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SEE 4433 POWER ELECTRONIC AND DRIVES GROUP 3 ASSIGNMENT 1 GROUP MEMBER: NAME : AMIR SHAHRIZAL BIN MOHD ZAHLAN MATRIX NO : SX091666EEJ03 NAME : AMMIRUL NIZAM BIN OTHMAN MATRIX NO : SX091675EEJ03 NAME : AZIAN BINTI YAHAYA MATRIX 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 2 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 etc Introduction : Statement of problem Design methodology : All calculation of parameters MATLAB/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. References Appendix: Datasheets of selected devices used in the design. 3 2.0 Problem Statement 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 4 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. 5 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 ) T (1−𝐷) = = D = = 𝑉𝑜 (𝑉𝑜−𝑉𝑑) −12𝑉 (−12𝑉−9𝑉) DT 1 𝑓 1 3000𝐾ℎ𝑧 = 3.33 µs = (0.571) (3.33µs) = 1.90 µs = 0.571 3.2.2 Voltage Switch Close Switch Open VL = Vd = 9V VL = Vo = -12V 3.2.3 Load Current P = IV Io = 𝑃𝑜 𝑉𝑜 ideal components Ps = Po = 5 12 = 0.42A 6 3.2.4 Converter Operating in Continuous Current Mode (CCM) down to 5% load Find RNEW with output current drop to 5% Io 5% RNEW = 5 = 100 𝑥 0.42𝐴 = 0.021A 𝑉𝑜 Io 5% = 571.4Ω 3.2.5 Minimum Inductance Value Lmin = = (1−D)2 xR 2𝑓 (1−0.571)2 x(571.4) 2(300𝐾ℎ𝑧) = 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 = = Vo (𝐷) ∆Vo𝑅𝑓 12 (0.571) (0.12)(571.4)(300𝐾ℎ𝑧) = 0.33 𝑥 10−6 𝑓 = 0.33 µf 3.2.8 Ripple Factor r ∆Vo 0.12 = Vo = = 0.001 = 1% 12 7 3.2.9 ∆IL, Imax & Imin (𝑉𝑑)(𝐷𝑇) ∆IL = Imax = Io 5% + Imin = Io 5% + L = ∆𝐼𝐿 2 ∆𝐼𝐿 2 9 𝑥 1.90 µ𝑠 175.3 µH = 0.021 + = 0.021 - = 97.5mA 97.5𝑚𝐴 2 97.5𝑚𝐴 2 8 = 69.8mA = 27.8mA 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: 9 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: Q2 IRF232 D1 V2 V1 9V 300kHz 15 V L1 175.3µH C2 0.33µF GND Figure 4.2.1 : Buck-Boost converter circuit drawing 10 R1 571.4Ω 5.0 Analysis & Simulation Results a. Simulation plots for output Voltage & Current converter operating in CCM mode down to 5% load 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 16 PASS 0.29W 11 b. Peak-to-Peak Output Ripple Voltage CH1: 100mV/Div: Output ripple Voltage Result Specification Peak to peak output ripple voltage 240 mV 12 Simulation Result Result Less than 200 mV PASS Analysis Perform analysis on the circuit to understand the operation each of the components used 1. Voltage and current across diode D1. XSC1 Ext T rig + _ B A + XCP1 Q2 IRF232 D1 V2 V1 9V 300kHz 15 V L1 175.3µH C2 0.33µF R1 571.4Ω GND CH1: 10V/Div: VR of D1 CH2: 200mA/Div: Id of D1 13 _ + _ 2. Voltage and current across inductor L1 XSC1 Ext T rig + _ B A + Q2 IRF232 XCP1 D1 V2 V1 9V 300kHz 15 V L1 175.3µH C2 0.33µF GND CH1:10V/Div: VL of L1 CH2:100mA/Div: IL of L1 14 R1 571.4Ω _ + _ 3. Relationship between duty cycle (Vgs) and Vds of MOSFET Q1. XSC1 Ext Trig + _ B A + Q2 IRF232 _ + _ D1 V2 V1 9V 300kHz 15 V L1 175.3µH C2 0.33µF R1 571.4Ω GND XCP1 CH1: 10V/Div: Vds of Q1 CH2: 10V/Div: Duty cycle (Vgs) of Q1 15 Ton Toff ON OFF ON OFF t VL SWITCH STATE 5.2.1.2 Summary of buck boost operation Typical waveform IL t ID t VD t IC t t Output ∆Q t Iload Vo 16 6.0 Discussion 7.0 8.0 Conclusion References 1. Daniel W.Hart, Power Electronic, International EdittionMcGraw.Hill. 2. SEE 4433: POWER ELECTRONICS AND DRIVES teaching module. 17