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“The Electronics Of Radio” NorCal 40B Transceiver Build Lab Notes: Problem 5A, 5B, & 5C

May 4, 2025 by KM1NDY

This is a continuation of my solutions to the problems contained within “The Electronics of Radio” text by David B. Rutledge. These posts are probably only useful if you also are following along with the book and building the transceiver.

[Update: The links to all problem solutions as I go through them will be posted here.]

First off, you can skip all of the stuff in purple. I went through this half page of the book for a few days trying to work things out, and a lot of this will seem like gibberish (because in some ways it is.)

Channel 2 of the signal generator is set to output a 5V 1KHz sine wave.

Channels 1 and 2 of the oscilloscope are each set at 200mV/div. The time scale is set for both at 100μSec/div.

Channel 1 of the oscilloscope is set up with a 10x probe connected to the positive output of the signal generator, and the ground lead of the probe on the ground of the signal generator. The oscilloscope measures a Vpp of 4.7V at the 1MΩ input setting.

Channel 2 of the oscilloscope is set up with a 6′ length of 50Ω RG8X coax attached to the center portion of a T-connector. One limb of the T-connector is terminated with a 50Ω load (toward the left in the picture below) and the other limb is connected to a 1x oscilloscope probe. The oscilloscope probe is attached to the positive output of the signal generator and the ground lead is connected to the ground of the oscilloscope.

Channel 2 of the oscilloscope measures a Vpp of 7.00V at the 1MΩ input setting. Of note, something is not working quite right with these onboard oscilloscope measurements. I think this voltage is actually 0.7Vpp.

Testing with a 5V 1kHz square wave shows a similar pattern. Note that both channels are being shown in these pictures which is why the signal is doubled.

A 1 mH inductor is added between the positive and negative terminals of the signal generator.

Only channel 2 of the oscilloscope is considered for the remainder of this excercise. This is the waveform shown in channel 2.

The following shows the t2 measurement (134 μs).

The peak-to-peak voltage is 1.3V.

Below the calculated peak-to-peak current through the inductor is 26mA.

Oops! 26mA is not right! I measured the RMS current (AC) is 36.88 mA. This is 104.3 mA peak-to-peak current.

Which exactly matches the 100mA current seen in LTspice (just read the top line of the graphed current below which is taken from the inductor, L1).

Which made me look at my calculations again and conclude that the 1.3Vpp measurement is not correct, and instead I stuck with the original 5Vpp signal generator output, for a current calculation of 100mA.

So, I am going to conclude this problem. I have actually made a bunch more measurements.

I am left confused about why I am measuring so many different voltages and currents. There are so many considerations it is quite hard to keep track of. But I did learn how to do AC+DC RMS voltage and current (combined readings) on my Fluke 45 multimeter. And how to use its dual display. I also ordered a proper BNC Tee connector and 50Ω load so I can set coax cable measurements up correctly on the oscilloscope. Although, to be honest, my scope has the option built in for a 50Ω input. Regardless, I am heading on to the next part of Problem 5 now though.

…Except I decided to build the circuit one more time… Instead of worrying about trying to add in a funky 50Ω termination that seemed unnecessary anyway, I went ahead and added a 47Ω resistor.

The oscilloscope showed the expected rounded square wave at a little under half the peak-to-peak voltage of the function generator, using the 10x setting on the probe.

What the author calls t2, the point where the signal rises to the zero voltage line (the hatched line in the center of the graticule), is approximately 10μs. Thus, t2 is approximately 50% of the voltage discharge/recharge, and is 0.693τ, where τ is the time constant of an RL circuit. The formula is τ = L/R = 1mH / 50Ω = 20μs. And 0.693τ = 0.693 x 20μs = 13.86μs. Which is a lot closer to the measured 10μs I finally got. And for what its worth, I was not able to get a cleaner view of this particular waveform on my oscilloscope.

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