DC/AC THREE – PHASE INVERTER
Student Name
Nguyễn Tuấn Minh
Đỗ Lê Trà My
Đào Đăng Khoa
Đào Gia Huy
Student ID
20202794
20200416
20202751
20202789
2
TABLE OF CONTENTS
1. Introduction
2. System Modeling
2.1. Circuit diagram
2.2. Block diagram
3. Control problems
3
I. INTRODUCTION
• An inverter (DC to AC inverter) is an electronic
device that converts direct current (DC) power
into alternating current (AC) power. Inverters are
commonly used in a variety of applications,
including renewable energy systems,
uninterruptible power supplies (UPS), electric
vehicles, and more.
4
I. INTRODUCTION
Powering AC Devices
and Appliances
Enhancing Power
Quality
WHY INVERTER IS
ESSENTIAL ?
Supporting Off-Grid
Power Systems
Providing Portable
Power
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I. INTRODUCTION
MODELING THE
PROBLEM
OVERVIEW AND DEFINE THE
PROBLEMS
DIVIDE INTO BASIC
SECTION AND
MODELING EACH
APPLYING PHYSIC LAWS TO FIND
MATHEMATICAL EQUATIONS
ADD CONTROLLER TO
ADJUST OUTPUT
CALCULATE THE COEFFICIENT OF
CONTROLLER
SIMULATION AND
CHECK THE RESULT
MODELING THE PROBLEM
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II. SYSTEM MODELING
Parameters:
• Input voltage: 𝑉𝐷𝐶 = 24𝑉𝐷𝐶
• Output voltage: 𝑉𝑜𝑢𝑡 = 12𝑉𝐴𝐶 − 50𝐻𝑧
• Max output current: 𝐼𝑜𝑢𝑡𝑚𝑎𝑥 = 2𝐴
• Switching frequency: 𝑓𝑠𝑤 = 5𝑘𝐻𝑧
• Output power: 𝑃 = 4𝑊
• LC filter: 𝐿𝑓 = 3.3𝑚𝐻, 𝑟𝐿 = 6.6 mΩ
𝐶𝑓 = 47𝑢𝐹
• R load: 𝑅 = 6Ω
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II. SYSTEM MODELING
Circuit diagram of three – phase inverter
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II. SYSTEM MODELING
BLOCK DIAGRAM
𝑉𝑟𝑒𝑓
𝐺𝑃𝐼 (𝑠)
𝐺𝑃𝐼 (𝑠)
Voltage PI
controller
Current PI
controller
𝐼𝑑𝑞
𝑑𝑞/𝛼𝛽
𝑎𝑏𝑐/𝑑𝑞
𝑉𝑑𝑞
𝑎𝑏𝑐/𝑑𝑞
𝑆𝑉𝑃𝑊𝑀
𝑇ℎ𝑟𝑒𝑒 − 𝑝ℎ𝑎𝑠𝑒
𝑖𝑛𝑣𝑒𝑟𝑡𝑒𝑟
𝐼𝑎𝑏𝑐
𝑉𝑎𝑏𝑐
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II. SYSTEM MODELING
Reference Frame Transformation
Three - phase to the Stationary Reference Frame (abc → 𝛂β)
Clark Transformation
10
II. SYSTEM MODELING
Reference Frame Transformation
Stationary Reference Frame to the Synchronous Reference (𝛂β → dq)
Park Transformation
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III. MATHEMATICAL MODEL
Circuit diagram of three – phase inverter
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III. MATHEMATICAL MODEL
Assume 𝑽𝑖 = 𝑣𝑖𝐴 𝑣𝑖𝐵 𝑣𝑖𝐵 𝑇 , 𝑰𝑖 = 𝑖𝑖𝐴 𝑖𝑖𝐵 𝑖𝑖𝐵 𝑇
𝑽𝐿 = 𝑣𝐿𝐴 𝑣𝐿𝐵 𝑣𝐿𝐵 𝑇 , 𝑰𝐿 = 𝑖𝐿𝐴 𝑖𝐿𝐵 𝑖𝐿𝐵 𝑇
Apply KCL and KVL at the LC output filter, we have:
𝑑𝑽𝐿
1
1
= 𝑰𝑖 − 𝑰𝐿
𝑑𝑡
𝐶𝑓
𝐶𝑓
𝑑𝑰𝑖 𝑟𝐿
1
1
+ 𝑰𝑖 = 𝑽𝑖 − 𝑽𝐿
𝑑𝑡 𝐿𝑓
𝐿𝑓
𝐿𝑓
((1)
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III. MATHEMATICAL MODEL
𝒂𝒃𝒄 reference frame  𝜶𝜷 reference frame by the following expression:
2𝜋
4𝜋
𝑗3
𝑗3
𝑗0
𝑿𝛼𝛽 = 𝑥𝑎 𝑒 + 𝑥𝑏 𝑒
+𝑥𝑐 𝑒
((2)
where 𝑿𝛼𝛽 = 𝑥𝛼 + 𝑗𝑥𝛽
(1) can be transformed to the following:
where 𝑽𝐿𝛼𝛽 = 𝑣𝐿𝛼
𝑽𝑖𝛼𝛽 = 𝑣𝑖𝛼
𝑑𝑽𝐿𝛼𝛽
1
1
= 𝑰𝑖𝛼𝛽 − 𝑰𝐿𝛼𝛽
𝑑𝑡
𝐶𝑓
𝐶𝑓
𝑑𝑰𝑖𝛼𝛽 𝑟𝐿
1
1
+ 𝑰𝑖𝛼𝛽 = 𝑽𝑖𝛼𝛽 − 𝑽𝐿𝛼𝛽
𝑑𝑡
𝐿𝑓
𝐿𝑓
𝐿𝑓
𝑣𝐿𝛽 𝑇 , 𝑰𝐿𝛼𝛽 = 𝑖𝐿𝛼 𝑖𝐿𝛽 𝑇
𝑣𝑖𝛽 𝑇 , 𝑰𝑖𝛼𝛽 = 𝑖𝑖𝛼 𝑖𝑖𝛽 𝑇
((3)
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III. MATHEMATICAL MODEL
𝜶𝜷 reference frame  𝒅𝒒 reference frame by the following expression:
𝑿𝑑𝑞 = 𝑥𝑑 + 𝑗𝑥𝑞 = 𝑿𝛼𝛽 𝑒 −𝑗𝜃
((4)
𝑡
𝜔 𝜏 𝑑𝜏 + 𝜃𝑜 is the transformation angle, 𝜔 is the angular frequency
0
where 𝜃 𝑡 =
(𝜔 = 2𝜋𝑓)
(3) can be transformed to the following:
where 𝑽𝐿𝑑𝑞 = 𝑣𝐿𝑑
𝑽𝑖𝑑𝑞 = 𝑣𝑖𝑑
𝑑𝑽𝐿𝑑𝑞
1
1
+ 𝑗𝜔𝑽𝐿𝑑𝑞 = 𝑰𝑖𝑑𝑞 − 𝑰𝐿𝑑𝑞
𝑑𝑡
𝐶𝑓
𝐶𝑓
𝑑𝑰𝑖𝑑𝑞 𝑟𝐿
1
1
+ 𝑰𝑖𝑑𝑞 + 𝑗𝜔𝑰𝐿𝑑𝑞 = 𝑽𝑖𝑑𝑞 − 𝑽𝐿𝑑𝑞
𝑑𝑡
𝐿𝑓
𝐿𝑓
𝐿𝑓
𝑣𝐿𝑞 𝑇 , 𝑰𝐿𝑑𝑞 = 𝑖𝐿𝑑 𝑖𝐿𝑞 𝑇
𝑣𝑖𝑞 𝑇 , 𝑰𝑖𝑑𝑞 = 𝑖𝑖𝑑
((5)
𝑖𝑖𝑞 𝑇
15
III. MATHEMATICAL MODEL
(5) can be written as
1
1
𝑣𝐿𝑑 = 𝜔𝑣𝐿𝑞 − 𝑖𝐿𝑑 + 𝑖𝑖𝑑
𝐶𝑓
𝐶𝑓
1
1
𝑣𝐿𝑞 = −𝜔𝑣𝐿𝑑 − 𝑖𝐿𝑞 + 𝑖𝑖𝑞
𝐶𝑓
𝐶𝑓
𝑟𝐿
1
1
𝑖𝑖𝑑 + 𝑖𝑖𝑑 = − 𝑣𝐿𝑑 + 𝜔𝑖𝑖𝑞 + 𝑣𝑖𝑑
𝐿𝑓
𝐿𝑓
𝐿𝑓
𝑟𝐿
1
1
𝑖𝑖𝑞 + 𝑖𝑖𝑞 = − 𝑣𝐿𝑞 − 𝜔𝑖𝑖𝑑 + 𝑣𝑖𝑞
𝐿𝑓
𝐿𝑓
𝐿𝑓
((6)
((7)
((8)
((9)
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III. MATHEMATICAL MODEL
From (8), (9), we have
𝑑𝑣𝑖𝑑
𝐿𝑓
+ 𝑟𝐿 𝑖𝑖𝑑 = −𝑣𝐿𝑑 + 𝐿𝑓 𝜔𝑖𝑖𝑞 + 𝑣𝑖𝑑
𝑑𝑡
𝑣𝑖𝑞
𝐿𝑓
+ 𝑟𝐿 𝑖𝑖𝑞 = −𝑣𝐿𝑞 − 𝐿𝑓 𝜔𝑖𝑖𝑑 + 𝑣𝑖𝑞
𝑑𝑡
((10)
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IV. CONTROL STRATEGY
BLOCK DIAGRAM
𝑉𝑟𝑒𝑓
𝐺𝑃𝐼 (𝑠)
𝐺𝑃𝐼 (𝑠)
Voltage PI
controller
Current PI
controller
𝐼𝑑𝑞
𝑑𝑞/𝛼𝛽
𝑎𝑏𝑐/𝑑𝑞
𝑉𝑑𝑞
𝑎𝑏𝑐/𝑑𝑞
𝑆𝑉𝑃𝑊𝑀
𝑇ℎ𝑟𝑒𝑒 − 𝑝ℎ𝑎𝑠𝑒
𝑖𝑛𝑣𝑒𝑟𝑡𝑒𝑟
𝐼𝑎𝑏𝑐
𝑉𝑎𝑏𝑐
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IV. CONTROL STRATEGY
4.1. INNER CURRENT LOOP
𝑣𝐿
𝑖𝑖𝑟𝑒𝑓
𝐺𝑃𝐼 (𝑠)
𝐺𝑆𝑉𝑃𝑊𝑀 (𝑠)
𝐺𝑝𝑙𝑎𝑛𝑡 (𝑠)
𝑖𝑖
((10)
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IV. CONTROL STRATEGY
Output voltage references (𝑣𝑖𝑑,𝑟𝑒𝑓 , 𝑣𝑖𝑞,𝑟𝑒𝑓 ) are modified by means of adding one decoupling
term and one feed-forward voltage as
𝑣𝑖𝑑,𝑟𝑒𝑓 = −𝑣𝐿𝑑 + 𝜔𝐿𝑓 𝑖𝑖𝑞 + 𝑣𝑖𝑑
((11)
𝑣𝑖𝑞,𝑟𝑒𝑓 = −𝑣𝐿𝑞 − 𝜔𝐿𝑓 𝑖𝑖𝑑 + 𝑣𝑖𝑞
The current control loop after modification can
be rewritten as
𝑑𝑖𝑖𝑑
𝐿𝑓
+ 𝑟𝐿 𝑖𝑖𝑑 = 𝑣𝑖𝑑,𝑟𝑒𝑓
𝑑𝑡
𝑑𝑖𝑖𝑞
𝐿𝑓
+ 𝑟𝐿 𝑖𝑖𝑞 = 𝑣𝑖𝑞,𝑟𝑒𝑓
𝑑𝑡
𝐿𝑑
𝑖𝑑
𝑖𝑑
1
𝑟𝐿 + 𝐿𝑓 𝑠
𝑖𝑑
((12)
𝑖𝑞
𝑖𝑞
1
𝑟𝐿 + 𝐿𝑓 𝑠
𝑖𝑞
𝐿𝑞
20
IV. CONTROL STRATEGY
Laplace transform:
𝐿𝑓 𝑠𝐼𝑖𝑑 𝑠 + 𝑟𝐿 𝐼𝑖𝑑 𝑠 = 𝑉𝑖𝑑,𝑟𝑒𝑓 (𝑠)
𝐿𝑓 𝑠𝐼𝑖𝑞 𝑠 + 𝑟𝐿 𝐼𝑖𝑞 (𝑠) = 𝑉𝑖𝑞,𝑟𝑒𝑓 (𝑠)
The filter (plant) transfer functions Gplant s
((13)
:
𝐼𝑖 𝑠
1
𝐺𝑝𝑙𝑎𝑛𝑡 =
=
=
𝑉𝑖 (𝑠) 𝑟𝐿 + 𝐿𝑓 𝑠
1
𝑟𝐿
𝐿𝑓
1+ 𝑠
𝑟𝐿
21
IV. CONTROL STRATEGY
4.2. OUTER VOLTAGE LOOP
𝑖𝐿
𝑣𝑟𝑒𝑓
𝐺𝑃𝐼 (𝑠)
𝑖𝑟𝑒𝑓
1
𝑖𝑖
𝐺(𝑠)
𝑣𝑖
22
IV. CONTROL STRATEGY
From (6), (7), we have
𝑑𝑣𝐿𝑑
𝐶𝑓
= 𝐶𝑓 𝜔𝑣𝐿𝑞 − 𝑖𝐿𝑑 + 𝑖𝑖𝑑
𝑑𝑡
𝑑𝑣𝐿𝑞
𝐶𝑓
= −𝐶𝑓 𝜔𝑣𝐿𝑑 − 𝑖𝐿𝑞 + 𝑖𝑖𝑞
𝑑𝑡
((14)
Similarly,
𝑑𝑣𝑖𝑑
𝐶𝑓
= 𝑖𝑖𝑑
𝑑𝑡
𝑑𝑣𝑖𝑞
𝐶𝑓
= 𝑖𝑖𝑞
𝑑𝑡
Laplace Transform
𝐶𝑓 𝑠𝑉𝑖𝑑 𝑠 = 𝐼𝑖𝑑 (𝑠)
𝐶𝑓 𝑠𝑉𝑖𝑞 𝑠 = 𝐼𝑖𝑞 (𝑠)
𝑉𝑖𝑞 𝑠
𝑉𝑖𝑑 𝑠
1
⇒𝐺 𝑠 =
=
=
𝐼𝑖𝑑 (𝑠) 𝐼𝑖𝑞 (𝑠) 𝑠𝐶𝑓
23
THANK YOU !
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