Design and Implementation of On-board Electrical Power ... - OUFTI-1

The power losses in the inductor are equal to I 2 avR L , where R L is the series resistance ofthe inductor. They represent a loss of 1% efficiency ifR L < 0.01 V outI out= 0.01R. (5.10)We now have all the elements necessary to design the inductor.follows:The procedure is as• The inductance is given by Eq. 5.9, with maximum V in and maximum I out .• The peak current in the inductor is given by Eq. 5.7, with maximum V in and maximumI out .• An acceptable inductor series resistance is given by Eq. 5.10.Capacitor designFrom the integration of i C = C dv Cdtis found to beand Fig. 5.3, the voltage ripple on the capacitor ∆v CUsing the value of ∆i L given by Eq. 5.6, we have∆v C = ∆i LT s8C . (5.11)∆v C = V outηV inV in − V out8f 2 s LC , (5.12)C = V outηV inV in − V out8f 2 s L∆v C. (5.13)This formula gives the value of the capacitor C corresponding to a chosen voltage ripple∆v C .”Boost” converter [16]PrincipleThe power stage of a boost converter includes the same components than a Buck converter,but the position of the switch and the inductor are inverted (Fig. 5.4). The switch is inposition 1 during a time interval DT s (phase 1) and in position 2 during an interval D ′ T s(phase 2), with D ′ = 1 − D.In phase 1, the voltage v L across the inductor is equal to the input voltage V in . Theinductor is being charged of magnetic energy. The current in the inductor is increasing andthe slope of the waveform is given by v L (t) = L di L(t)dt. The variation of i L over the phase 1 is52