Overly simple explanation alert!!
In very simplistic terms, a boost converter circuit looks like this: -
To the left is the power source (48 volts) and to the right is your load. In between is a switch (usually a MOSFET) that "shorts" the inductor to 0V for a short period of time then the switch goes open circuit.
The inductor will accumulate energy during the period it is grounded and, when the switch goes open circuit, that energy is released into the capacitor and load via the diode. This process cycles at many, many times per second.
My explanation is a bit of a simplistic approach but, if you look how energy is accumulated and transferred, it should help you understand how the values of L and C are calaculated. Using your numbers, you have a load that needs to have 100 volts across it. Say that load is 1kohm - this means it needs to be "continuously" fed 10 watts to maintain the 100 volts.
But, if the switch permanently remained open circuit (and the diode were perfect), the load inevitably receives a constant 48 volts (a power of 2.3 watts). So, the power needed to be generated by switching-action is not 10 watts but about 8 watts i.e. there is a constant background power of 2 watts feeding the load and propping it up. This is a tad simplistic because maybe up to 50% of the time the inductor is shorted out thus not delivering that "background" power to the load. You could make an argument for assuming the background power is more like 1 watt on a 50:50 switching cycle. Moving on...
Let's say the switch operates 10,000 times per second - that tells us that the energy transferred per switch operation is: -
Energy per cycle is \$\dfrac{8W}{10,000}\$ = 800 \$\mu\$J.
Let's say the inductor is 100 \$\mu\$H and see how the numbers stack-up. The energy contained in an inductor is: -
Energy = \$\dfrac{L I^2}{2}\$ so, to get 800 \$\mu\$J, a current of 4 amps is needed.
How long does the switch need to be closed for to get 4 amps through the inductor? The switch closes and the current ramps up at a rate determined by the power source voltage and the inductor's value: -
V = \$L\dfrac{di}{dt}\$ is the formula to use.
Power voltage is 48 volts and this divided by 100 \$\mu\$H = 0.48 amps per \$\mu\$s. We need 4 amps so this means the switch needs to be "on" for about 8.3 \$\mu\$s. This informs us that the duty cycle is about 8.3% (10 kHz switching frequency).
Remember, this was just me throwing numbers together to help you follow the process and I think that a more appropriate duty cycle would be closer to 50% - this is just a simplistic look at a simple boost converter.
If a 1mH inductor was chosen, a peak current of 1.265 amps is needed to create a stored energy of 800 \$\mu\$J. V/L implies a current rate of 48mA per \$\mu\$s and therefore the on period of the switch needs to be about 26 \$\mu\$s.
How big should the output capacitor be? This is primarily a question of controlling the ripple voltage across the load. Let's say that ripple should be 1V p-p - a simple approximation is to assume the load always takes a constant 0.1 amps i.e. there is 100 volts across a 1kohm resistor. The voltage droops in the 26 \$\mu\$s that the inductor is being charged so knowing that....
I = C \$\dfrac{dV}{dt}\$ we can see how much capacitance is needed.
I = 0.1 amps and dv/dt is 1 volt per 26 \$\mu\$s. Hence C = 2.6 \$\mu\$F
Remember, this is a very simplistic look at how a theoretical boost converter would work.
Best Answer
With 12 volts, 220 uH and 31 kHz you can analyse the energy transferred per cycle. At 50% duty, the time period is 16.13 us and this allows the current in the inductor to ramp up to 880 mA. This current and 220 uH implies a stored energy of 85 uJ. Barring losses, this energy is transferred 31,000 times per second to the load and output capacitor.
This is a power delivery of 2.64 watts at a duty of 50%.
If the load is 2.2 kohm and you want 120 volts across it then it requires 6.55 watts. So clearly you need more like 80% duty to achieve your aims. But this relies on perfect components with minimal losses and that won't happen in reality.
It's likely you are pushed significantly over the 80% duty requirement and this starts to be bothersome for boost converters. I see you have 4 inductors in series/parallel so, maybe just use two in parallel to reduce the inductance to 110 uH and the current will ramp to 1760 mA at 50% duty. This will then store an energy of 170 uJ and deliver a power of 5.28 watts i.e. a significant step closer.
But you have to choose inductors that don't get close to saturation at this higher current.