TABLE OF CONTENTS We have seen that the AC component (I AC = ?I /2) is derivable from the voltseconds law. From the basic inductor equation V =L dI/dt, we get
So the current swing I PP ? ?I can be intuitively visualized as "voltseconds per unit inductance". If the applied voltseconds doubles, so does the current swing (and AC component). And if the inductance doubles, the swing (and AC component) is halved.
Let us now consider the DC level again. Note that any capacitor has zero average (DC) current through it in steady-state, so all capacitors can be considered to be missing altogether when calculating DC current distributions. Therefore, for a buck, since energy flows into the output during both the on-time and off-time, and via the inductor, the average inductor current must always be equal to the load current. So
On the other hand, in both the boost and the buck-boost, energy flows into the output only during the off-time, and via the diode. Therefore, in this case, the averag e diode current must be equal to the load current. Note that the diode current has an average value equal to I L when it is conducting (see the dashed line passing through the center of the down-ramp in the upper half of Figure 3.3). If we calculate the average of this diode current over the entire switching cycle, we need to...
