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# DC to AC converter (Inverter) - PowerPoint PPT Presentation

0. T/2. T. 0.5V s. C. D. Q 1. R L. I L. t. V s. V g 1. V L. Q 2. D. 0.5V s. C. Dead band = 1 μ s. 0. T/2. T. V g2. v g1. DC to AC converter (Inverter). Single phase. t. v g2. V L , I L. n=∞ v L = Σ n=1,3,5,.. (2 V s ) / (n Π ) × sin(n ω t). V s /2. V n. t. 0.

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## PowerPoint Slideshow about ' DC to AC converter (Inverter)' - paiva

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Presentation Transcript

T/2

T

0.5Vs

C

D

Q1

RL

IL

t

Vs

Vg1

VL

Q2

D

0.5Vs

C

0

T/2

T

Vg2

vg1

DC to AC converter (Inverter)

Single phase

t

vg2

VL, IL

n=∞

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

Vs/2

Vn

t

0

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

vL

Vs/2

0.5Vs

C

D1

Q1

LL

RL

IL

t

0

T/2

T

Vs

Vg1

VL

Q2

D2

-Vs/2

0.5Vs

C

Q1

Q2

Q1

Vg2

iL

t

D1

Q1

D2

Q2

Heavily inductive load (RL → 0)

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

Θn= tan-1(nωL / R)

HFn Harmonic Factor for the nth harmonic

HFn= (VLn) / VL1 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

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

D1

Q1

LL

RL

IL

Vs

Vg1

VL

• 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)

Q2

D2

0.5Vs

C

Vs= 48V

R= 2.4 Ω

Vg2

iQ1

iQ2

t

t

0

T/2

T

0

T

T/2

Peak current in each transistor = 24/2.4 = 10A

Average current in each transistor = 5 A

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

VL3= 21.6/3 = 7.2 V

VL5= 21.6/5 = 4.32 V

VL7= 21.6/7 = 3.086 V

d) Vh= ( 242 – 21.62 )0.5 = 10.46 V

THD = 10.46 / 21.6 = 0.4843

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

T/2

T

D

D

Q1

Q3

RL

IL

Vs

Vg1

Vg3

VL

Q2

D

Q4

D

t

Vg4

Vg2

0

T/2

T

Vg1, Vg2

The H-bridge single phase inverter

t

Vg3, Vg4

VL, IL

n=∞

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

Vs

Vn

t

0

T/2

T

-Vs

Q1, Q2

Q1, Q2

Q3, Q4

VLrms= Vs

VL1= (4Vs) / (√2Π)

1

2

3

4

5

6

7

8

n

Harmonic contents in the output voltage

D

Q1

Q3

RL

IL

Vs

Vg1

Vg3

VL

Q2

D

Q4

D

Vg4

Vg2

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

• 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

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 ×{ [ 14.4/32]2+ [8.64/52]2

+[6.17/72] 2}0.5

= 1/43.2 ×{ 1.6 + .3456

+0.1259+ .. }0.5

= 0.033 (same)

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)

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.

D

Q1

Q3

C

L

R

IL

Vs

Vg1

Vg3

VL

Q2

D

Q4

D

Vg4

Vg2

• 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.

D3

D5

D1

Q1

Q5

Q3

R

c

b

a

a

Vg5

Vg3

Vg1

Q2

Q4

Q6

D6

D2

D4

R

Vs

Vg2

Vg6

Vg4

b

R

c

120o conduction

Three-phase inverters

180o conduction

120o conduction

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

1 in another lower leg

vG1

ωt

60o

vG2

ωt

60o

vG3

ωt

60o

vG4

ωt

60o

vG5

ωt

60o

vG6

ωt

60o

For 60o ≤ ωt < 120o

For 120o ≤ ωt < 180o

For 0 ≤ ωt < 60o

R

R

R

a

a

a

Vs

R

Vs

R

R

b

b

b

n’

n’

n’

Vs

R

R

R

c

c

c

For 240o ≤ ωt < 300o

For 300o ≤ ωt < 360o

For 180o ≤ ωt < 240o

R

R

R

a

a

a

R

R

b

b

R

Vs

b

n’

n’

Vs

n’

Vs

R

R

c

c

R

c

vab

Vs

0.5Vs

CV

CV

CV

CV

ωt

CV

CV

CV

CV

60o

- 0.5Vs

vbc

-Vs

ωt

60o

vca

ωt

60o

van’

ωt

60o

vbn’

0.5Vs

ωt

60o

-0.5Vs

Vcn’

ωt

60o

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

vG1

ωt

60o

vG2

ωt

60o

vG3

ωt

60o

vG4

ωt

60o

vG5

ωt

60o

vG6

ωt

60o

R

a

R

a

n’

a

n’

R

b

R

R

b

b

n’

R

c

R

c

c

R

Vs

Vs

Vs

R

R

R

R

b

For 60o ≤ ωt < 120o

For 120o ≤ ωt < 180o

For 0 ≤ ωt < 60o

For 240o ≤ ωt < 300o

For 300o ≤ ωt < 360o

For 180o ≤ ωt < 240o

R

a

n’

a

R

n’

a

b

n’

R

b

c

R

c

R

c

R

Vs

Vs

Vs

vab

Vs

CV

CV

ωt

CV

CV

60o

vbc

-Vs

ωt

60o

vca

ωt

60o

van’

⅔Vs

⅓Vs

ωt

60o

vbn’

ωt

60o

Vcn’

ωt

60o

VL

Vs

δ

3Π/2

ωt

Π/2

0

Π

δ

-Vs

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)

Voltage control techniques of single phase inverters

Multiple pulse width modulation

Single pulse width modulation

VL

Vs

δ

δ

δ

3Π/2

7Π/6

11Π/6

ωt

Π/6

Π/2

Π/3

2Π/3

4Π/3

0

5Π/6

5Π/3

Π

αm=2

δ

δ

δ

-Vs

P= # of pulses per half cycle

P=3

Decreases DF significantly

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

VLrms= Vs √(δ/Π)

VLrms= Vs √ (pδ/Π)

δ = M T/ (2p)

Where M is the amplitude modulation index 0 ≤ M ≤ 1

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 > 1 over-modulation

if MF is an odd number, quarter-wave symmetry

is obtained and no even harmonics are present

in the output voltage.

ωt

α1

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

0

-Vs

MA

0

1