Power Electronics
Single Phase AC-AC Converters
Dr. Firas Obeidat
Table of contents
1 • Introduction
2 • The Single Phase AC Voltage Controller
3
• The Single Phase AC Voltage Controller - Resistive Load
4 • The Single Phase AC Voltage Controller - RL Load
Introduction
An ac voltage controller is a converter that controls the voltage, current, and average power delivered to an ac load from an ac source.
The phase-controlled ac voltage controller has several practical uses including light-dimmer circuits and speed control of induction motors.
In a switching scheme called phase control, switching takes place during every cycle of the source, in effect removing some of the source waveform before it reaches the load.
Integral-cycle control, the source is connected and
disconnected for several cycles at a time.
The Single Phase AC Voltage Controller
Basic Operation
For the single phase AC voltage controller shown, electronic switches are shown as parallel thyristors (SCRs). This SCR arrangement makes it possible to have current in either direction in the load. This SCR connection is called antiparallel or inverse parallel because the SCRs carry current in opposite directions. A triac is equivalent to the antiparallel SCRs.
Other controlled switching devices can be used instead of SCRs.
Load current contains both positive and negative half-cycles. An analysis identical to that done for the controlled half-wave rectifier can be done on a half cycle for the voltage controller. Then, by symmetry, the result can be extrapolated to describe the operation for the entire period.
S1 conducts if a gate signal is applied during the positive half-cycle of the source. S1 conducts until the current in it reaches zero.
A gate signal is applied to S2 during the negative half-cycle of the source, providing a path for negative load current.
vs
L o a d
+ vsw -
S1
S2
+
-
+
- io
vo
The Single Phase AC Voltage Controller
Basic Operation
Basic observations about this controller
The SCRs cannot conduct simultaneously.
The load voltage is the same as the source voltage when either SCR is on. The load voltage is zero when both SCRs are OFF.
The switch voltage vsw is zero when either SCR is ON and is equal to the source voltage when neither is ON.
The average current in the source and load is zero if the SCRs are on for equal time intervals. The average current in each SCR is not zero because of unidirectional SCR current.
The rms current in each SCR is 1/√2 times the rms load current if the SCRs are on for equal time intervals.
vs
+ vsw -
S1
S2 +
-
+
- io
vo
R
The Single Phase AC Voltage Controller - Resistive Load
𝑣𝑠 𝜔𝑡 = 𝑉𝑚𝑠𝑖𝑛𝜔𝑡
𝑣𝑠 𝜔𝑡 = 𝑉𝑚𝑠𝑖𝑛𝜔𝑡 𝛼 < 𝜔𝑡 < 𝜋 𝑎𝑛𝑑 𝜋 + 𝛼 < 𝜔𝑡 < 2𝜋
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Let the voltage source be
Output voltage is
The rms load voltage is
𝑉𝑜,𝑟𝑚𝑠 = 1
𝜋 (𝑉𝑚𝑠𝑖𝑛𝜔𝑡)2 𝑑𝜔𝑡
𝜋
α
= 𝑉𝑚
2 1 −𝛼
𝜋 + 𝑠𝑖𝑛2𝛼 2𝜋
The Single Phase AC Voltage Controller - Resistive Load
The power factor of the load is
The rms current in the load and the source is 𝐼𝑜,𝑟𝑚𝑠 = 𝑉𝑜,𝑟𝑚𝑠
𝑅 = 𝑉𝑚
𝑅 2 1 − 𝛼
𝜋 + 𝑠𝑖𝑛2𝛼 2𝜋
𝑝𝑓 = 𝑃
𝑆 = 𝑉𝑜,𝑟𝑚𝑠2 𝑅
𝑉𝑠,𝑟𝑚𝑠𝐼𝑠,𝑟𝑚𝑠 = 𝑉𝑜,𝑟𝑚𝑠2 𝑅
𝑉𝑠,𝑟𝑚𝑠(𝑉𝑜,𝑟𝑚𝑠 )𝑅 = 𝑉𝑜,𝑟𝑚𝑠 𝑉𝑠,𝑟𝑚𝑠 =
𝑉𝑚
2 1 − 𝛼𝜋 + 𝑠𝑖𝑛2𝛼 2𝜋 𝑉𝑚 2
𝑝𝑓 = 1 − 𝛼
𝜋 + 𝑠𝑖𝑛2𝛼 2𝜋
The pf=1 for α=0, which is the same as for an uncontrolled resistive load, and the power factor for α>0 is less than 1.
The average source current is zero because of half-wave symmetry.
𝐼𝑠,𝐴𝑣𝑔 = 𝐼𝑜,𝐴𝑣𝑔 = 0
The Single Phase AC Voltage Controller - Resistive Load
The average SCR current is 𝐼𝑆𝐶𝑅,𝐴𝑣𝑔 = 1
2𝜋
𝑉𝑚𝑠𝑖𝑛𝜔𝑡
𝑅 𝑑𝜔𝑡
𝜋
𝛼
= 𝑉𝑚
2𝜋𝑅 (1 + 𝑐𝑜𝑠𝛼)
𝐼𝑆𝐶𝑅,𝑟𝑚𝑠 = 1
2𝜋 (𝑉𝑚𝑠𝑖𝑛𝜔𝑡
𝑅 )2 𝑑𝜔𝑡
𝜋
α
= 𝑉𝑚
2𝑅 1 − 𝛼
𝜋 + 𝑠𝑖𝑛2𝛼 2𝜋 The rms SCR current is
𝐼𝑆𝐶𝑅,𝑟𝑚𝑠 = 𝑉𝑚
2 2𝑅 1 − 𝛼
𝜋 + 𝑠𝑖𝑛2𝛼
2𝜋 = 𝐼𝑜,𝑟𝑚𝑠 2
π
α 2π 2π+α 3π ωt
iS1
vs/R
Example: The single-phase ac voltage controller has a 120-V rms 60-Hz source.
The load resistance is 15 Ω. Determine (a) the delay angle required to deliver 500 W to the load, (b) the rms source current, (c) the rms and average currents in the SCRs, (d) the power factor.
𝑉𝑜,𝑟𝑚𝑠 = 𝑉𝑚
2 1 − 𝛼
𝜋 + 𝑠𝑖𝑛2𝛼
2𝜋 = 2 × 120
2 1 − 𝛼
𝜋 + 𝑠𝑖𝑛2𝛼 2𝜋 𝑃 = 𝑉𝑜,𝑟𝑚𝑠2
𝑅 → 𝑉𝑜,𝑟𝑚𝑠2 = 𝑃𝑅 = 500 × 15 = 7500 𝑉𝑜,𝑟𝑚𝑠2 = 1202(1 − 𝛼
𝜋 + 𝑠𝑖𝑛2𝛼 2𝜋 )
∴ 7500 = 14400(1 − 𝛼
𝜋 + 𝑠𝑖𝑛2𝛼 2𝜋 ) 𝛼
𝜋 − 𝑠𝑖𝑛2𝛼
2𝜋 = 0.479 2𝛼 − 𝑠𝑖𝑛2𝛼 = 3.01
𝛼 = 1.54 rad = 88.1𝑜 (a)
The Single Phase AC Voltage Controller - Resistive Load
The Single Phase AC Voltage Controller - Resistive Load
(b)
𝐼𝑜,𝑟𝑚𝑠 = 𝑉𝑜,𝑟𝑚𝑠
𝑅 = 86.6
15 = 5.77 A
𝑉𝑜,𝑟𝑚𝑠2 = 7500 → 𝑉𝑜,𝑟𝑚𝑠 = 86.6 From (a)
𝐼𝑆𝐶𝑅,𝐴𝑣𝑔 = 𝑉𝑚
2𝜋𝑅 1 + 𝑐𝑜𝑠𝛼 = 2 × 120
2𝜋15 1 + 𝑐𝑜𝑠88.1 = 1.86 A
𝐼𝑆𝐶𝑅,𝑟𝑚𝑠 = 𝐼𝑜,𝑟𝑚𝑠
2 = 5.77
2 = 4.08 A (c)
𝑝𝑓 = 𝑃
𝑆 = 𝑃
𝑉𝑠,𝑟𝑚𝑠𝐼𝑠,𝑟𝑚𝑠 = 1 − 𝛼
𝜋 + 𝑠𝑖𝑛2𝛼
2𝜋 = 500
120 × 5.77 = 0.722 (d)
The Single Phase AC Voltage Controller - RL Load
vs
+ vsw -
S1
S2 +
-
+
- io
vo
R
L
When a gate signal is applied to S1 at 𝜔t=α in single phase AC voltage controller with RL load, Kirchhoff’s voltage law for the circuit is expressed as.
𝑉𝑚𝑠𝑖𝑛𝜔𝑡 = 𝑅𝑖𝑜 𝑡 + 𝐿𝑑𝑖𝑜 𝑡 𝑑𝑡
The solution for current in this equation (as obtained in the controlled half wave rectifier section) is
𝑖𝑜 𝜔𝑡 = 𝑉𝑚
𝑍 sin 𝜔𝑡 − 𝜃 − sin (𝛼 − 𝜃)𝑒(𝛼−𝜔𝑡) 𝜔𝜏 𝛼 ≤ 𝜔𝑡 ≤ 𝛽
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
where
𝑍 = 𝑅2 + (𝜔𝐿)2 𝜃 = 𝑡𝑎𝑛−1(𝜔𝐿 𝑅 )
The extinction angle 𝛽 is the angle at which the current returns to zero, when 𝜔𝑡= 𝛽,
𝑖𝑜 𝛽 = 𝑉𝑚
𝑍 sin 𝛽 − 𝜃 − sin (𝛼 − 𝜃)𝑒(𝛼−𝛽) 𝜔𝜏
(1)
(2)
The Single Phase AC Voltage Controller - RL Load
A gate signal is applied to S2 at 𝜔t=π+α, and the load current is negative but has a form identical to that of the positive half- cycle.
The above equation must be solved numerically for 𝛽.
The angle (𝛽-α) is called the conduction angle 𝛶.
𝛶 =𝛽-α
In the interval between π and 𝛽 when the source voltage is negative and the load current is still positive, S2 cannot be turned on because it is not forward biased.
The gate signal to S2 must be delayed at least until the current in S1 reaches zero, at 𝜔t= 𝛽. The delay angle is therefore at least 𝛽- π.
α≥𝛽- π
The Single Phase AC Voltage Controller - RL Load
When α=θ, equation (2) becomes sin 𝛽 − 𝜃 = 0
which has a solution 𝛽 − 𝜃 = π
Therefore
𝛶 = 𝜋 When α=θ
When 𝛶=𝜋, one SCR is always conducting, and the voltage across the load is the same as the voltage of the source. The load voltage and current are sinusoids for this case, and the circuit is analyzed using phasor analysis for AC circuits. The power delivered to the load is continuously controllable between the two extremes corresponding to full source voltage and zero.
The expression of rms load current is
𝐼𝑜,𝑟𝑚𝑠 = 1
𝜋 𝑖𝑜2(𝜔𝑡) 𝑑𝜔𝑡
𝛽
α
= 1 𝜋
𝑉𝑚
𝑍 sin 𝜔𝑡 − 𝜃 −𝑉𝑚
𝑍 sin (𝛼 − 𝜃)𝑒(𝛼−𝜔𝑡) 𝜔𝜏
2
𝑑𝜔𝑡
𝛽
α
The Single Phase AC Voltage Controller - RL Load
The average output voltage can be found as
An other way to find the expression of rms load current is
𝐼𝑜,𝑟𝑚𝑠 = 1
𝑅2 + (𝜔𝐿)2 𝑉𝑚
2 1
𝜋(𝛽 − 𝛼 − 1
2𝑠𝑖𝑛2𝛽 + 1
2𝑠𝑖𝑛2𝛼) Power absorbed by the load is determined from
𝑃 = 𝐼𝑜,𝑟𝑚𝑠2𝑅
The power factor of the load is 𝑝𝑓 = 𝑃
𝑆 = 𝑉𝑜,𝑟𝑚𝑠𝐼𝑜,𝑟𝑚𝑠
𝑉𝑠,𝑟𝑚𝑠𝐼𝑠,𝑟𝑚𝑠 = 𝑉𝑜,𝑟𝑚𝑠
𝑉𝑠,𝑟𝑚𝑠 = 1
𝜋(𝛽 − 𝛼 − 1
2𝑠𝑖𝑛2𝛽 + 1
2𝑠𝑖𝑛2𝛼) 𝑉𝑜,𝑟𝑚𝑠 = 1
𝜋 (𝑉𝑚𝑠𝑖𝑛𝜔𝑡)2 𝑑𝜔𝑡
𝛽
α
= 𝑉𝑚 2
1
𝜋(𝛽 − 𝛼 − 1
2𝑠𝑖𝑛2𝛽 + 1
2𝑠𝑖𝑛2𝛼)
The Single Phase AC Voltage Controller - RL Load
Each SCR carries one-half of the current waveform, making the average SCR current
𝐼𝑆𝐶𝑅,𝐴𝑣𝑔 = 1
2𝜋 𝑖𝑜(𝜔𝑡) 𝑑𝜔𝑡
𝛽
α
= 1 2𝜋
𝑉𝑚
𝑍 sin 𝜔𝑡 − 𝜃 − sin (𝛼 − 𝜃)𝑒(𝛼−𝜔𝑡) 𝜔𝜏 𝑑𝜔𝑡
𝛽
α
The rms current in each SCR is 𝐼𝑆𝐶𝑅,𝑟𝑚𝑠 = 𝐼𝑜,𝑟𝑚𝑠
2
The average load current is zero, 𝐼𝑜,𝐴𝑣𝑔 = 0
Example: For the single-phase voltage controller with RL load, the source is 120Vrms at 60 Hz, and the load is a series RL combination with R=20Ω and L=50mH. The delay angle α is 90. Determine (a) an expression for load current for the first half-period, (b) the rms load current, (c) the rms SCR current, (d) the average SCR current, (e) the power delivered to the load, and (f) the power factor.
(a)
The Single Phase AC Voltage Controller - RL Load
The extinction angle is determined from the numerical solution of i(β)=0 in the above equation.
The Single Phase AC Voltage Controller - RL Load
(b)
(c)
(d) (e) (f)
𝐼𝑜,𝑟𝑚𝑠 = 1
27.5 × 120 × 1
𝜋(3.83 − 1.57 − 1
2𝑠𝑖𝑛440 + 0) = 3.27 A 𝐼𝑜,𝑟𝑚𝑠 = 1
𝑅2 + (𝜔𝐿)2 𝑉𝑚
2 1
𝜋(𝛽 − 𝛼 − 1
2𝑠𝑖𝑛2𝛽 − 1
2𝑠𝑖𝑛2𝛼) Or
