Step 1: You briefly short the inductor. Inductors cannot instantaneously change current, so they will linearly ramp up the current over time. This builds up a magnetic field in the inductor.
Step 2: Stop shorting the inductor. Inductors cannot instantaneously change current, so the now-built-up magnetic field continues pushing current into the switching node. The magnetic field and the inductor current linearly ramp down over time.
The forced current will push charge onto the parasitic capacitance of the switch (from switch to ground), the inductor itself (from inductor output to inductor input), and the reverse diode capacitance (anode to cathode).
Since capacitor voltage is charge over capacitance, once enough charge is forced onto the capacitance at the switch node, eventually the voltage from the switch node to the output capacitance is high enough to turn the diode on in forward conduction. The rest of the inductor current is forced into the output capacitance until the remaining magnetic field in the inductor is depleted.
Step 3) Repeat very fast to reduce inductor size and ripple current required (100s of kHz or MHz speed). Vary the duration for which the inductor is shorted in step 1 according to how much charge you need to put on the output capacitor. You could figure this out open-loop by noting that output current at the high voltage side is in charge per second, output voltage is equal to charge over output capacitance, calculating the time taken for the ramp to grow and decay, etc. Or you could design a closed loop control scheme that looks at the output voltage and converts it to shorted duration for you (this is what most integrated circuit boost converters do).
In summary, you dump current into a inductor to build up a magnetic field, then you use the inductor's magnetic field to yeet current up over a large voltage difference.
I know how boost converters work but reading your great explanation made me realize how similar they are to an impact wrench. Ever wonder how an impact wrench creates such massive torque?
Spin up a fly wheel then let it hit the dogs greatly amplifying the torque through stored kinetic energy similar to an inductor being dumped.
I always like how electricity can be compared to mechanical and hydraulic systems, it's not always perfect but there is obviously a lot of overlap between voltage, current, pressure, flow, torque and rpm. Power is the common thread.
Shorting the inductor temporarily converts some of the energy in the stored electric field of the input capacitor to "stored" magnetic field in the inductor (not quite, since it's only present when current is flowing, but close enough). Roughly the same amount of energy is eventually converted back into stored electric field in the output capacitor, minus the losses from parasitic capacitances/resistances and radiated emissions.
To keep things simple, imagine a lossless boost converter. In terms of power-in and power-out, a lossless converter has the same input and output power. If the converter output is doing real work (resistive load), the output power is equal to the output voltage times the load current. Therefore, at a lower input voltage, the converter sees a much higher input current than the load current - it has to, because power-in equals power-out. If it intuitively feels like you're pulling more energy out of the input capacitor than you're pushing into the output capacitor because of the temporary shorting of the inductor, remember - it's not lost to real work, just cleverly exploited by the transformation to magnetic field to losslessly overcome the difference in potential energy between the input charge at low voltage and the output charge at high voltage.
Energy in cap: E = 1/2 * C * V^2
C = Q / V
E = Q * V / 2
Energy in inductor: E = 1/2 * L * I^2
I = dQ/dt
I don't know how to write out the integral notation on HN, so you can fill in the blanks (sorry) - integrate the inductor current ramp up and ramp down portions, set equal to input charge pulled and output charge pushed, observe that energy is conserved with less charge at higher voltage on the output.
Step 2: Stop shorting the inductor. Inductors cannot instantaneously change current, so the now-built-up magnetic field continues pushing current into the switching node. The magnetic field and the inductor current linearly ramp down over time.
The forced current will push charge onto the parasitic capacitance of the switch (from switch to ground), the inductor itself (from inductor output to inductor input), and the reverse diode capacitance (anode to cathode).
Since capacitor voltage is charge over capacitance, once enough charge is forced onto the capacitance at the switch node, eventually the voltage from the switch node to the output capacitance is high enough to turn the diode on in forward conduction. The rest of the inductor current is forced into the output capacitance until the remaining magnetic field in the inductor is depleted.
Step 3) Repeat very fast to reduce inductor size and ripple current required (100s of kHz or MHz speed). Vary the duration for which the inductor is shorted in step 1 according to how much charge you need to put on the output capacitor. You could figure this out open-loop by noting that output current at the high voltage side is in charge per second, output voltage is equal to charge over output capacitance, calculating the time taken for the ramp to grow and decay, etc. Or you could design a closed loop control scheme that looks at the output voltage and converts it to shorted duration for you (this is what most integrated circuit boost converters do).
In summary, you dump current into a inductor to build up a magnetic field, then you use the inductor's magnetic field to yeet current up over a large voltage difference.