You have the right idea for a basic unregulated supply. A transformer, four diodes, and as large a cap as you can manage will serve well enough for a lot of purposes, but isn't appropriate for all.
There are two main problems with such a unregulated supply. First, the voltage is not known well. Even with ideal components, so that the AC coming out of the transformer is a fixed fraction of the AC going in, you still have variations in that AC input. Wall power can vary by around 10%, and that's without considering unusual situations like brownouts. Then you have the impedance of the transformer. As you draw current, the output voltage of the transformer will drop.
Second, there will be ripple, possibly quite significant ripple. That cap is charged twice per line cycle, or every 8.3 ms. In between the line peaks, the cap is supplying the output current. This decreases the voltage on the cap. The only way to decrease this ripple in this type of design is to use a bigger cap or draw less current.
And don't even think about power factor. The power factor a full wave bridge presents to the AC line is "not nice". The transformer will smooth that out a little, but you will still have a crappy power factor regardless of what the load does. Fortunately, power factor is of little concern for something like a bench supply. Your refrigerator probably treats the power line worse than your bench supply ever will. Don't worry about it.
Some things you can't do with this supply is run a anything that has a tight voltage tolerance. For example, many digital devices will want 5.0 V or 3.3 V ± 10%. You're supply won't be able to do that. What you should probably do is aim for 7.5 V lowest possible output under load, with the lowest valid line voltage in, and at the bottom of the ripples. If you can guarantee that, you can use a 7805 regulator to make a nice and clean 5 V suitable for digital circuits.
Note that after you account for all the reasons the supply voltage might drop, that the nominal output voltage may well be several volts higher. If so, keep the dissipation of the regulator in mind. For example, if the nominal supply output is 9 V, then the regulator will drop 4 V. That 4 V times the current is the power that will heat the regulator. For example, if this is powering a digital circuit that draws 200 mA, then the dissipation in the regulator will be 4V x 200mA = 800mW. That's will get a 7805 in free air quite hot, but it will probably still be OK. Fortunately, 7805 regulators contain a thermal shutdown circuit, so they will just shut off the output for a while instead of allowing themselves to get cooked.
if the input voltage is so high, why can the transformer in a
switching power supply be smaller than that of a traditional (not
switching) transformer (which first drops and then rectify the 220 V
AC signal)?
Because the operating frequency is so much higher than 50 or 60Hz. With a higher operating frequency, the inductance of the primary can be proportionately smaller requiring fewer turns and a smaller core size because core saturation is caused by ampere-turns. It's not unusual for a switcher to run at 200kHz i.e. 4000 times higher than 50 Hz.
It's all about core saturation, even for a laminated mains transformer - the current in the primary when the secondary is unloaded is the current that saturates the core. The magnetizing inductance of the primary is proportional to primary turns squared so if you double the turns you quadruple the inductance and quarter the magnetization current. So amps have gone down by 4, turns have doubled but, importantly, ampere-turns have halved.
(Generality alert) This is why a 230V mains transformer needs about 1000 turns on the primary - it needs to maintain a primary inductance of about 10H. This then limits current to about 73mA and ampere-turns will be about 73 At. For a given core size, the mean length of the mag field might be 300mm so, mag field strength, H will be about 244 At/m. At this field strength iron and ferrite and other such transformer core materials are seeing a peak flux density that is on the cusp of beginning to cause core saturation. A bigger core means less saturation. More turns means less saturation. Higher frequency means less saturation. Weigh this against the down sides. More turns = more cost and more copper losses. Bigger cores means bigger product and more cost. higher frequency means smaller core, fewer turns, smaller cost BUT cost of control electronics to run this higher frequency has to be considered.
why greater voltages would require an isolation? That is: why flyback
converters are used instead of buck-boost converters in such cases?
It's a legislative safety thing as mentioned on page 8
Best Answer
Yes that’s what it means. Planes often use 400 Hz generators and the plane’s frame becomes “Earth ground”.
As always, Gnd just means 0V anywhere in the universe and Earth bonded implies 0V to earth ground with some impedance <100 Ohms each to another grid location with an earth ground, unless in dry remote areas with poor earth bonding.
Earth is used by all 3 pin instruments to bypass EMI line filter leakage currents only and provide a good low impedance reference for scope probe ground leads.Sometimes DSO’s will put a plastic cap or an inductor between 0Vdc and Earth gnd pin for various reasons to isolate earth ground common mode RF noise (L), or block AC current but bypass RF to a relatively low impedance AC gnd using a cap only.
Engineers do use DSO’s on planes for R&D work.