Many times on the VHF Reflector I see folks asking for the equations to build a Power Divider. Well, not only the equations for power dividers but everything else used in Ham Radio and Electronics in general. I have to wonder if folks these days have no reference material or access to them at all. So, I thought I’d do the research for those of little reference and maybe turn it into a Think Tank article.
To start let’s find out just what constitutes a Power Divider? Well as the name implies it is a device that divides power (equally) between two output ports from one input port or combines two antennas as an array and presents them as a 50 ohm load to a transmitter/receiver combination or transceiver. The Divider/Combiner can be a quarter wave section of transmission line or it can be a can half wavelength section. First off several things need to be understood about transmission line theory and they are; A quarter wave section of line is like a matching transformer. It will transform a high impedance to a low impedance from one end to the other. A half wavelength section of line will present the input impedance to the output end. The surge impedance of a transmission line is the characteristic impedance of the line and remains the same no matter what impedance it is transforming. A 50 ohm transmission line has a 50 ohm impedance along its entire length. If the instantaneous impedance say 100 ohms is introduced atone end of a half wavelength of line it will be 100 ohms a half wavelength or 180 electrical degrees later. This may not seem relevant right now but will when we see that an impedance at the ends of a couple of joining lines must be assumed in order to present a desired terminal impedance. Sometimes for clarity we call the surge impedance the DC impedance and the instantaneous impedance the AC impedance.
Building Power Dividers is by no means Rocket Science by any stretch of the imagination. A lot of common sense goes into the construction of a Power Divider. Many do not understand the preservation of phase in an antenna system, and this can get you in more trouble than the attempt to stack a pair of like antennas is worth. When stacking a pair or a quad array of like antennas the transmission lines between the antennas and the Power Divider must be equal lengths. They do not have to be an even number of half wavelengths (or quarter wavelengths) contrary to a few ‘old wives tales’ that have been floating around Ham Radio circles for years. However; there is a situation in stacking a pair of antennas where the stacking harness is made up of a ¼ wavelength section and a ¾ wavelength section, which negates the use of a Power Divider. This isn’t important at this time and may be discussed later in the text.
Power Dividers generally are of the ‘two way’ or the ‘four way’ variety (a three way is possible but not commonly used). The ‘two’, ‘four’ or (‘three’) refers to the ports of division. If we want to feed two like antennas we use a ‘two’ way and feeding four we use a ‘four’ way. We sometimes designate the two types as ‘2X’ and ‘4X’. The two varieties can exist as a quarter wavelength device or a half wavelength device. I prefer the half wavelength device because a 4 way can then be made of 50 ohm transmission line. I’ll show proof of this later. The body of the Power Divider can be constructed of round or square material, however; the general equation for coaxial transmission line must then be modified to compensate for the geometric configuration. The center conductor of a Power Divider can be made of round, square or rectangular material, but again the general equation must be modified to compensate for the change in material geometry. The general equation used to determine the surge impedance of coaxial transmission line that has a round center conductor and a round outer conductor is as follows:
Zo = 138 Log (D/d)
Log is always to base 10, (ln is natural log).
Zo is the surge impedance, generally 50 ohms,
D = inside diameter of the outer conductor,
d = outside diameter of the inter conductor.
Any dielectric other than air requires that the constant 138 be modified by dividing it by the square root of the dielectric constant of the insulating material.
Most common materials (aluminum, copper, brass), in air, will have a velocity factor of about 96%. The devices themselves are fairly low Q and broad banded enough so that free space equations for lengths may be used. However; for use in VHF/UHF devices the constant 5600/f (MHz) will yield the length in inches of a half wavelength of material taking into account a velocity factor of 96%. Something else that must be understood in the mathematical design of a Power Divider or Impedance Converter (Bazooka) is the geometric mean of two numbers. To find the geometric mean of two numbers they are multiplied together and then the square root of the product is taken. I mention the Bazooka only as a teaching device to demonstrate geometric mean.
A Bazooka would be used if there was ever a need to convert a 75 ohm line to a 50 ohm system. A quarter wavelength (at the frequency of interest) of transmission line would be used. That piece of transmission line would have a certain surge impedance in order to cause a minimum of mismatch between the input and output impedance’s. The product of the two values (50 and 75) is 3,750. The square root of this number is 61.24, and this is the number that would be used as Zo in the general equation for coaxial transmission lines. The derivation of the general equation is required in order to acquire the proper D/d ratio which in this case is 2.78. In other words the inside diameter of the outer conductor must be 2.78 times the outside diameter of the inter conductor. For example, if the inter conductor were be a piece of 3/32” brazing rod, the outer conductor would have to have an inside diameter of 0.2585” in this case at hand. A piece of ¼” copper pipe would suffice and could be used as the discrepancy between 0.2583 and 0.2500 would only be 3.25% and not an error worth considering. When building these devices try to work around standard or commonly available values. The Bazooka is then fitted with a coax connector of choice and the other end attached to the coax cable of interest that needs to be transformed to the new higher or lower impedance. Or made to transformer to transform 50 to 75 ohms or vise versa.
Now let’s see about building a quarter wavelength, 2X Power Divider for use at 144.2 MHz. The communications system in which this device is intended to be used is most likely going to be 50 ohms. The two antennas that will be stacked as an array will also most likely have terminal impedance’s of 50 ohms. First determine the length of material needed for the quarter wave section by (5600/144.2)/2 = 19.4175 (19.4”). The two antennas impedance’s (50 ohms) are being paralleled at one end of the Power Divider and the resultant impedance at that end will be 25 ohms. Since the device is to be used in a 50 ohm system the other end of the Power Divider will need to be 50 ohms. The impedance of the Power Divider must be the geometric mean of the two numbers (25 and 50), which is 35.36 ohms. We turn the crank on the general equation and find the required ratio of the conductors must be 1.804:1. The Power Divider inner conductor is going to be 19.5” long and therefore the outer conductor can be an inch longer on both ends making it 21.5”. The added inch on each end will give room for mounting the coax connectors and allow the center pins of the connectors to reside at points that are the distance or length of the inner conductor. At that length since the device will be out in the weather you might want to make it somewhat substantial. How about 5/8” (0.625”) copper pipe? To find the outside diameter, of the inner conductor we simply divide 0.625 by the acquired ratio (1.804:1) and find the it is 0.3465. This is real close to 11/32 (0.3438) and brass tubing in 64ths of an inch is readily available in most hobby shops and hardware stores. Remember; Pipe is measured by inside diameter and tubing by outside diameter. The percentage difference between 11/32” tubing and the mathematical statement of 0.3465 is about ¾ of a percent and will change the impedance from the desired 35.36 ohms to 35.83 ohms, a difference of <1.29%. Again, not enough to really worry about.
The 2X Power Divider could have been made with a half wavelength of material so that the antennas would be connected to both ends and fed in the middle. At the middle, the Power Divider needs to present a 50 ohm terminal impedance to a transmission line, if in fact the system is 50 ohms. The inner conductor is essentially two, one-quarter wavelength sections in series. To obtain 50 ohms of the two basically paralleled conductors, 100 ohms at that point must be assumed. The two, one-quarter wavelength sections that are joined in the center have to be the geometric mean of 50 ohms and 100 ohms or 70.7 ohms. Does this number look familiar? Yes, it is near that of 75 ohm transmission line. Could a piece of 75 ohm transmission line be used as a Power Divider? Yes, absolutely! The VSWR incurred by the difference of 70.7 to 75 ohms is 1.06:1, again not enough to worry about. If constructing the Power Divider of 75 ohm transmission line be sure to take into account the velocity of propagation of whatever line is used. If a constructed Power Divider is more desirable, then just turn the crank on the equations and fit the pieces together.
The 4X Power Divider can be built two ways. The phasing lines of all four antennas can be attached to one end of a quarter wavelength device or the device can be a half wavelength long and two each of the four antennas attached to each end. And the feed point will be in the middle of the device. In the first case, one end will of course be 50 ohms. The other end will be the result of paralleling four 50 ohm antennas (transmission lines). The result will be the impedance, at one end, of the device that will be 12.5 ohms. The geometric mean of 12.5 and 50 is 25 ohms. A possible problem arises here as the impedance of a transmission line gets smaller the outside diameter of the inner conductor will get larger or closer to the inside diameter of the outer conductor. Let’s consider using 5/8” copper pipe as the outer conductor. The inner conductor will have to have a 0.4110” outer diameter. This leaves only a tenth of an inch spacing between the two conductors. A little warping, a little condensation and there’s going to be trouble. It would be better (but not much), if it is desirable to use a quarter wavelength section to parallel two quarter wavelengths of 50 ohm hard line (Heliax or coax). Then gin up the 4X, end fitting on one end of the paralleled lengths and the same on the 50 ohm end.
Myself, I would make the 4X Power Divider of a half wavelength of material, placing two coax fittings at each end and feed the device in the center. In this manner, the geometric mean of the end and center impedance is the square root of the product of 25 and 100 ohms, or a 50 ohm result. As I mentioned previously a 4X device can be made from 50 ohm transmission line or again turn the crank on the equations, select the values for the material needed, tack them all together and have a 4X power Divider that way. By the way, the spacing between inter and outer conductors then if 5/8” copper pipe is used for the outer conductor is on the order of 180 thousandths of an inch or almost twice the distance of that of the quarter wavelength section.
I mentioned the Dividerless phasing harness in the beginning; let me explain. If it is desirable to stack a pair of like antennas that have a 50 ohm terminal impedance and require on the order of 5/8 wavelength spacing this scheme works well. An example of these would be a pair of 2M Big Wheels. Anyway the coax used will be one wavelength long at the frequency of interest taking the velocity factor of the coax into consideration. This coax must be 75 ohm line; Belden 8213, RG-11, RG-6, Cable TV slick side, aluminum jacketed, hard-line. Whatever the dimension comes out to be, divide it by 4 and cut that dimension or 1/4th off one end. Install four connectors of choice and place a triple female “T” adapter between the two prepared lengths. Attach the antennas to the coax ends and feed with 50 ohm line. Make sure the two antennas are 180 degrees out of phase, i.e. in the case of Gamma matched antennas, the Gamma matches must be on opposite sides of the two stacked antennas.
From: The Rocky Mountain VHF+ Newsletter THINK TANK August 1999
Dave W6OAL –
Olde Antenna Laboratory 41541 Dublin Drive Parker, CO 80138
