BEKE 3673 INDUSTRIAL POWER ELECTRONICS ASSIGNMENT 2 4 BEKG S1 DR AZZIDDIN BIN MOHAMAD RAZALI No Name Matric no 1 ARIF HAIKAL BIN ABDUL HALIM B011910032 2 MOHAMMAD IRSYAD SYAHMI BIN MOHD ZAMRI B011910141 Youtube Video Link : https://youtu.be/mX8PR0j2YQA
PART 1 1.0 PROJECT BACKGROUND In this assignment, a switch mode power supply (SMPS) circuit needs to be designed. Given the input is a single-phase AC 240 , 50 Hz and converter rated output is 200W with variable DC output 0 to 60V. The converter output voltage ripple must not exceed 5% at steady state. 2.0 SWITCH MODE POWER SUPPLY DESIGN The rectifier is needed to convert from AC to DC because the input is in AC. Since the voltage is high, the AC voltage needs to step down using a transformer before the conversion into the rectifier. A capacitor is used to smoothed out the rectified DC Voltage and will be used as the supply voltage for buck converter. Assuming non-ideal IGBT with = 1 and =0.8 Figure 1: SMPS Circuit Design 3.0 CALCULATION The calculation started from the AC input side to determine the design parameter.
Using formula : Peak AC Input, 1, = √2 =√2 (240) =339.41V The buck converter output maximum is 60V. Assuming the input voltage for buck converter needed is 65V or more. Hence; ≥ 65 is the output voltage for rectifier. Hence, the input voltage for rectifier can be calculated using this formula: Peak voltage for input rectifier, = + 2 ≥ 65 + 2 ≥ 65 + 2(0.8) ≥ 66.6V To find transformer ratio: Transformer ratio, n=1 2 = 339.41 66.6 =5.096≅ 5 Using n=5 To find transformer output peak voltage, 2 = 1 5 = 339.41 5 = 67.882V Rectifier output peak voltage, = 67.882 − 2(0.8) =66.282V Then, the maximum duty ratio to output 60V can be calculated using this formula: = − − (1 − )
= (− − ) − D= + −+ D= 60+0.8 66.282−1+0.8 = 0.92 For continuous current mode (CCM), , > 0 Since ∆, = −− (1 − ) , = − ∆ 2 Hence , = − − 2 (1 − )T>0 = − − R 2 (1 − ) Let switching frequency f=20khz Load resistance R=10Ω = 60−0.8 60 ( 10 2(20) )(1-0.92) =19.73μH Let L=300μH To calculate output capacitance Ripple factor, r = ∆
Peak to peak ripple voltage, ∆= ∆ 8C = + 8LC (1-D) 2 ∆ = (1 + )(1 − ) 8 2 Since r≤ 0.05 Hence (1+ )(1−) 8 2 ≤ 0.05 C≥ (1+ )(1−) 8 2(0.05) C≥ (1+ 0.8 60 )(1−0.92) 8(300μ)(20) 2(0.05) =1.69x10−6F =1.69 μF We chose 25 μF as the capacitor. The parameter for the switch mode power supply (SMPS) as below: IGBT conduction voltage, 1V Diode conduction voltage, 0.8V Buck converter input voltage, 66.282V Transformer turns ratio,n 5 Output resistance, R 10Ω
Buck converter inductance, L 300μH Buck converter capacitance,C 25μF Switching frequency,f 20kHz Maximum duty ratio,D 0.92 ∴Theoritical inductor current ripple, maximum and minimum inductor current inductor current ripple ∆ , = | −− (1 − )| =| −60−0.8 (300μ)(20) (1 − 0.92)| =0.811A Maximum inductor current = 0 + ∆ 2 = 60 10 + 0.811 2 =6.4055A Minimum inductor current = 0 - ∆ 2 = 60 10 − 0.811 2 =5.5945A Average inductor current = 6.4055+5.5945 2 =6A ∆0 0 = (1 + 0.8 60)(1 − 0.92) 8(300μ)(20) 2 (25) =0.00342=0.342%
Simulation Results From the simulation, the result of the voltage of buck converter Vs, output voltage Vo, inductor current iL, and diode current iD.
Figure 2 Vs against time Figure 3 iL against time
Figure 4 Graph of diode current against time Figure 5 Output voltage Vo against time From Figure 2, , = 66.0159 + 65.9658 2 = 65.991 From Figure 3, For average inductor current,
, = 6.40661 + 5.8131 2 = 6.12 For ripple inductor current, Δ = 6.40661 − 5.8131 = 0.5935 From figure 5, For average output voltage and ripple, , = 61.1664 + 60.9565 2 = 61.0614 Δ = 61.1664 − 60.9565 61.0614 = 0.0034 = 0.34% Table 1 Comparison between theoretical and simulation results Parameters Theoretical Simulation Buck voltage, Vs 66.282V 65.991 V Output voltage, Vo 60V 61.0614 V Output voltage ripple, r 0.342% 0.34% Inductor current, 6A 6.12 A , 6.4055A 6.40661 A , 5.594A 5.8131 A Current ripple,Δ 0.811A 0.5935 A
SMPS WTIH PI CONTROLLER Using the same SMPS circuit, a PI controller is added to allow feedback control of the output voltage. To design the PI controller, the transfer function of each component is needed. Transfer function for the buck converter, () () = ( 2 + ( 1 ) + 1 ) () () = 65.991 7.5 ( 2 + ( 1 4000) + 1 200000000) Transfer function for PWM, () () = 1 1 = 1 Transfer function for PI controller, () () = + The design in Figure is used in Simulink to obtain bode plot and response of the system for PI controller Figure 6 Block diagram of the system
Figure 7 Block diagram for PI controller Figure 8 Block diagram for PWM Figure 9 Block diagram for buck converter
Using = 0.5 and = 5000 for initial simulation, the result obtained by bode plot and pole-zero map in figures below. Figure 10 Bode plot for = 0.5 and = 5000 Figure 11 Pole-zero map for = 0.5 and = 5000 − , = 68.4 2 = 10.89
ℎ = −185 − (−180) = −5° The cut-off frequency and phase margin does not fulfil the condition of the stable control loop. Cut-off frequency need to between 30° to 60°. Then the cut-off frequency needs to be ¼ or 1/5 of switching frequency. Therefore, the initial condition needs to change. To obtain phase margin 45°, = 20 log = 20 log 5000 = 73.98 Shifting down, = 73.98 − 39.5 = 34.48 = 20 log = 10 34.48 20 = 52.97 = ( ) = 0.5 5000 (52.97) = 0.0053
By using the value of = 0.0053 and = 52.97 Figure 12 Bode plot for = 0.0053 and = 52.97 Figure 13 for = 0.0053and = 52.97 − , = 14.1 2 = 2.244
ℎ = −123 − (−180) = 57° Based on the calculations above, this has shown that the condition of stable control has been fulfilled. Applying the PI controller into SMPS circuit below: Figure 14 SMPS circuit connect with PI controller Figure 15 Output voltage waveform
Figure 16 Inductor current waveform Calculation and analysis, From Figure, , = 61.163 + 58.8805 2 = 60.02 Δ = 61.163 − 58.8805 60.02 = 0.038 = 0.38% From figure, , = 6.7996 + 5.353 2 = 6.0763 Δ = 6.7996 − 5.353 = 1.4466 From this simulation, the overshoot of output voltage is 63.675 V, and the overshoot of inductor current is 9.6123 A YouTube video link: