Lecture 9 Dr Oday A Ahmed ACAC Voltage Controller Converter Cycloconverter Indirect Matrix Converter direct Matrix Converter Or AC Choppers ACAC Voltage Controller Converter There are two different types of thyristor control used in practice to control the ac power ID: 716901
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Slide1
AC-ACConverters
Module II
Lecture 9Slide2Slide3
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)
Slide10Slide11Slide12
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 harmonicsSlide13Slide14
SolutionSlide15Slide16Slide17Slide18Slide19
Single Phase Full Wave Ac ChopperSlide20
Dr. Oday A. AhmedSlide21Slide22Slide23Slide24Slide25
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.Slide29Slide30
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.Slide32Slide33Slide34Slide35
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. Slide36Slide37
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
Slide46Slide47
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
Slide53
Dr. Oday A. Ahmed
Solution Continue for aSlide54
Dr. Oday A. Ahmed
We can note above that at
β
=197.42 the LHS
RHS
Solution Continue for aSlide55Slide56