REPORT ON THREE PHASE INVERTER 120 DEG. MODE

1. EE 560: Power Electronics Converter
Three Phase Inverter 120o mode
Name : Akash Vishwakarma
Roll No. : 224102101 20th Spt. 2022
1 Objective:
A three phase inverter is operating in 120o conduction mode and supplying a load of impedance R
= 10 Ω, L = 15 mH in each phase. Assume input voltage (Vdc) = 380 V, switching frequency (fs)
= 50 Hz. Load is star connected.
2 Three Phase Inverter Design:
2.1 Output Voltage (For any Load):
• Expression of Output Line Voltage(Six Stepped wave):
vph(t) =
∞
X
n=1,5,7,11,13−−
3Vin
nπ
Sin(nwt + ϕn) (1)
• Expression of Output Phase Voltage(Quasi Square wave):
vL(t) =
∞
X
n=1,5,7,11,12−−
2Vin
nπ
Sin(
nπ
3
)Sin(nwt +
π
6
) (2)
• nth harmonic Voltage:
vLn(t) =
3Vin
nπ
Sin(nwt + ϕn) (3)
vphn(t) =
2Vin
nπ
Sin(
nπ
3
)Sin(nwt) (4)
• Fundamental Output Voltage component:
vL1(t) =
3Vin
π
Sin(wt +
π
3
) = 362.87Sin(wt +
π
3
)
vph1(t) =
√
3Vin
π
Sin(wt +
π
6
) = 209.5Sin(wt +
π
6
)
• Output RMS Voltage:
VL =
1
√
2
Vin = 268.7V
Vph =
1
√
6
Vin = 155.13V
• Fundamental Output rms Voltage is:
VL1 =
3Vin
√
2π
= 256.6V
Vph1 =
√
3Vin
√
2π
= 148.14V
1

2. 2.2 Load Current (For any load)
• Expression for load current:
iL(t) = iph(t) =
∞
X
n=1,5,7,11,12−−
2Vin
nπ|Zn|
Sin(
nπ
3
)Sin(nwt +
π
6
− ϕn) (5)
• nth harmonic Current
iLn(t) = iphn(t) =
2Vin
nπ|Zn|
Sin(
nπ
3
)Sin(nwt +
π
6
− ϕn)
• where Zn is nth order impedance
Zn = R + jnwL (6)
• Fundamental Load Current Component is:
io1(t) = 32.83sin(wt − 25.23o
)
• Load current (RMS):
Iphr =
p
Iph1 + Iph5 + Iph37 − −− = 23.5A
2.3 Total Harmonic Distortion and Distortion Factor in Output Voltage
• Distortion Factor
g =
VL1
VLr
=
Vph1
Vphr
= 0.95 (7)
• Total Harmonic Distortion
THD =
r
1
g2
− 1 ∗ 100 = 31% (8)
2

3. 2.4 Simulation model of Three Phase Inverter 120o
mode
3

4. 3 Waveforms and Results:
4