Design and analysis of Two inductor boost converter

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.

The MLZ series of inductors is not designed for power conversion (buck, boost, etc.), but instead for power supply filtering. This means that a) the windings are not optimized for low skin losses, and b) the magnetic core is not optimized for low core loss. From a circuit design perspective, this means the boost regulator will be less efficient.

If efficiency is not what you're after -- if you simply care for functionality and compactness -- then I don't see any problem using MLZ (or other "ferrite choke") inductors, as long as your circuit doesn't overheat. In practice, this means sticking to low power levels. Just ensure the current rating covers your needs (with healthy margin), and evaluate the power supply thoroughly.