1. DC to AC converter (Inverter)
vg1
DC to AC converter (Inverter)
Single phase t
T/2 T
vg2 t
T/2 T
Dead band = 1 μs Q1 Q2
Vg2
Vg1 Vs C RL D
0.5Vs VL IL
VL, IL
n=∞
vL=Σn=1,3,5,.. (2 Vs) / (nΠ) × sin(nωt)
Vs/2 Vn t
T/2 T
-Vs/2 Q1 Q2 Q1
VLrms= Vs / 2
VL1= (2Vs) / (√2Π) 1 2 3 4 5 6 7 8 n
Harmonic contents in the output voltage

2. Heavily inductive load (RL → 0)
T/2 T
Vs/2
-Vs/2 vL Q1 Q2 iL D1 D2
Heavily inductive load (RL → 0) Q1 Q2
Vg2
Vg1 Vs C RL D1 D2
0.5Vs VL IL LL
R-L load
iL= Σ (2Vs) / [ nΠ √ (R2+n2ω2L2) ] × sin (nωt - Θn)
Θn= tan-1(nωL / R)

3. Performance Parameters
HFn Harmonic Factor for the nth harmonic
HFn= (VLn) / VL for n > 1
THD 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

4. a) The rms value of the load fundamental voltage. b) The output power.
Vs= 48V
R= 2.4 Ω
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 = (2Vs) / Π × sin ( ωt)
VL1rms = ( 2 × 48 ) / ( Π × √2 ) = 21.6 V
VLrms= 0.5 Vs = 24 V
PL= (VLrms)2 / R = 242 / 2.4 = 240 W c)
iQ1 t
T/2 T
iQ2
Peak current in each transistor = 24/2.4 = 10A
Average current in each transistor = 5 A
d) Vh= ( 242 – )0.5 = V
THD = / 21.6 =
VL3= 21.6/3 = 7.2 V
VL5= 21.6/5 = 4.32 V
VL7= 21.6/7 = V
DF = 1/21.6 ×{ [ 7.2/32]2+ [4.32/52]2
+[3.086/72] 2}0.5
= 1/21.6 ×{
}0.5 = Q1 Q2
Vg2
Vg1 Vs C RL D1 D2
0.5Vs VL IL LL

5. The LOH = 3rd harmonic
HF3= 1/3 =
DF3= /32 = note that VL3= which is > so LOH =3

6. The H-bridge single phase inverter
Vg1, Vg2
The H-bridge single phase inverter Q3 Q2
Vg2
Vg3 Vs RL D VL IL Q1 Q4
Vg1
Vg4 t
T/2 T
Vg3, Vg4 t
T/2 T
Dead band = 1 μs
VL, IL
n=∞
vL=Σn=1,3,5,.. (4 Vs) / (nΠ) × sin(nωt) Vs Vn t
T/2 T
-Vs
Q1, Q2
Q3, Q4
Q1, Q2
VLrms= Vs
VL1= (4Vs) / (√2Π) 1 2 3 4 5 6 7 8 n
Harmonic contents in the output voltage

7. a) The rms value of the load fundamental voltage. b) The output power.
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. Q3 Q2
Vg2
Vg3 Vs RL D VL IL Q1 Q4
Vg1
Vg4
a) vL1 = (4Vs) / Π × sin ( ωt)
VL1rms = ( 4 × 48 ) / ( Π × √2 ) = 43.2 V
VLrms= Vs = 48 V
PL= (VLrms)2 / R = 482 / 2.4 = 960 W c)
Vs= 48V
R= 2.4 Ω
iQ1, iQ2
iQ3, iQ4 t t
T/2 T
T/2 T
Peak current in each transistor = 48/2.4 = 20A
Average current in each transistor =10 A
DF = 1/43.2 ×{ [ 14.4/32]2+ [8.64/52]2
+[6.17/72] 2}0.5
= 1/43.2 ×{
}0.5
= (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 – )0.5 = V
THD = / 43.2 =
(same)

8. Voltage is twice that of the 2-transistor circuit.
LOH = 3rd harmonic
HF3 = 1/3
DF3= 1/(3×32) = (same) note that VL3= 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.

9. The H-bridge inverter shown in figure has an
RLC load with R=10Ω, L=31.5mH, C=112μF.
The inverter frequency is 60 Hz and the dc input
Voltage is Vs=220V.
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. Q3 Q2
Vg2
Vg3 Vs R D VL IL Q1 Q4
Vg1
Vg4 L C

10. Three-phase inverters
120o conduction
Three-phase inverters
180o conduction
120o conduction Vs Q1 Q4
Vg1
Vg4 D1 D4 a Q3 Q6
Vg3
Vg6 D3 D6 b Q5 Q2
Vg5
Vg2 D5 D2 c R
@ any time only 2 transistors are conducting: 1 in an upper leg
1 in another lower leg

11. ωt
60o
vG1
vG2
vG3
vG4
vG5
vG6

12. For 0 ≤ ωt < 60o Vs a b c For 60o ≤ ωt < 120on’
For 240o ≤ ωt < 300o
For 300o ≤ ωt < 360o

13. ωt
60o
vab
vbc
vca
van’
vbn’
Vcn’ CV Vs
0.5Vs
- 0.5Vs
-Vs
-0.5Vs

14. 180o conduction ( 3 transistors are conducting at any time)
vG1
vG2
vG3
vG4
vG5
vG6
180o conduction ( 3 transistors are conducting at any time)

15. For 0 ≤ ωt < 60o For 60o ≤ ωt < 120o For 120o ≤ ωt < 180o Vsa b c n’ R Vs a b c n’ R Vs a b c n’
For 240o ≤ ωt < 300o
For 300o ≤ ωt < 360o
For 180o ≤ ωt < 240o R n’ a R a R R b R n’ b a n’ R c b R c c R R R Vs Vs Vs

16. ωt
60o
vab
vbc
vca
van’
vbn’
Vcn’ CV Vs
-Vs
⅓Vs
⅔Vs

18. Voltage control techniques of single phase inverters
Single pulse width modulation
Multiple pulse width modulation Π 2Π ωt VL
Π/2
3Π/2 δ
-Vs Vs VL Vs δ δ δ
7Π/6
3Π/2
11Π/6 ωt
Π/6
Π/3
Π/2
2Π/3
5Π/6 Π
4Π/3
5Π/3 2Π
αm=2 δ δ δ
-Vs
P= # of pulses per half cycle
P=3
vL= Σn=1,3,5,.. (4Vs / nΠ) sin(nδ/2) sin(nωt) ∞
Decreases DF significantly
VLrms= Vs √(δ/Π)
vL= Σn=1, 3, ..Σm=1{4Vs /(nΠ) sin{ nδ/4 [ sin n(αm+3δ/4) – sin n(Π+αm+δ/4) ] }× sin(nωt) ∞ 2p
VLrms= Vs √ (pδ/Π)
δ = M T/ (2p)
Where M is the amplitude modulation index 0 ≤ M ≤ 1

19. Sinusoidal Pulse Width ModulationAc 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 > 1 over-modulation

20. if MF is an odd number, quarter-wave symmetry
is obtained and no even harmonics are present
in the output voltage. α1 ωt α2
For a 3-phase inverter, MF should be an odd
triplen number
180o- α1
180o – α2
SPWM reduces greatly the DF U1 Vs
<1 ωt
over-modulation MA
-Vs 1