The old world way of doing it. Based on the formulas posted earlier.

RF Amp and antenna input circuit

The input should be matched to the antenna and the amp input. I will use a bandpass filter for 80 meters. First the page from the book.

and the results.

The sweep.

The radio less the LO

Years ago my friend Andy built the 40/80. It is a good example of the dual conversion as presented in the block diagram. What I have in mind this time is a JFET design rather than the MOSFET. I bought a box of K2539's for $5 and I tend to use them first. I also have some J112 and J175 to try. My design tool of choice is LTspiceIV. Other tools may give good results but LTspice is an old friend. The next contender is QUCS "Quite Unerversal Circuit Simulator". I have used QUCS just enough to see the possibilty. It is available is several packages for Linux, Windows and Mac. Now things to consider: Should I use a DBM or active detector? Single conversion or dual conversion? Thrown together with a minimum of parts. etc.... look at the 40/80 here: ***** https://radio.radiotrician.org/2018/11/dbm-and-dual-gate-mixer.html***** It uses a DBM and dual gate mxer. DBM maybe but no dual gates this time.

Bandspread vs range continued from previous post

If you observed the effects of adjusting the Vcap ratio by adjusting the low end capacitance and the high end by adjusting the "trimmer" It should be easy to understand the importance of making the adjustments in order. The 600KC adjustment sets the bandwith while centering the range over the correct dial point. The 1400KC adjustment sets the high scale to the correct point. You must work between the two to get the proper operation of the set. Both effect both ends of the scale so the 1400KC could be adjusted without first setting the 600KC and then you are looping back and forth with to wide a band. You can set the dial to 1400KC and adjust either one adding the low range adjustment while reducing the high range adjustment and hold the output steady. Anywho here are a couple of excert from the service manual.

The alignment procedure on one sheet.

Setting the bandset for a LC tuning circuit.

As simple as I can state it the tuning range is determined by the CIRCUIT high to low capacitance. When you look at a schematic you may se the Vcap range is 0pfd to 360pfd but that is just the Vcap you will also find trimmers and strays in the circuit. The sim's work through a series showing the hi-lo tuning with a Vcap and strays. Once the circuit is set and the range determined I adjust the strays to show the effect. Then I use the same circuit and adjust the inductor. You should be able to see the ratio of hi-lo capacitance sets the tuning range while adjusting the inductor shifts the band being covered. One way of looking at it is the band being covered is definrd by a VEE. By adjusting the capacitance you widen or reduce the VEE's width. Adjusting the inductor will slide the VEE across the band to establish the center point. We can hit a brick wall when aligning a set by making the wrong adjustment first. Consider this: the range is defined by the (VcapHI plus strays) / (VcapLO plus strays). Look at the schematic, do you see capacitors in the tuning circuit? You must add them to the Vcap values to establish the range. Next time you find an impossible to align set consider this: set the dial to the low point and find the frequency being received. set the dial to the high point and find what frequency is being received. Divide the measured high by the measured low and square it to determine the actual tuning range. Divide the dial high by the dial low and square it to determine the desired tuning range. If they are not close it is time to examine the extra components in the circuit. So the goal is to set the width of the VEEs arms and the center point of the range. Hummm? Makes me wonder about that "set the dial to 1400kc and adjust...". Oh well now for some screen shots.

Charging and discharging a coil to power a joule thief.

Notice the time delay when power is switched on. 

The CEMF in the coil delays the current buildup. 

The current continues to build as long as power is applied. 

When the input falls the current decays just as it increased when it was applied.

The voltage (green) reverses polarity.

The current (red falls just as it rose in the same polarity.

What you need to see here is the EMF produces current which produces CEMF opposes the current. There is a time delay as current builds and as current falls.

The inductor opposes change in current. This delay in rise and storage then delays the decay as it releases the energy back into the circuit.

How does the voltage reverse in the green waveform? The current is unidirectional and the polarity is determined by seeing the coil as a source or sink. 

(sinking)

As it is drawing power from the power source the bottom terminal is negative and the top terminal is positive. 

(sourcing)

As it returns energy to the circuit the top terminal is negative and the bottom is positive.

With this in mind look at the joule thief.

If the 2N3904 is switched on the coil is directly across the battery. The top is positive and the bottom is negative.

As the field builds in the coil it couples to the base coil and turns the 2N3904 off.

As the field collapses the current is maintained with the top of the coil negative and the bottom positive.

As you can see the diode "sees" the battery voltage and the coil voltage in series and the voltage is the sum off the two.

This is how can a "dead" cell powers a LED that requires between 2 and 3 volts.

As we know there is no such thing as a free energy (outside nuke). What we have here is energy conversion. The low voltage is stepped up as the current is maintained for a short period of time.

Power is a function of time. We dissipate 1mw of power in 1 millisecond or 1 microsecond. To produce the power the short pulse would be higher voltage or current. 

A tip of the hat to the ladies and thanks for the project idea. I can see a couple of joule thiefs in making.