buck or boost converter derivation of output voltage

Hello,

I have been deriving the operation of ideal buck and boost converters using Fundamentals of Power Electronics by Erickson and Maksimovic and have a few questions.

For either the ideal buck or boost, when you solve for the inductor voltage (or current) how do you know that the DC component of the output voltage will be the same when the switch is open and when the switch is closed. Do you just assume a constant output DC voltage because that's what you want to attain?

What I mean is that if I call the DC output voltage V1 during the switch open phase and V2 during the switch closed phase, how do I prove that V1 = V2 in one complete switching cycle?

I guess this also leads into the question of how does one know the inductor current will eventually enter a steady state?

Thanks! Any help is much appreciated!

In most switching regulators you come across two components , Inductor and Capacitor the LC circuit form a second order filter to filter out high switching frequency noise .

The switching function of the switch can be explained as follows.

when a switch is closed ,
a voltage differential exists on the inductor and a current rises given by the equation V=-L*dIL/dt
a portion of which flows through the load and a portion flows through the output capacitor given by the equation Ic = C*dVc/dt
this gives rise to an energy inside the inductor which stores energy during conduction.

when the switch is opened ,
the sudden change in current in the inductor changes the polarity of the voltage w.r.t to the input supply and the output capacitor making a diode in the buck converter freewheel or a boost converter to conduct this charges the capacitor to and energy possessed by the inductor.

So it is not just the voltage , current will also play an important role in a switching converter.

Current never is in a steady state in a switching converter but a small ripple current is needed to make use of the Inductor and Capacitor to regulate.

A small ripple Voltage will always be present on the output .

yes. For simpler calculation it is assumed that output DC voltage is constant.