chap 4 - inverters

8/2/2019 Chap 4 - Inverters

1/60

Chapter 4Chapter 4

INVERTERSINVERTERS

(DC-AC)(DC-AC)


2/60

IntroductionIntroduction

Convert DC power to AC power at a desired outputConvert DC power to AC power at a desired output

voltage or current and frequency.voltage or current and frequency.

Definition:

General block diagramof inverter


3/60

Introduction (cont)Introduction (cont)

Static circuits no moving partsStatic circuits no moving parts

Converts DC to AC power by switching the DC inputConverts DC to AC power by switching the DC input

voltage in a pre-determined sequence to generatevoltage in a pre-determined sequence to generateAC voltageAC voltage

Output waveform is not purely sinusoidalOutput waveform is not purely sinusoidal


4/60


Classified according to:Classified according to:

number of phases

use of power semiconductor devicescommutation principles

output waveforms


5/60


Induction motor drives, traction, standby power supplies,Induction motor drives, traction, standby power supplies,

and uninterruptible ac power supplies (UPS).and uninterruptible ac power supplies (UPS).

Application:


6/60


(1)(1) Voltage source inverter (VSI)Voltage source inverter (VSI)

3 types of inverter:

DC voltage source isconstant.

A large capacitor across DCsource to stable the output.

Output current is dependent

on the load.Widely used in industry


7/60


(2) Current source inverter (CSI)(2) Current source inverter (CSI)

Current source is DC power

supply.Output current is defined,

based on the gating pattern.

Output voltage is dependent

on the load.


8/60


(3) Current regulated inverter(3) Current regulated inverter

The current regulated

inverters are becomingpopular especially for speed

control of AC motors.

In this category, there is a

current sensing circuit that

senses the actual value of

the current at every instant.


9/60

Single-Phase Half-Bridge InverterSingle-Phase Half-Bridge Inverter

Uses 2 semiconductor for switches.

S1 and S2 connect and disconnect alternately.

The combination of 2 switches provides the 4

states.

Basic principles:

State S1 S2 Output

1 + - +E

2 - - 0

3 - + -E

4 + + 0


10/60

Single-Phase Half-Bridge Inverter (cont)Single-Phase Half-Bridge Inverter (cont)

State 1 & 3 repeatedState 1 & 3 repeatedalternately willalternately willproduced square-produced square-Wave AC voltage.Wave AC voltage.

State 2 & 4 makes thestep-wave or quasi-

quare-wave is obtained.


11/60

Half-Bridge VSIHalf-Bridge VSI


Output

waveform

with R load

Output

waveform

with RL load


12/60

If S1 or S2 are closed (TIf S1 or S2 are closed (TONON), the half-wave average output), the half-wave average output

voltage, Vvoltage, VO(ave)O(ave);;

T

Tcycledutyd

where

EdT

TE

T

TEV

ON

ONONaveO

==

===

;

222/

)(

The RMS output voltage, VO(RMS);

EdV RMSO 2)( =



13/60


For the resistive load, the half-wave average outputFor the resistive load, the half-wave average outputcurrent, Icurrent, IO(ave)O(ave);;

R

VI

aveO

aveO

)(

)( =

The average current in the switch, iS1 and iS2 = IO(ave) / 2

The average power absorbed by the load, PL;

R

dE

R

VP

RMSO

L

22)( 2==


14/60

Example 4.1Example 4.1

A single-phase half-wave inverter with E=100V,A single-phase half-wave inverter with E=100V,

d=50% and resistive load, R=1d=50% and resistive load, R=1..

(i) Find the average load current.(i) Find the average load current.

(ii) Find the average switch current.(ii) Find the average switch current.

(iii) find the power delivered to the load.(iii) find the power delivered to the load.


15/60

Solution 4.1Solution 4.1

(i)(i) VVO(ave)O(ave) = 2Ed = 2(100)(0.5) = 100V= 2Ed = 2(100)(0.5) = 100V

IIO(ave)O(ave) = V= VO(ave)O(ave) / R = 100/1 = 100A/ R = 100/1 = 100A

(ii)(ii) Average current in the switch = IAverage current in the switch = IO(ave)O(ave)/2 =100/2 = 50A/2 =100/2 = 50A

(iii)(iii) The RMS output voltage,The RMS output voltage,

VEdV RMSO 100)100)(5.0*2(2)( ===

Power load, PPower load, PLL;;

kWRVPRMSOL

101/100/ 22 )( ===


16/60

In time domain, the totalIn time domain, the totalRMS value of the loadRMS value of the loadoutput voltage,output voltage,

The instantaneous outputvoltage is;

The fundamental rms

output voltage is;

Single-Phase Half-Bridge Inverter (cont)


17/60



18/60



19/60



20/60



21/60

Solution 4.2 (cont)Solution 4.2 (cont)


22/60


Frequency of AC output;Frequency of AC output;

Tf

1=

2 methods to make the AC output voltage closer tosinusoid:

(i) use a filter circuit on the output side but increase the

power losses, cost and weight.

(ii) use a Pulse Width Modulation (PWM) as aswitching scheme to modify the shape of output

voltage.


23/60

In practical, a dead time as shown below is required in S1In practical, a dead time as shown below is required in S1

and S2 to avoid shoot-through faults.and S2 to avoid shoot-through faults.



24/60

Single-Phase Full-Bridge InverterSingle-Phase Full-Bridge Inverter

Full-Bridge VSIFull-Bridge VSI

Outputwaveform

with R load

Output

waveform

with RL load


25/60

Single-Phase Full-Bridge Inverter (cont)Single-Phase Full-Bridge Inverter (cont)


26/60



27/60



28/60



29/60



30/60



31/60



32/60



33/60



34/60

Three-Phase InverterThree-Phase Inverter


35/60


36/60


37/60

Total Harmonic DistortionTotal Harmonic Distortion


38/60


39/60


40/60

Spectrum of square-waveSpectrum of square-wave


41/60

Fourier series analysis for square wave inverterFourier series analysis for square wave inverter


42/60


43/60



44/60



45/60


46/60


47/60

Pulse Width Modulation (PWM)Pulse Width Modulation (PWM)


48/60

Pulse Width Modulation (PWM) .contPulse Width Modulation (PWM) .cont


49/60

Pulse Width Modulation (PWM) .contPulse Width Modulation (PWM) .cont


50/60


51/60


52/60


53/60


54/60


55/60


56/60


57/60


58/60


59/60


60/60

Sekian.Sekian.

chap 4 - inverters

Documents