(3) When both the switch pairstDLJ本T本+T.are off (ton<t<ton+△): D, and D,N211Vaconduct. i, through D, and D2NDdecrease linearly.木T(a)Vo, = 0N2Va/N10Vi= -V.(10 -32)VlLTiDl=ip2 =$2T/2Lo1(b)
(3) When both the switch pairs are off (t o n<t<to n+Δ): D 1 and D 2 1 conduct. iL through D and D 2 decrease linearly. 2 D 1 D 2 = iL i = i (10 −32) oi vL = − Vo v = 0 voi 0 t N 2 V d/N 1 ton Ts /2 iL Vo R Io (b) t 0 D 1 N1 iL Io (a) T 1 + N2 u 2 - N2 D2 L +vo i - +Vd - +Vo - +v1 - T 2 T 3 T 4
1.3Comparison of the full-bridge converter(FB) with the half-bridge converter (HB) ☆For identical input and output voltages and powerratings requires:N2= 2( 2)(10-34)(N,)HB(N,)FBThe switch currents Iare:(10 -35)(I)HB = 2(IT)FBIn large power ratings, it may be advantageous to usea full-bridge converter over a half-bridge converter
1.3 Comparison of the full-bridge converter (FB) with the half-bridge converter (HB) ☆ For identical input and output voltages and power ratings requires: (10 −34) 1 HB 1 FB = 2 N N N2 N2 (10 −35) The switch currents IT are: (IT )HB = 2(IT )FB In large power ratings, it may be advantageous to use a full-bridge converter over a half-bridge converter