5KVA Inverter Boost H-Bridge

8/12/2019 5KVA Inverter Boost H-Bridge

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Bajopas Vol. 2 Number 1 June, 2009

6

Bayero Journal of Pure and Applied Sciences, 2(1): 6 - 13Received: November, 2008

Accepted: February, 2009

5KVA POWER INVERTER DESIGN AND SIMULATION BASED ON BOOSTCONVERTER AND H-BRIDGE INVERTER TOPOLOGY

*Musa, A. and G.S.M. GaladanciDepartment of Physics, Bayero University, PMB 3011, Kano

*Correspondence author

ABSTRACTFive (5) kVA power inverter was designed and simulated base on two topologies; Boost converterand Half-bridge inverter topology. A 555 timer IC was used as the control at fixed frequencies of 25kHz and 50 Hz for the two stages. The results of the simulation were obtained. The graphs for bothstages were plotted and the results show a significant increase in the voltage and duty cycle. Thewave form of the output gives a square wave form.

Keywords: Power inverter, Simulation, Topology, Converter

INTRODUCTIONThe World demand for electrical energy is constantlyincreasing and conventional energy resources arediminishing and even threatened to be depleted. Withthe increasing popularity of alternative power sources,such as solar and wind, the need for static inverters toconvert dc energy stored in batteries to conventionalac form has increased substantially. This conversioncan either be achieved by transistors or by SCRs(silicon controlled rectifiers). For lower and mediumpower outputs, transistorised inverters are suitable butfor high power outputs, SCRs should be used.Transistor inverters are useful in wide variety ofapplications. They provide power to the complicated

electronic systems of orbiting satellites and coolastronaut suits(http;//Guru/PEInverters/introduction/pdf/inv.pdf,2006).

However, MOSFETs (Metal OxideSemiconductor Field Effect Transistor) and IGBTs(Insulated Gate Bipolar Transistor) are goodcandidates for the switches in dc to ac conventionaltechniques. With the development electronic devicesand circuits, newer equipment often have convertersthat allow the user to change the voltage andfrequency applied to the equipment .Power electronicsalso provides the industries with effective methods tosave energy and improve performance (Sen, 2002).

Power electronic devices can be divided in totwo categories: Single component and hybrid

component. The single component is composed of onesolid-state device, and hybrid component is acombination of several single componentsmanufactured as a single block. The single devices areloosely divided into three type :(1) the two layerdevices such as diode , (2) the three layer devicessuch as transistor , and (3) the four layer devices suchas the Thyristor . The hybrid components includewider range of devices such as the Insulated GateBipolar Transistor (IGBT), Static Insulated Transistor(SIT), and Darlington Transistor (DT). Solid statedevices are main building components of anyconverter. Their function is mainly to mimic themechanical switches by connecting and disconnecting

electrical loads, but at very high speed. When themechanical switch is open (off), the current throughthe switch is zero and the voltage across its terminal isequal to the source voltage. When the switch is closed(on), its terminal voltage is zero and its current isdetermined by load impedance (Mukund, 2006) .

Design Stage for fly Boost Converter and PowerInverterBoost converter designBased on the dc-dc converter topology; Boostconverter circuit design is shown with duty cycle (D),output input gain with pulse width modulation to drivethe converter at 25 kHz shown in Figure 1.0

Figure 1.0: Boost converter schematic diagram

PWM &

V i


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The output to the input voltage gain of the inverter is given by (Simon and Oliva, 2005).

2.0

and 2.1

where , is the duty cycle of switching converter that is

2.2

where switched off time for the transistor

switched on time for the transistor and

which is define by the following relation

2.3 a

or 2.3 b

However, with 555-timer ICs an inexpensive regulated SMPS is constructed, feedback and pulse widthmodulation were shown in Figure 2.0. If the desired inverter frequency and the voltage are known, the powerconsumed by the switching on and off of the converter can be estimated by the following equation(http://supertex.com/pdf/app.note/AN-36.pdf, 2008).

2.4

GND

DI S

OUT RST

VCC

THR

CON

TRI

2

0

0

3

5

0

6

7

Fig. 2 DC-DC Converter

0

1

0

BatteryCin

CTRD

RC

RFB

555

ZFB

L

Qsw

D

Cout

0

4 Vout

Determine operating frequency, Duty cycle, and InductorNeglecting switch resistance, inductor losses, and other parasitic, the relationship between these

parameters can be approximated by the following

2.5

Selection of a converter frequency is very important,since many applications require certain frequencies forEMI reasons. Higher switching frequencies allow theuse of smaller inductors but lead to high switching

losses. Conversely, lower frequencies can reduceswitching losses but require large inductors. Converterin the range of 20 kHz-100 kHz are generally suitable.Duty cycle is calculated as follows.

2.6

The equation above yield duty cycles greater than 100%. For the inductor rating, peak inductor current can beapproximated using the following equation

2.7

Select Q sw and DFor switching transistor Q sw, the most importantparameters are break down voltage, on resistance,peak current, and power dissipation. For the diode,the important parameters are reverse break down

voltage, peak repetitive current, and average forwardcurrent and reverse recovery time. Since peak inductorcurrent also flows through the switch and the diode, itmay be used to rate these components as well.

2.8

Maximum average current will occur at minimum input voltage

2.9

Figure 2.0 DC-DC Boost converter


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Average power dissipation in the switch can be estimated from the following equation.

3.0

Where R sw = switch on resistance

Select C in and C out Input capacitance C in functions as an input bypasscapacitor to reduce the effective source impedance

and also EMI by restricting high frequency currentpaths to short loops.. For the best performance C in impedance should be less than 1 .

3.1

Where

Output capacitance C out stores high voltage energyand also reduces EMI by restricting high frequencycurrent paths to stop short loops. The value of C out

must largely dependent on the desired ripple voltageon Vout . Generally, ripple (as a fraction of outputvoltage) of about 10% is adequate.

3.2

Where I out = output current to the inverter,

Ripple = and , both C in and C out should be high frequency

types

Select Timing Components ,

Timing components , determine

nominal converter frequency and maximum duty cycle.Selection of these components is an iterative process.

The ratio sets the maximum possible duty

cycle, while , together determine

nominal output frequency. Maximum duty cycle canalso be calculated using the following equation.

3.3

The ratio must be greater than for proper operation of the 555 timer. If less, timing capacitor

voltage will be unable to discharge to and output of the 555 will remain low. Nominal converter

frequency can be calculated using the following equation.

3.4

Select Feedback components

Output voltage is determine by the zener voltage plus an amount of bias voltage needed to vary duty cycle ofthe timing

3.5

The amount bias will vary depending on load and input voltage. The extreme limits of bias voltage are given inequations 3.6 and 3.7.

3.6

3.7

and

If is too low, it will prevent timing capacitor voltage from rising to as required for nominal

operation of the 555, resulting in switch staying on and the current rising to destructive levels. To prevent

this from occurring, the ratio of must always be greater than two.

3.8


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Power Inverter stage design However, in this part H-bridge inverter topology wasadopted. The half bridge inverter in the simplest formis shown is Figure 3.0 H-bridge inverter topology haslower number of switches and simplecontrol.(http://Guru/PEInverter/Half-bridge/pdf/inv.pdf, 2006). The duration of the deadband should be large enough to allow the switch that

is turned off to close before the other switch, to startconducting. The advantage of H-bridge invertertopology has lower number of switches and simplecontrol (Emadi and Stoyan, 1992). The differential

equation for the current from 0=t to 2T t = governing the current and voltage flows of half-bridgeinverter may be written as

( ) ( ) sV t Ridt

t di L =+ 3.9

Its general solution is of the form

( ) t s Ae R

V t i += 4.0

Where , the time constant for R-L load, is R

L= , Applying the initial condition, that is

( ) maxmin0 I I i == when 0=t , we get max I RVs

A = .

Hence, the expression for the current in the circuit during 0=t and 2T t = is

( ) ( ) t t e I e R

Vst i =

max1 4.1

The current attains its maximum value at 2T t = and is obtained from (4.1) as

( ) 2max2max 1 T e I e RV

I T s =

Rearranging the term we get

+

=

2

2

max

1

1T

T

e

e

R

Vs I 4.2

Hence the current in the circuit during 0=t and 2T t = is

( ) ( )

t

T

T st s e

e

e

R

V e

R

V t i

+=

2

2

11

1 4.3

Figure 3.0 (i) Half-bridge inverter with dual supply, and (ii) single supply

MATERIALS AND METHODSTwo methods were employed for the inverter designedand simulation; the computer simulation with NIMultisim 10. Software and the circuit analysis using

the above equations. The design topology which wasadopted is shown in Fig.4 with two stages; boostconverter and half-bridge inverter, and their controls.


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Figure 4.0 DC-AC Inverter circuit designRESULTS:Based on the simulation and the equations above the results were plotted using origin 50 with the followinggraphs:

0.0 0.2 0.4 0.6 0.8 1.0

0

2

4

6

8

10

Buck Converter Boost Converter Buck-Boost Converter

V out /V in

Duty cycle(%)

Figure 5.0 Graph of commonly converters compared analytically with equation 2.0 which shows the boost

converter has high gain.

0.88 0.89 0.90 0.91 0.92 0.93 0.94

8

10

12

14

16

18

Boost Converter

V out /V in

Duty Cycle

Figure 6.0 Graph of the gain to the duty cycle from the simulation result

1 2 3 4 5 6 7 8 9 10

0.0

0.2

0.4

0.6

0.8

1.0

B

D ( d u t y c y c l e )

RC

/RD

ratio

Figure 7.0 Graph of maximum duty cycle with ratio of timing resistors

Feedbac

DC-DC (BoostConverter

Control555-timer IC

DC-AC (H-Bridge

Control555-timer IC

1st Stage 2 nd Stage


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However, the nominal converter frequency waschosen to be 25 kHz which gave an inductor of 500 Hand inductor peak current I L(pk) using equations (2.6)and (2.7) with design power level of 700mW.Maximum duty is 70% at 6.0volt correspond to R C /R D

ratio of 3.4 with equation 10; the R C value was 40.5k which indicated R D is 11.9k . For maximum regulation,R FB should be greater than 81.0k since R C was40.5k then R FB is slightly greater than 81.0k whichapproximately 90.0k

0 20 40 60 80 100 120 140 160 1800

10

20

30

40

50

60

70

R C /R D =3 R C /R D =4 R C /R D =5 R C /R D =6 R C /R D =7 R C /R D =8 R C /R D =9 R C /R D =10

F r e q e n c y (

k H z )

R C kohms

Figure 8.0 Graph of nominal Converter frequency for C T = 1nF

2 3 4 5 6

20

40

60

80

100

120

140

160

180

200

R C /R D =3 R C /R D =4 R C /R D =5 R C /R D =6 R C /R D =7 R C /R D =8 R C /R D =9 R C /R D =10

V B I A S

( v o l t )

R FB /R C

Figure 9.0 Graph of Bias voltages plot from equation (3.7)

However, for maximum regulation R FB are slightly greater than two as equation (3.8). Since R C is 41.0k than R FB must be greater than 82 k but 90.0 k was selected.

Figure 10.0 Converter output voltage from the simulation

Figure 11.0 Output wave form for the two AND-gate


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Figure 12.0 Inverter output voltage from the simulationTable 1.0 Requirement for Comparison

12k

90k

41k

7uF

GND

DI S

OUT RST

VCC

THR

CON

TRI

500uH

2

0

0

3

5

1nF0

6

702BZ2.2

VN10LF

4.7uF

Fig. 2 DC-DC Converter

0

1

6-12v Cin

CTRD

RCRFB

555

ZFB

L

Qsw

D

Cout

VoutBAV21

XSC1

A B

Ext Trig+

+

_

_ + _

0

0

8

U1

LM555CM

GND

1

DI S7

OUT 3RST 4

VCC

8

THR6

CON5

TRI2

R124k

C2330nF

R210k

R3

42k Key=A

50%1

0 0

U3A74HC113D_4V

1J3

~1Q 6

1CLK 1

1K 2

1Q 5

~1PR

4

U4

3554AM

6

5

7

2

1

2

R4

10k

R5100k Key=A 50%

6

7

00

5

8

U2A

7408NU2C

7408N

10

11

9

U5A

MCT62

1

8

7

U5B

MCT64

3

5

6

R6

7k

R7

7k

3

12

0

13

14

0

XSC1

A B

Ext Trig+

+

_

_ + _

15

16

17

18

0

0

T1

TS_POWER_25_TO_1

12k

90k

41k

7uF

GND

DI S

OUT RST

VCC

THR

CON

TRI

500uH

1nF

02BZ2.2

VN10LF

4.7uF6-12v Cin

CTRD

RC RFB

555

ZFB

L

Qsw

D

Cout

VoutBAV21

0

22

0

27

26

0

25

24

0

0

23

0

20

4

Q3

IRFP250

Q2

IRFP250

XMM1

U7LM7812CT

LI NEVREG

COMMON

VOLTAGE

0 28

2119

S/no. Parameter Required Achieved1 Vout (converter) 48.0 V 100V2 Vout (inverter) 220V 220V3 P out (converter) 700mW 0.15W4 P out (inverter) 5kVA 2.09kVA5 Converter wave form Modulated Modulated6 Inverter wave form Modified Sine wave Square wave with

distortion

7 I out (inverter) 22.7 A 9.53A8 I Lpk 330mA 185.32mA9 Inverter Freq. 50Hz 48.6Hz10 Converter Freq. 25kHz 24.6kHz

Figure 14. DC-DC/DC-AC power Inverter stage

TS25T1 powerTransformer

Figure. 13 DC-DC converter stages with the nominal calculated values


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SUMMARY, DISCUSSION AND CONCLUSIONThe selection of the above topology simulated andworked with the nominal calculated values. It alsoshows that from table 1.0 the frequencies for bothconverter and inverter were close to the requiredvalues together with the wave forms.

In conclusion, the circuits when simulatedperformed much better than the expected result dueto calculation assumptions and approximations. Itshows that the duty cycle of the converter changecompletely as seen from the Fig. 6 when comparedwith general topologies in Fig.5. It is also shown fromFig. 11 output voltage high than expected.

REFERENCEShttp://Guru/PEInverters/introduction/pdf/inv.pdf 24th

February, 2006PC Sen. (2002), Power Electronics . Tata McGraw-Hill

publishing company limited, New Delhi, India,726p

Mukund R. Patel (2006). Wind and Solar PowerSystem: Design, Analysis, and Operation (2 nd edition). CRC press, Taylor & Francis, U.S.A221-224pp

Simon ANG and Alejandro Oliva (2005). PowerSwitching Converters, (2 nd edition) CRCpress, Taylor & Francis, 32-38pp

http://supertex.com/pdf/app-note/AN-36.pdf 23thJanuary, 2008.

http://Guru/PEInverte rs/Half-bridge/pdf/inv.pdf 24th February, 2006

Ali Emadi Abdolhosein Nasiri and Stoyan B Bekiarou(1992). Uninterruptable Power Supplies& ActiveFilter .53p

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5KVA Inverter Boost H-Bridge