“The Electronics Of Radio” NorCal 40B Transceiver Build Lab Notes: Problem 16A, 16B, 16C, 16D, 16E
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.]
The circuit is constructed for part A.
The leads are placed for the function generator and oscilloscope.
A.
C2 is adjusted for maximum output through the subcircuit, which is 330 mV.
Pprimary = Vprimary2 / R = 0.52 / 50 = 5 mW
Psecondary = Vsecondary2 / R = 0.332 / 1500 = 72.6 µW
Power loss ratio = 10 * log10(Psecondary / Pprimary ) = 10 * log10(72.6 µW / 5 mW) = -18.38 dB
Next is to find the bandwidth. The center frequency is 7 MHz, and the output voltage at this point is 330 mV. The pictures above and below show the upper and lower 3dB cut-off frequencies with a voltages equivalent to 0.707 * fc , i.e. 233 mV . The bandwidth therefore is 6.577 MHz to 7.517 MHz.
I measured the capacitance (17.5 pF) of the oscilloscope probe to help with the next part.
B.
The following link shows how to model a transformer the correct way with LTSpice. Very helpful.
https://www.analog.com/en/resources/media-center/videos/5579254291001.html
Zpri / Zsec = (Npri / Nsec)2 = Lpri / Lsec = 1500 / 50
In the circuit below, C2 at a capacitance of 15pF displays the largest S21 peak around 7MHz. Decreasing the capacitance of C2 increases the frequency of the maximum peak, while decreasing the capacitance of C2 decreases the frequency.
I reran the circuit simulation with a narrower frequency range and changed the voltage scale to decibels. I also changed the capacitance of C2 to 15.5pF as it produced a peak closer to the 7.0MHz center frequency. The FLower at 3dB is 6.7MHz.
The FUPPER at 3dB is 7.3MHz. The 3dB bandwidth is 6.7MHz to 7.3MHz.
C.
A note about the “20MHz BW LIMIT” toggle switch in the vertical control section of the oscilloscope. This is essentially a low pass filter, blocking signals with frequencies greater than approximately 20MHz. I wish I knew about this a LONG time ago. No more radio stations on my test leads!
This is the a waveform without the 20MHz BW LIMIT deployed:
This is the same waveform with the 20MHz BW LIMIT deployed:
Note: One thing about the oscilloscope is that I am not always certain when you add a termination or not. In this case I am adding a 50 ohm termination. In doing so, I do not know if I should correct my results or not. I am going to leave the results as is, accepting these answers may be wrong.
The jumper is connected as described (see R2).
Loss (dB) = 20 x log10 (Vout / Vin) = 20 x log10 (150mV / 500mV) = -10.5dB
Now the book says if the loss is greater than 7dB, then something is not working correctly. If I were to double the voltage output given the voltage divider effect of a 50 ohm termination, I would get
Loss (dB) = 20 x log10 (Vout / Vin) = 20 x log10 (300mV / 500mV) = -4.44dB
Well, I like that answer better…
Ok, let’s try again…
Function Generator: 5V 7MHz signal
Directly connected to oscilloscope with 10x probe into 1MΩ scope termination.
Oscilloscope measurement = 4.86V
Now lets put this through the circuit…
Function Generator: 5V 7MHz signal
Function generator connected to circuit and outputted to oscilloscope with 10x probe into 1MΩ scope termination.
Oscilloscope measurement = 6.12V
So, it appears my circuit is showing a larger voltage than the input voltage when I set it up like this… hmmm… There is still something here I am not getting, but I am going to move on.
D.
This is the waveform of the 2.8MHz VFO image after passing through the RF filter. It is supposed to be irregular. It is also with the 1MΩ termination starting from 20Vpp on the Function Generator.
With the 50Ω termination at the oscilloscope and 10x oscilloscope probe, the waveform is much smaller and about one quarter of the amplitude at 25mV.
Ri = 20 log (Vrf / Vvi) dB = 20 log (20 /0.025) = 58.1dB
E.
The following is the circuit design with the additional LC series resonant component added. Of note, this is for the NorCal 40A.
The simulation shows an VFO image (i.e., fvi = 2.8MHz) voltage of 382µV.
Therefore the image rejection ratio
Ri = 20 log (Vrf/Vvi) dB = 20 log (0.500/0.000382) dB = 62.3dB
But!!! The NorCal 40B has a slightly different circuit with a resistor to ground and a potentiometer between the LC circuit and RF filter. It is shown below. The potentiometer (nominally labeled as R4 & R5) is arbitrarily set at 500Ω each. Of note, the potentiometer acts as a voltage divider and functions as the RF gain knob on the NorCal 40B.
The simulation shows an VFO image (i.e., fvi = 2.8MHz) voltage of 236µV.
Ri = 20 log (Vrf/Vvi) dB = 20 log (0.500/0.000236) dB = 66.5dB
Or if I change the voltage divider to 1 ohm and 999 ohms (to ground):
Ri = 20 log (Vrf/Vvi) dB = 20 log (0.500/0.000386) dB = 62.2dB
Or if I change the voltage divider to 999 ohm and 1 ohms (to ground):
Ri = 20 log (Vrf/Vvi) dB = 20 log (0.500/0.000094) dB = 74.5dB