Hello Dario.
Sorry for the delay... but my spare time is very limited...
Unfortunately I've not a current probe for the scope so I've modified a little bit the schematics inserting a 1ohm resistor series to the inductor. I've then used two scope channels to measure the voltage across the resistor terminals. I've also placed a multimeter measuring current flowing either on the input side, either on the output side.
In the attached picture you can see the results.
Channel1, the yellow line, is the output side voltage (more or less 5V). Channel 2, the cyan line, is the input voltage (an average of 31.4V).
Channel3 is the "hot" resistor terminal (the one attached to the inductor), Channel4, the green line, is the resistor on the capacitor side (i.e. it's attached to the same position of CH1 but the sensitivity on this channel is 1V). The red line is the difference between Ch3 and Ch4.
With a 1ohm resistor I can imagine to read Amps instead of Volts. Effectively we can see a sawtooth with an averaged value supposed to be a bit less than a half of Ampere.
The Frequency measurement on Ch3 reports that the switching is running at 24.3kHz. It's probably something I can modify changing the capacitor value connected to the switching IC... Probably increasing the frequency will rise up the efficiency... but I've to investigate about.
On the multimeter snapshots, below the scope picture, are the DC and AC current measurements at the input side (on the left) and at the output side (on the right).
With this collected data I can calculate:
Pin = Vin x Iin = 31.40.10 = 3.14W
Pout = Vout x Iout = 5.40.43 = 2.32W
having an efficiency of E = Pout/Pin *100 = 73.8%
I think is an interesting value, do you? We have to consider that there are some power losses introduced by the 1ohm resistor...
Thanks!
Marco Signorini.
