Cycloconverter
Single-phase
Cycloconverter
Rectifier: From AC
Variable DC voltage.
Chopper: From constant DC voltage
Inverter: From DC
variable DC voltage.
variable magnitude and variable frequency
AC voltage.
There are ways to convert from AC
AC
• Two stages: AC to DC and then DC to AC.
• One stage: cycloconverter
Input: AC
(Constant V & f)
Cycloconverter
AC - AC
Output: AC
Adjustable Voltage with
Lower frequency
The ac power at one frequency is converted directly to another frequency generally
lower than the input supply frequency without any intermediate dc stage.
Applications: Cement mill drives, Ship propulsion drives, Rolling mill drives,
starting synchronous motors, etc.
Construction
A typical cycloconverter consists of one or more pairs of back-to-back connected controlled
rectifiers. By the controlled opening and closing of the thyristors, it is possible to fabricate
output voltage waveforms having a fundamental component of the desired output frequency.
Negative or N Converter:
generate negative cycle
Positive or P Converter:
generate positive cycle
Thyristor
𝑷𝟏
𝑷𝟑
𝑵𝟏
𝑵𝟑
𝑵𝟒
𝑵𝟐
L 𝑽𝑳𝒐𝒂𝒅
O
A
D
𝑽𝒎 𝒔𝒊𝒏𝝎𝒕
𝑷𝟒
𝑷𝟐
Cycloconverter types based on
the input and output phases
Single-phase to
single-phase
Bridge
Center tap
Three-phase to
Three-phase
Three
pulse
bridge
Six
pulse
bridge
Three-phase to
single-phase
Three
pulse
Six
pulse
Required
Operation Method
Magnitude of 𝑽𝒊𝒏𝒑𝒖𝒕 = Magnitude of 𝑽𝒐𝒖𝒕𝒑𝒖𝒕
𝒏=
Firing angle 𝜶𝒑 = 𝟎
𝒇𝒊𝒏
𝒇𝒐
𝒇𝒐𝒖𝒕 = 𝟎. 𝟓 𝒇𝒊𝒏𝒑𝒖𝒕
𝒏=𝟐
n is the number of half supply cycle. It is integer.
Positive or P Converter
(working)
Negative or N Converter
(disabled or not working)
𝑷𝟏 & 𝑷𝟐 𝑶𝑵
𝑷𝟏
𝑷𝟑
L
O
A
D
𝑷𝟒
𝑷𝟐
For the first one cycle
and negative parts)
𝑵𝟑
𝑵(positive
𝟏
of
input voltage,
the
positive Converter operates
supplying current to the
load. In the next cycle, the
negative converter operates
supplying current to the load
in the reverse direction.
𝑵𝟒
𝑵𝟐
Operation Method
Note that when one of the converters operates
the other one is disabled, there is no current
circulating between the two rectifiers.
Positive or P Converter
(Working)
Negative or N Converter
(disabled or not working)
𝑷𝟑 & 𝑷𝟒 𝑶𝑵
𝑷𝟏
𝑷𝟑
L
O
A
D
𝑷𝟒
𝑷𝟐
𝑵𝟏
𝑵𝟑
𝑵𝟒
𝑵𝟐
Operation Method
𝑵𝟏 & 𝑵𝟐 𝑶𝑵
Positive or P Converter
(disabled or not working)
𝑷𝟏
Negative or N Converter
(working)
𝑷𝟑
L
O
A
D
𝑷𝟒
𝑷𝟐
𝑵𝟏
𝑵𝟑
𝑵𝟒
𝑵𝟐
Operation Method
𝑵𝟑 & 𝑵𝟒 𝑶𝑵
Positive or P Converter
(disabled or not working)
𝑷𝟏
Negative or N Converter
(working)
𝑷𝟑
L
O
A
D
𝑷𝟒
𝑷𝟐
𝑵𝟏
𝑵𝟑
𝑵𝟒
𝑵𝟐
𝒄𝒖𝒓𝒓𝒆𝒏𝒕
𝒇
𝟓𝟎
N = 𝒇𝒊𝒏 = 𝟐𝟓 = 𝟐
𝒐
𝒄𝒖𝒓𝒓𝒆𝒏𝒕
Disadvantages:
N = 1,2,3,…
𝒇
𝟓𝟎
N = 𝒇𝒊𝒏 = 𝟏𝟎 = 𝟓
𝒐
Single-phase to single-phase: Bridge
R- Load
Single-phase to single-phase: Bridge
𝑽𝒐 (𝒓𝒎𝒔) =
𝟐𝑵 𝝅
න 𝑽𝒎 𝒔𝒊𝒏 𝒘𝒕 𝟐 𝒅𝒘𝒕
𝟐𝝅𝑵 𝜶𝑷
𝜶𝒑
=
=
𝑉𝑂(𝑟𝑚𝑠)
𝑽𝒎
𝝅
𝑽𝒎
𝒔𝒊𝒏 𝟐 𝝅
𝝅−
𝟐
𝟏
𝒔𝒊𝒏 𝟐 𝝎𝒕
𝝎𝒕 −
𝝅
𝟐
𝟐
𝜶𝒑
𝟏
𝟐 𝝅
Zero
−
𝜋
𝒔𝒊𝒏 𝟐𝜶𝒑
𝜶𝒑 −
𝟐
𝑽𝒐 (𝒓𝒎𝒔)
𝑰𝑶 (𝒓𝒎𝒔) =
𝑹
𝑷𝒐𝒖𝒕 = 𝑰𝟐𝒐(𝒓𝒎𝒔) ∗ 𝑹
Single-phase to single-phase: Bridge
Inductive load
RL - Load
𝜷 𝒂𝒏𝒈𝒍𝒆 𝒊𝒔 𝒏𝒐𝒕 𝒈𝒊𝒗𝒆𝒏
𝑽𝒐 (𝒓𝒎𝒔) =
𝜷 𝒂𝒏𝒈𝒍𝒆 𝒊𝒔 𝒈𝒊𝒗𝒆𝒏
𝒔𝒊𝒏 𝟐𝜶𝒑
𝟏
𝝅 − 𝜶𝒑 +
𝝅
𝟐
𝟐
𝑽𝒎
𝛽
𝛽
𝜶𝒑
=
𝜷
𝑽𝒐 (𝒓𝒎𝒔) =
𝑽𝒎
𝟏
𝟐 𝝅
𝑽𝒎
𝜷
𝟏
𝒔𝒊𝒏 𝟐 𝝎𝒕
𝝎𝒕 −
𝝅
𝟐
𝟐
𝜶𝒑
𝒔𝒊𝒏 𝟐 𝜷
𝜷−
𝟐
−
𝒔𝒊𝒏 𝟐𝜶𝒑
𝜶𝒑 −
𝟐
Single-phase to single-phase: Bridge
Inductive load
RL - Load
𝑰𝒐 (𝒓𝒎𝒔) =
The rms value of the load current
𝑹𝟐 + 𝑿𝟐𝑳
𝑰𝒊𝒏 (𝒓𝒎𝒔) = 𝑰𝒐 (𝒓𝒎𝒔) =
The converter current
𝑰𝒐 (𝒓𝒎𝒔) =
𝑽𝒐(𝒓𝒎𝒔)
𝑰𝟐𝒑 (𝒓𝒎𝒔) + 𝑰𝟐𝑵 (𝒓𝒎𝒔)
𝑰𝒑 (𝒓𝒎𝒔) = 𝑰𝑵 (𝒓𝒎𝒔)
𝟐 ∗ 𝑰𝟐𝑵 (𝒓𝒎𝒔) 𝑰𝒐 (𝒓𝒎𝒔) = 𝟐 𝑰𝑵 (𝒓𝒎𝒔) 𝑰𝑵 (𝒓𝒎𝒔) =
𝑰𝑵 (𝒓𝒎𝒔) = 𝑰𝑷 (𝒓𝒎𝒔)
𝑰𝑵 (𝒓𝒎𝒔) =
Thyristor current
𝟐 ∗ 𝑰𝟐𝒕𝒉𝒚.
𝑰𝒐(𝒓𝒎𝒔)
𝟐
Output power
𝟐
𝑰𝟐𝒕𝒉𝒚.𝑵𝟏 (𝒓𝒎𝒔) + 𝑰𝟐𝒕𝒉𝒚.𝑵𝟑 (𝒓𝒎𝒔)
𝑰𝒕𝒉𝒚.𝑵𝟏 = 𝑰𝒕𝒉𝒚.𝑵𝟑 = 𝑰𝒕𝒉𝒚.
𝑰𝑵 (𝒓𝒎𝒔) =
𝑰𝒐(𝒓𝒎𝒔)
= 𝟐 𝑰𝒕𝒉𝒚.
𝑷𝒐𝒖𝒕 = 𝑽𝒐(𝒓𝒎𝒔) ∗ 𝑰𝒐 𝒓𝒎𝒔 ∗ 𝒄𝒐𝒔 ∅
𝑰𝒐(𝒓𝒎𝒔)
𝑰𝒕𝒉𝒚. =
𝟐
Single-phase to single-phase: Bridge (MATLAB: 4 Pulse gene.)
Periodic time of
the output wave:
𝟏 Τ 𝒇𝒐
Firing angle 𝜶𝒑 = 𝟎
Single-phase to single-phase: Bridge (MATLAB: 4 Pulse gene.)
Firing angle 𝜶𝒑 = 𝟎
𝒄𝒖𝒓𝒓𝒆𝒏𝒕
𝒄𝒖𝒓𝒓𝒆𝒏𝒕
𝒄𝒖𝒓𝒓𝒆𝒏𝒕
𝒄𝒖𝒓𝒓𝒆𝒏𝒕
Periodic = 1/𝒇𝟎 = 1/12.5 = 0.08
𝒇
𝟓𝟎
N = 𝒊𝒏 =
= 𝟒 (even number)
𝒇
𝟓𝟎
N = 𝒇𝒊𝒏 = 𝟐𝟓 = 𝟐
𝒐
(even number)
𝒇𝒐
𝟏𝟐.𝟓
Phase delay P12 = 0 Phase delay P34 = 0.01
Phase delay N12 = 0.04 Phase delay N34 = 0.05
Single-phase to single-phase: Bridge (MATLAB: 4 Pulse gene.)
Pulse is on
Output = 1
Single-phase to single-phase: Bridge (MATLAB: 4 Pulse gene.)
Single-phase to single-phase: Bridge (MATLAB: 4 Pulse gene.)
Firing angle 𝜶𝒑 = 𝟎
Periodic = 1/𝒇𝟎
= 1/10 = 0.1
𝒇𝒊𝒏
𝟓𝟎
N = 𝒇 = 𝟏𝟎 = 𝟓 (𝒐𝒅𝒅 𝒏𝒖𝒎𝒃𝒆𝒓)
𝒐
Phase delay P12 = 0
Phase delay P34 = 0.01
Phase delay N12 = 0.06
Phase delay N34 = 0.05
Single-phase to single-phase: Bridge (MATLAB: 4 Pulse gene.)
Periodic = 1/𝒇𝟎
= 1/10 = 0.1
𝒇
𝟓𝟎
N = 𝒇𝒊𝒏 = 𝟏𝟎 = 𝟓 (𝒐𝒅𝒅 𝒏𝒖𝒎𝒃𝒆𝒓)
𝒐
Phase delay P12 = 0
Phase delay P34 = 0.01
Phase delay N12 = 0.05
Phase delay N34 = 0.06
Single-phase to single-phase: Bridge (MATLAB: 2 Pulse gene.)
Single-phase to single-phase: Bridge (MATLAB: 2 Pulse gene.)
Single-phase to single-phase: Bridge (MATLAB: 2 Pulse gene.)
Single-phase to single-phase: Bridge (MATLAB: 2 Pulse gene.)
Single-phase to single-phase: Bridge (MATLAB: 2 Pulse gene.)
Single-phase to single-phase: Bridge (MATLAB: 3 Pulse gene.)
𝑷𝑵𝟏𝟐
𝑷𝑵𝟑𝟒
𝟏
𝒇𝟎
𝑷𝟓
Single-phase to single-phase: Bridge (MATLAB: 3 Pulse gene.)
Periodic = 1/𝒇𝟎
= 1/10 = 0.1
𝒇𝒊𝒏
N= 𝒇 =
𝒐
𝟓𝟎
= 𝟓 (𝒐𝒅𝒅 𝒏𝒖𝒎𝒃𝒆𝒓)
𝟏𝟎
Disadvantages:
N = 1,2,3,…
Single-phase to single-phase: Bridge (MATLAB: 3 Pulse gene.)
Periodic = 1/𝒇𝟎
= 1/10 = 0.1
𝒇𝒊𝒏
N= 𝒇 =
𝒐
𝟓𝟎
= 𝟓 (𝒐𝒅𝒅 𝒏𝒖𝒎𝒃𝒆𝒓)
𝟏𝟎
Disadvantages:
N = 1,2,3,…
Single-phase to single-phase:
Center tap
Single-phase to single-phase: Center tap (step down)
𝒇𝒊𝒏 = 𝟓𝟎 𝑯𝒛, 𝒇𝒐𝒖𝒕 = 𝟐𝟓 𝑯𝒛
𝑭𝒓𝒐𝒎 𝟎 → 𝝅 𝑷𝟏 𝐢𝐬 𝐨𝐧
𝑷𝟏
𝒇𝒓𝒐𝒎 𝝅 → 𝟐𝝅 𝑷𝟐 𝒊𝒔 𝒐𝒏
𝑷𝟏
𝑷𝟐
Single-phase to single-phase: Center tap (step down)
𝑵𝟐
𝑭𝒓𝒐𝒎 𝟐𝝅 𝒕𝒐 𝟑𝝅 → 𝑵𝟐
𝑭𝒓𝒐𝒎 𝟑𝝅 𝒕𝒐 𝟒𝝅 → 𝑵𝟏
𝑵𝟐
𝑵𝟏
Single-phase to single-phase: Center tap (step down)
Single-phase to single-phase: Center tap (step down)
𝑷𝟏 → 𝑷𝟐 → 𝑵𝟐 → 𝑵𝟏
𝑷𝟏
𝑷𝟐
𝑵𝟐
𝑵𝟏
𝑷𝟑
Single-phase to single-phase: Center tap (step up)
𝒇𝒊𝒏 = 𝟓𝟎 𝑯𝒛,
𝑨𝒕 𝒛𝒆𝒓𝒐 → 𝑷𝟏
𝑨𝒕 𝝅Τ𝟐 → 𝑵𝟐
𝒇𝒐𝒖𝒕 = 𝟏𝟎𝟎 𝑯𝒛
𝑷𝟏
𝑷𝟏
𝑵𝟐
Single-phase to single-phase: Center tap (step up)
𝒇𝒊𝒏 = 𝟓𝟎 𝑯𝒛,
𝑷𝟏
𝑨𝒕 𝝅 → 𝑷𝟐
𝑨𝒕 𝟑𝝅Τ𝟐 → 𝑵𝟏
𝒇𝒐𝒖𝒕 = 𝟏𝟎𝟎 𝑯𝒛
𝑷𝟏
𝑵𝟐
𝑵𝟐
𝑷𝟐
𝑷𝟐 𝑵𝟏
Single-phase to single-phase: Center tap (step up)
Single-phase to single-phase: Center tap (step up)
𝑷𝟏 → 𝑵𝟐 → 𝑷𝟐 → 𝑵𝟏
𝑷𝟏
𝑵𝟐
𝑷𝟐
𝑵𝟏
𝑷𝟑
Single-phase to single-phase: Problems
(Center tap)
Solution
Single-phase to single-phase: Problems
(a)
𝟏
𝒇𝒐𝒖𝒕
𝟏
= 𝟐𝟓 = 𝟎. 𝟎𝟒
Single-phase to single-phase: Problems
(b)
(c)
(d)
Single-phase to single-phase: Problems
(d) Input power factor when the harmonic content take in consideration
Solution
(a)
Single-phase to single-phase: Problems
𝑷𝟏
(a)
𝑷𝟐
𝑵𝟐
𝑵𝟏
𝟏
𝒇𝒐𝒖𝒕
𝟏
= 𝟏𝟔.𝟔 = 𝟎. 𝟎𝟔
𝑷𝟑
Single-phase to single-phase: Problems
(b)
𝑰𝒐 =
𝟏𝟎𝟏. 𝟔
= 𝟏𝟓. 𝟓 𝑨
𝟔. 𝟓𝟐
𝑰𝒐(𝒓𝒎𝒔)
𝑰𝒕𝒉𝒚. =
= 𝟕. 𝟕𝟗 𝐀
𝟐
(c)
𝑰𝒊𝒏 = 𝑰𝒐 = 𝟏𝟓. 𝟓 𝑨
𝑽𝑨𝒓𝒂𝒕𝒊𝒏𝒈 = 𝑬 × 𝑰𝒊𝒏 = 𝟐𝟑𝟎 × 𝟏𝟓. 𝟓 = 𝟑𝟓𝟔𝟓 𝑽𝑨
𝑷𝒐 = 𝑬𝒐 × 𝑰𝒐 𝐜𝐨𝐬 𝜽 = 𝟏𝟎𝟏. 𝟔 × 𝟏𝟓. 𝟓 × 𝐜𝐨𝐬 𝟒𝟎 = 𝟏𝟐𝟎𝟔 𝑾
Single-phase to single-phase: Problems
𝑷. 𝑭𝒊𝒏 =
𝑷𝑶
𝟏𝟐𝟎𝟔
=
= 𝟎. 𝟑𝟑𝟖𝟒
𝑽𝑨
𝟑𝟓𝟔𝟓
(d)
This proves that the harmonic content in the output voltage reduces the
power-factor.
Single-phase to single-phase: Analysis
when p-converter operating alone, the output voltage is positive half-cycles with
average output voltage 𝑉𝑑𝑐1 = 𝑉𝑑𝑜 𝑐𝑜𝑠𝛼 . Also when n-converter operating alone,
the output voltage is negative half-cycles with average output voltage
𝑉𝑑𝑐2 = −𝑉𝑑𝑜 𝑐𝑜𝑠𝛼 = −𝑉𝑑𝑐1
𝑻𝒐𝒕𝒂𝒍 𝑯𝒂𝒓𝒎𝒐𝒏𝒊𝒄
𝑽 = 𝑽𝑭𝒖𝒏𝒅. + 𝑽𝒉𝒂𝒓𝒎. + 𝑽𝒅𝒄
𝑯𝒂𝒓𝒎𝒐𝒏𝒊𝒄 ↑ → 𝑭𝒖𝒏𝒅. ↓ → 𝑻𝑯𝑫 ↑ → 𝑷. 𝑭 ↓
In order to reduce the harmonic and increase the fundamental
component, Lookup table for P converter , and lookup table for N
converter are used to apply different angle
Single-phase to single-phase: : Analysis
Single-phase to single-phase: : Analysis
Single-phase to single-phase: : Analysis
900
360
900
360
Single-phase to single-phase: : Analysis
0
270 90
00 1260
Single-phase to single-phase: Problems
Single-phase to single-phase cycloconverter “ Bridge type” with input
voltage 220 V, input frequency 50 Hz, and the output frequency is 12.5
Hz. The firing angle of P converter and N converter equal 36 and 90,
respectively . The load is resistance and it equals 10 ohm.
(a) Calculate the rms of the output voltage and the current.
(b) Sketch the input and output voltage waveforms
© If the different fringing angle technique is used, where the firing
angles of 𝑃12 , 𝑃34 , 𝑃12 , 𝑎𝑛𝑑 𝑃34 𝑒𝑞𝑢𝑎𝑙 27, 72, zero, and 126,
respectively. Calculate the rms output voltage and current.
(d) If the different fringing angle technique is used, where the firing
angles of 𝑃12 , 𝑃34 , 𝑃12 , 𝑎𝑛𝑑 𝑃34 𝑒𝑞𝑢𝑎𝑙 27, 72, zero, and 90,
respectively. Calculate the rms output voltage and current.
Solution
Single-phase to single-phase: Problems
(a)
𝑽𝒐 (𝒓𝒎𝒔) =
𝟐𝑵 𝝅
න 𝑽𝒎 𝒔𝒊𝒏 𝒘𝒕 𝟐 𝒅𝒘𝒕
𝟐𝝅𝑵 𝜶𝑷
𝑽𝒐 (𝒓𝒎𝒔) =
𝝅
𝝅
𝟏
𝟐 ∗ න 𝑽𝒎 𝒔𝒊𝒏 𝒘𝒕 𝟐 𝒅𝒘𝒕 + 𝟐 ∗ න 𝑽𝒎 𝒔𝒊𝒏 𝒘𝒕 𝟐 𝒅𝒘𝒕
𝝅∗𝟒
𝟑𝟔
𝟗𝟎
𝑽𝒐 (𝒓𝒎𝒔) =
𝑽𝟐𝒎 𝟐
𝒔𝒊𝒏𝟐𝝎𝒕
𝟐
𝒔𝒊𝒏𝟐𝝎𝒕
∗ 𝝎𝒕 −
+ ∗ 𝝎𝒕 −
𝝅∗𝟒 𝟐
𝟐
𝟐
𝟐
𝟑𝟔
𝟗𝟎
𝝅
𝑽𝒐 (𝒓𝒎𝒔) =
𝑽𝟐𝒎
𝝅∗𝟒
𝝅
𝟑𝝅
𝒔𝒊𝒏𝟐 ∗ 𝟑𝟔
𝝅
𝒔𝒊𝒏𝟐 ∗ 𝟗𝟎
𝝅−
+
+ 𝝅−
+
𝟏𝟓
𝟐
𝟐
𝟐
𝟏
𝑽𝒐 (𝒓𝒎𝒔) = 𝟐𝟐𝟎 𝟐
𝝅∗𝟒
𝟑𝝅
𝒔𝒊𝒏𝟐 ∗ 𝟑𝟔
𝝅
𝒔𝒊𝒏𝟐 ∗ 𝟗𝟎
𝝅−
+
+ 𝝅−
+
𝟏𝟓
𝟐
𝟐
𝟐
Single-phase to single-phase: Problems
𝑽𝒐 (𝒓𝒎𝒔) = 𝟐𝟐𝟎 𝟐
(b)
𝟏
𝝅∗𝟒
𝝅−
𝟑𝝅
𝒔𝒊𝒏𝟐 ∗ 𝟑𝟔
𝝅
𝒔𝒊𝒏𝟐 ∗ 𝟗𝟎
+
+ 𝝅−
+
𝟏𝟓
𝟐
𝟐
𝟐
= 𝟏𝟖𝟕. 𝟒 𝑽
Single-phase to single-phase: Problems
(c)
𝝅
𝑽𝒐 (𝒓𝒎𝒔) =
𝟏
𝝅∗𝟒
𝝅
න 𝑽𝒎 𝒔𝒊𝒏 𝒘𝒕 𝟐 𝒅𝒘𝒕 + න 𝑽𝒎 𝒔𝒊𝒏 𝒘𝒕 𝟐 𝒅𝒘𝒕 +
𝟐𝟕
𝝅
𝟕𝟐
𝝅
න 𝑽𝒎 𝒔𝒊𝒏 𝒘𝒕 𝟐 𝒅𝒘𝒕 + න
𝟎
𝑽𝒐 (𝒓𝒎𝒔) = 𝟐𝟐𝟎 𝟐
𝟏
𝝅∗𝟖
= 184.7 V
𝑽𝒎 𝒔𝒊𝒏 𝒘𝒕 𝟐 𝒅𝒘𝒕
𝟏𝟐𝟔
𝟑𝝅
𝒔𝒊𝒏𝟐 ∗ 𝟐𝟕
𝟓𝝅
𝒔𝒊𝒏𝟐 ∗ 𝟕𝟐
+
+ 𝝅−
+
+
𝟐𝟎
𝟐
𝟏𝟐. 𝟓
𝟐
𝒔𝒊𝒏𝟐 ∗ 𝟎
𝟑𝝅
𝒔𝒊𝒏𝟐 ∗ 𝟏𝟐𝟔
𝝅−𝟎 +
+ 𝝅−
+
𝟐
𝟒. 𝟐𝟖𝟓
𝟐
𝝅−
Single-phase to single-phase: Problems
Single-phase to single-phase: Problems
Single-phase to single-phase: Problems
(d)
𝝅
𝑽𝒐 (𝒓𝒎𝒔) =
𝟏
𝝅∗𝟒
𝝅
න 𝑽𝒎 𝒔𝒊𝒏 𝒘𝒕 𝟐 𝒅𝒘𝒕 + න 𝑽𝒎 𝒔𝒊𝒏 𝒘𝒕 𝟐 𝒅𝒘𝒕 +
𝟐𝟕
𝟕𝟐
𝝅
න 𝑽𝒎 𝒔𝒊𝒏 𝒘𝒕 𝟐 𝒅𝒘𝒕 + න 𝑽𝒎 𝒔𝒊𝒏 𝒘𝒕 𝟐 𝒅𝒘𝒕
𝟎
𝑽𝒐 (𝒓𝒎𝒔) = 𝟐𝟐𝟎 𝟐
𝟏
𝝅∗𝟖
= 195.8 V
𝝅
𝟗𝟎
𝟑𝝅
𝒔𝒊𝒏𝟐 ∗ 𝟐𝟕
𝟓𝝅
𝒔𝒊𝒏𝟐 ∗ 𝟕𝟐
𝝅−
+
+ 𝝅−
+
+
𝟐𝟎
𝟐
𝟏𝟐. 𝟓
𝟐
𝒔𝒊𝒏𝟐 ∗ 𝟎
𝝅
𝒔𝒊𝒏𝟐 ∗ 𝟗𝟎
𝝅−𝟎 +
+ 𝝅−
+
𝟐
𝟐
𝟐
Single-phase to single-phase: Problems
Alpha = 27,72,0,90
(d)
Single-phase to single-phase: Application
Single-phase to single-phase: Application
Single-phase to single-phase: Application
Single-phase to single-phase: Application
REFERENCES
1
“Power electronics devices, circuits, and applications” Muhammad H. Rashid
(4th edition)
2
“SPEED CONTROL OF SINGLE AND THREE PHASE INDUCTION MOTOR USING FULL
BRIDGE CYCLOCONVERTER” 2018 IEEE conference
3
https://www.youtube.com/watch?v=ia21tsfasnc&t=2448s
4
https://www.youtube.com/watch?v=l4CfEB61fMc&list=PLgwJf8NK2e5Lyu3Hj_EJms-boPTpRbpn
5
https://www.youtube.com/watch?v=LudX641uR4g&list=PLSWRPBzGkib8X_Kd71
mlqrKLtJKhQqwYO&index=3
5
https://www.youtube.com/watch?v=d6GJ1svbP9I&t=3s
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