DC to AC converter (Inverter)

Single phase

Q1

Q2

Vg2

Vg1Vs

C

C

RL

D

D

0.5Vs

0.5Vs

VL

IL

vg1

t

vg2

0 T/2 T

t0 T/2 T

Dead band = 1 μs

t0 T/2 T

Vs/2

-Vs/2

VL, IL

Q1 Q2 Q1

n=∞

vL=Σn=1,3,5,.. (2 Vs) / (nΠ) × sin(nωt)

Vn

n1 2 3 4 5 6 7 8

Harmonic contents in the output voltage

VLrms= Vs / 2

VL1= (2Vs) / (√2Π)

Q1

Q2

Vg2

Vg1Vs

C

C

RL

D1

D2

0.5Vs

0.5Vs

VL

ILLL

Heavily inductive load (RL → 0)

t0 T/2 T

Vs/2

-Vs/2

vL

Q1 Q2 Q1

t

iL

D1 Q1 D2 Q2

R-L load

iL= Σ (2Vs) / [ nΠ √ (R2+n2ω2L2) ] × sin (nωt - Θn)

Θn= tan-1(nωL / R)

Performance Parameters

HFn Harmonic Factor for the nth harmonic

HFn= (VLn) / VL1 for n > 1

THD Total Harmonic Distortion

∞

THD = 1 / VL1 ( Σ V2n ) 0.5

n=2, 3, 4,…

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 amplitudeIs ≥ 3% of the fundamental

The harmonic voltage Vh

∞

Vh= ( Σ VLn2 ) 0.5 = ( VLrms

2 – VL12 )0.5

n= 3, 5, 7, ..

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

b) VLrms= 0.5 Vs = 24 V PL= (VLrms)2 / R = 242 / 2.4 = 240 W

c)

iQ1

t0 T/2 T

iQ2

t0 T/2 TPeak 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

VL3= 21.6/3 = 7.2 VVL5= 21.6/5 = 4.32 VVL7= 21.6/7 = 3.086 V

DF = 1/21.6 ×{ [ 7.2/32]2+ [4.32/52]2

+[3.086/72] 2}0.5 = 1/21.6 ×{ 0.64 + 0.02986 +0.004+ .. }0.5

= 0.038

Q1

Q2

Vg2

Vg1Vs

C

C

RL

D1

D2

0.5Vs

0.5Vs

VL

ILLL

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

The H-bridge single phase inverterVg1, Vg2

t

Vg3, Vg4

0 T/2 T

t0 T/2 T

Dead band = 1 μs

t0 T/2 T

Vs

-Vs

VL, IL

Q1, Q2 Q3, Q4

n=∞

vL=Σn=1,3,5,.. (4 Vs) / (nΠ) × sin(nωt)

Vn

n1 2 3 4 5 6 7 8

Harmonic contents in the output voltage

VLrms= Vs

VL1= (4Vs) / (√2Π)

Q3

Q2

Vg2

Vg3Vs

RL

D

DVL

IL

Q1

Q4

Vg1

Vg4

D

D

Q1, Q2

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 = (4Vs) / Π × sin ( ωt)

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

t0 T/2 T

iQ3, iQ4

t0 T/2 TPeak 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)

VL3= 43.2/3 = 14.4 VVL5= 43.2/5 = 8.64 VVL7= 43.2/7 = 6.17 V

DF = 1/43.2 ×{ [ 14.4/32]2+ [8.64/52]2

+[6.17/72] 2}0.5

= 1/43.2 ×{ 1.6 + .3456 +0.1259+ .. }0.5

= 0.033 (same)

Q3

Q2

Vg2

Vg3Vs

RL

D

DVL

IL

Q1

Q4

Vg1

Vg4

D

D

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 anRLC load with R=10Ω, L=31.5mH, C=112μF.The inverter frequency is 60 Hz and the dc inputVoltage is Vs=220V.a) Express the instantaneous load current in Fourrier series.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

Q2

Vg2

Vg3Vs

RD

DVL

IL

Q1

Q4

Vg1

Vg4

D

D

L C

Three-phase inverters120o conduction

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

R

R

a

b

c

@ any time only 2 transistors are conducting: 1 in an upper leg 1 in another lower leg

ωt60o

ωt60o

vG1

vG2

ωt60o

vG3

ωt60o

vG4

ωt60o

vG5

ωt60o

vG6

For 0 ≤ ωt < 60o

R

R

R

Vs

a

b

c

For 60o ≤ ωt < 120o

R

R

R

Vs

a

b

c

For 120o ≤ ωt < 180o

R

R

RVs

a

b

c

For 180o ≤ ωt < 240o

R

R

R

Vs

a

b

c

n’ n’ n’

n’

For 240o ≤ ωt < 300o

R

R

R

Vs

a

b

c

n’

For 300o ≤ ωt < 360o

R

R

RVs

a

b

c

n’

ωt60o

ωt60o

vab

vbc

ωt60o

vca

ωt60o

van’

ωt60o

vbn’

ωt60o

Vcn’

CV

CV CV

CV CV

CV CV

CV

Vs

0.5Vs

- 0.5Vs

-Vs

0.5Vs

-0.5Vs

ωt60o

ωt60o

vG1

vG2

ωt60o

vG3

ωt60o

vG4

ωt60o

vG5

ωt60o

vG6

180o conduction ( 3 transistors are conducting at any time)

For 0 ≤ ωt < 60o For 60o ≤ ωt < 120o For 120o ≤ ωt < 180o

For 180o ≤ ωt < 240o For 240o ≤ ωt < 300o

R

R

Vs

R a

b

c

n’

For 300o ≤ ωt < 360o

R

R

Vs

a

R b

cn’

R

Vs

a

R bR c

n’

R

Vs

aRb

R

c

n’

R

Vs

a

Rb

R

c

n’

Vs

R

R

Ra

bc

n’

R

ωt60o

ωt60o

vab

vbc

ωt60o

vca

ωt60o

van’

ωt60o

vbn’

ωt60o

Vcn’

CV

CV

Vs

-Vs

CV

CV

⅓Vs

⅔Vs

Voltage control techniques of single phase inverters

Single pulse width modulation

0Π 2Π

ωt

VL

Π/2

3Π/2

δ

δ

-Vs

Vs

vL= Σn=1,3,5,.. (4Vs / nΠ) sin(nδ/2) sin(nωt) ∞

VLrms= Vs √(δ/Π)

Multiple pulse width modulation

0 Π 2Π

ωt

VL

Π/2

3Π/2

-Vs

Vs

Π/3 2Π/3Π/6 5Π/6

δδ

δ

7Π/6 11Π/6

4Π/3 5Π/3

P= # of pulses per half cycleP=3

δ δδ

Decreases DF significantly

αm=2

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

Sinusoidal Pulse Width Modulation

Carrier waveform

Reference waveform

Ar

Ac

MA = Amplitude Modulation Index

Ar

MA = _______

Ac

MF = Frequency Modulation Index

carrier frequencyMF = --------------------------- (= 5) reference frequency

fC = carrier frequency

fR = reference frequency

0 ≤ MA ≤ 1

If MA > 1 over-modulation

ωt

ωt

-Vs

α1 α2

180o- α1

180o – α2

Vs

if MF is an odd number, quarter-wave symmetryis obtained and no even harmonics are presentin the output voltage.

For a 3-phase inverter, MF should be an oddtriplen number

SPWM reduces greatly the DF

MA0

U1

0 1

<1

over-modulation