“The Electronics Of Radio” NorCal 40B Transceiver Build Lab Notes: Problem 10
This continues a series of blog posts on David Rutledge’s text, “The Electronics of Radio”, that I am studying while building the NorCal 40B transceiver. This series of posts will not be a review of the book, nor is it a assembly manual. Rutledge presents a series of problems at the each chapter that aid in understanding electronics and building the 40M QRP CW transceiver. I am going to try to go through all of these problems and document them here. All of these are titled similarly, so search for them that way. For what its worth, most people will want to skip these posts, they are really for my own self-education on electronics and may not make a lot of sense unless you have Rutledge’s book.
[The links to all problem solutions as I go through them will be posted here.]
Problem 10 addresses Transmission Line Theory. This particular topic is extremely complex and not something I am particularly interested in taking a deep dive into right now. I will be skipping Problems 11 and 12 (which also deal with transmission lines), and thankfully getting back into the transceiver build with Problem 13.
A.
I did not have the equipment necessary to establish the setup required in part A, so I modified it.
I used two 15-foot lengths of RG58c/u coax terminated on all four ends with a male BNC, and attached to one another with a barrel connector, for a total length of 30 feet (9.144 meters).
My Koolertron 60MHz function generator was set to deliver a 1MHz pulse with a width of 50ns and an amplitude of 5Vpp, representing a 5% duty cycle. Notice Channel 1, the bottom settings.
Yes, the waveforms look horrible. The bottom waveform is the input pulse and has twice the amplitude of the top waveform (which is the output of the coax, and also terminated with a 50 ohm load unlike the input pulse).
The delay between the arrival of the input signal at the oscilloscope and the output from the 9.144 meter coaxial cable is 78.15ns.
The propagation velocity = (length of wire) / (delay) = 117 x 106 m/s.
The velocity as a fraction of speed of light = (117 x 106 m/s) / (3.00 x 108 m/s) = 39%.
Of note, the published nominal velocity of propagation of RG58c/u is 66%.
B.
I plugged my 40M off-center fed dipole antenna into the t-connector of the oscilloscope.
The primary pulse and its various reflected waveforms are shown below with a periodicity of 1000ns, i.e., 1 MHz.
The delay from the primary pulse to the reflected pulse is measured at 131 ns.
Just to check, the pulse signal without the antenna attached is shown below. A tee-connector from the signal generator is being used to split the pulse between the antenna and the oscilloscope.
The length of the antenna feedline = delay x velocity = 131ns * ( 117 x 106 m/s) = 15.3 meters, i.e., 50.2 feet. Holy cow! My feedline IS 50 feet long!!!! However, before I get too excited, the velocity factor of RG8X (my shack’s actually coax) is actually nominally about 79%, so…
Let’s try this again…
antenna feedline length = [(reflection delay) / 2] * velocity = (131ns/2) * [0.8 * (3 x 108 m/s)] = 15.72 meters, or 51.6 feet. Wooo Hooo!!!! I do actually think THIS calculation is correct!!!
The reflection delay is divided by two, because essentially the wave needs to travel to the end of the wire (i.e., one length) and then back to the oscilloscope (i.e., two lengths). This means that the delay that we are measuring is for 100 feet of coax cable (since I already know how long that coax is). When I find the velocity by taking 80% of the speed of light, and multiply that by half of the time of the delay, I get an antenna feedline length of right around 50 feet!!!! Yay!
C.
First, this problem called for a 1:1 transformer with a 1Ω parallel resistor across the output. This is to avoid shorting the input signal to the ground of the oscilloscope. I covered this idea in a prior post so its interesting that it is coming up here. I built up a 1:1 transformer on a mix 43 toroid and added a 1 ohm resistor across the windings on one side.
I added a 1MHz 5Vpp with a 5% duty cycle (i.e., 50 nanoseconds) pulse into the black side of the transformer.
Below you can see the pulsed input wave (single peaked waveform) juxtaposed with the multi-peaked output signal. Note that the input terminals of the oscilloscope are fitted with “T” connectors, with the probe in one end and a 50Ω termination in the other.
Below you can see the 540 mV wave, measured with a 1x probe.
Now the output signal (with the oscilloscope probe across the red wires, shows a messy waveform with a peak-to-peak voltage of 15 mV. This voltage apparently corresponds to the amperage of the circuit. I suppose that the oscilloscope with a 5Vpp output into a 50 ohm load would have a 100mA current. Or a 540Vpp output into a 50 ohm load would have a 10.8mA current. This is actually not too far off from my measurements! I got to admit, I find it hard to keep track of the myriad of different voltage dividers that play a role in the measurements actually seen on the oscilloscope.
Now the book calls for constructing a particular circuit [Figure 4.16 in the book], which is still testing the 30 feet of RG58c/u (at least in my case) coaxial cable. The same
Initially, the coax is terminated at its far end with a 50Ω load. The same 1MHz 5Vpp 50ns signal is applied to the circuit.
Measured below is the voltage (750 mV).
And allegedly, the voltage across the isolation transformer and 1Ω resistor (which is 50mV) corresponds to the current, 50mA. I guess to calculate Z0 (the characteristic impedance of the cable), then I would just
I guess to calculate Z0 (the characteristic impedance of the cable), then I would just use Ohm’s law.
R = V / I = .750V / 0.05A = 15Ω. Well…I know that the coax actually has a characteristic impedance of 50Ω. I have to be honest, I really do not know how to sort this all out. There are too many variables, and I am at baseline not even quite sure I am constructing the circuit correctly. And I really do not think my makeshift transformer is working exactly as I would like it to. I am going to just be glad I could calculate a resistance that had the correct order of magnitude and leave it at that.
D.
Part D asks us to remove the 50Ω load from the end of the coax and leave an open circuit.
Interestingly, the voltage across the coax becomes very large
Measurements show that the peak-to-peak waveform across the coaxial cable is now 2.5V. The waveform now appears to be modulated with a higher frequency sinusoid; this is probably due to wave reflections from the now open end.
The trace that indicates current now shows a slight decrease to 40mV, or 40mA.
E. This answer is clearly wrong, but I am posting it anyway. The inductance and capacitance is off by orders of magnitude compared to the published specs.
That’s all for now!
KM1NDY