SEE 4433

POWER ELECTRONIC AND DRIVES

GROUP 3ASSIGNMENT 1

GROUP MEMBER:

NAME : AMIR SHAHRIZAL BIN MOHD ZAHLANMATRIX NO : SX091666EEJ03

NAME : AMMIRUL NIZAM BIN OTHMANMATRIX NO : SX091675EEJ03

NAME : AZIAN BINTI YAHAYAMATRIX NO : SX080513EEJ03

LECTURER:PM DR AWANG BIN JUSOH

CONTENTS PAGES

1. Quesstion

2. Problem Statement

3. Design Methodology“Calculation of all parameters

4. Simulation

5. Conclusion

6. Reference

7. Appendix

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1.0 Question

Problem (c)

Assuming that you working in Cellular Phone Company and your following task is to design a simple mobile converter to charge the phone's battery. The converter has the technical specification as follows:

Input voltage = 9 V Output voltage = -12 V Power rating = 5 W Peak-to-peak output ripple voltage < 2% (or 240mVpp) Switching frequency = 300 kHz Converter operating in CCM down to 5% load Assume all components are ideal

Establish the design parameter for this converter topology. Using MATLAB/Simulink/PSpice simulation package, conduct the computer simulation to verify and confirm the design that have been done. Submit a report with the following elements:

Cover page : Title, Group No and members etcIntroduction : Statement of problemDesign methodology : All calculation of parametersMATLAB/Simulink/PSpice simulation: Procedures, circuits, results and discussion. In the discussion, compare the theoretical design values and the obtained simulation results.

Conclusion: Summarize of work and result in few sentences.ReferencesAppendix: Datasheets of selected devices used in the design.

2.0 Problem Statement3

To design a simple mobile converter to charge the phone's battery with the specification Input

Voltage = 9 V and Output Voltage = -12 V with the output voltage always negative and the mobile

converter are operating in Continuous Current Mode (CCM) down to 5% load. The suitable

converter circuit to produce this output is Buck Boost Converter.

There are many types of dc-dc converter which is buck (step down) converter, boost (step up)

converter and buck-boost (step up-step down) converter. A Buck Boost Converter is a type of switch

mode converter that combine the principles of the Buck Converter and Boost Converter in a single

circuit. With the combination of these two regulator designs, it is possible to have a regulator circuit

that can cope with a wide range of input voltages both higher and lower than that needed by the

circuit

Figure a: buck-boost equivalent circuit

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Figure b: buck-boost operating state

The output voltage is of the opposite polarity as the input. This is a switched-mode power

supply with a similar circuit topology to the boost converter and the buck converter. The output

voltage is adjustable based on the duty cycle of the switching transistor. One possible drawback of

this converter is that the switch does not have a terminal at ground; this complicates the driving

circuit. Neither drawback is of any consequence if the power supply is isolated from the load circuit

(if, for example, the supply is a battery) as the supply and diode polarity can simply be reversed.

The switch can be on either the ground side or the supply side.

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3.0 Design Methodology

3.1 Specification given:

Input voltage = 9 V Output voltage = -12 V Power rating = 5 W Peak-to-peak output ripple voltage < 2% (or 240mVpp) Switching frequency = 300 kHz Converter operating in CCM down to 5% load Assume all components are ideal

3.2 Design Calculation

3.2.1 Duty Cycle

Vo = Vd ( D

(1−D)) T = 1f

= 1

3000 K h z= 3.33 µs

D = Vo

(Vo−Vd) DT = (0.571) (3.33µs) = 1.90 µs

= −12 V

(−12 V−9V ) = 0.571

3.2.2 Voltage

Switch Close Switch Open

VL = Vd = 9V VL = Vo = -12V

3.2.3 Load Current

P = IV ideal components Ps = Po

Io = PoVo

= 512

= 0.42A

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3.2.4 Converter Operating in Continuous Current Mode (CCM) down to 5% load

Find RNEW with output current drop to 5%

Io 5% = 5

100x 0.42 A = 0.021A

RNEW = Vo

I o 5 % = 571.4Ω

3.2.5 Minimum Inductance Value

Lmin = (1−D)2 xR

2 f

= (1−0.571 )2 x(571.4)

2(300 Khz)

= 175.3 µH

3.2.6 Output Voltage Ripple

∆Vo = Vo(D)RCf

= ∆Vo p-p / 2

= 0.24

2 = 0.12V

3.2.7 Capacitance, C

C = V o(D)∆ V o Rf

= 12(0.571)

(0.12 ) (571.4 )(300 Khz)= 0.33 x10−6 f= 0.33 µf

3.2.8 Ripple Factor

r = ∆ V oV o

= 0.1212

= 0.001 = 1%

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3.2.9 ∆IL, Imax & Imin

∆IL = (Vd )(DT )

L =

9 x1.90 µs175.3 µH

= 97.5mA

Imax = Io 5% + ∆ IL

2 = 0.021 +

97.5 mA2

= 69.8mA

Imin = Io 5% + ∆ IL

2 = 0.021 -

97.5 mA2

= 27.8mA

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4.0 Procedure

a. Begin by first opening up MultiSim.

b. To create file/Open a schematic file

c. As a default a blank file named “Curcuit 1” is opened up on the workspace. To save a

schematic under a different name simply click on File/Save As on the toolbar and enter the

name of your choice. To open an existing file click on File/Open on the toolbar and select

the file to open.

1. Placing components:On the toolbar select Place/Component, the following window will appear:

Figure 4.1 : Selecting a component

From this list, find all components needed to suit in buck boost converter circuit such as VDC,

transistor( Mosfet) , diode, inductor, resistor, capacitor, clock voltage and ground components.

Make the circuit for buck converter using the following parts:

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2. Editing the component

a. Once placed in the Mutisim window.

b. Arrange the components and wiring each other by clicking at the components terminal pin as

per buck boost converter circuit.

c. Set the parameter of the component by left double click on the component. Set parameter for

each of the components based on the specification given and the calculation done in the

previous section.

d. The schematic of the circuit as per below:

Figure 4.2.1 : Buck-Boost converter circuit drawing

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D1

R1571.4Ω

L1175.3µH

GND

C20.33µF

V19 V

V2

300kHz 15 V

Q2IRF232

5.0 Analysis & Simulation Results

a. Simulation plots for output Voltage & Current converter operating in CCM mode down to 5% load

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Specification Simulation Result Error

(%)

Result

Output Voltage -12 V -13.0 V 8.3 PASS

Output Current 0.021 A 0. 0228 A 8.6 PASS

Output Power 0.25W 0.29W 16 PASS

b. Peak-to-Peak Output Ripple Voltage

Result

Analysis

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CH1: 100mV/Div: Output ripple Voltage

Specification Simulation Result Result

Peak to peak output ripple voltage 240 mV Less than 200 mV PASS

D1

R1571.4Ω

L1175.3µH

GND

C20.33µF

V19 V

V2

300kHz 15 V

XSC1

A B

Ext Trig+

+

_

_ + _

XCP1Q2

IRF232

D1

R1571.4Ω

L1175.3µH

GND

C20.33µF

V19 V

V2

300kHz 15 V

XSC1

A B

Ext Trig+

+

_

_ + _

XCP1Q2IRF232

Perform analysis on the circuit to understand the operation each of the components used

1. Voltage and current across diode D1.

2. Voltage and current across inductor L1

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CH1: 10V/Div: VR of D1

D1

R1571.4Ω

L1175.3µH

GND

C20.33µF

V19 V

V2

300kHz 15 V

XSC1

A B

Ext Trig+

+

_

_ + _

XCP1Q2IRF232

3. Relationship between duty cycle (Vgs) and Vds of MOSFET Q1.

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CH1:10V/Div: VL of L1

CH2:100mA/Div: IL of L1

D1

R1571.4Ω

L1175.3µH

GND

C20.33µF

V19 V

V2

300kHz 15 V

XSC1

A B

Ext Trig+

+

_

_ + _

XCP1

Q2IRF232

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CH1: 10V/Div: Vds of Q1

CH2: 10V/Div: Duty cycle (Vgs) of Q1

SWIT

CH S

TATE Ton Toff

ON OFF ON OFF

VL

ILID

VD

IC

∆Q

Out

put

IloadVo

6.0 Discussion

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5.2.1.2 Summary of buck boost operation Typical waveform

t

t

t

t

t

t

t

7.0 Conclusion

8.0 References

1. Daniel W.Hart, Power Electronic, International EdittionMcGraw.Hill.2. SEE 4433: POWER ELECTRONICS AND DRIVES teaching module.

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