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

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..

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

Báo cáo Đồ án Điện tử công 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

TABLE OF CONTENTS

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 4

1.2.1 Principle 8

1.2.2 Calculate schematic parameters 9

CHAPTER 2: DESIGN CIRCUIT 2.1 Power circuit valve: 11

2.2 Calculation of choosing IGBT 11

2.3 IGBT protection 13

2.4 Cooling calculation for IGBT 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 23

Nowadays, with the rapid development of high-capacity half-sell technology,power converters using power semiconductor components have been widely used inindustry and life to meet daily needs the higher the society In practical use ofelectricity, 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 practicalapplications such as aircraft, ship, train drive systems During the period of studyand research, is studying and researching the Power Electronics subject and itsapplications in the fields of modern production systems So in order to master thetheory and apply that knowledge in practice, we received the subject project with thetopic: "Designing single-phase reverse-voltage circuits" With the assigned topic, wehave applied our knowledge to learn and research the theory, especially we delvedeeply into design calculations for product completion

Under the enthusiastic guidance of teacher Nguyen Thi Diep together with theefforts of the team members, we have completed our project However, due to thelimited time and knowledge, the shortcomings are inevitable when implementing thisproject So I hope to receive many comments from teachers and friends to the topic ismore complete

- Current independent inverter: convert DC to AC.

- Resonance independent inverter: when active it always forming an RLC resonantoscillating loop

- The load of the independent reciprocating is the AC device can be 1-phase or phase so it can also be manufactured in two forms: 1-phase inverter 3-phase inverter

Regarding the semiconductor structure, IGBT is nearly identical to the MOSFET, thedifference is that it has an extra layer connected to the Collector to create a pnpsemiconductor structure between the Emiter (similar to the original pole) and theCollector (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 alarge enough value Then, forming a channel with electromagnetic particles likeMOSFET, the electrons move towards the pole C, pass the P-N junction to createColector current

- The switching time of the IGBT is faster than conventional transistors, about0.15ms late when open, about 1ms late when locking Very small IGBT control powerusually 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 currentmode is much lower than Mosfet However, due to this structure, the switching time ofthe IGBT is slower than that of Mosfet, especially when locked The figure shows theequivalent structure of IGBT with Mosfet and a Tranzitor p-n-p Notation line throughIGBT 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 fullydischarged, so the current i1 = 0, but i2 will not decline rapidly due to the amount ofelectrical 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 currentlasting 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 voltageincreases from 0 to the value Ug During the delay time when Io is turned on, the controlsignal charges the capacitor Cgc, causing the voltage between the control terminal andemite to increase exponentially from 0 to the Uge (3 to 5v) value Only from there didMosfet unfold in the IGBT structure The current between the colecto-emite increasesaccording 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 Ugedetermines the value of current Io through the colecto Because diode is still conductingthe 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, thevoltage 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 IGBTline of the IGBT The Uce voltage starts to decrease IGBT switches the working pointacross 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-emiteresistance to the Ron value when fully saturated.

Uce = IoRon

After the opening time Ton when the capacitor Cgc has completed discharge, thevoltage between the control pole and the emito continues to increase exponentially withthe time constant CgcRg to the final value Ug

-Clock process

E, Basic parameters of IGBT valve

Table 1.IGBT’s parameters

No Parameter Parameters’s

1 Voltage Disruptive

Disruptive voltage GE U(BR)GẺ When short circuit GSMaximum

voltage between

CE terminals

resistance to connect between terminal G and E

current IC At the specified temperature of

the shellMaximum pulse

drain current ICM With specified pulse width

Voltage GE

regulationResistor DS when

generation capacity

According to the regulation

Lock delay time tD(OFF)Drain terminal

current increase time

tR

Drain terminal current decrease time

Total charge of the Gate terminal QG According to the regulation

set in the middle

RThj.A No heat sink

Thermistor establishes

"shell-heatsink"

RThj.S

The transient thermistor between transients "np-shell"

ZThj.C With current

pulses have specified time

Maximum permissible temperature resistance at

"pn" transients

TJ(max) Both negative

and positive temperatures

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

Figure1.7 IGBT bridge circuit 1.2.2 Principle:

Valves are divided into two groups, Tr1, Tr2 and Tr3, Tr4, operating at all loadlines Valves of the same group control together (open or lock together) but the twogroups are controlled in the left state, controlling one group having to lock the othergroup and vice versa.

Single-phase voltage independent inverse is a special case of symmetric 1-wayvoltage 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 inparallel because the current had to flow in the old direction (due to the inductive load), sothe current could not go backwards through Tr3, Tr4, but forced flow through the diodesparallel 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 intime, the current cannot return to zero, so only D3, D4 lead this whole interval until thevalve 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 currentwill reverse, the source E will power again for the download By the time T, Tr1, Tr2opened 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, thecurrent 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

• The average value is zero (= 0)

• After a half cycle with equal but opposite values:

• After a period of repeating the state:

1.2.3 Calculate schematic parameters:

According to the basic harmonic method:

With coefficient k = 1

Input voltage for base load:

U1(t )= 4 E

π sin ωtt=1,273 E sinθ=U 1 m sin θ (1.4)

Basic load current :

Power consumption from the given source:

P dθ=EI dθ=E 2 I 1 m

π cos φ

(1.11) Load’s power:

2)

CHAPTER 2: DESIGN CIRCUIT

2.1 Power circuit valve:

Figure 2.1 Power circuit2.2 Calculation of choosing IGBT

- Voltage applied to the valve: U = 310V

- Considering the load as resistive, we have the current through the valve

500

2.3A 220

be the first priority when calculating the selection and the capacity is the working voltage

of the valve Uv; effective current flows through the valveI VRMS and the average current

flowing through the valve I VAV

In which the selected valve voltage shall satisfy the condition U v

=(1.6÷2)U ngmax

=>U v=¿ 2×310 = 620V

The flow rate of the flow valve is selected depending on the cooling condition If thepublic semiconductor and is only cooled by natural convection radiator, the ability towithstand electrical brittle is only 25 ÷ 30% of the rated current written on the valve If

the power semiconductor valve is cooled by a radiator and has a cooling fan, the currentcapacity is 50 ÷ 70% of the rated current written on the valve If the industrialsemiconductor valve is cooled by a radiator and has a cooling solution, the currentresistance 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 : U V = 600V và I V = 10A

Based on the above calculation we can choose IGBT : FGA25N120AN

Features ofFGA25N120AN:

- Fast switching speed

- Low saturation voltage: VCE (sat) = 2.5 V, IC = 25A

- High input impedance

VGE(th) Threshold gate-emitter

VCC= 600 V, Ic=

25A,

RG= 10Ω , VGE=15V,

TC = 25°C

VCE(sat) Saturated

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

- 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

Figure 2.6: Standard heatsink

2.5 Amplified control signals for IGBTs

To amplify the IGBT control signal, there are 3 options:

3.1 Structure of inverter control circuit:

Figure 3.1 Inverter control circuit’s structure

- Pulse amplifier: increase the power to open/close the valve

3.2 -phase inverter control circuit:

Figure 3.2 1-phase inverter control structureControl circuit for this type just has 1 step to generate rectangle pulse for Ufx, afterthat it’s goes through the frequency division step to make sure lead area valves arecompletely 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

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

MOSFETS

Go to

Pulseamplifier

Determinelead area

Pulsedistribute

Pulse

generation

PulseamplifierPulse

distribution

Pulse

amplifier

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 therectifier 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

3.4 Stitch creating serrated.

Hình 3.6: Circuit Stitch creating serrated

select OA species TL082 The control angle range is 168 degrees

⇒Capacitor C launch time: tp = 168× 10× 10−3

⇒ 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)

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 thetiming of the control pulse ⇒ Determine the control angle α

The comparison stage can be done with an element such as a transistor, or an OAalgorithm amplification

- We use the OA element because it allows to ensure the highest accuracy is to usededicated 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