I am teaching myself about boost converters, and I understand the operating principles- to a point. Specifically, I understand what happens to the conductor during the switch-closed portion of the cycle. My understanding:
As the switch-closed circuit has basically no load, the voltage source will continue to drive the current through the inductor higher and higher, at a rate that depends on the inductance. This is what the inductance equation dictates, V=L(di/dt). Thus the voltage across the inductor will remain at the voltage provided by the source during the switch-closed portion of the cycle. The inductor accumulates energy.
Then, the switch opens, and the load "downstream" of the inductor suddenly becomes much higher, because now the circuit includes whatever resistance is offered by the high-voltage component the converter is hooked up to.
So now the inductor sees a drop in current because of this new load, and responds with a voltage in the direction of the current to compensate. This voltage is proportional to the rate at which the current drops when the switch opens.
Now to my question: what is that rate? How do I find it?
In order for the system to be in steady state, the current must, in the remaining portion of the cycle, drop to what it was at the beginning of the last cycle, before the switch closes again. If the current doesn't drop fast enough, then the inductor won't dissipate all the energy it gained during the switch-closed portion, and so it will gain more and more energy every cycle. That doesn't seem ideal.
But I feel like I'm missing some key piece of information about the switch-open portion that would let me solve for what the output voltage (or at least di/dt) will be. I know how to find what it should be- it's whatever it takes to go from Imax to Imin in the allowed time. But I don't know how to prove that it will be, if indeed it will be. And if it's not guaranteed to be, what variables (load resistance, duty cycle, etc.) would I have to manipulate to get it there?