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 and design the current-mode converter, Proc. 11th Int. Power Electronics Conf., Dallas, Texas, 1984, 1–11. 59 Appendix Substituting the duty ratio law of (13) into (3) yields: [ ˙x] =               a1 a2 0 0 0 0 a3 a4 0 a5 0 a5 0 0 a1 a2 0 0 0 a5 a3 a4 0 a5 0 0 0 0 a1 a2 0 a5 0 a5 a3 a4                             iL1 vo1 iL2 vo2 iL3 vo3               +               b1 0 0 b2 0 b3 0 0 b4 b5 0 b1 0 b2 0 0 b3 0 b4 b5 0 0 b1 b2 0 0 0 b3 b4 b5                          ˆvc1 ˆvc2 ˆvc3 ˆvg ˆio            (17) where: a1 = −He TD M c , a2 = −DL + Kr TD R iMc , a3 = DC + 3LHe TD 2McRC , a4 = −1 RC − 3LKr TD 2RiMcRC , a5 = −1 CR , b1 = 1 TD R iMc , b2 = 1 L + (Kf − Kr) TD R iMc , b3 = −3L TD 2RiMcRC , b4 = 3L(Kr − Kf ) TD 2RiMcRC , b5 = 1 C The set of equations given by (17) represents the converter of Fig. 1 when all current loops are closed. Transfer-function relations between small changes in any of the system states and small changes in the control input, the source voltage, or the load current can be obtained as usual by introducing Laplace transforms and converting into transfer function form. The above procedure is repeated for n modules, and the following general set of small-signal transfer functions are derived with n as a variable: vo vc = b3s − a1b3 + a3b1 s2 − s(a1 + a4 − a5) + a1(a4 − a5) − a2a3 × s2 − s(a1 + a4 + (n − 2)a5) + a1(a4 + (n − 2)a5) − a2a3 s2 − s(a1 + a4 + (n − 1)a5) + a1(a4 + (n − 1)a5) − a2a3 (18) (19) vo vg = b4s + a3b2 − a1b4 s2 − s(a1 + a4 + (n − 1)a5)+ a1(a4 + (n − 1)a5) − a2a3 vo io = 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 by breaking the loop at point (A) in Fig. 5, leaving the current loops of modules 2 and 3 closed. Equation (17) becomes: [ ˙x] = d dt               iL1 vo1 iL2 vo2 iL3 vo3               = Ati               iL1 vo1 iL2 vo2 iL3 vo3               + Bti            ˆd1 ˆvc2 ˆvc3 ˆvg ˆio            (18) where: Ati =               0 −D L 0 0 0 0 D C a5 0 a5 0 a5 0 0 a1 a2 0 0 0 a5 a3 a4 0 a5 0 0 0 0 a1 a2 0 a5 0 a5 a3 a4               (19) Bti =               Vg D L 0 0 1 L 0 −3Vg D 2 CR 0 0 0 b5 0 b1 0 b2 0 0 b3 0 b4 b5 0 0 b1 b2 0 0 0 b3 b4 b5               (20) Transfer function ˆiL1 / ˆd1 is then derived, and the currentloop gain (Ti1 ) is found using the relationship: (Ti1 ) = ˆ iL1 ˆd1 R iFmHe (21) where Fm and He are given by (6) and (10), respectively.

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