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Chapter 1
(Part 3)
By
Dr. Ayman Yousef
Single phase
AC voltage controllers
Single-phase full-wave
AC voltage controllers
with R-L load
Power Electronics II
Dr. Ayman Yousef
Single-phase full-wave AC voltage
controllers with R-L load
 The thyristor T1 is forward biased during the positive half cycle of input
supply and conduct when suitable pulse applied on its gate ωt = α.
 The thyristor T2 is forward biased during the negative half cycle of input
supply and conduct when suitable pulse applied on its gate ωt = π + α .
Power Electronics II
Dr. Ayman Yousef
Single-phase full-wave AC voltage
controllers with R-L load
 Due to the inductance in the load, the load
current io flowing through T1 would not
fall to zero at ωt = π , when the input
supply voltage starts to become negative.
 The thyristor T1 will continue to conduct
the load current until all the inductive
energy stored in the load inductor L is
completely utilized and the load current
through T1 falls to zero at ωt = β =
extinction angle.
 The thyristor T1 conducts from ωt = α to β . The conduction angle of T1 is
δ = (β − α ), which depends on the delay angle α and the load impedance
angle φ.
Power Electronics II
Dr. Ayman Yousef
Single-phase full-wave
AC voltage controllers
with R-L load
In case of  > +
 The problem, when (T2) is fired at t = + ,
while (T1) is still conducting due to load
inductance, (T2) cannot be turned on. This will
result in asymmetric output voltage and current
waveforms.
 The solution of this problem, continuous
gating signals with a duration of (-) should
be used .
 Another problem, when the current in (T1) falls
to zero, (T2) would be turned on. But, a
continuous gating signal results in high
switching losses of thyristors and requires a
larger isolating transformer in the gating
circuit.
 In practice, a train of pulses with short
durations, are normally used to overcome these
problems.
Power Electronics II
Dr. Ayman Yousef
Waveforms of single phase full wave ac
voltage controller with RL load
In case of (α < φ)
 The load voltage and current can be
sinusoidal if the firing delay angle (α) is
less than the load angle (φ).
In case of (α > φ)
 If (α) is greater than (), which is
usually the case, the load current would
be discontinuous and non-sinusoidal.
 In discontinuous load current operation
occurs for α > φ and β < (π +α )
i.e., (β −α ) < π , conduction angle < π .
Waveforms of single phase
full wave ac voltage controller
with RL load for α > φ.
Power Electronics II
Dr. Ayman Yousef
Analysis of Single phase full wave
AC voltage controller with inductive load
 Assuming that thyristor (T1) is triggered at ωt =α , The load current which flows
through the thyristor T1 during ωt =α to β can be found from the equation:
 The solution of the above differential equation gives the general expression for
the thyristor (load) current, during this period, is in the form:
Power Electronics II
Dr. Ayman Yousef
Analysis of Single phase full wave
AC voltage controller with inductive load
 The value of the constant A1 can be determined from the initial condition. i.e. initial
value of load current iO = 0 , at ωt = α . Hence from the equation for Io equating io
to zero and substituting ωt = α
 Substituting the value of constant A1 from the above equation into the expression
for io , we obtain
Power Electronics II
Dr. Ayman Yousef
Analysis of Single phase full wave
AC voltage controller with inductive load
 The extinction angle β , depends upon the load inductance and is defined as the
value of ωt at which the load current io falls to zero and T1 is turned off, and it can
be estimated by using the condition that io = 0 , at ωt = β
 The extinction angle β can be determined from this equation by using the iterative
method of solution (trial and error method).
 In the exponential term the value of α and β should be substituted in radians.
 After β is calculated, we can determine the thyristor conduction angle δ.
Power Electronics II
Dr. Ayman Yousef
Analysis of Single phase full wave
AC voltage controller with inductive load
 In case of α > φ, the type of operation will be discontinuous load current, and we
get β < (π +α), and greater than π radian or 180o . Then, the range of extinction
angle β
 Conduction angle δ increases as α is decreased for a known value of β.
 Maximum thyristor conduction angle δ = (β − α ) = π radians = 180o for α ≤ φ .
 After β is calculated, we can determine the thyristor conduction angle δ.
Power Electronics II
Dr. Ayman Yousef
Analysis of Single phase full wave
AC voltage controller with inductive load
Power Electronics II
Dr. Ayman Yousef
Analysis of Single phase full wave
AC voltage controller with inductive load
Power Electronics II
Dr. Ayman Yousef
Ex .3: A single-phase full-wave AC voltage controller supplies an RL
load. The supply rms voltage is 120 V, 60 Hz. The load has R=2.5W and
L=6.5 mH, the firing delay angles of thyristors are equal: α1= α2= π/2.
Solution
 = tan-1 (L/R) = tan-1 (2πx60x6.5x10-3/2.5) = 44.4
 In order to determine the conduction angle δ, firstly, we must find the extinction
angle β from the given equation
Determine:
(a) the conduction angle of the thyristor T1
(b) the rms output voltage
(c) the rms thyristor current
(d) the average thyristor current
(e) the rms output current
(f) the input power.
Power Electronics II
Dr. Ayman Yousef
 By solving this equation numerically using trial and error method , and because
α > φ, the extinction angle () can be determined with the given range
  220
2
1
o 2
2β
sin
2
2α
sin
)
(
π
1
V





















 s
V
v
25
.
90
2
1
o 2
2x220
sin
2
2x90
sin
)
90
220
(
π
1
120
V 




















Power Electronics II
Dr. Ayman Yousef
Power Electronics II
Dr. Ayman Yousef
Harmonic analysis of output
voltage and current with RL load
Power Electronics II
Dr. Ayman Yousef
Harmonic analysis of output voltage
and current with RL load
The fundamental component of load voltage
The nth component of load voltage
n = 3, 5, 7, ………,
Power Electronics II
Dr. Ayman Yousef
Harmonic analysis of output voltage and
current with RL load
The RMS of load voltage from Fourier series
Harmonic analysis of output voltage
and current with RL load
The RMS of the load current obtained from Fourier series
Power Electronics II
Dr. Ayman Yousef
Harmonic analysis of output voltage and
current with RL load
Active power
Reactive power
Power Electronics II
Dr. Ayman Yousef

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Ch1 AC voltage controllers_Part 3 (1).pdf

  • 1. Chapter 1 (Part 3) By Dr. Ayman Yousef Single phase AC voltage controllers
  • 2. Single-phase full-wave AC voltage controllers with R-L load Power Electronics II Dr. Ayman Yousef
  • 3. Single-phase full-wave AC voltage controllers with R-L load  The thyristor T1 is forward biased during the positive half cycle of input supply and conduct when suitable pulse applied on its gate ωt = α.  The thyristor T2 is forward biased during the negative half cycle of input supply and conduct when suitable pulse applied on its gate ωt = π + α . Power Electronics II Dr. Ayman Yousef
  • 4. Single-phase full-wave AC voltage controllers with R-L load  Due to the inductance in the load, the load current io flowing through T1 would not fall to zero at ωt = π , when the input supply voltage starts to become negative.  The thyristor T1 will continue to conduct the load current until all the inductive energy stored in the load inductor L is completely utilized and the load current through T1 falls to zero at ωt = β = extinction angle.  The thyristor T1 conducts from ωt = α to β . The conduction angle of T1 is δ = (β − α ), which depends on the delay angle α and the load impedance angle φ. Power Electronics II Dr. Ayman Yousef
  • 5. Single-phase full-wave AC voltage controllers with R-L load In case of  > +  The problem, when (T2) is fired at t = + , while (T1) is still conducting due to load inductance, (T2) cannot be turned on. This will result in asymmetric output voltage and current waveforms.  The solution of this problem, continuous gating signals with a duration of (-) should be used .  Another problem, when the current in (T1) falls to zero, (T2) would be turned on. But, a continuous gating signal results in high switching losses of thyristors and requires a larger isolating transformer in the gating circuit.  In practice, a train of pulses with short durations, are normally used to overcome these problems. Power Electronics II Dr. Ayman Yousef
  • 6. Waveforms of single phase full wave ac voltage controller with RL load In case of (α < φ)  The load voltage and current can be sinusoidal if the firing delay angle (α) is less than the load angle (φ). In case of (α > φ)  If (α) is greater than (), which is usually the case, the load current would be discontinuous and non-sinusoidal.  In discontinuous load current operation occurs for α > φ and β < (π +α ) i.e., (β −α ) < π , conduction angle < π . Waveforms of single phase full wave ac voltage controller with RL load for α > φ. Power Electronics II Dr. Ayman Yousef
  • 7. Analysis of Single phase full wave AC voltage controller with inductive load  Assuming that thyristor (T1) is triggered at ωt =α , The load current which flows through the thyristor T1 during ωt =α to β can be found from the equation:  The solution of the above differential equation gives the general expression for the thyristor (load) current, during this period, is in the form: Power Electronics II Dr. Ayman Yousef
  • 8. Analysis of Single phase full wave AC voltage controller with inductive load  The value of the constant A1 can be determined from the initial condition. i.e. initial value of load current iO = 0 , at ωt = α . Hence from the equation for Io equating io to zero and substituting ωt = α  Substituting the value of constant A1 from the above equation into the expression for io , we obtain Power Electronics II Dr. Ayman Yousef
  • 9. Analysis of Single phase full wave AC voltage controller with inductive load  The extinction angle β , depends upon the load inductance and is defined as the value of ωt at which the load current io falls to zero and T1 is turned off, and it can be estimated by using the condition that io = 0 , at ωt = β  The extinction angle β can be determined from this equation by using the iterative method of solution (trial and error method).  In the exponential term the value of α and β should be substituted in radians.  After β is calculated, we can determine the thyristor conduction angle δ. Power Electronics II Dr. Ayman Yousef
  • 10. Analysis of Single phase full wave AC voltage controller with inductive load  In case of α > φ, the type of operation will be discontinuous load current, and we get β < (π +α), and greater than π radian or 180o . Then, the range of extinction angle β  Conduction angle δ increases as α is decreased for a known value of β.  Maximum thyristor conduction angle δ = (β − α ) = π radians = 180o for α ≤ φ .  After β is calculated, we can determine the thyristor conduction angle δ. Power Electronics II Dr. Ayman Yousef
  • 11. Analysis of Single phase full wave AC voltage controller with inductive load Power Electronics II Dr. Ayman Yousef
  • 12. Analysis of Single phase full wave AC voltage controller with inductive load Power Electronics II Dr. Ayman Yousef
  • 13. Ex .3: A single-phase full-wave AC voltage controller supplies an RL load. The supply rms voltage is 120 V, 60 Hz. The load has R=2.5W and L=6.5 mH, the firing delay angles of thyristors are equal: α1= α2= π/2. Solution  = tan-1 (L/R) = tan-1 (2πx60x6.5x10-3/2.5) = 44.4  In order to determine the conduction angle δ, firstly, we must find the extinction angle β from the given equation Determine: (a) the conduction angle of the thyristor T1 (b) the rms output voltage (c) the rms thyristor current (d) the average thyristor current (e) the rms output current (f) the input power. Power Electronics II Dr. Ayman Yousef
  • 14.  By solving this equation numerically using trial and error method , and because α > φ, the extinction angle () can be determined with the given range   220 2 1 o 2 2β sin 2 2α sin ) ( π 1 V                       s V v 25 . 90 2 1 o 2 2x220 sin 2 2x90 sin ) 90 220 ( π 1 120 V                      Power Electronics II Dr. Ayman Yousef
  • 15. Power Electronics II Dr. Ayman Yousef
  • 16. Harmonic analysis of output voltage and current with RL load Power Electronics II Dr. Ayman Yousef
  • 17. Harmonic analysis of output voltage and current with RL load The fundamental component of load voltage The nth component of load voltage n = 3, 5, 7, ………, Power Electronics II Dr. Ayman Yousef
  • 18. Harmonic analysis of output voltage and current with RL load The RMS of load voltage from Fourier series
  • 19. Harmonic analysis of output voltage and current with RL load The RMS of the load current obtained from Fourier series Power Electronics II Dr. Ayman Yousef
  • 20. Harmonic analysis of output voltage and current with RL load Active power Reactive power Power Electronics II Dr. Ayman Yousef