I don't fully follow that theory. I am making 3 versions of this circuit: one 7.5v to 12v at about 20mA, one 7.5v to 40v at 6mA, and also a buck from 7.5v to 5v at about 150mA. I don't understand the relationship between current and inductance - does higher current require higher inductance or lower? I can't find that video, and I haven't any equipment to plunder. I could easily wind one if I knew the dimensions and wire guage and number of turns. Or if I buy one do I need to specify inductance, saturation current, or what?
Some suggestions:
The maximum voltage on the switch node (collector) of the MC34063 is 40, so I don't think you will achieve your 7.5 to 40V directly - but as it only requires 6mA you could use a charge pump /voltage doubler on the output for this.
The 7.5 to 12V converter only needs 20mA, IMO whilst the MC34063 is cheap and available, it isn't well suited to very low output currents, and is somewhat over kill. There are plenty of cheap chinese boost converter modules available for a couple of US dollars that will be more efficient, and have lower noise output than any MC34063 circuit.
Again, the 7.5V to 5V would be about 66% efficient using a linear regulator. Using an MC34063 for a relatively small step down wouldn't fair much better.
A handy design tool that makes a for a good
starting point is here:
http://www.nomad.ee/micros/mc34063a/Note the 'Ipeak' is the peak current in the inductor - any inductor you use must have a saturation current greater than this. Its DC resistance will impact on efficiency, but for the most part isn't particularly critical, just make sure if the Ipeak is say 400mA, you get one with a Isat of >600mA, preferably with a larger margin for error. Designing switching converters isn't always straight forward, not only do parts have to meet certain specs (like using inductors designed for power electronics, rather than filtering) but also, the layout can be critical, keeping high current paths short and low impedance - don't run long thin wires for the switching path. Copying a reference schematic is a good starting point, but also look at the PCB layout and component specs.
If your 7.5V supply isn't coming from a battery - where you want to maximise battery life - then efficiency isn't as critical, especially at such low currents, I would suggest a linear regulator for the 5V, and a single boost converter to convert 7.5V to 15V, then use a 12V linear regulator to get a *clean* 12V from this 15V line. Also, adding a voltage tripler (5 cheap diodes, and 5 cheap caps) to the boost converter would get you 3*15 (minus diode drops) = ~42V @ <10mA. Similar to this setup:
http://www.clivemaxfield.com/area51/do-not-delete/max-4-john-01.jpgThat would mean you only need one boost converter and one linear regulator (cheap) to produce all three rails. Less parts, so it'll be cheaper, smaller, and less to go wrong, but it would require a bit of experimenting with the charge-pump bit. If you decide to give it a try I'll try and run some tests in LTspice to give you an idea of component values.