vg
DC to AC converter
(Inverter)
1
Single phase
t
0
T/2
T
0
T/2
T
vg
2
0.5Vs
C
RL
V
VL
s
0.5Vs
C
D
IL
Q1
Vg1
D
t
Q2
Dead band = 1 μs
Vg2
n=∞
vL=Σn=1,3,5,.. (2 Vs) / (nΠ) × sin(nωt)
VL, IL
Vs/2
Vn
0
T/2
t
T
-Vs/2
Q1
Q2
VLrms= Vs / 2
1 2 3 4 5 6 7 8 n
Harmonic contents in the output voltage
VL1= (2Vs) / (√2Π)
Q1
Heavily inductive load (RL → 0)
0.5Vs
C
RL
V
s
LL
VL
0.5Vs
C
D1
IL
vL
Vs/2
0
Q1
t
T
-Vs/2
Q1
Vg1
D2
T/2
Q2
Q1
iL
Q2
Vg2
t
R-L load
D1 Q1
D
2
iL= Σ (2Vs) / [ nΠ √ (R2+n2ω2L2) ] × sin (nωt - Θn)
Θn= tan-1(nωL / R)
Q2
Performance Parameters
HFn Harmonic Factor for the nth harmonic
HFn= (VLn) / VL1
THD
for n > 1
Total Harmonic Distortion
The harmonic voltage
Vh
∞
THD = 1 / VL1 ( Σ V2n ) 0.5
n=2, 3, 4,…
∞
Vh= ( Σ VLn2 ) 0.5 = ( VLrms2 – VL12 )0.5
n= 3, 5, 7, ..
DF
Distortion Factor
∞5
DF = 1 / VL1 [ Σ ( VLn / n2 )2 ] 0.5
n=2, 3, …
The Distortion Factor of the nth harmonic = VLn / ( VL1 n2)
for n > 1
Lowest Order Harmonic LOH
is the harmonic component that is the closest to the fundamental and its amplitude
Is ≥ 3% of the fundamental
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.
0.5Vs
C
RL
V
s
D1
LL
Vg1
VL
0.5Vs
Q1
IL
Q2
D2
C
Vg2
a) vL1 = (2Vs) / Π × sin ( ωt)
Vs= 48V
R= 2.4 Ω
VL1rms = ( 2 × 48 ) / ( Π × √2 ) = 21.6 V
b) VLrms= 0.5 Vs = 24 V
PL= (VLrms)2 / R = 242 / 2.4 = 240 W
c)
iQ1
iQ2
t
0
T/2
T
0
Peak current in each transistor = 24/2.4 = 10A
Average current in each transistor = 5 A
d) Vh= ( 242 – 21.62 )0.5 = 10.46 V
THD = 10.46 / 21.6 = 0.4843
T/2
T
t
DF = 1/21.6 ×{ [ 7.2/32]2+ [4.32/52]2
+[3.086/72] 2}0.5
VL3= 21.6/3 = 7.2 V
= 1/21.6 ×{ 0.64 + 0.02986
VL5= 21.6/5 = 4.32 V
+0.004+ .. }0.5
VL7= 21.6/7 = 3.086 V
= 0.038
The LOH = 3rd harmonic
HF3= 1/3 = 0.3333
DF3= 0.3333/32 = 0.03703
note that VL3= 0.3333 which is > 0.03 so LOH =3
Vg1, Vg2
The H-bridge single phase inverter
Q1
D
IL
RL
D
s
Q4
VL
D
T/2
T
0
T/2
T
Vg3, Vg4
Q3
Vg1
V
t
0
Vg3
D
Q2
Vg4
t
Vg2
Dead band = 1 μs
VL, IL
n=∞
vL=Σn=1,3,5,.. (4 Vs) / (nΠ) × sin(nωt)
Vs
Vn
0
T/2
T
t
-Vs
Q1, Q2
Q3, Q4
VLrms= Vs
1 2 3 4 5 6 7 8 n
Harmonic contents in the output voltage
VL1= (4Vs) / (√2Π)
Q1, Q2
Calculate:
a) The rms value of the load fundamental voltage.
b) The output power.
c) The average and peak current in the transistor. V
d) The THD, DF, the HF and DF of the LOH.
s
D
Q1
IL
D
RL
Q3
Vg1
Vg3
VL
Q4
Q2
D
D
Vg4
Vg2
a) vL1 = (4Vs) / Π × sin ( ωt)
Vs= 48V
R= 2.4 Ω
VL1rms = ( 4 × 48 ) / ( Π × √2 ) = 43.2 V
b) VLrms= Vs = 48 V
PL= (VLrms)2 / R = 482 / 2.4 = 960 W
c)
iQ1, iQ2
iQ3, iQ4
t
0
T/2
T
0
Peak current in each transistor = 48/2.4 = 20A
Average current in each transistor =10 A
d) Vh= (482 – 43.22 )0.5 = 20.92 V
THD = 20.92 / 43.2 = 0.4843
(same)
T/2
T
t
DF = 1/43.2 ×{ [ 14.4/32]2+ [8.64/52]2
+[6.17/72] 2}0.5
VL3= 43.2/3 = 14.4 V
= 1/43.2 ×{ 1.6 + .3456
VL5= 43.2/5 = 8.64 V
+0.1259+ .. }0.5
VL7= 43.2/7 = 6.17 V
= 0.033 (same)
LOH = 3rd harmonic
HF3 = 1/3
DF3= 1/(3×32) = 0.03703 (same)
note that VL3= 14.4 which is > 0.03×VL1 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.
The H-bridge inverter shown in figure has an
RLC load with R=10Ω, L=31.5mH, C=112μF.
D
D
Q1
L C
R
IL
The inverter frequency is 60 Hz and the dc input
V
V
Voltage is Vs=220V.
VL
s
D
Q
a) Express the instantaneous load current in
4
D
Fourrier series.
V
b) 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.
Q3
g1
g4
Vg3
Q2
Vg2
120o conduction
Three-phase inverters
180o conduction
120o conduction
R
a
Q1
V
s
a
Vg1
Q4
Vg4
D1
Q3
b
Vg3
Q6
D4
Vg6
D3
Q5
Q2
Vg2
R
c b
Vg5
D6
D5
D2
R
c
@ any time only 2 transistors are conducting: 1 in an upper leg
1 in another lower leg
vG
1
60o
ωt
vG
2
60o
ωt
vG
3
60o
ωt
vG
4
60o
ωt
vG
5
60o
ωt
vG
6
60o
ωt
For 60o ≤ ωt < 120o
For 0 ≤ ωt < 60o
R
R
R
b
s
n’
b
R
c
R
n’
V
s
For
180o
≤ ωt <
For 240o ≤ ωt < 300o
240o
c
b
R
n’
s
R
R
c
R
R
a
R
n’
V
s
c
n’
For 300o ≤ ωt < 360o
a
a
b
b
R
R
V
a
V
s
R
c
R
a
a
V
For 120o ≤ ωt < 180o
R
b
R
c
R
V
s
n’
vab
Vs
CV
60o
vbc
0.5Vs
CV
CV
CV
- 0.5Vs
CV
CV
CV
CV
ωt
-Vs
ωt
60o
vca
ωt
60o
van’
ωt
60o
vbn’
0.5Vs
60o
Vcn’
60o
ωt
-0.5Vs
ωt
180o conduction ( 3 transistors are conducting at any time)
vG1
60o
ωt
vG2
60o
ωt
vG3
vG4
60o
60o
ωt
ωt
vG5
vG6
60o
60o
ωt
ωt
For 60o ≤ ωt < 120o
For 0 ≤ ωt < 60o
a
c
R
R
R
a
R
n’
For
b
a
n’
b
R
R
b
b
n’
R
R
c
c
R
Vs
Vs
180o
For 120o ≤ ωt < 180o
≤ ωt <
240o
n’
R
Vs
For 240o ≤ ωt < 300o
R
a
R
b
c
c
For 300o ≤ ωt < 360o
a
n’
a
R
R
c
R
R
n’
R
R
R
Vs
Vs
Vs
b
vab
Vs
CV
60o
vbc
CV
ωt
CV
CV
-Vs
60o
ωt
vca
60o
ωt
van’
⅓Vs
⅔Vs
60o
ωt
vbn’
60o
ωt
Vcn’
60o
ωt
Voltage control techniques of single phase inverters
Multiple pulse width modulation
Single pulse width modulation
VL
VL
Vs
Vs
δ
δ
3Π/2
0
Π/2
Π
δ
7Π/6
ωt
0 Π/6 Π/3 Π/2 2Π/3 5Π/6 Π
2Π
αm=2
δ
-Vs
δ
3Π/2
4Π/3
δ
11Π/6
5Π/3
δ
ωt
2Π
δ
-Vs
P= # of pulses per half cycle
P=3
∞
vL= Σn=1,3,5,.. (4Vs / nΠ) sin(nδ/2) sin(nωt)
VLrms= Vs √(δ/Π)
∞
Decreases DF significantly
2p
vL= Σn=1, 3, ..Σm=1{4Vs /(nΠ) sin{ nδ/4 [ sin n(αm+3δ/4) – sin n(Π+αm+δ/4) ] }× sin(nωt)
VLrms= Vs √ (pδ/Π)
δ = M T/ (2p)
Where M is the amplitude modulation index 0 ≤ M ≤ 1
Sinusoidal Pulse Width Modulation
Ac
Ar
Reference waveform
MA = Amplitude Modulation Index
Ar
Carrier waveform
MA = _______
Ac
MF = Frequency Modulation Index
carrier frequency
MF = --------------------------reference frequency
0 ≤ MA ≤ 1
If MA > 1 over-modulation
(= 5)
fC = carrier frequency
fR = reference frequency
if MF is an odd number, quarter-wave symmetry
is obtained and no even harmonics are present
in the output voltage.
α1 α
ωt
180o2- α1
For a 3-phase inverter, MF should be an odd
triplen number
180o – α2
SPWM reduces greatly the DF
U1
Vs
<1
ωt
over-modulation
0
0
1
MA
-Vs