DC to AC converter (Inverter)

1
vg1
DC to AC converter (Inverter)
Single phase
t
vg2
VL, IL
n8 vLSn1,3,5,.. (2 Vs) / (n?)
sin(n?t)
Vs/2
Vn
t
0
T/2
T
-Vs/2
Q1
Q2
Q1
VLrms Vs / 2 VL1 (2Vs) / (v2?)
n
1
2
3
4
5
6
7
8
Harmonic contents in the output voltage
2
Heavily inductive load (RL ? 0)
R-L load
iL S (2Vs) / n? v (R2n2?2L2) sin (n?t -
Tn) Tn tan-1(n?L / R)
3
Performance Parameters
HFn Harmonic Factor for the nth harmonic
HFn (VLn) / VL1 for n gt 1
THD Total Harmonic Distortion
The harmonic voltage Vh
8 THD 1 / VL1 ( S
V2n ) 0.5 n2, 3, 4,
8 Vh ( S VLn2 ) 0.5 ( VLrms2 VL12
)0.5 n 3, 5, 7, ..
DF Distortion Factor
85 DF 1 / VL1 S (
VLn / n2 )2 0.5 n2,
3, The Distortion Factor of the nth harmonic
VLn / ( VL1 n2) for n gt 1
Lowest Order Harmonic LOH
is the harmonic component that is the closest to
the fundamental and its amplitude Is 3 of the
fundamental
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The LOH 3rd harmonic HF3 1/3 0.3333 DF3
0.3333/32 0.03703 note that VL3 0.3333
which is gt 0.03 so LOH 3
6
Vg1, Vg2
The H-bridge single phase inverter
t
Vg3, Vg4
VL, IL
n8 vLSn1,3,5,.. (4 Vs) / (n?)
sin(n?t)
Vs
Vn
t
0
T/2
T
-Vs
Q1, Q2
Q1, Q2
Q3, Q4
VLrms Vs VL1 (4Vs) / (v2?)
n
1
2
3
4
5
6
7
8
Harmonic contents in the output voltage
7
Calculate a) The rms value of the load
fundamental voltage. b) The output power. c) The
average and peak current in the transistor. d)
The THD, DF, the HF and DF of the LOH.

• a) vL1 (4Vs) / ? sin ( ?t)
• VL1rms ( 4 48 ) / ( ? v2 ) 43.2 V
• VLrms Vs 48 V
• PL (VLrms)2 / R 482 / 2.4 960 W
• c)

Vs 48V R 2.4 O
iQ1, iQ2
iQ3, iQ4
t
t
0
T/2
T
0
T
T/2
Peak current in each transistor 48/2.4
20A Average current in each transistor 10 A
DF 1/43.2 14.4/322 8.64/522
6.17/72 20.5 1/43.2
1.6 .3456 0.1259 ..
0.5 0.033 (same)
VL3 43.2/3 14.4 V VL5 43.2/5 8.64 V VL7
43.2/7 6.17 V
d) Vh (482 43.22 )0.5 20.92 V THD
20.92 / 43.2 0.4843
(same)
8
LOH 3rd harmonic HF3 1/3 DF3 1/(332)
0.03703 (same) note that VL3 14.4 which is gt
0.03VL1 so LOH 3 The quality of the output
voltage is the same as for the 2-transistor
circuit however, the H bridge inverter the output
power is 4 times higher and the fundamental
output Voltage is twice that of the 2-transistor
circuit.
9

• The H-bridge inverter shown in figure has an
• RLC load with R10O, L31.5mH, C112µF.
• The inverter frequency is 60 Hz and the dc input
• Voltage is Vs220V.
• Express the instantaneous load current in
• Fourrier series.
• Calculate the rms load current at the fundamental
• frequency.
• c) Calculate the THD of the load current.
• d) Calculate the total power absorbed by the
  load as well as the fundamental power.
• e) Calculate the average dc current drawn from
  the supply.
• f) Calculate the rms and the peak current of
  each transistor.

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120o conduction
Three-phase inverters
180o conduction
120o conduction
_at_ any time only 2 transistors are conducting 1
in an upper leg
1 in
another lower leg
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For 60o ?t lt 120o
For 120o ?t lt 180o
For 0 ?t lt 60o
For 240o ?t lt 300o
For 300o ?t lt 360o
For 180o ?t lt 240o
R
a
a
n
R
n
a
b
n
R
b
c
R
c
R
c
R
Vs
Vs
Vs
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Voltage control techniques of single phase
inverters
Multiple pulse width modulation
Single pulse width modulation
VL
Vs
d
d
d
3?/2
7?/6
11?/6
?t
?/2
?/6
0
?/3
2?/3
5?/6
4?/3
5?/3
?
2?
am2
d
d
d
-Vs
P of pulses per half cycle P3
Decreases DF significantly
vL Sn1,3,5,.. (4Vs / n?) sin(nd/2) sin(n?t)
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VLrms Vs v(d/?)
VLrms Vs v (pd/?)
d M T/ (2p)
Where M is the amplitude modulation index 0 M
1
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Sinusoidal Pulse Width Modulation
Ac
Ar
Reference waveform
MA Amplitude Modulation Index Ar MA
_______ Ac MF Frequency
Modulation Index carrier frequency MF
--------------------------- ( 5)
reference frequency
Carrier waveform
fC carrier frequency fR reference frequency
0 MA 1 If MA gt 1 over-modulation
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if MF is an odd number, quarter-wave symmetry is
obtained and no even harmonics are present in the
output voltage.
?t
a1
a2
For a 3-phase inverter, MF should be an
odd triplen number
180o- a1
180o a2
SPWM reduces greatly the DF
U1
Vs
lt1
?t
over-modulation
0
-Vs
MA
0
1