IGBT-ĐỒ-ÁN-ĐIỆN-TỬ-CÔNG-SUẤT-nghịch lưu áp 1 pha

TRƯỜNG ĐẠI HỌC ĐIỆN LỰC
KHOA KỸ THUẬT ĐIỀU KHIỂN & TỰ ĐỘNG HĨA

Báo cáo Đồ án Điện tử cơng suất
Tên đề tài: Thiết kế bộ nghịch lưu áp 1 pha
Giảng viên hướng dẫn:

Nguyễn Thị Điệp

Sinh viên thực hiện :

Phạm Trung Tuấn Anh

Lớp: CLC D12 CNTD

HÀ NỘI, 10/ 2020

1


TABLE OF CONTENTS

2


Page
Chapter 1 :Introduction
1.1 Inverters and applications
1.1.1:Define

3

1.1.2 . Classify:

3

1.1.3. Application:

3

1.1.2 IGBT
1.2.1 Principle
1.2.2 Calculate schematic parameters

4
8
9

CHAPTER 2: DESIGN CIRCUIT
2.1. Power circuit valve:

11

2.2. Calculation of choosing IGBT

11

2.3. IGBT protection
2.4. Cooling calculation for IGBT

13

14

2.5. Amplified control signals for IGBTs

15

CHAPTER3: DESIGN THE CONTROL CIRCUIT
3.1 Structure of inverter control circuit

17

3.2 -phase inverter control circuit:

17

3.3 Synchronous

18

3.4 . Stitch creating serrated

19

3.5 The comparison stage

20

3.6 Stitch amplification creates pulses

21

CHAPTER 4: Complete circuit simulation

1

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OPENING
Nowadays, with the rapid development of high-capacity half-sell technology,
power converters using power semiconductor components have been widely used in
industry and life to meet daily needs. the higher the society. In practical use of
electricity, we need to change the frequency of the supply, inverters are widely used
in electric drive, in induction heating devices, in lighting equipment …
The inverters are inverters that convert direct to AC indirectly with great practical
applications such as aircraft, ship, train drive systems .. During the period of study
and research, is studying and researching the Power Electronics subject and its
applications in the fields of modern production systems. So in order to master the
theory and apply that knowledge in practice, we received the subject project with the
topic: "Designing single-phase reverse-voltage circuits". With the assigned topic, we
have applied our knowledge to learn and research the theory, especially we delve
deeply into design calculations for product completion.
Under the enthusiastic guidance of teacher Nguyen Thi Diep together with the
efforts of the team members, we have completed our project. However, due to the
limited time and knowledge, the shortcomings are inevitable when implementing this
project. So I hope to receive many comments from teachers and friends to the topic is
more complete.

2

Chapter 1: Introduction
1.1 . Inverters and applications:
3.6.1 Define:
Independent inverter is a device that changes or converts direct current (DC) to
alternating current (AC) with fixed or variable frequency.
3.6.2 Classify:
- Voltage power independent inverter: allow to change from direct voltage (E) to
alternating voltage with characteristic as same as grid voltage.
- Current independent inverter: convert DC to AC.
- Resonance independent inverter: when active it always forming an RLC resonant
oscillating loop.
- The load of the independent reciprocating is the AC device can be 1-phase or 3phase so it can also be manufactured in two forms: 1-phase inverter 3-phase inverter.
1.1.3. Application:
- Power supply
- Adjusting AC motor speed
- Electricity delivery
- Metallurgical and energy source converters
.

I.1.2 IGBT

3


A Structure:

Regarding the semiconductor structure, IGBT is nearly identical to the MOSFET, the
difference is that it has an extra layer connected to the Collector to create a pnp
semiconductor structure between the Emiter (similar to the original pole) and the

Collector (similar to the drain), without is nn as in mosfet. Thus, IGBT can be considered
as equivalent to a p-n-p transistor with base current controlled by a MOSFET.
B, Sign

C, Working Principle
- Polarize for IGBT star UCE> 0, then on pole G a control voltage Uge> 0 with a
large enough value. Then, forming a channel with electromagnetic particles like
MOSFET, the electrons move towards the pole C, pass the P-N junction to create
Colector current.
- The switching time of the IGBT is faster than conventional transistors, about
0.15ms late when open, about 1ms late when locking. Very small IGBT control power

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usually open as a control voltage of + -15V. To open normally + 15V signal, lock signal
-15V
D, Valve opening conditions
- Due to the n-p-n structure, the forward voltage between C and E in IGBT current
mode is much lower than Mosfet. However, due to this structure, the switching time of
the IGBT is slower than that of Mosfet, especially when locked. The figure shows the
equivalent structure of IGBT with Mosfet and a Tranzitor p-n-p. Notation line through
IGBT consists of two components: i1 line through Mosfet, i2 stream through Tranzitor.
The Mosfet part in IGBT can be locked quickly if the charge between G and E is fully
discharged, so the current i1 = 0, but i2 will not decline rapidly due to the amount of
electrical accumulation in (equivalent to the base of the structure. pnp structure) can only
be lost due to the self-neutralization of the charge. This appears the area of the current
lasting when the IGBT is locked.
- Diagram of testing an IGBT course:

-Open process:
The IGBT opening process is similar to this one in Mosfet when the input voltage
increases from 0 to the value Ug. During the delay time when Io is turned on, the control
signal charges the capacitor Cgc, causing the voltage between the control terminal and
emite to increase exponentially from 0 to the Uge (3 to 5v) value. Only from there did
Mosfet unfold in the IGBT structure. The current between the colecto-emite increases
according to the linear rule from 0 to the load current Io during time Tr. During Tr time,
the voltage between the control terminal and the emite increases to the value Uge
determines the value of current Io through the colecto. Because diode is still conducting
the load current of Io, the voltage Uce is still pinned up to the 1-way voltage Udc. Next,
the opening process takes place in 2 phases T1 and T2. During these two phases, the
voltage between the control terminal remains the same Uge to maintain the current Io,
because the control current is purely the discharge current Cgc capacitor. IGBT still work
in linear mode. In the first stage, the lock and recovery process of the diode takes place.
Due to the regenerative current of the diode Do create a current pulse above the IGBT

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line of the IGBT. The Uce voltage starts to decrease. IGBT switches the working point
across the linear mode region to the saturation region. Stage 2 continues the process of
decreasing resistance in the resistive region of the colecto leading to the colecto-emite
resistance to the Ron value when fully saturated.
Uce = IoRon.
After the opening time Ton when the capacitor Cgc has completed discharge, the
voltage between the control pole and the emito continues to increase exponentially with
the time constant CgcRg to the final value Ug.

-Clock process

E, Basic parameters of IGBT valve

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Table 1.IGBT’s parameters
No
1

2

3

4

5

6

Parameter
Voltage

Current

Control

Amplification

Power

Active mode

Parameters’s
name
Disruptive
voltage CE
Disruptive
voltage GE
Maximum
voltage between
CE terminals

Symbol

Maximum drain
current

IC

Maximum pulse
drain current

ICM

Maximum GE
voltage
Voltage GE
threshold
Conductivity

UGE

Resistor DS when
led
Maximum heat
generation
capacity
Loss of energy
when opening
Loss of energy
when locking
Open delay time

RCE(on)

Lock delay time

tD(OFF)

Drain terminal
current increase
time
Drain terminal
current decrease
time
Input port
capacitance

tR

7

Measurement
mode

U(BR)CE
U(BR)GẺ
UCE

UGE(th)
gFS

PC

When short circuit
GS
When there is
resistance to
connect between
terminal G and E
At the specified
temperature of
the shell
With specified
pulse width
When short circuit
CE
When UCE>0
According to the
regulation

With specified
shell temperature

EON
EOFF
tD(ON)
According to the
regulation

tF
CISS

CISS= CGE+ CGC


7

Temparature

Output port
capacitance
Capacitance
switch
Total charge of
the Gate terminal
The thermistor
set in the middle
of the transient
"np-shell"
Transition-setting

thermistor "npenvironment"
Thermistor
establishes
"shell-heatsink"

COSS

COSS= CGC+ CCE

CRSS

CRSS= CGC

QG

According to the
regulation
With standard
heat sink

The transient
thermistor
between
transients "npshell"
Maximum
permissible
temperature
resistance at "pn"
transients

ZThj.C

With current
pulses have
specified time

TJ(max)

Both negative
and positive
temperatures

RThj.C

RThj.A

No heat sink

RThj.S

-Analysis of independent single-phase voltage source inverter diagram, bridge
diagram:

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Figure1.7 IGBT bridge circuit

1.2.2 Principle:
Valves are divided into two groups, Tr1, Tr2 and Tr3, Tr4, operating at all load

lines. Valves of the same group control together (open or lock together) but the two
groups are controlled in the left state, controlling one group having to lock the other
group and vice versa.
Single-phase voltage independent inverse is a special case of symmetric 1-way
voltage pulse hashing.
* According to the figure above:
T: the period of the output voltage.
In the range t = 0 ÷ T / 2: Tr1, Tr2 are open and Tr3, Tr4 lock so Ut = E.
Current from source E through Tr1 through load, Tr2 and then on source.
At the time of T / 2 two valves Tr1, Tr2 locked, Tr3, Tr4 were controlled open in
parallel because the current had to flow in the old direction (due to the inductive load), so
the current could not go backwards through Tr3, Tr4, but forced flow. through the diodes
parallel to them.
In the range t = 0 ÷ T: Tr1, Tr2 closed and Tr3, Tr4 open so Ut = -E.
The accumulated inductance energy is returned to the source:
+ If the accumulated energy at Lt inductance is large enough and can not dissipate in
time, the current cannot return to zero, so only D3, D4 lead this whole interval until the
valve opens at the beginning of the cycle.
+ If the accumulated energy is not large enough, the bundle will be dissipated before
T and the current reaches 0, however because Tr3, Tr4 are open now, the load current will
reverse, the source E will power again. for the download. By the time T, Tr1, Tr2 opened
again, the current was negative, so it could not be reversed immediately, forcing it to flow
through diodes D1, D2, only when the energy on the inductance ends, the current will
return to a positive value.
In this method the current always works in continuous mode.
Output voltage satisfies the conditions of AC conditioning voltage:
• The output voltage has two positive and negative signs

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• The average value is zero (= 0)
• After a half cycle with equal but opposite values:
u(t)=-u(t)+u(t+T/2)
• After a period of repeating the state:
u (t) = u (t + T)

(1.2)

(1.3)

1.2.3 Calculate schematic parameters:
According to the basic harmonic method:
With coefficient k = 1
Input voltage for base load:

Basic load current :

(1.4)

(1.5)

With

(1.6)

and

(1.7)

Average current across the valve:
(1.8)
(1.9)

The mean value of the single-stroke current is equal to the mean flow through the valve:
(1.10
)
Power consumption from the given source:
(1.1

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1)
Load’s power:
(1.1
2)

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CHAPTER 2: DESIGN CIRCUIT
2.1. Power circuit valve:

Figure 2.1 Power circuit
2.2. Calculation of choosing IGBT
- Voltage applied to the valve: U = 310V
- Considering the load as resistive, we have the current through the valve
I

P 500

 2.3A
U 220

Consider that the selected power valve must be based on current and voltage
parameters in the circuit. Specifically, when calculating the capacity valves, they must
meet the conditions specified by the manufacturer. In which the parameters usually must
be the first priority when calculating the selection and the capacity is the working voltage
of the valve Uv; effective current flows through the valve and the average current flowing
through the valve .
In which the selected valve voltage shall satisfy the condition.
=(1.6ữ2)
=> 2ì310 = 620V.
The flow rate of the flow valve is selected depending on the cooling condition. If the
public semiconductor and is only cooled by natural convection radiator, the ability to
withstand electrical brittle is only 25 ÷ 30% of the rated current written on the valve. If

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the power semiconductor valve is cooled by a radiator and has a cooling fan, the current
capacity is 50 ÷ 70% of the rated current written on the valve. If the industrial
semiconductor valve is cooled by a radiator and has a cooling solution, the current
resistance can reach 100% of the rated current on the valve. save naturally. So we have:
I = (25÷30%) IVRMS
IVRMS =(2.3*100)/25=9,2A



Select valve yes :
= 600V và = 10A.
Based on the above calculation we can choose IGBT : FGA25N120AN
Features of FGA25N120AN:
- Fast switching speed
- Low saturation voltage: VCE (sat) = 2.5 V, IC = 25A
- High input impedance

2.3: IGBT FGA25N120AN
Symbo
l
VCES
VGES
IC

ICM

Description
Lock voltage collectoremitter

1200

Uni
t
V

Voltage gate-emitter

± 20

V

25

A

75

A

125

W

One-way colltter current
(TC=100oC)
Current collectter peak
repeat

PD

Maximum power
dissipation (TC=100oC)

TJ

Transition temperature

Symbo
l

VGE(th)

FGA25N120AN

Description
Threshold
voltage

o

-55 đến +150
FGA25N120
AN

gate-emitter

13

Unit
5.5

C

Sym
bol
V


VCE(sat)
td(on)

td(off)
tr

Saturated
collector- VCC= 600 V, Ic=
emitter voltage
25A,
Delay opening time
RG= 10Ω , VGE=
Late closing time
15V,
Growth time

2.5

V

60

ns

170

ns

60

ns

TC = 25°C

Table 2.1: Some parameters of FGA25N120A

2.3. IGBT protection
-Usually IGBTs are used in high frequency switching circuits, from 2 to tens of kHz. At
such high switching frequencies, malfunctions can destroy the element very quickly and
lead to total equipment failure. The most common fault is overcurrent due to short circuit
from the load side or from faulty elements due to fabrication or assembly. The IGBT
current can be disconnected by bringing the control voltage to a negative value. However,
overcurrent can take the IGBT out of saturation mode, leading to a sudden increase in
heat generation capacity, destroying the element after a few switching cycles. On the
other hand, when locking the IGBT again for a very short time when a very large current
leads to too great a current rate, causing overvoltage on the collector, emiter, immediately
puncturing the element. Besides, there were also unexpected incidents and interference
effects. Therefore, we must calculate the protection for the semiconductor valves when
the failure occurs ... For short circuit protection and overload current use Aptomat or fuse.
- Principle of choosing this device is according to the current with
Ibv = (1,11,3)Ilv.
-The protective current of Aptomat must not exceed the short circuit current of the
transformer.
From above, we choose the fuse to protect with:
Ibv = (1,11,3)Ilv= 1.3*2.3=2.99 (A)
We choose fuse 3A to protect over current for the IGBT.

2.4. Cooling calculation for IGBT

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Semiconductors are very sensitive to temperature. If working, the temperature of the
laminate surface is higher than the permissible temperature Tjm ,it can damage the

semiconductor device. So the calculation of radiant heat for the joint is very necessary:
+ When calculating the diagram of thermal isotherms shown as follows:
Inside:
Tj:is the temperature of the joint.
Tv: The temperature of the case of the semiconductor device.
Tr: The temperature of the radiator fins.
Ta: The air temperature of the working environment.
Rjv: Thermal resistance between the coupling face and the semiconductor housing
Rvt: The heat resistance between the cover and the diffuser.
Rra: Heat resistor fins and ambient air.

Figure 2.4: Diagram of thermal isotherms

+ The temperature is transferred from the hot to the cold zone, the heat capacity
transferred is proportional to the wrong heat and inversely proportional to the thermistor
Rth.
∆P=
Inside T1 is the hot zone temperature, T2 is the cold zone temperature, the refractory
temperature

Rth = Rjv + Rvr + Rra is calculated by

- In heat problems often give us know Tjm, Ta, Rth, ∆P. It is required to determine whether
to be cooled by natural convection or by how much the fan must be cooled m/s.

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Figure 2.5:
a)Vol-ampere characteristics

b ) Curve representing radiator blade heat and cooling fan speed
c) Curve representing the radiator fins resistance and the medium

With the above data, we choose heat dissipation by convection radiator.
Have T1 = 155 oC, T2 = 30 oC, ∆P = 125 => Rth=1oC/W
So we can choose the type of heatsink below:

Figure 2.6: Standard heatsink

2.5. Amplified control signals for IGBTs

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To amplify the IGBT control signal, there are 3 options:
1. Pulse transformer
2. Dedicated IC
3. Transistor
-Amplification by pulse transformers is capable of isolating, but difficult in usage and
fabrication.
- Transistors amplification is more compact than a pulse transformer, but only for small
power circuits.
- Amplifying with a dedicated IC for this circuit uses IC IR2110 that both responds to
large frequencies and is quite easy to use without requiring in-depth knowledge.

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CHAPTER3: DESIGN THE CONTROL CIRCUIT
3.1 Structure of inverter control circuit:

Pulse
generation

Pulse
distribute

Determine lead
area

Pulse
amplifier

Go to
MOSFETS

Figure 3.1 Inverter control circuit’s structure
 Blocks’s function :
- Pulse generation block: to make sync signal for hole system with the frequency is
proportional to the fundamental harmonic of out voltage
- The pulse distribute block: distribute pulse signal into individual valve following the
working order of working priciple.
- Determine lead area block: make the valve working with specifically control
method.
- Pulse amplifier: increase the power to open/close the valve.
3.2 -phase inverter control circuit:

Pulse amplifier
Pulse
generation

Pulse
distribution

Pulse amplifier

Figure 3.2 1-phase inverter control structure
Control circuit for this type just has 1 step to generate rectangle pulse for U fx, after
that it’s goes through the frequency division step to make sure lead area valves are
completely equal and reverse phase. Before the power being amplified, we need to make
a open delay to prevent short circuit of two valves in line.

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With it’s application, this kind of control circuit can be very easy to use by using
simple pulse-number technique, with the oscillate generation step and voltage divide step
using Flip Flop circuit at binary counting mode with Uoscillator=2*Uout.

3.3. Synchronous
- Select the synchronous circuit two half cycle:

Figure 3.4: Synchronous circuit diagram.
The two half-cycle rectifier circuit has a midpoint using diodes D1, D2 and the
rectifier load is resistor R0. The rectifier voltage Ucl after being created is brought to the
(+) pole of the Opam to compare with 0 (because the (-) pole of the opam is grounded).
If Ucl> 0 then Udb is equal to the saturation voltage (Ubh).
If Ucl> 0 then Udb is equal to the negative saturation voltage (-Ubh).
The point of intersection of Ucl and 0 is the transition point of the output voltage.

Figure 3.5: Oscillation graph synchronous circuit diagram.

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3.4. Stitch creating serrated.

Hình 3.6: Circuit Stitch creating serrated.
Activities:
+ When Udb < 0 then D3 guide; so
UR4 = Udb Udb = UC1
When the voltage reaches the threshold of Dz voltage diode, it will keep the output
voltage at this voltage stabilizer (if there is no Dz ⇒ UC increased to + Udb).
+ When Udb > 0 then D3 block Capacitor is launched UC dropped to 0 and Dz keep
UC in value - 0,7.
- Calculation:
Cycles: T = 1 / f = 0,02 (s) = 20 (ms).
select OA species TL082. The control angle range is 168 degrees.
Capacitor C launch time: tp = = 9,33 (ms).
Choose the voltage regulator diode BZX79C has UDZ = 10 (V).
Select capacitors C = 220 (nF).
select R6 = 51k serial variable resistor P1 = 8k.
Time capacitor C loaded: tn = T/2 – tp = 10 – 9,33 = 0,67 (ms).
The saturation voltage of the OA: Udb = E – 1,5 = 12 – 1,5 = 10,5 (V).
So select R4 = 1 (k).

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Figure 3.7: Oscillation graph Circuit Stitch creating serrated.
3.5. The comparison stage.

Function: Compare the control voltage with the restraint voltage to determine the
timing of the control pulse ⇒ Determine the control angle α
The comparison stage can be done with an element such as a transistor, or an OA
algorithm amplification.
- We use the OA element because it allows to ensure the highest accuracy is to use
dedicated OA coparator, with low cost, without complicated adjustment.

- Comparison using two-door OA:

Hình 3.8: Comparison circuit.
The two voltages to be compared are applied to two different poles of the OA.

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In the above case Uđk = U +, Utua = U If UDC> Ura ⇒ Ura = + Ubh.
If Uđk Valve selection calculation:
Select Opam type TL082.
Select resistance R1 = 10k, R2 = 10k.
Udk = 4 (V).

Hình 3.9: Oscillation graph comparison circuit.

3.6. Stitch amplification creates pulses.
- Task: Create pulses to open Thyristor, pulses to open Thyristor require:
+ Enough capacity.
+ There is a vertical slope, usually rectangular pulse.
+ Isolation between the control circuit and the force circuit ⇒ Using a pulse
transformer.

- Several ways to amplify pulses.
Direct: Isolating between the control circuit and the force circuit is not allowed.
+ Coupling through optical elements: Only withstand loads of tens of mA
⇒ Insufficient capacity to open the force valve.
By amplifying pulse: Most commonly used today, Easy to isolate between the control
circuit and the force circuit, transmitting pulses in the form of beam pulses.
⇒ Choose how to amplify by pulse transformer.

22


Amplification by a pulse transformer:

Figure 3.12: Amplifier circuit.
- Operation: Input voltage is a beam pulse voltage, rectangular shape, need to
open 2 thyristors, when there is a pulse, there is current I5 so there is current
flowing through the pulse transformer. This current will induce the secondary of
the control pulse transformer. Use positive pulse because positive pulse energy is
taken from source E, and negative pulse is discharged by the energy of inductor
coil, this energy is small.

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