Page 73 - 2024-Vol20-Issue2

69 | Shukir

Fig. 1. Power system equivalent circuit diagram.

filters in addition to switches losses to improve the efficiency boost converter. Applying Kirchhoff’s current law(KCL) to
and the power factor of the inverter. A twin inverter providing the DC side.
more than one line to feed loads is used to reduce filters and
switches losses. In order to obtain a stable output voltage un- Id.c = (D · IB - Ic) · ?c (3)
der various conditions, the control system is designed based
on a fuzzy logic controller, which is tolerant of inaccurate Id.c = (D · IB - 2p fsbCb(VB · D + D · IB · 2p fsbLb) · ?c)·?c (4)
data, can model nonlinear functions of arbitrary complexity,
and relies on natural language, so the number and variety of Applying (KVL) to the AC side
engineering applications of fuzzy logic have increased signifi-
cantly. Gaussian membership functions are used because of v + L f di - VL = 0 (5)
their smoothness and brevity of notation, which is a popular dt
method for defining fuzzy sets, and has the advantage of be-
ing smooth and non-zero at all points [9]. The synchronous di (6)
converter is used instead of transformer to prevent the con- v + L f dt = VL
verter capacitor from short circuit, as the two switches are
never operated together, and the power losses in the mosfet v = va.c = Vr.m.s (7)
are less than the losses in the diode [10]. Schematic diagram
of the conversion of DC power into AC power, mathematical di
model, principle operation of the twin inverter, determination Vr.m.s + L f dt = VL
of converter parameters, and the construction of the fuzzy
logic controller are presented in this paper.

II. MATHEMATICAL MODEL Vr.m.s = Vvd.c · m · ?inv (8)
2 (9)
Fig. 1 demonstrated the equivalent circuit of the power system.
The mathematical representation of the power system is as
following [11]; Kirchhoff’s voltage law (KVL) applied to the
DC side (the sum of voltage in a closed loop is zero)

Vd.c = (VB · D +VL) · ?c (1) Vr.m.s = 0.707(VB · D + D · IB · 2p fsbLb) · ?c · m · ?inv (10)

where: VB is the battery voltage, D represents the duty cycle, Ir.m.s = vId.c · m · ?inv (11)
and ?c is the boost converter efficiency. 2

Vd.c = (VB · D + D · IB · 2p fsbLb) · ?c (2)

Here, IB is the battery current, fsb represents the switching Ir.m.s = 0.707 (D · IB - 2p fsbCb(VB · D + D · IB · 2p fsbLb) · ?c)
frequency of the boost converter, and Lb is the inductor of the · ?c · m · ?inv