. This elevation angle translates to loss of performance at distances of 350 and 700 miles (1 or 2 hop E skip) and distances of 550 and 1100 miles for F layer skip. These numbers vary quite a bit because they depend on the height of the ionospheric layer layers. These heights vary with the time of day and sun activity. Variations in the F layer height could cause this distance of reduced performance for a horizontal antenna one full wavelength above the ground to vary form about 400 to 800 miles.
If you are most interested in these distances be sure your antenna is not a full wavelength above the ground. A horizontal antenna half as high (one-half wavelength) will have its maximum performance at these distances and (all other things being equal) could actually be 10 dB stronger than the higher antenna!
If you do not have an antenna for these bands yet, I suggest building a simple dipole fed with coax. For lengths of up to 100 feet, RG 8X is not a bad choice. The loss for 100 feet of RG8 X should be less than 2 dB at 15 meters. Less loss will require larger and heavier coax. I feel it is most important to get the dipole outdoors if possible. It does not have to be real high as long as it is away from any nearby objects, especially metalic objects. On 15 meters I consider 20 to 25 feet high. That height puts the antenna a halfwave above the ground. Any antenna that is a halfwave above the ground is frequently a better general purpose antenna than one twice as high. I will go into more detail about antenna patterns at a later date. For now let's just say that I think the magic number for the height of a horizontal antenna (not a vertical antenna) is one half wavelength!
That's 16 feet for 10meters, 22 feet for 15 meters and 33 feet for 20 meters. Antennas that are in the clear, away from any nearby objects work very well at these heights. One problem may be 16 feet on 10 meters. It may be hard to get a 10 meter antenna in the clear at only 16 feet. In that case the best solution is to double the height to 32 feet. The same goes for the other bands. Doubling the height to one full wavelength will give added performance at long distances but will produce a reduction in performance at an intermediate distance. I am most interested in DX on 15 meters, so I chose to put my 15 meter dipole at about 44 feet. That height is acheivable because I have some nice tall trees I can use for support. I chose to do the same for my 10 meter dipole. If I ever put up a tower again I expect to only go about 35 feet. I have had my 20 meter beam at 70 feet and if I put it back up I think it will do almost as well at a halfwave as at one full wave. I also will not have the null at a vertical angle of 30 degrees.
All horizontal antennas will exhibit a decrease in performance or null at an elevation angle of 30 degrees. This elevation angle translates to loss of performance at distances of 350 and 700 miles (1 or 2 hop E skip) and distances of 550 and 1100 miles for F layer skip. These numbers vary quite a bit because they depend on the height of the ionospheric layer layers. These heights vary with the time of day and sun activity. Variations in the F layer height could cause this distance of reduced performance for a horizontal antenna one full wavelength above the ground to vary form about 400 to 800 miles.
If you are most interested in these distances be sure your antenna is not a full wavelength above the ground. A horizontal antenna half as high (one-half wavelength) will have its maximum performance at these distances and (all other things being equal) could actually be 10 dB stronger than the higher antenna!
Verticals if you can not get horizontal antennas "high"
I have recommended that horizontal antennas be a half wave above ground for best performance. For the lower bands, this states to be a problem for most of us. While a half wave on 20 meters is only 33 feet, on 40 this becomes 66 feet and on 80 meters this is now 120 or 130 feet! Most of us can only have "low" dipoles on 80 or 160 meters. In order to have a good performing antenna on these bands, especially for DX, we have to either put up a very high tower or use some form of vertical. It is amazing how much improvement a simple bent vertical or inverted L antenna can be over a dipole at the same "low" heigth.
The low dipole does a great job for distances up to maybe 500 miles on 160 or 80 meters. If you want a reasonably strong signal at greater distances, some form of vertical is porbably your answer.
I was able to hang a half square antenna for 80 meters between two pine trees. I was not able to get the vertical elements the desired 66 feet high so I bent the bottom ends. This does not affect the performance much at all because, in this antenna, the two ends at the ground are open. (Not connected to anything). What do we know about the open ends of an antenna? The current is zero at the ends. So the last 5 or 10 feet on a 80 meter antanna wire dies not cary much current. If there is not much current we find that there is not much radiation. Most of the radiation from an antenna is from the high or higher current areas on the antenna. The maximum height of my half square antenna is only about 50 feet. For working DX, say Europe for examply, I would need an 80 meter dipole about 130 feet high to equal the 50 foot high half square! The antenna mocking programs such as EZNEC predict this, and I can say that my experience with this antenna has verified that prediction time and again.
The low dipole does a great job for distances up to maybe 500 miles on 160 or 80 meters. If you want a reasonably strong signal at greater distances, some form of vertical is porbably your answer.
I was able to hang a half square antenna for 80 meters between two pine trees. I was not able to get the vertical elements the desired 66 feet high so I bent the bottom ends. This does not affect the performance much at all because, in this antenna, the two ends at the ground are open. (Not connected to anything). What do we know about the open ends of an antenna? The current is zero at the ends. So the last 5 or 10 feet on a 80 meter antanna wire dies not cary much current. If there is not much current we find that there is not much radiation. Most of the radiation from an antenna is from the high or higher current areas on the antenna. The maximum height of my half square antenna is only about 50 feet. For working DX, say Europe for examply, I would need an 80 meter dipole about 130 feet high to equal the 50 foot high half square! The antenna mocking programs such as EZNEC predict this, and I can say that my experience with this antenna has verified that prediction time and again.
Saturday, February 27, 2010
More about the ImpedanceConcept
The concept of impedance is very important in antennas and transmission lines. Remember that impedance is the ratio of voltage to current. Let's look at what happens when we apply a radio frequency voltage to one end. When the voltage is first applied, a certain amount of current will flow. What ever the ratio voltage to current turns out to be is called the characteristic impedance of the transmission line. If we apply 10 volts and find that initially .2 amps flows, then 10/. 2 = 50. The impedance of the line is 50 ohms. If the line was 186,000 miles long we would be able to measure this .2 amp current for about two seconds. After however long it takes the applied voltage and current to reach the end and be reflected this current may change. If the transmission line is connected to an impedance equal to the characteristic impedance, then there will be no reflection and the current remains at its initial value. You will also hear the term surge impedance used interchangeably with characteristic Impedance. Usually you will hear it shortened to just impedance. Most of us will be using coax for transmission line and it will probably have an impedance of 50 ohms or somewhere close to that. Coax is also frequently found with 75-ohm impedance. In order to get as much of your power as possible into the antenna so it can be radiated, you need to have the antenna impedance as close to the transmission line's impedance as possible. If the antennas impedance does not match that if the line, you can use a simple matching network. The main idea is to present a load to the transmission line so there is a minimum mismatch and minimum reflected voltage and current where the line connects to the antenna.
If there is a reflection it causes standing waves to be set up on the transmission line. Usually I like to see the standing wave ratio less than 2:1.
The higher you go in frequency, the more important it is to have a low SWR. We will discuss SWR in more detail at a later date.
If there is a reflection it causes standing waves to be set up on the transmission line. Usually I like to see the standing wave ratio less than 2:1.
The higher you go in frequency, the more important it is to have a low SWR. We will discuss SWR in more detail at a later date.
Saturday, February 13, 2010
Two halves does not always make a whole
Antennas do not have to be resonant to work well, but it does make feeding them a little less complicated in general. When learning the basics it may make it easier to study resonant antennas first. It is my intention to stick with resonant antennas for a while longer. There may be exceptions as there is with anything.
The shortest resonant antenna is a half wave long. Remember a half wave in free space is determined by the formula 492/frequency in MHz. When confined to a wire, the speed changes and there are other effects that appear to change the speed. The sum total of all these effects is to reduce the length by about 5%. So you change the formula to 468 if you are using most kinds of wire. That means that a half wave of wire is determined by dividing 468 by the frequency in MHz. So for example on 1.850 MHZ (160 meter band) the length of a half wave of wire is 468/1.85 = 252.9 feet or about 253 feet. The formula for a half wave in a wire is not exact. Although the length depends primarily on the speed of the radio signal in the wire, the type of insulation and its thickness makes a big difference. There are also effects caused by the wire being close to the ground or any other conductor, even wood or tree branches. There is an effect due to the insulator at the end of the wire. Some of these effects, especially due to the insulation, are more pronounced the higher the frequency. This formula is also not good if the antenna is more than one half wave long. The formula was reduced from 492 to 468 for a one half wavelength wire that has two ends connected to an insulator. If the antenna is say, one wavelength long, there are still only two ends, but there is effectively a half wavelength of wire in the middle that has no open ends. That has an effect on the calculation. It turns out that a large part of the reason that a half wave wire antenna is physically shorter than a half wave in free space is due to the insulators at the two ends or the “end effects”. The way I see it, the ends will either have an insulator or be hanging loose. In either case there is an end effect of one sort or another. If you connect two half wave lengths of wire together, you will find that together they are shorter than they should be for full wave resonance. The calculation of each half wave was based on each length having two ends. Remember, a full wavelength of wire will be slightly longer than two half-wave lengths added together. This is not a big deal, but something to keep in mind if you are constructing a wire antenna longer than a half wave long.
The shortest resonant antenna is a half wave long. Remember a half wave in free space is determined by the formula 492/frequency in MHz. When confined to a wire, the speed changes and there are other effects that appear to change the speed. The sum total of all these effects is to reduce the length by about 5%. So you change the formula to 468 if you are using most kinds of wire. That means that a half wave of wire is determined by dividing 468 by the frequency in MHz. So for example on 1.850 MHZ (160 meter band) the length of a half wave of wire is 468/1.85 = 252.9 feet or about 253 feet. The formula for a half wave in a wire is not exact. Although the length depends primarily on the speed of the radio signal in the wire, the type of insulation and its thickness makes a big difference. There are also effects caused by the wire being close to the ground or any other conductor, even wood or tree branches. There is an effect due to the insulator at the end of the wire. Some of these effects, especially due to the insulation, are more pronounced the higher the frequency. This formula is also not good if the antenna is more than one half wave long. The formula was reduced from 492 to 468 for a one half wavelength wire that has two ends connected to an insulator. If the antenna is say, one wavelength long, there are still only two ends, but there is effectively a half wavelength of wire in the middle that has no open ends. That has an effect on the calculation. It turns out that a large part of the reason that a half wave wire antenna is physically shorter than a half wave in free space is due to the insulators at the two ends or the “end effects”. The way I see it, the ends will either have an insulator or be hanging loose. In either case there is an end effect of one sort or another. If you connect two half wave lengths of wire together, you will find that together they are shorter than they should be for full wave resonance. The calculation of each half wave was based on each length having two ends. Remember, a full wavelength of wire will be slightly longer than two half-wave lengths added together. This is not a big deal, but something to keep in mind if you are constructing a wire antenna longer than a half wave long.
Sunday, February 7, 2010
Sketching Current on the Antenna
I may be repeating a few things as we go along, but I think the important principles of radio and antenna theory need to be reinforced. I have found over the years that current, voltage, impedance and standing waves were among the most misunderstood concepts in radio or electronics. I think everyone has heard the term SWR (Standing Wave Ratio). I have been talking about standing waves on antennas. These are both standing waves of voltage and standing waves of current.
The ratio of voltage to current is the impedance at that point. When dealing with transmission likes, the common SWR is not a voltage to current ratio, but a ratio of two voltages measured at two different places on a line (VSWR). It could also be the ratio of two currents measured at different places on the line (ISWR). In fact there are numerous methods that can be used to measure transmission line SWR. I will eventually get around to explaining that. Right now I want to get the concept of antenna current and voltage down first. That really comes first and is a direct cause of the transmission line SWR.
There are a couple things that we need to know about antennas. First we want them to radiate as high a percentage of the power we supply to them as possible. (We want the transmission line to deliver as high a percentage of the transmitter power to the antenna as possible.) In order for the antenna to be able to accept all the power from the feed line we must have an antenna whose impedance matches that of the feed line. There are numerous physical things that will determine this ‘feed point” impedance. Almost all of these things are within our control, but we must have a basic understanding of them and how they work in order to control them. Otherwise it is just by trial and error.
The most important thing is to know approximately what the current and voltages are on an antenna. In order to start we need to pick a frequency and determine the wavelength and half wavelength. I always sketch the antenna then note some ballpark information about the current standing wave at various points along the wire. What do we know about the current at an open end? (It will be zero) Then I put a dot a quarter wave back from every end. What do we know about current at this point? (Current will be a maximum.) Continue on away from the end another quarter wave (if the wire is long enough). What do we know about the current here? It will be zero or a minimum. The reason it is not zero is that some of it is radiated as it travels along the wire to and from the ends, and when it crosses “fresh” current, there is not enough of it to completely cancel out the “fresh” current.
You should be able to sketch a half wave wire and draw a smooth curve representing the current. It should look like the first half of a sine wave. The curve representing current should start at zero on one end, rise like a sine wave to a maximum in the center and down to zero at the other end. If the wire is one wave long, we draw that current curve the same for the first half wave (going from left to right) then the second half wave is drawn below the line down to a maximum negative value, then back up to zero at the end. The current above the curve represents positive current and below the line represents negative current. It does not really matter which way the first arrow goes, what does matter is that the current in the first half wave flows one way and the current in the second half wave flows the opposite way. Another way to illustrate can be seen if you draw a long line from left to right. Divide it into four equal segments. So you can follow what I am doing, number the segments left to right 1, 2, 3 and 4. Over segments 2 and 4 draw an arrow pointing right. Over segments 1 and 3 draw an arrow pointing left. The currents in segments 2 and 4 are in phase; the currents in 1 and 3 are also in phase with each other but out of phase with currents in segments 2 and 4. To start to analyze any antenna, you can draw a sketch. Divide the wire into half wave segments and draw arrows over each half-wave segment, being sure to change the direction of the arrows in each adjacent half-wave segment.
Once you have drawn the sketch, you immediately know where the current is zero (at the ends) and where the current is a maximum (between the zero points). You also know that the points where the impedance is low are at the same place where the current is a maximum. These are the most likely places that you would want to connect a coaxial feed line. (There are exceptions, but we will deal with the exceptions one at a time later. For now, be advised that there seems to always be an exception to everything)
The impedance is different at different points on the wire. The voltage to current ratio at the ends is very high. Voltage is at its highest. We say the current is zero. You cannot really divide something like 100 or 200 volts by zero. You have to divide by almost zero, and get a close answer. In math or physics we always seem to run into problems where we want to divide by zero. If you divide by zero you get infinity. Not a good answer. So what we do is usually try and figure out what the limit is as we approach zero. We divide by smaller and smaller numbers to approximate dividing by zero. You might think of it a instead of dividing by zero, dividing by 1/1000 amps or maybe 1/1,000,000 amps. Sometimes you get close enough to the right answer that no one can ever tell it is not exactly the right number! In our case it is almost always good enough to say that the impedance at the end of an antenna wire is very high and leave it at that. I do. Funny things happen in the real world when you try and work with impedances like at the end of a wire where they are real high. There are situations where things become unstable and unpredictable. There are cases where impedances rise very fast to an extremely high inductive value and then like you would flip a switch, they change to a very high capacitive value. For most of us, we need only know that such conditions do exist, recognize when they may exist, understand that it is best to avoid them! I will try and help you with that. I just rambled some. Let me back up and start again.
The impedance is different at different points on the wire. The voltage to current ratio is very high at the ends and this ratio goes down as you move away from the end. The impedance reaches a minimum at the middle of each half-wave section. You can connect cut the wire and connect a feed line anywhere. In the center you will get a good match to 50 or 75 ohm coax. Somewhere between the center and the end you may find a point that is a close match for 300-ohm line. No matter where you choose to connect a transmission line on any length wire there will be an impedance at that point that is acting like either a pure resistance, or a resistance in series with a coil (inductor) or capacitance. If the impedance is not a good match for the chosen feed line, you have several choices. First you can change the point of connection. That sometimes works. Second you can change the type of feed line. Third you can add a matching circuit or matching network to transform the impedance of the antenna to the value of the transmission line. Fourth you can change the length of the antenna (longer or shorter). Lastly, you can do nothing. That means you accept the additional loss due to the mismatch. We normally do not accept much of a mismatch when using coax. However, when using open wire or parallel wire transmission line, typically 300 or 600-ohm lines, we find that the increased losses due to a rather large mismatch are not significant and we can live with that small loss.
The ratio of voltage to current is the impedance at that point. When dealing with transmission likes, the common SWR is not a voltage to current ratio, but a ratio of two voltages measured at two different places on a line (VSWR). It could also be the ratio of two currents measured at different places on the line (ISWR). In fact there are numerous methods that can be used to measure transmission line SWR. I will eventually get around to explaining that. Right now I want to get the concept of antenna current and voltage down first. That really comes first and is a direct cause of the transmission line SWR.
There are a couple things that we need to know about antennas. First we want them to radiate as high a percentage of the power we supply to them as possible. (We want the transmission line to deliver as high a percentage of the transmitter power to the antenna as possible.) In order for the antenna to be able to accept all the power from the feed line we must have an antenna whose impedance matches that of the feed line. There are numerous physical things that will determine this ‘feed point” impedance. Almost all of these things are within our control, but we must have a basic understanding of them and how they work in order to control them. Otherwise it is just by trial and error.
The most important thing is to know approximately what the current and voltages are on an antenna. In order to start we need to pick a frequency and determine the wavelength and half wavelength. I always sketch the antenna then note some ballpark information about the current standing wave at various points along the wire. What do we know about the current at an open end? (It will be zero) Then I put a dot a quarter wave back from every end. What do we know about current at this point? (Current will be a maximum.) Continue on away from the end another quarter wave (if the wire is long enough). What do we know about the current here? It will be zero or a minimum. The reason it is not zero is that some of it is radiated as it travels along the wire to and from the ends, and when it crosses “fresh” current, there is not enough of it to completely cancel out the “fresh” current.
You should be able to sketch a half wave wire and draw a smooth curve representing the current. It should look like the first half of a sine wave. The curve representing current should start at zero on one end, rise like a sine wave to a maximum in the center and down to zero at the other end. If the wire is one wave long, we draw that current curve the same for the first half wave (going from left to right) then the second half wave is drawn below the line down to a maximum negative value, then back up to zero at the end. The current above the curve represents positive current and below the line represents negative current. It does not really matter which way the first arrow goes, what does matter is that the current in the first half wave flows one way and the current in the second half wave flows the opposite way. Another way to illustrate can be seen if you draw a long line from left to right. Divide it into four equal segments. So you can follow what I am doing, number the segments left to right 1, 2, 3 and 4. Over segments 2 and 4 draw an arrow pointing right. Over segments 1 and 3 draw an arrow pointing left. The currents in segments 2 and 4 are in phase; the currents in 1 and 3 are also in phase with each other but out of phase with currents in segments 2 and 4. To start to analyze any antenna, you can draw a sketch. Divide the wire into half wave segments and draw arrows over each half-wave segment, being sure to change the direction of the arrows in each adjacent half-wave segment.
Once you have drawn the sketch, you immediately know where the current is zero (at the ends) and where the current is a maximum (between the zero points). You also know that the points where the impedance is low are at the same place where the current is a maximum. These are the most likely places that you would want to connect a coaxial feed line. (There are exceptions, but we will deal with the exceptions one at a time later. For now, be advised that there seems to always be an exception to everything)
The impedance is different at different points on the wire. The voltage to current ratio at the ends is very high. Voltage is at its highest. We say the current is zero. You cannot really divide something like 100 or 200 volts by zero. You have to divide by almost zero, and get a close answer. In math or physics we always seem to run into problems where we want to divide by zero. If you divide by zero you get infinity. Not a good answer. So what we do is usually try and figure out what the limit is as we approach zero. We divide by smaller and smaller numbers to approximate dividing by zero. You might think of it a instead of dividing by zero, dividing by 1/1000 amps or maybe 1/1,000,000 amps. Sometimes you get close enough to the right answer that no one can ever tell it is not exactly the right number! In our case it is almost always good enough to say that the impedance at the end of an antenna wire is very high and leave it at that. I do. Funny things happen in the real world when you try and work with impedances like at the end of a wire where they are real high. There are situations where things become unstable and unpredictable. There are cases where impedances rise very fast to an extremely high inductive value and then like you would flip a switch, they change to a very high capacitive value. For most of us, we need only know that such conditions do exist, recognize when they may exist, understand that it is best to avoid them! I will try and help you with that. I just rambled some. Let me back up and start again.
The impedance is different at different points on the wire. The voltage to current ratio is very high at the ends and this ratio goes down as you move away from the end. The impedance reaches a minimum at the middle of each half-wave section. You can connect cut the wire and connect a feed line anywhere. In the center you will get a good match to 50 or 75 ohm coax. Somewhere between the center and the end you may find a point that is a close match for 300-ohm line. No matter where you choose to connect a transmission line on any length wire there will be an impedance at that point that is acting like either a pure resistance, or a resistance in series with a coil (inductor) or capacitance. If the impedance is not a good match for the chosen feed line, you have several choices. First you can change the point of connection. That sometimes works. Second you can change the type of feed line. Third you can add a matching circuit or matching network to transform the impedance of the antenna to the value of the transmission line. Fourth you can change the length of the antenna (longer or shorter). Lastly, you can do nothing. That means you accept the additional loss due to the mismatch. We normally do not accept much of a mismatch when using coax. However, when using open wire or parallel wire transmission line, typically 300 or 600-ohm lines, we find that the increased losses due to a rather large mismatch are not significant and we can live with that small loss.
Friday, February 5, 2010
Slight review and trap dipole principle
In a resonant half wave dipole, the voltage and current enter the antenna at the center, travel to the ends and get reflected. The current reflection is out of phase so the currents at the ends cancel. There is always zero current at the ends. (There is always zero current at the end of a an open wire. The voltage is also reflected but is not reversed in polarity like the current, so the voltage at the end is pretty much twice the original voltage. Due to the fact that the voltage and current vary with time, the actual currents and voltages that are set up on the antenna (these are called standing waves) varies or oscillates between two values. At the center the current varies at any given instant between a maximum positive value and a maximum negative value. As you move toward either end the current varies between two limits of decreasing value until at the end the two values go to zero. The voltage and current at the antenna terminals will be in phase. That means when the voltage peaks the current peaks. When the voltage is zero the current is zero. This is just like current and voltage in a pure resistor. When the radio frequency voltage is first applied at the transmitter end of a feed line current will flow (in a ratio of voltage to current determined by the characteristic impedance of the line) Lets look at the current first. The current reaches the terminals then flows to the end, gets reflected (out of phase) and flows back to the center. When it gets to the center it is exactly in phase with the current from the next cycle (remember this is only true for a resonant condition). These two currents then simply add together. Now lets think about the voltage. The voltage signal enters the antenna and travels down the wire to the end where it is reflected but the phase is not reversed. By the time it gets back to the terminals, the voltage is out of phase with the voltage of the next cycle. This reflected voltage tends to cancel the incoming voltage much like happens to current at the ends of the antenna. We should call the voltage and current that comes up the transmission line the initial voltage and current. We can also call it the incident voltage and current. These are the voltages and currents existing before things get muddy due to the reflections, standing waves and all that. The current and voltage waves that are set up on the antenna are standing waves of voltage and current. Initial voltage and current are in phase. Like they would be in a pure resistance. If the antenna is resonant the resulting current will still be in phase with this terminal voltage. The resonant antenna will act like a pure resistance. Now if the antenna is long or short, the resulting standing waves set up on the antenna will be a bit different and will cause the resulting current at the antenna terminals to be slightly out of phase with the applied voltage which will cause the antenna to not act like a pure resistance. The antenna will act like it has a coil or capacitor connected in series with it, even though it does not. It just acts like it does. That’s why we say the antenna is capacitive if it is short and inductive if it is long.
We can take a center fed dipole that is resonant and make it longer. We can also take a center fed dipole and add a coil to both sides. If we pick the right size coil, you will not be able to tell the difference at the antenna terminals. The same goes for making the antenna shorter or adding a capacitor. You will not be able to tell the difference. You can take a short antenna and add a coil to make it “appear” to be longer. You can take a long antenna and add a capacitor and make it appear shorter. This is the principle behind trap antennas. The trap is designed so that it makes the antenna appear to be longer or shorter so that the voltage and current at the terminals will be in phase and of a low enough impedance to be a good match to a coax transmission line.
We can take a center fed dipole that is resonant and make it longer. We can also take a center fed dipole and add a coil to both sides. If we pick the right size coil, you will not be able to tell the difference at the antenna terminals. The same goes for making the antenna shorter or adding a capacitor. You will not be able to tell the difference. You can take a short antenna and add a coil to make it “appear” to be longer. You can take a long antenna and add a capacitor and make it appear shorter. This is the principle behind trap antennas. The trap is designed so that it makes the antenna appear to be longer or shorter so that the voltage and current at the terminals will be in phase and of a low enough impedance to be a good match to a coax transmission line.
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