SMALL- AND LARGE-SIGNAL MODELLING OF A MODULAR BOOST-DERIVED DC-DC CONVERTER FOR HIGH-OUTPUT VOLTAGE APPLICATIONS

M.R.D. Al-Mothafar

References

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  11. [11] V.G. Bello, Using the SPICE2 CAD package to simulate anddesign the current-mode converter, Proc. 11th Int. PowerElectronics Conf., Dallas, Texas, 1984, 1–11.59AppendixSubstituting the duty ratio law of (13) into (3) yields:[ ˙x] =a1 a2 0 0 0 0a3 a4 0 a5 0 a50 0 a1 a2 0 00 a5 a3 a4 0 a50 0 0 0 a1 a20 a5 0 a5 a3 a4iL1vo1iL2vo2iL3vo3+b1 0 0 b2 0b3 0 0 b4 b50 b1 0 b2 00 b3 0 b4 b50 0 b1 b2 00 0 b3 b4 b5ˆvc1ˆvc2ˆvc3ˆvgˆio(17)where:a1 =−HeTD Mc, a2 =−DL+KrTD RiMc,a3 =DC+3LHeTD 2McRC, a4 =−1RC−3LKrTD 2RiMcRC,a5 =−1CR,b1 =1TD RiMc, b2 =1L+(Kf − Kr)TD RiMc,b3 =−3LTD 2RiMcRC, b4 =3L(Kr − Kf )TD 2RiMcRC, b5 =1CThe set of equations given by (17) represents theconverter of Fig. 1 when all current loops are closed.Transfer-function relations between small changes in anyof the system states and small changes in the control input,the source voltage, or the load current can be obtained asusual by introducing Laplace transforms and convertinginto transfer function form.The above procedure is repeated for n modules, andthe following general set of small-signal transfer functionsare derived with n as a variable:vovc=b3s − a1b3 + a3b1s2 − s(a1 + a4 − a5) + a1(a4 − a5) − a2a3×s2− s(a1 + a4 + (n − 2)a5) + a1(a4 + (n − 2)a5) − a2a3s2 − s(a1 + a4 + (n − 1)a5) + a1(a4 + (n − 1)a5) − a2a3(18)(19)vovg=b4s + a3b2 − a1b4s2 − s(a1 + a4 + (n − 1)a5)+ a1(a4 + (n − 1)a5) − a2a3voio=b5(s − a1)s2 − s(a1 + a4 + (n − 1)a5)+ a1(a4 + (n − 1)a5) − a2a3(20)Current-loop gain of (say, module 1) is obtained bybreaking the loop at point (A) in Fig. 5, leaving the currentloops of modules 2 and 3 closed. Equation (17) becomes:[ ˙x] =ddtiL1vo1iL2vo2iL3vo3= AtiiL1vo1iL2vo2iL3vo3+ Btiˆd1ˆvc2ˆvc3ˆvgˆio(18)where:Ati =0 −DL0 0 0 0DCa5 0 a5 0 a50 0 a1 a2 0 00 a5 a3 a4 0 a50 0 0 0 a1 a20 a5 0 a5 a3 a4(19)Bti =VgD L 0 0 1L 0−3VgD 2CR0 0 0 b50 b1 0 b2 00 b3 0 b4 b50 0 b1 b2 00 0 b3 b4 b5(20)Transfer function ˆiL1/ ˆd1 is then derived, and the current-loop gain (Ti1) is found using the relationship:(Ti1 ) =ˆiL1ˆd1RiFmHe (21)where Fm and He are given by (6) and (10), respectively.

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