microsoft powerpoint - chapter2_parta1 [compatibility mode]

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SCHMY-MMU, Preliminary Meeting 23rd August 2002

5

Chapter 2

Line-Commutated Diode Rectifiers

ObjectiveTo understand the operation of

Uncontrolled Rectifier

To understand the operation of

Controlled Rectifier

Ultimately

Control of Speed of a DC Motor




5

Why I should Learn this chapter?

In Industry DC Power Source needed

e.g To run a DC Motor

WHY DC Supply is not used?

E.g. Battery

Reliability Problem

Forces Engineer to use Power

Converter to supply DC load

using AC Power Source

DC

SUPPLY

Constant DC

Voltage

DC

MOTOR

Needs Variable DC voltage

for motoring

(Speed Variation)

AC

SUPPLY Constant AC

Voltage & Frequency

Converters Fed DC Drives




5

Presents Box Office MoviePresents Box Office Movie

DC MotorDC MotorAPPLICATIONAPPLICATION

Staring:Staring:

ArmatureArmature

Speed & TorqueSpeed & Torque

Special Appearance:Special Appearance:

Variable Speed DrivesVariable Speed Drives

Ok Fine…..Why we should learn two different types of Rectifier?

Uncontrolled Rectifier (Natural Commutating Rectifier)

Controlled Rectifier (Phase Controlled Rectifiers)

Diode

V

t

Constant DC Voltage AC Power Source

Thyristor/SCR

V

t

Variable DC Voltage AC Power Source




5

Important Points which Student has to ConcentrateFlow for This Chapter

Type of Rectifier

Uncontrolled (Part A) Controlled (Part B)

Type of Uncontrolled Rectifier

Half Wave

Full Wave

AC Source

Single Phase

Three Phase

Type of Load

R Load

RL Load

Output Waveform

Average and RMS Values

Rectifier Performance Parameters e.g Efficiency, Form Factor

Type of Controlled Rectifier

Half Wave

Full Wave

AC Source

Single Phase

Three Phase

Type of Load

R Load

RL Load Is there any input

inductance, LS ?

YES/NO

Part A: Uncontrolled Rectifier

Line Frequency Diode Rectifier




5

Animation on How a Diode Works?

How a Diode Works ?

DIODE is ONONLY If

ANODE gets more

positive voltage

Than Cathode

How a Diode Works (cont.)?

Practical Diode

Ideal Diode

+ 0.7V -

+ 0V -

For our explanation and

analysis

We assume An Ideal

Diode




5

A Simple Circuit [Resistive load]

ππππ====ωωωωωωωω

ππππ==== ∫∫∫∫

ππππ rms,S

0rms,Savg,o

V2)t(d)tsin(V2

2

1V

V d or V Out

i

V diode

2

V)t(d)]tsin(V2[

2

1V

rms,S

0

2

rms,Srms,o ====ωωωωωωωωππππ

==== ∫∫∫∫ππππ

Average Output Voltage

r.m.s Output Voltage

A single phase half wave uncontrolled

Rectifier is as shown in figure

below.

Draw the output voltage, Voltage

across Diode and current.

[Show for at least 2 complete cycles.]

Calculate average and rms output

voltage

Given, input power source; 240V,

50Hz and load resistor , 15 Ω ΩΩ Ω .

Exercise 1 Exercise 1 Exercise 1 Exercise 1




5

A Simple Circuit (R-L Load)

Current continues to flows fora while even after the input

voltage has gone negative

[ ] [ ])cos(12

V2tcos2

V2

)t(d)tsin(V22

1)t(d)t(v

2

1V

rms,s

0

rms,s

0 rms,s

2

0 oavg,o

β−π

=ω−π

=

ωωπ

=ωωπ

=

β

βπ

∫∫

2

2

22

2

1

0

2 β−β

π=ωω

π= ∫

β sinV )t ( d )]t sin( V [ V rms , s

rms , srms ,o



A Simple Circuit (R-L Load) [ cont.]




5

A Simple Circuit (R-L Load) [ cont.]

t

t

t1

vL

vdiode

Area A

Area B

when diode conducts

when diode does not

conducts

t

i

vs

vo

vR

vR

vo

i

t1 t2 t3

T

0 t4

ββββvs

peak

Single Phase Half wave

Uncontrolled Rectifier with RL

Load.

Simulation




5

A Simple Circuit (Load has a dc back-emf)

Current begins to

flow when the

input voltage

exceeds the dc

back-emf

Current continues

to flows for a

while even after

the input voltagehas gone below

the dc back-emf

Single-Phase Diode Rectifier Bridge

Large capacitor at the dc output for filtering and energy

storage

Source Inductance

Which affects current

commutation




5

Diode-Rectifier Bridge Analysis

Two simple (idealized) cases to begin with

Waveforms with a purely resistive load and a

purely dc current at the output




5

In both cases, the dc-side voltage waveform is the same

Output voltage and current

Wave forms

Input Voltage and Current

Wave forms

∫∫ π

π==

0

2

0

1

2

1dwt )wt ( vdt )t ( v

/ T V o

T

oavg .o

rms ,S rms ,S

V .V

9022=

π=

∫∫

π

π

==0

22

0

22

1

2

1 )wt ( d )]wt sin( V [ dt )]t ( v[

/ T

V rms ,S

T

orms ,o

rms ,S rms ,S rms ,S

V ][ V

)wt ( d )wt ( sinV

=π

π=

π= ∫

π

2

22

0

2

∫π

π=

02

1 )wt ( d )wt sin( V rms ,S

Single Phase Full Wave Rectifier with Purely Resistive






5

Diode-Rectifier Bridge Input Current

•Idealized case with a purely dc output current

•Problem created to the power source

Mr. Fourier

Harmonics ofInput Current

Effect of DC-Side Current on THD, PF and DPF

• Very high THD at low current values

Bigger LS

So it is important to add

inductance in series at

source




5

Diode-Rectifier Bridge Analysis with AC-Side Inductance

Output current is assumed to be purely dc

Line Inductance

Constant Load Current

(Continuous Conduction)

Understanding Current Commutation

Half Wave Rectifier

Assuming inductance in this circuit to be zero




5

Understanding Current Commutation (cont.)

Inductance in this circuit is included

During the Commutation

After the Current

Commutation is completed

Current Commutation Waveforms

Shows the volt-seconds needed to commutate current




5

Current Commutation in Full-Bridge Rectifier

• Shows the necessary volt-seconds


Note the current loops for analysis

During Current

Commutation




5


[[[[ ]]]]ucos1V

d2

sinV2V m

u

md ++++

ππππ====θθθθ

ππππθθθθ

==== ∫∫∫∫ππππ

(((( )))) (((( )))) (((( ))))

)u2sin2

1

u(

1

1VV

dsinV

dsinV

V

S)rms(d

u

2

2

m

u

2

m2

)rms(d

−−−−ππππ−−−−====

θθθθθθθθππππ

====θθθθππππ

θθθθ==== ∫∫∫∫∫∫∫∫

ππππππππ



A single phase full wave diode bridge rectifier as shown in figure

below is connected to an inductive load of 2H in series with a 10 Ω ΩΩ Ω

resistor from a 240V, 50Hz A.C. mains supply. If the line

inductance is 10mH, calculate ;

a) The average output voltage and current supplied to the load.

Ignore

the voltage drops across the diodes.

b) The commutation overlap angle

c) The output r.m.s voltage.

Exercise 2 Exercise 2 Exercise 2 Exercise 2




5

Rectifier with a dc-side voltage

Part A1 of Chapter 2

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