AC-AC Converters Module II - PowerPoint Presentation

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AC-ACConverters

Module II

Lecture 9Slide2
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Dr. Oday A. Ahmed

AC-AC Voltage Controller Converter

Cycloconverter

Indirect Matrix Converter

direct Matrix ConverterSlide4

Or AC Choppers

AC-AC Voltage Controller Converter

There are two different types of thyristor control used in practice to control the ac power

flow

 On-Off control

 Phase control

Only Phase control will be considered in this lecture.

The ac voltage controllers are classified:

Single Phase AC Controllers.

Three Phase AC Controllers.

Uni

-directional or half wave ac controller

Bi-directional or full wave ac controller.Slide5

Applications

of Ac Choppers

Lighting / Illumination control in ac power circuits. Slide6

Induction heating.

Applications

of Ac Choppers

Electromagnetic induction is unique because it actually generates heat inside the material that is heated, has an immediate effect.

Compared to other heating techniques, it takes less time to heat and is more efficient and accurate.Slide7

Transformer tap changing

Applications

of Ac Choppers

On load tap changers

Tap changers offer variable control to maintain the supply voltage between certain limits. About 96% of all power transformers today above 10MVA incorporate on load tap changers as a means of voltage regulation.Slide8

Speed control of induction motors

Applications

of Ac ChoppersSlide9

Half wave AC phase controller (Unidirectional Controller)

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the half wave ac controller has the drawback of limited range RMS output voltage control.

can result in the problem of core saturation of the input supply transformer.

Ac power flow to the load can be controlled only in one half cycle.

gives limited range of RMS output voltage control

the half wave ac controller has input DC component and even order harmonicsSlide13
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SolutionSlide15
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Single Phase Full Wave Ac ChopperSlide20

Dr. Oday A. AhmedSlide21
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We can notice from the figure, that we obtain a much better output control

characteristic by using a single phase full wave ac voltage controller.

No input DC component and No even order harmonics , WHY?Slide26

Performance ParametersSlide27

Performance ParametersSlide28

Problem

What is the Average current through Triac in the following circuit?

Solution

In the case of a single phase full wave ac voltage controller circuit using a Triac with

resistive load, the average current

=0. Because the Triac conducts in both the half cycles and the current is alternating and we obtain a symmetrical thyristor current waveform which gives an average value of zero on integration.Slide29
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Dr. Oday A. Ahmed

single phase full wave ac controller with a common cathode configurationSlide31

due to the need of two power diodes the costs of the devices increase.

there are two power devices conducting at the same time the voltage drop across the ON devices increases and the ON state conducting losses of devices increase and hence the efficiency decreases.Slide32
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if there is a large inductance in the load circuit, thyristor T1 may not be turned OFF at the zero crossing points, in every half cycle of input voltage and this may result in a loss of output control.

This would require detection of the zero crossing of the load current waveform in order to ensure guaranteed turn off of the conducting thyristor before triggering the thyristor in the next half cycle, so that we gain control on the output voltage.

three power devices conducting together at the same time there is more conduction voltage drop and an increase in the ON state conduction losses and hence efficiency is also reduced. Slide36
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Dr. Oday A. Ahmed

Phase control

V

o

=V

s

V

o

=-V

s

V

o

=0Slide38

T

1

conducts from

ωt

=α to

ωt

=β,

T

1

then conduction angle δ=( β- α).

δ depends on α and the load impedance angle ϕSlide39

for α>ϕ and β< π+ α

ϕ

Discontinuous load current operation occurs Slide40

for α>ϕ and β< π+ αSlide41

For very large load inductance ‘L’ the SCR may fail to commutate

and the load voltage will be a full sine wave

for

α

ϕ

A continuous load current and the output voltage waveform appears as a continuous sine wave identical to the input supply voltage waveform

We lose the control on the output voltage and thus we obtain:Slide42

The load current which flows through the thyristor during

ωt

=α to β

Output Current for (Inductive Load)

The solution of the above differential equation gives the general expression for the output load current which is of the formSlide43

At

ωt

=α,

i

o

=0Amp, hence,

By substituting

ωt

=α: then,

which results in the instantaneous output current equal to:Slide44

Calculate Extinction Angle β

At

ωt

=β,

i

o

=0Amp, hence,Slide45

β can be determined from this transcendental equation by using

the iterative method of soluti

on (

trial and error method

).

For δ<π, for β< π+ α the load current waveform appears as a discontinuous current waveform.

When α is decreased and made equal to the load impedance angle α

ϕ, we obtain from the expression for

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Performance ParametersSlide48

Load CurrentSlide49

Dr. Oday A. Ahmed

EXAMPLE

SolutionSlide50

Dr. Oday A. Ahmed

Solution Continue for aSlide51

Dr. Oday A. Ahmed

Solution Continue for a

We will use trail and error to find the solution:Slide52

Dr. Oday A. Ahmed

Solution Continue for a

To find the right angle for

β

the LHS must be = RHS of following equation:

As shown above for

β

=180 the LHS

RHS, then we will select another angle for

β

until find solution

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Dr. Oday A. Ahmed

Solution Continue for aSlide54

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We can note above that at

β

=197.42 the LHS

RHS

Solution Continue for aSlide55
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