Chapter 10-SPICE for Power Electronics and Electric Power-Book, Lecture notes of Power Electronics

Download Chapter 10-SPICE for Power Electronics and Electric Power-Book and more Lecture notes Power Electronics in PDF only on Docsity! 330 SPICE for Power Electronics and Electric Power, Second Edition EXAMPLE 10.1 FINDING THE PERFORMANCE OF A HALF-BRIDGE RESONANT- PULSE INVERTER WITH BJT SWITCHES A half-bridge resonant inverter is shown in Figure 10.1(a). The control voltages are shown in Figure 10.1(b). The DC input voltage is 100 V. The output frequency is fo = 5 kHz. The load resistance is R = 1 Ω, and the load inductance is L = 50 µH. Use PSpice to (a) plot the instantaneous output current io and the instantaneous input supply currently is and (b) calculate the Fourier coefficients of the output current io. The BJT parameters are IS = 2.33E–27, BF = 13, CJE = 1PF, CJC = 607.3PF, and TF = 26.5NS. SOLUTION The values of gate voltage Vg and base resistance Rb must be such that the transistors are driven into saturation at the expected load current. The PSpice schematic with The switching frequency {FREQ} and the duty cycle {DUTY_CYCLE} are defined as variables. The model parameters for the BJTs and the freewheeling diodes are as follows: FIGURE 10.1 Half-bridge resonant inverter. (a) Circuit, (b) control voltages. + − + − + − VY 4 µF 4 µF is C1 C2 L 5 6 7 8 3 9 10 R Rb2 vg1 vg1 vg2 Rb1 D1 Q1 Q2D2 (a) Circuit (b) Control voltages 1 0 4 2 50 µH 1 Ω 15 Ω 15 Ω 0 V VX ioVs 100 V 0 80 100 200 t (µs) t (µs)80 100 180 200 0 40 V vg2 40 V Resonant-Pulse Inverters 331 .MODEL QMOD NPN(IS=6.83E-14 BF=13 CJE=1pF CJC=607.3PF TF=26.5NS) for IGBTs .MODEL DMD D(IS=2.22E-15 BV=1200V CJO=0PF TT=0US)) for BJTs The listing of the circuit file is as follows: FIGURE 10.2 PSpice schematic for Example 10.1. Example 10.1 Half-bridge resonant inverter SOURCE
VS 1 0 DC 100V .PARAM DUTY_CYCLE=80 FREQ=5kHz Vg1 8 3 PULSE (0 40 0 1ns 1ns {{Duty_Cycle}/(200* {Freq})-2ns} {1/{Freq}}) Vg2 10 0 PULSE 0 40 {1/(2*{Freq})} 1ns 1ns {{Duty_Cycle}/(200*{Freq})-2ns} {1/{Freq}} CIRCUIT

Rb1 8 7 15 Rb2 10 9 15 VY 1 2 DC 0V ; Voltage source to measure supply current VX 3 6 DC 0V ; Measures load current R 6 5 1 L 5 4 50UH C1 2 4 4UF C2 4 0 4UF D1 3 2 DMOD ; Diode D2 0 3 DMOD ; Diode .MODEL DMOD D(IS=2. 2E–15 BV=1200V CJO=0 TT=0) ; Diode model parameters C1 4 uF QMOD Q2 8 Vx 0 V +− DMD D2 Vg1+− 9 + − E1 E GAIN = 10 C2 4 uF DMD D1 L 50 uH − − + + + − E2 E GAIN = 10 10 11 12 2 3 QMOD Q1 4 I Rb1 150 5 R 1 Vy 0 V + Vg2+− 6 Rb2 150 7 Vs 100 V + − Parameters: FREQ = 5 kHz DUTY_CYCLE = 80 − 334 SPICE for Power Electronics and Electric Power, Second Edition fo = 29.3 kHz. Use PSpice to (a) plot the instantaneous output current io and the instantaneous input supply current is and (b) calculate the Fourier coefficients of the output current io. Use voltage-controlled switches to perform the switching action. SOLUTION The inductor Ls acts as a current source. It is generally necessary to adjust the on time of the switches to match the resonant frequency of the circuit. The PSpice with proper dot signs. The switching frequency can be varied by changed by changing the parameter {Freq}. The model parameters for the switch are as follows: .MODEL SMD VSWITCH (RON=1M ROFF=10E6 VON=1V VOFF=0V) The listing of the circuit file is as follows: FIGURE 10.5 Parallel resonant inverter. (a) Circuit, (b) control voltages. + − + − + − 1 2 3 5 7 8 6 M M R 4 12 12 0.1 Ω 0.1 Ω 1.5 kΩ 0.5 mH 0.5 mH 0 V 0 V L C 0.01 µF Vy Ls Rs L1 L2 R1 L3 S1 R2 S2 vg2 Rg2 10 MΩ vg1 vg2 10 MΩRg1 vg1 4 mH 2 µH 0.1 CE 0.01 µF 13 13 0 9 10 11 Vx io 0 Vs 100 V (a) Circuit (b) Control voltages 0 17 34 t (µs) 0 17 34 t (µs) 20 V 20 V Resonant-Pulse Inverters 335 Note the following: (a) The PSpice pzplots of the instantaneous output current I(VX) and the current (b) The Fourier coefficients of the output current are as follows: Example 10.2 Push–pull resonant inverter SOURCE
VS 1 0 DC 100V .PARAM Freq=29.4kHz Vg1 12 0 PULSE (0 10V 0 1ns 1ns {1/(2*{Freq})-2ns} {1/{Freq}}) Rg1 12 0 10MEG Vg2 13 0 PULSE (0 10V {1/(2*{Freq})} 1ns 1ns {1/(2* {Freq})-2ns} {1/{Freq}}) Rg2 13 0 10MEG CIRCUIT

VX 9 10 DC 0V ; Measures load current VY 1 2 DC 0V ; Voltage source to measure supply current RS 3 4 0.1 LS 2 3 4MH CE 5 6 0.01UF L1 5 7 0.5MH R1 7 4 0.1 L2 4 8 0.5MH R2 8 6 0.1 L3 9 0 3.5MH K12 L1 L2 0.9999 K14 L1 L4 0.9999 K24 L2 L4 0.9999 L 10 11 2UH C 10 11 0.01UF R 11 0 1.5K S1 6 0 12 0 SMOD ; Voltage-controlled switch S2 5 0 13 0 SMOD ; Voltage-controlled switch .MODEL SMOD VSWITCH (RON=0.01 ROFF=10E+6 VON=1V VOFF=0MV) ANALYSIS


.TRAN 0.1US 120US ; Transient analysis .PROBE ; Graphics post-processor .OPTIONS ABSTOL = 1.00N RELTOL = 0.01 VNTOL = 0.1 ITL5=50000 .FOUR 29.3KHZ I(VX) ; Fourier analysis .END 336 SPICE for Power Electronics and Electric Power, Second Edition Note: Because of the constant-current inductor Ls, the input current remains approx- imately constant within a certain amount of ripples. The output voltage V(L3:L1) will depend on the turns ratio Ns/Np = (L3/L1)∫. 10.3 ZERO-CURRENT SWITCHING CONVERTERS (ZCSC) A power device of a ZCSC is turned on and off at zero current by using an LC- resonant circuit. The device remains on and provides a path for completing the FIGURE 10.6 PSpice schematic for Example 10.2. FOURIER COMPONENTS OF TRANSIENT RESPONSE I(VX) DC COMPONENT = 9.019181E−05 Harmonic No Frequency (Hz) Fourier Component Normalized Component Phase (Deg) Normalized Phase (Deg) 1 2.930E+04 7.433E–02 1.000E+00 5.243E+01 0.000E+00 2 5.860E+04 2.302E–02 3.098E–01 1.345E+02 8.202E+01 3 8.790E+04 8.330E–03 1.121E–01 −4.895E+01 −1.014E+02 4 1.172E+05 4.527E–03 6.091E–02 1.299E+02 7.745E+01 5 1.465E+05 2.641E–03 3.553E–02 −4.305E+01 −9.548E+01 6 1.758E+05 1.825E–03 2.456E–02 1.424E+02 8.993E+01 7 2.051E+05 1.300E–03 1.749E–02 −2.515E+01 −7.759E+01 8 2.344E+05 1.083E–03 1.457E–02 1.535E+02 1.011E+02 9 2.637E+05 8.259E–04 1.111E–02 −1.467E+01 −6.710E+01 TOTAL HARMONIC DISTORTION = 3.387136E+01 PERCENT + − + - SMD S1 12 R 1.5 k * 5 7 Vs 100 V + − Ce 0.01 uF 1 R1 0.1 Ls 4 mH Vy 0 V + + − − + - SMD S2 2 Parameters: Freq = 29.4 kHz 3 Rs 0.1 Vg2 10 V + − L3 3.5 mH L1 0.5 mH 4 Vg1 10 V +- L 2 uH K K123 Coupling = 0.9999 K_Linear L3 L1 L2 11 C 0.01 uF * 9 R2 0.5 mH 10 6 Vx 0 V +- 8 13 L2 0.1 Resonant-Pulse Inverters 339 The listing of the circuit file is as follows: The PSpice plots of the gate voltage V(9), the instantaneous capacitor voltage V(2,4), the capacitor current I(C), the diode voltage V(4), and the instantaneous power is 13.06 W. Example 10.3 ZCSC SOURCE
VS 1 0 DC 15V .PARAM Freq=8.33kHz Duty_Cycle=62.5 Rg 9 0 10MEG ; Control voltage Vg 9 0 PULSE (0 1 0 1ns 1ns {{Duty_Cycle}/(100* {Freq})-2ns} {1/{Freq}}) CIRCUIT

VY 1 2 DC 0V ; Voltage source to measure supply current VX 6 0 DC 0V ; Measures load current VN 7 3 DC 0V ; Measures the current-controlled switch R 5 6 55 LE 4 10 150UH RE 10 5 0.01 CE 5 0 20UF L 3 4 10UH C 2 4 20UF ; Initial condition DM 0 4 DMOD ; Diode .MODEL DMOD D (IS=2.2E−15BV=1200VCJO=0TT=0) ; Diode model parameters S1 2 8 9 0 SMOD ; Voltage-controlled switch .MODEL SMOD VSWITCH (RON=0.001ROFF=10E+6VON=10VVOFF=5V) W1 8 7 VN IMOD ; Current-controlled switch .MODEL IMOD ISWITCH (RON=1E+6 ROFF=0.01 ION=0 IOFF=1UA) ; Model parameters ANALYSIS


.TRAN 0.1US 400US ; Transient analysis .PROBE ; Graphics post-processor .OPTIONS ABSTOL = 1.00NRELTOL = 0.1VNTOL = 0.1ITL5=50000 .END 340 SPICE for Power Electronics and Electric Power, Second Edition EXAMPLE 10.4 FINDING THE PERFORMANCE OF AN L-TYPE ZERO-CURRENT SWITCHING RESONANT INVERTER voltage is shown in Figure 10.11(b). SOLUTION The circuit file is similar to that of Example 10.3, except that the capacitor C is connected across the diode Dm. The statement for C is changed to C 4 0 20UF ; No initial condition The PSpice plots of the instantaneous capacitor voltage V(2,4), the inductor current I(L), and the diode voltage V(4), and the instantaneous power of the IGBT switch 10.3, i.e., 13.06 W. FIGURE 10.10 Plots for Example 10.3. 20 W 10 V 20 V −10 V 0 W V (Z1:C, Z1:E)* IC(Z1) V (C:1, C:2) V (E1:3, E1:4) V (Dm:2) I (C) 0 V 0 V 0.6 ms 0.7 ms 0.8 ms Time 0.9 ms 1.0 ms 5 A 0 A −5 A −10 A 10 A SELL>> 0 V Switch power loss Capacitor voltage Gate voltage Capacitor current Diode voltage Resonant-Pulse Inverters 341 10.4 ZERO-VOLTAGE SWITCHING CONVERTER (ZVSC) A power device of a ZVSC is turned on when its voltage becomes zero because of the resonant oscillation. At zero voltage, the resonant current becomes maxi- mum. The device remains on and supplies the load current. The switching period must be long enough to complete the resonant oscillation. EXAMPLE 10.5 FINDING THE PERFORMANCE OF A ZERO-VOLTAGE SWITCHING INVERTER The DC input voltage is Vs = 15 V. The switching frequency is fs = 2.5 kHz. Use PSpice to plot the instantaneous capacitor voltage vc, the inductor current iL, the diode current vDm, and the load voltage vL. Use a voltage-controlled switch to perform the switching action. FIGURE 10.11 L-type ZCSC. (a) Circuit, (b) control voltage. FIGURE 10.12 PSpice schematic for Example 10.4. + − + − 1 2 8 9 S1VY vg L LB Re W1 Vn Dm Vx vg 0 V Ce 220 µF C 20 µF (a) Circuit (b) Control voltage Rg 10 MΩ 7 3 4 10 5 R 50 6 0 0 V Vs = 15 V 20 V 0 75 120 t (µs) 0 V 10 µH 150 µH 0.01 Ω 12 Le 150 uH −+ + − E1 E GAIN = 10 C 20 uF R 5Vs 15 V + − Vx 0 V + - 1 Z1 IXGH40N60 Re 0.01 Rg 250 2 Parameters: FREQ = 8.33 kHz DUTY_CYCLE = 62.5 Ce 20 uF 7 L 10 uH DMD Dm 3 Vn 0 V + Vg1 + − 4 10 5 9 Vy 0 V + 11 −− 344 SPICE for Power Electronics and Electric Power, Second Edition The PSpice plots of the instantaneous capacitor voltage V(2,4), the inductor current FIGURE 10.15 PSpice schematic for Example 10.5. FIGURE 10.16 Plots for Example 10.5. Vg +− 6 Parameters: Freq = 2.5 kHz DUTY_CYCLE = 30 Re 0.01 R 50 Le 150 uH 10 9 DMD D1 L 20 uH 2 Rg Ce 220 uF C 20 uF Vy 0 V + Vx 0 V + −− + + − E1 E Gain = 10 7 1 8 3 Vs 15 V + − − Z1 IXGH40N60 DMD Dm 54 1 k 11 V 10 V 9 V SEL>> 8 V V (Ce:2) 4.0 A −4.0 A 0 A 0.4 ms 0.6 ms 0.8 ms 1.0 ms Time 1.2 ms 1.4 ms 1.6 ms I (L) Inductor current Diode voltage Switch voltage Output voltage 20 V 0 V V (C:1, C:2) V (Dm:2) 346 SPICE for Power Electronics and Electric Power, Second Edition 10.5.2 EXPERIMENT RI.2 Single-Phase Full-Bridge Resonant Inverter FIGURE 10.17 Single-phase half-bridge resonant inverter. Objective To study the operation and characteristics of a single-phase full-bridge resonant (transistor) inverter. Applications The resonant inverter is used to control power flow in high-frequency applications, AC and DC power supplies, input stages of other converters, etc. Textbook Apparatus Warning See Experiment RI.1 Experimental procedure Set up the circuit as shown in Figure 10.18. Repeat the steps in Experiment RI.1. Report See Experiment RI.1. FIGURE 10.18 Single-phase full-bridge resonant inverter. R = 10 Ω 1 Ω Rs = 1 kΩ C2 D2 Vs 100 V C1 D1 Q1 Q2 Rs Rs Cs Vg1 Vg2Cs Cs = 0.1 µF 4 µF 4 µF 50 µH A AV V V R L Vs 100 V C 220 µF A AV Vg1 Q1 Rs R D1 D4 D2 D3 Cs Cs = 0.1 µF C L Rs Rs = 1 kΩCs Rs Cs Rs Cs Q4 Q2 Q3 Vg4 Vg2 Vg3 V V 1 Ω 4 µF 50 µH Resonant-Pulse Inverters 347 10.5.3 EXPERIMENT RI.3 Push–Pull Inverter 10.5.4 EXPERIMENT RI.4 Parallel Resonant Inverter Objective Applications applications, AC and DC power supplies, input stages of other converters, etc. Textbook Apparatus Warning See Experiment RI.1. Experimental procedure Report See Experiment RI.1. FIGURE 10.19 Push–pull inverter. Objective To study the operation and characteristics of a single phase push–pull parallel resonant (transistor) inverter. Applications The parallel resonant inverter is uded to control power flow in high-frequency applications, AC and DC power supplies, input stages of other converters, etc. Textbook See Reference 1, Section 8.3. Apparatus See Experiment RI.1. Warning See Experiment RI.1. Experimental procedure Report See Experiment RI.1. Vs 100 V Ce 200 µF A A V Vg1 Q1 D1D2 Cs = 0.1 µF Rs Rs = 1 kΩ Cs Rs Cs Q2 Vg2 V Vo RL = 10 Ω Resonant-Pulse Inverters 351 10.2 with the following specifications: DC supply voltage, Vs = 100 V Load resistance, R = 1 Ω Load inductance, L = 50 µH Load capacitance, C = 4 µF Output frequency should be as high as possible (a) Determine the ratings of all components and devices under worst-case conditions. (b) Use SPICE to verify your design. (c) Provide a cost estimate of the circuit. 10.3 specifications: DC supply voltage, Vs = 100 V Load resistance, R = 100 Ω Peak value of load voltage, Vp = 140 V Output frequency, fo = 1 kHz (a) Determine the ratings of all components and devices under worst-case conditions. (b) Use SPICE to verify your design. (c) Provide a cost estimate of the circuit. 10.4 following specifications: DC supply voltage, Vs = 100 V Load resistance, R = 1 kΩ Load inductance, L = 2 µH Load capacitance, C = 0.1 µF Peak value of load voltage, Vp = 140 V (a) Determine the ratings of all components and devices under worst-case conditions. (b) Use SPICE to verify your design. (c) Provide a cost estimate of the circuit. It is required to design the single-phase full-bridge resonant inverter of Figure 10.18 It is requir d to design the push–pull inverter of Figure 10.19 with the following It is required to design the parallel resonant inverter of Figure 10.20 with the 352 SPICE for Power Electronics and Electric Power, Second Edition 10.5 DC supply voltage, Vs = 20 V Load resistance, R = 100 Ω Average output voltage, V(DC) = 10 V with ±5% ripple (a) Determine the ratings of all components and devices under worst-case conditions. (b) Use SPICE to verify your design. (c) Provide a cost estimate of the circuit. 10.6 DC supply voltage, Vs = 20 V Load resistance, R = 100 Ω Average output voltage, V(DC) = 10 V with ±5% ripple (a) Determine the ratings of all components and devices under worst-case conditions. (b) Use SPICE to verify your design. (c) Provide a cost estimate of the circuit. 10.7 DC supply voltage, Vs = 20 V Load resistance, R = 100 Ω Average output voltage, V(DC) = 10 V with ±5% ripple (a) Determine the ratings of all components and devices under worst-case conditions. (b) Use SPICE to verify your design. (c) Provide a cost estimate of the circuit. 10.8 Use PSpice to find the worst-case minimum and maximum DC output voltages Vo(max) and Vo(min) for Problem 10.1. Assume uniform tolerances of ±15% for all passive elements and an operating temperature of 25°C. It is required to design the ZCSC of Figure 10.21 with the following specifications: It is required to design the ZCSC of Figure 10.22 with the following specifications: It is required to design the ZVSC of Figure 10.23 with the following specifications: