eab3023 l11 students

8/3/2019 EAB3023 L11 Students

1/13

EAB 3023 POWER ELECTRONICS 1: DC-AC INVERTER

Page 1

EAB/EBB 3023: POWER ELECTRONICS 1

LECTURE SLIDES #11: DC-AC (INVERTER)

Lecturer: Dr. Zuhairi Baharudin

Room: 22-03-09

Phone: 05-3687810

email: [email protected]


2/13


Learning Objectives

At the end of the class, you should be able to:

Define DC-AC inverter and state its application

Identify type of Inverter

Explain the operation of half-bridge and full-bridge inverter


3/13


Contents

Introduction

Half bridge inverter, R, L load

Half bridge inverter, R, L load with feedback/flyback diode

Full bridge inverter, R, L load with feedback/flyback diodeAverage transistor and diode current

Inductive load analysis


4/13


Introduction

An inverter is an electrical device that converts direct

current (DC) to alternating current (AC)

The applications of inverter such as:

Solar photovoltaic

UPS

HVDC power transmission

AC motor speed control


5/13


Square wave Inverter

The basic fundamental of inverter

It converts a DC voltage to a square wave AC voltage

Although, the output is nonsinusoidal, it may be an

adequate AC signal for some applications


6/13


Half-Bridge: R, Load

S1 and S2 are switched on and off

alternatively at a 50% duty cycle


7/13


Half-Bridge: R, L, Load

When S1 ON, current will slowly rise

D1 and D2 depending on

the value of L time constant,


8/13


Half-Bridge: R and L load, with feedback/flyback diode

0


9/13


Full-Bridge: R, load


10/13


Full-Bridge: R,L, load


11/13


20)( / TtAeR

Vti

tdc

o!

X

From Kirchhoffs Voltage law during the interval 0 to T/2,

dt

tdiLtRiV

o

odc

)()( !

The output current is expressed as

Where A is a constant evaluated from the

Initial condition andR

L!X

At t=T/2, the voltage becomes Vdc and the current can be written as

TtT

BeR

Vti

Ttdc

o

!

2)( /)2/( X

Where B is a constant evaluated from the

Initial condition

The formulation for the output current


12/13


Evaluating the output current, at t=0,

min

0)0( IAeR

Vi

dc

o!! Thus,

R

VIA

dc!

min

20)()( /min TteR

VI

R

Vti tdcdco !

X

Finally the output current, can be written as

Likewise, evaluating the output current, at t=T/2,

max

0)2/( IBe

R

VTi

dc

o!

! Thus,

R

VIB

dc!

max

Thus,the output current, can be written as

TtT

eR

VI

R

Vti

Ttdcdc

o

!

2)()( /)2/(

max

X

The formulation for the output current


13/13


Solving forImax and Imin,

-

!!

X

X

2/

2/

minmax

1

1

s

s

T

T

dc

e

e

R

VII

Power absorbed can be determined fromRIP

rms

2!

where the rms current is given by

!!2

0

2

0

2)(

2)(

1s

T

s

T

s

rmsdtti

Tdtti

TI

dteR

VI

R

V

TI

sT

tdcdc

s

rms

22

0

/min )(2 -

! X

If the switches are ideal, the power supplied by the source must be the same as

Absorbed by the load:indcdcIVP !

Imin, Imax and power

eab3023 l11 students

Documents