# Matching an AM transmitter to a tower

This was a fun little project I was involved with last year.  Diplexing two AM stations to a single tower.  This particular tower was brand new, replacing an older tower that was rusting from the inside out.  As such, it had slightly different characteristics than the old tower, so it was a whole new project.  Fortunately, the old Antenna Tuning Units (ATU) were made by Kintronics, so they had plenty of head room on both side of the circuit for matching purposes.

The replacement tower is up, the new unipole antenna has been installed and now it is time to match the transmitters to the tower. This involves using some math. At some point in radio history, someone decided that all transmitters should have an output impedance(Z) of 50Ω (ohms). Impedance in an alternating current (AC) circuit (all radio frequency is AC) is like resistance in a direct current (DC) circuit. The only difference is impedance requires the use of the Z axis to calculate. You remember 9th grade algebra and the Cartesian Coordinate graphing system, the X and the Y axis. Looking down on the X and Y axis, the Z axis would be stick straight up, which makes it a three dimensional problem.

Station number one, broadcasts on 980 kHz at 5,000 watts. The tower is 240 feet tall which is close to 1/4 wave length, nearly ideal for an AM station. Using a Delta Operating Impedance Bridge (OIB-1) and a Potomac SD-31 frequency Generator, I measured the impedance at the base of the tower. On 980 khz it is 74Ω with +j160 reactance. Using the ohms law pie chart, anything can be  figured out about electricity:

Math pie chart

Therefore, the base current will be I=√(P/R) =√(5000/74) = 8.3 amps. The voltage will be E=√PxR =√(5000×74) = 608 Volts.

A bit about reactance; it is noted by using the letter j, which indicates it is an imaginary number. Basically in an AC circuit, there is inductance and capacitance. They are the reciprocal of each other, sort of (this could get into a long, long post if I have to explain the roles of inductance and capacitance in and AC circuit). Reactance is an undesired inductive or capacitive component that has to do with the lead or lag time between the voltage wave form peak and the current wave form peak. In standard utility company parlance it is know as the “Power Factor”. In RF circuits it causes inefficient power transfer and it needs to be canceled. A +j value indicates that the reactance is inductive, and therefore needs to be canceled out with a capacitor. A -j value indicates the reactance is capacitive and needs to be canceled out with an inductor.

Then there is the difference in impedance, the transmitter and transmission line is 50Ω and the tower is 74Ω. Enter the antenna tuning unit (ATU). The ATU matches the base impedance of the tower through the use of a T network:

To determine the value of each leg of the T network, we need to employ math again. Here is where the details will catch up with you. Remember, there are two stations on this tower, a 980 kHz and a 1430 kHz. We need to make two T networks, one for each station. There are a few characteristics of a T network that can be used to our advantage here. A T network can also function as a low pass or high pass filter depending on the relationship between capacitance and inductance. In an inductive circuit the phase is advanced and in a capacitive circuit the phase is retarded. If we can make the phases of the two stations 180° opposing, this makes an excellent start to a filter network. Therefore, one station should be +90° and the other should be -90°.

So, on 980 kHz we want to match 50Ω to 74Ω with a +90° phase shift. Simple. Each leg of the T network needs to have the following value:

Z(leg)=√Z(antenna) x Z(transmitter) or Z=√(50 x 74) = √3700 = 60.8Ω

This is a highly simplified diagram that does not show the pass/reject filters employed between the ATU and the tower to properly combine both stations onto one antenna. That would be an extensive topic that I am not even sure I could adequately describe here:

So each leg needs to have an impedance of 60.8Ω. The input leg is inductive, the ground leg is capacitive and the output leg is inductive. Remember, the output leg is already inductive by +j163. The inductive reactance needs to be canceled out, but some of it can be used in the T network. To make the output network match the rest of the T network, it will need 102.4Ω capacitive reactance (163-60.8=102.4Ω). To calculate these values, we use the L and C formulas which are 980 KHz = .98 mHz):

C = 1/(2π f (mHz)Xc) or 1/(6.28 x .98 x 60.8) or 0.00267 uf (ground leg, 60.8Ω)

C= 1/(2π f (mHz)Xc) or 1/(6.28 x .98 x 102.4) or 0.00159 uf (output leg, 102.4Ω)

and

L= Xl / (2π f(mhz) or 60.8 / (6.28 x .98) or 9.88 uH (input leg, 60.8Ω)

This combination should get us close to the Z 50Ω impedance that the transmitter is looking for.

The next frequency is 1430 kHz with a power of 10,000 watts. This frequency should be retarded by -90 degrees, so the input will be capacitive with in inductive leg to ground and a capacitive output. The tower measures 165Ω with -j105. Perfect!

Again, the current and voltage at the base of the tower on this frequency will be I=√(P/R) = 7.78 amps and E=√PR = 1,285 volts.

Z= √(50×165) = √8250 = 90.82Ω

L = Xl / (2π f(mHz) = 90.82 / (6.28 x 1.43) = 10.11 uH (ground leg, 90.82Ω)

C= 1 / (2π f(mHz) Xc) = 1 / (6.28 x 1.43 x 90.82) = 0.00122 uf (input leg, 90.82Ω)

and

C= 1 / (2π f(mHz) Xc) = 1 / (6.28 x 1.43 x (-j105-90)) = 0.0074 uf (output leg, 75.8Ω) See comments below.

Since the current and voltage for both stations are additive (with slight variations due to phasing on the two frequencies) the total current at the tower base will be 8.3 amps + 7.8 amps = 16.1 amps and the total voltage will be 608 volts + 1285 volts = 1,893 volts. Now you know why there is a fence around the bottom of the tower!

That is the theoretical part.  Using the OIB-1 and the generator, I tuned leg to ground to give the approximate valued noted above.  The inductive legs are easier to tune than the capacitive legs.  Since the value of each component is stamped on the name plate, I was able to estimate where the tap should go.  The capacitors are fixed, so they required some series/parallel connections to get the values close.

After all that, the transmitters are turned on and the system is measured under power.  Everything was pretty close, but a little bit of final tuning was required.

Once the transmitters were happy with the match, I did an full impedance sweep of both frequencies and recalculated the base currents for each station.  Then all of the harmonics and additive frequencies were checked to make sure that any spurious emissions were below the FCC required maximums.  That involved driving about 1 mile away and using a Potomac Instruments FIM-41.

The Frequencies measured were:

• 530 kHz, lower intermod product
• 1880 kHz, upper intermod product
• 1960 kHz, 980 second harmonic
• 2940 kHz, 980 third harmonic
• 2860 kHz, 1430 second harmonic
• 4290 kHz, 1430 third harmonic

And there you have it, that is how an AM transmitter is coupled to the base of a transmitting tower.

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### 9 comments to Matching an AM transmitter to a tower

• J. Aegerter

I happen to like Phasetek over Kintronics. Their equipment is top notch and their prices are lower.

• Dan Eggers

I had an insulated 197 ft tower for KTAM AM in College Station, TX, but they took that down and put up a 400 ft tower sitting on the ground. It has a loop of wire going up the side or something, and what I was really looking for was an explanation about how a grounded tower is excited, or is it just a seperate loop or something? –Dan

• Paul Thurst

Dan, sounds like a folded unipole antenna, sometimes known as a skirted tower. wire going up the side is likely attached to the tower at some point. Google folded unipole for more information

• pattymor

Please let me know how is it possible for the Output capacitor @1430Khz to be 75.8 Ohm?. Seems like something in the Math is not matching (-j105-90.82) = -j195.82; So, 0.0074uf @ 1.43Mhz is around 15.04 Ohm; far from output leg 75.8Ω which is 1,468pf @1.43Mhz.

• Paul Thurst

Great googly moogly, I’ll now have to dig out the file and try to figure out what I was doing here.

Update: I think I see what is wrong here, the math should look like this: -j105+90.82= -14.18 ohms, which needs to be canceled out with an inductor, since the output leg is looking for 90.82 ohms capacitive resistance, which works out to be 1.65 uH.

• Sean Richardson

Hey Paul,

Great article! I’ve read it several times….have built a spreadsheet based upon it now.

Curious if you could outline the formula/calculations for the output leg on the correction you made above? I’m coming up with -1.577 uh on that leg… a little different than what you’ve got.. not sure what I’m doing wrong.

• pattymor

Sean,

No much to get confused over the output leg, Paul had already show how to calculate the value of L from its XL impedance. For a little more understanding remember capacitors provide negative J and coils +J; so -105.J +14.18J Ohms = -90.82J capacitive resistance part of the tower’s base which is required to balance the output leg.

L= Xl / (2π f(mHz) = 14.18 / (6.28 x 1.43) = 1.58 uH (Series Output leg, think Paul twisted values from 1.56uh to 1.65uh )

• greg white

I need to figure out the impedance for a 200 ft tower (2 tower array) at 1220kc.

• Paul Thurst

What you need is an impedance brige.

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