“The Electronics Of Radio” NorCal 40B Transceiver Build Lab Notes: Problem 13
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.]
This particular problem deals with the harmonic filter just proximal to the antenna output. It is important to note that the schematic in “The Electronics of Radio” is actually for the NorCal 40A — not the 40B! The harmonic filter is different between the two, and this caused me quite a headache initially.
The book instructs you to add in 2 toroidal inductors and 3 capacitors (and the BMC jack).
It then instructs you to a send a 10Vpp 7MHz sine wave across C45 and measure with an oscilloscope out the BMC antenna port.
So first I tested my signal generator first (channel 2).
With the coaxial cable directly from the signal generator into the oscilloscope, a 10Vpp signal was seen.
The signal amplitude was halved when a 50Ω termination at the o-scope input was used (which is what the instructions called for, and yes, I know I have a built-in 50Ω termination in this scope).
The harmonics seen on the spectrum analyzer are pretty brutal, particularly the 2nd (14 MHz) and 3rd (21 MHz) which are about -30dBm and -24dBm down from the 7MHz fundamental respectively.
I send the 10Vpp 7MHz signal through C45 of the harmonic filter and measure it out of the BMC port.
I measure the signal on the oscilloscope through the 50Ω termination; a puny 7.7mV!
Despite the tremendous attenuation of the fundamental, the harmonics are suppressed >30dBm.
The filter appeared to work, but the attenuation simply seemed wrong. I increased the signal input frequency to 14MHz as instructed , and the fundamental was also around -40 dBm, while the harmonics were also about -75dBm. Clearly something was wrong — the filter should have rejected the 14MHz signal, not passed it at the same amplitude as the 7MHz signal it was specifically designed to pass.
A careful comparison of the schematics for the NorCal 40A and NorCal 40B solved the issue. The NorCal 40B has a harmonic filter design composed of five capacitors and two toroidal inductors. I soldered C44 and C24 onto the board to complete the filter and preceded to the first question.
A.
A 10Vpp 7MHz sine wave across the filter outputs a 3.25Vpp signal on the oscilloscope (with a 50Ω termination).
And looks like this on the spectrum analyzer.
A 10Vpp 14MHz sine wave across the filter outputs a 47mVpp signal on the oscilloscope (with a 50Ω termination).
And looks like this on the spectrum analyzer.
For the 7MHz 10Vpp signal:
Loss = 20log(10/V) dB = 20log(10/3.25) = 9.76 dB
For the 14MHz 10Vpp signal:
Loss = 20log(10/V) dB = 20log(10/47m) = 46.56 dB
This demonstrates that the 2nd harmonic (14MHz) has a loss of 46.56 dB through the filter, compared to the fundamental (7MHz) which has a loss of 9.76 dB. The goal is for the harmonics to be greater than 30dB below the fundamental to avoid spurious emissions.
B.
Inductance Formula:
L = AL * N2
AL = Inductance Constant (note: given as 4 nH/turn2)
N = # of turns
Inductance of L7:
L = AL * N2
L= 4n * 152 = 0.9μH
Inductance of L8:
L = AL * N2
L= 4n * 122 = 0.576 μH
C.
I used the nanoVNA to assess s-parameters of the filter.
Note, this is a terrible set up, with a bnc-to-alligator clip transmission line from CH0 (i.e., port 1). The Open-Short-Load calibration is done with the alligator clips. CH2 (i.e., port 2) does use the PCB antenna jack.
[S21] and [S11] are shown below. In essence, all of the harmonics pass through the filter at >40dB below the 7MHz signal (as shown by [S21]).
D.
Input impedance ( S11|Z| ) is shown below.
For the 1st through 4th harmonic, it is easier just to read the results from the chart.
Below is the Multisim simulation of the circuit with their VNA function in place. The VNA analysis assumes a 50Ω and I have added a 50Ω load to simulate an antenna with a matched impedance.
The frequency sweep is from 1MHz to 35MHz to account for the first four harmonics. The S11 curve shows the lowest return loss at around 6.44MHz, whereas the S21 curve shows the highest insertion gain at 6.44 MHz. This is quite idealized, and pretty much looks nothing like what I see with my attempt to measure this with the nanoVNA.
Next I tried an LTSpice simulation. I used this video for instruction on how to conduct S-parameters. I have to say, I also used Google’s “AI Mode” to walk me through some of the nuances of LTSpice, and it was quite informative!
The key to performing S-parameter analysis in LTSpice is to use the .net command which has the voltage out, i.e. I(Rout) and the voltage in (V1) as variables, and in that particular order.
If this is set up correctly, simply right click on the trace area and choose “Add trace”.
Now a whole bunch of parameters, including S-parameters, are selectable! This is pretty cool.
And finally! The S11 and S21 traces of the NorCal 40B’s harmonic filter:
Interestingly, this is somewhat similar to the measured S11 and S21 of the actual device with the NanoVNA. I’ll repost for comparison…
E.
The average power output of the circuit 71.2 (???? mW) with 1V peak input voltage.
By varying some of the capacitor and inductor values, I was able to nearly double the power output as instructed (to 137.89 ???mW). The units are not clear to me.
F.
I used this Chebyshev low pass filter design tool for this next problem.
14MHz is about -30dB on the S21 trace. This circuit would need to be tweaked to really work decently, but for now this is all I am going to do on this problem.
Well that’s it! On to the next problem soon…
KM1NDY