Our club president, Greg K3EA, and I had been discussing the quality and usefulness of the material published in various available sources about transmission lines. Instead of adding to the confusion, we decided to demonstrate some of the important properties of transmission lines with a series of experiments. We packaged these experiments into a seminar for the club (Delaware Valley Radio Association, W2ZQ). This blogpost is a summary of the experiments and the conclusions.
One of our objectives was to keep the seminar accessible to all Amateur Radio operators. So we have focused entirely on practical experiments with one exception. It is necessary to decompose some of the impedances encountered in these experiments into their resistive and reactive components. We just presented the results with a bit of explanation.
It is important to note that these experiments, to keep them simple, are for one type of cable (RG8X), two lengths of cable, one frequency, and three different resistive terminations. Changing any of these parameters will change the results but the general ideas still apply. The bottom line of these experiments is to show that terminating a transmission line in other than its characteristic impedance changes the input impedance of the line. As a result, the power delivered to the input of the line is equal to the power delivered to the load and lower than it would be if the line was terminated in its characteristic impedance.
The material for the experiment are:
- A QRP radio for a signal generator
- A "short" length of coax cable
- A 6.3 meter length of coax cable
- Three load resistors, 50 ohms, 25 ohms, and 112 ohms
- A current probe (described below)
- An oscilloscope
To keep things simple, we stuck to one frequency (14.25 MHz, in the middle of the 20 meter band, wavelength 17.3 meters in the RG8X coax) and one type of cable, RG8X (50 ohms, 79% velocity factor). For all practical purposes, a 6.3 meter length of RG8X coax cable in the middle of the 20 meter band can be considered lossless.
The experiments are:
- QRP radio driving a short length of cable terminated in 50 ohms
- QRP radio driving the 6.3 meter length of cable terminated in 50 ohms
- QRP radio driving the 6.3 meter length of cable terminated in 25 ohms
- QRP radio driving the 6.3 meter length of cable terminated in 112 ohms
In each experiment, the voltage at the radio, the voltage at the load, and the current through the coax are measured. Power delivered to the load is calculated using the voltage at the load and voltage and current at the radio. We also used the phase difference between the voltage and current at the radio end of the cable to estimate the resistive and reactive components of the input impedance.
First, let's review the current probe construction. Below is a picture of the current probe. It is essentially a one turn primary and an 11 turn secondary (it was supposed to be a 10 turn secondary but I got carried away). The primary is a short piece of coax cable with the shield grounded on one side to provide an electrostatic shield between the primary and the secondary (grounding it on both side would short out the transformer). The 100 ohm resistor (blue color with a red sleeve over the lead in the picture) is across the secondary. The voltage across the resistor is the secondary current times 100. The current through the primary is the secondary current times 11.
Experiment 1: QRP radio driving a short piece of coax cable terminated in 50 ohms:
Results:
In this first picture (voltage across the load), we are putting 20 volts peak to peak across a 50 ohm load.
So the power delivered to the load is:
If you noticed, the length of my short coax cable is not very short relative to the wavelength of the signal, so let's look at the delay through my short coax.
As you can see, the delay is about 8 nanoseconds. We can also calculate the delay based on the velocity of the electromagnetic signal in the RG8X coax:
1.8 m / 236,800,000 m/s = 7.6 ns
Since we have a current probe, we can also calculate the power delivered to the load using the current. In the photo below, the yellow trace is the voltage across the load the green trace is the voltage across the 100 ohms current sensing resistor across the secondary of the current measurement transformer.
So, we can calculate the power using the current measurement.
3.76 volts peak to peak / 2 = 1.88 volts peak
1.88 volts peak / 100 ohms = 18.8 mA peak (through the secondary)
18.8 mA x 11 = 206.8 mA peak = 0.2068 A peak (through the primary)
We can calculate the power now:
Which is the same (within the accuracy of the experiment) the same as the power we calculating using the voltage data.
This establishes the baseline and now we can proceed to the next experiment:
Experiment 2: QRP radio driving 6.3 meters of coax cable terminated in 50 ohms:
Results:
In the picture below the yellow trace is the voltage measured at the input of the coax cable and the green trace is the voltage measured at the load. They are not exactly 20 volts because I am using my Ten Tec Omni D as the signal generator and at low power levels adjustment is a bit tricky. Also, the voltage at the load is slightly less than voltage at the source because of some likely mismatch between the cable and the termination (and some of it because of my cheap scope probes).
But essentially, the power delivered at the input of the cable is the power delivered to the load.
We can also look at the time delay between the yellow and the green signal which is approximately 33 ns. We can calculate the time delay in the same way that we did earlier and we obtain 26.6 ns. There are a number of reasons for this difference:
- The difference between the cable specifications and the actual cable wave velocity
- Measurement error (e.g. probe mismatch)
- Stray inductance and capacitance
- Some combination of above.
But in general you note that when a cable is terminated in its characteristic impedance, other than time delay, it acts as if it is not there.
We can also calculate the power delivered to the load using the current data. In the photo below, the yellow trace is the voltage across the load and the green trace is the voltage across the current sensing resistor.
Knowing that the voltage across the current sense resistor is 3.56 volts peak to peak from the scope measurement, we can calculate the power delivered to the coax in the same way we did earlier. The result is 1.02 watts, close to 1 watt as we would expect.
We can also calculate the magnitude of the input impedance of the line by dividing the voltage by current:
- 3.56 volts peak to peak across a 100 ohm resistor gives us 0.0356 A peak to peak in the secondary
- Multiplying it by 11 gives us 0.3916 A peak to peak in the primary
- Dividing 20 volts peak to peak by 0.3916 A peak to peak, we get 51.1 ohms which approximately the characteristic impedance of the line and the load impedance.
If you note, there is about 2 ns of delay between the current and the voltage waveform with the current leading voltage. This is a 10 degree phase shift which suggests a slight (about 9 ohms) capacitive component in the load. The resistive component is 50.3 ohms. Nothing is perfect.
Now that we have demonstrated (in one experiment but if we go to the theory, this is true of all lossless transmission lines) that when a transmission line is terminated in its characteristic impedance, the input impedance is the same as the load impedance and the power delivered from the source is equal to the power consumed by the load.
Experiment 3: QRP radio driving 6.3 meters of coax cable terminated in 25 ohms:
Results:
Below is a picture of the voltage waveform at the source (yellow) and at the load (green).
Note that the voltage at the source is 14.0 volts peak to peak and voltage at the load 12.16 volts. What is going on? Before we explore this, let's just calculate the power delivered to the load. We know that the load is 25 ohms, so we can calculate the power:
Let's just note that the power delivered to the load is now lower.
If we put the 50 ohm load back in place of the 25 ohm load, we will get the earlier image, so we know that there has not been a mistake. Before getting to the explanation, let's look at the voltage and current at the input of the line:
In this picture, like previous ones, the yellow trace is the voltage at the input of the line and the green trace is the voltage across the current sense resistor also at the input of the line. Note that the input voltage has increased to 15.2 volts, that is because the Ten Tec Omni D is not the most stable "signal generator". But as long as we stay consistent with our measurements and analysis, the results would be valid.
Also note that in this picture, the current waveform (the voltage proportional to it) leads the voltage waveform by 7 nano seconds or approximately 36 degrees. We would expect the load to have a reasonable capacitive component to it.
Following the process I explained earlier, we can calculate the magnitude of the input impedance. The magnitude of the input impedance is 29.3 ohms and it has a 23.7 ohms resistive component and a 17.2 ohms capacitive component (note that that the resistive and capacitive components are 90 degrees out of phase with each other so the magnitude is the hypotenuse of a right angle triangle and we can verify this using the Pythagorean theorem).
The resistive component of the load is where the power is dissipated so let's calculate the power delivered to it. Note that we get the 0.5192 A peak to peak by dividing the sense resistor voltage (4.72 volts peak to peak) by 100 ohms to get the secondary current and multiply it by 11 (the turns ratio).
So, the power delivered to the resistive portion of the load is equal to the power delivered to the load at the end of the line. This makes sense because of the principal of conservation of energy. The power delivered to the load has to come from somewhere (the source) and the power delivered by the source has to be dissipated somewhere (the load) and the two have to be equal to each other in a lossless line.
Before getting into the explanation, let's run our final experiment.
Experiment 4: QRP radio driving 6.3 meters of coax cable terminated in 112 ohms:
Results:
Here is the voltage at the source (yellow trace) and the voltage at the load (green trace). Note that these voltages are larger than the ones we observed in previous cases.
As we have done before, we can use the power delivered to the load using the voltage across the load:
In this case, the power is also less than when the line is terminated in 50 ohms. We can also calculate the power delivered at the input of the line from the data below where the yellow trace is the voltage across the line at the source and the green trace is the voltage across the 100 ohm current sense resistor.
We will calculate the magnitude of the impedance as we did before by calculating the current through the primary and dividing the voltage by this current. The magnitude of the impedance is 94.0 ohms. In the image below, I have expanded the horizontal scale so we can better see the lag between the voltage and the voltage across the current sense resistor. In this case, the voltage leads current by 3 ns (scale is 2 ns per major division). So we know that the input impedance of the line has an inductive component to it.
From this we can calculate (details left out) that the 94.0 ohms impedance is composed of a 90.6 ohms resistive and a 24.9 ohms inductive components (the triangle rule applies here also). Now we can calculate the power delivered to the resistive part of the input impedance.
And all our conclusions from the last experiment apply again.
Now the explanation.
When the line is terminated in its characteristic impedance, the current through the load is equal to the current in the line and the product of the voltage and current is equal to the power input to the line. That is what we saw with the 50 ohm load cases.
As the generator is turned on, the wavefront starts traveling down the transmission line. At this point, it sees the 50 ohms impedance of the line and the voltage and current waveforms act accordingly. When the wavefront reaches the end of the line terminated in a resistance lower than its characteristic impedance, the voltage across the load has to be lower than the voltage that is arriving the current to be the current that was injected into the line (the current can't do anything else). So, some of the voltage is reflected and travels back towards the source. The phase of this voltage has to one degree or another oppose the incident voltage to reduce it. When it arrives back at the source, it appears to the source as a generator that opposes (or aids) the source. The combination of these interactions is what we saw in these experiments. When the line is terminated in a resistance higher than its characteristic impedance, the voltage has to be higher to support the current provided by the line. So the phase of the reflected voltage has to be such that it aids the incident voltage. We saw both of these effects in the experiments.
The ratio of the reflected voltage to the incident voltage is called the reflection coefficient. Without going into a lot of the math, the reflection coefficient can be calculated by knowing the line and load impedances. That is what is shown in the following equations. The reflection coefficients for our two experiments are also calculated. Note that in the case of the 25 ohm load, the reflection coefficient is negative, hence it reduces the voltage across the load as we saw. In the case of the 112 ohm load, the reflection coefficient is positive, hence it increases the voltage across the load, also as we saw in the experiment.
We can also calculate the Voltage Standing Wave Ratio (VSWR commonly referred to as SWR) from the reflection coefficient:
Note that we are using the absolute value of the reflection coefficient in the definition of the SWR. So the sign is lost in the process. We can also calculate the absolute value of the reflection coefficient from the SWR, but not its sign.
With a little bit of algebra, it is easy to see the incident and reflected voltage in each case. For example, in the case of the 25 ohm load, the incident voltage is 18.23 volts and the reflected voltage is -6.10 volts (the sign is because of the phase shift that reduces the voltage across the load).
So how can you use this data:
- Match your antenna and transmission line as close as possible (200 ohm antenna and ladder line, ground mounted radials and 50 ohms coax cable etc.).
- Use the ARRL local club, and other expert resources to achieve this (in the case of DVRA, the antenna committee).
- Avoid ideas that you can't verify in some reliable way.
- Your radio in all likelihood is designed to work with a 50 ohm load. Find a matching scheme that matches your antenna to your radio
- Keep SWR at 2 or lower for the most part
