Page 74 - 2024-Vol20-Issue2

70 | Shukir

the operation of IGBT transistors within 360 degrees.
Since the current is divided between the two bridges the losses
will be reduced to a half as obvious in the basic formulas:
Losses in a single inverter (LSI) is:

LSI = I2(Rac + Rdc)inductor + I2RRC (15)

Losses in the twin inverter (LTI) is:

LTI = 2 I 2 I2 (16)
2 2 RRC
· (Rac + Rdc)inductor +

= 1 I 2 [(Rac + Rdc)inductor + RRC] (17)
2

Losses reduction leads to improved inverter efficiency (?inv)

Fig. 2. The twin inverter. ?inv = Po Po (18)
+ Plosses

and improves the power factor of the inverter (p.f)

(12) p.f = Po (19)
S

VL = 0.707(VB · D + D · IB · 2p fsbLb) · ?c · m · ?inv+ and reduces the amount of heat that power supplies have to
dissipate

Lf d (0.707 (D · IB - 2p fsbCb (VB · D+ (13) ?T ? I2 (20)
dt

D · IB · 2p fsbLb) · ?c) · ?c · m · ?inv) In addition, the increase in the current of nonlinear loads
such as computers, printers, variable speed drives, and power
IL = VL (14) conversion systems leads to an increase in harmonics, so
R nonlinear loads can be divided into groups and feed them
using the twin inverter to protect the system from increasing
In the above equations, Cb designates the capacitor of the the harmonics.
boost converter, Lf denotes the inductor of the filter, m sig-
nifies the modulation index, and ?inv represents the inverter IV. A CONVERSION FLOW FROM THE DC
efficiency. VOLTAGE TO AC VOLTAGE

III. THE PRINCIPLE OPERATION OF THE The power flow in the system is illustrated in Fig. 3, where:
TWIN INVERTER
Converter output voltage = 1 Vi · ?c (21)
The twin inverter consists of two bridges feeding two groups -D
of loads, each load tied to a leg from the first bridge and
a leg from the second bridge as depicted in Fig. 2. In the Converter output current = Ii · (1 - D) (22)
positive half cycle, current passes from the positive terminal z
of the source through g1-3 to load 1 and back to the negative ?c
terminal through g2-4, and from the positive terminal to the
second load through g1-1 and back to the negative terminal TABLE I.
through g2-2. In the negative half cycle, current from the IGTB TRANSISTORS IN THE ON STATE THROUGHOUT
positive terminal passes to the first load through g2-3 and
back to the negative terminal through g1-4, and current feeds 360° DEGREES
the second load from the positive terminal through g2-1 and
returns to the negative terminal through g1-2. Table I shows 0 - 180° 180° - 360°
g1-1, g2-2 g2-1, g1-2
g1-3, g2-4 g2-3, g1-4