Chapter 7-SPICE for Power Electronics and Electric Power-Book, Lecture notes of Power Electronics

Download Chapter 7-SPICE for Power Electronics and Electric Power-Book and more Lecture notes Power Electronics in PDF only on Docsity! 202 SPICE for Power Electronics and Electric Power, Second Edition are affected by the area factor are marked with an asterisk (*) in the descriptions of the model parameters. The diode is modeled as an ohmic resistance (value = RS/area) in series with an intrinsic diode. The resistance is attached between node NA and an internal anode node. [(area) value] scales IS, RS, CJO, and IBV, and defaults to 1. IBV and BV are both specified as positive values. The DC characteristic of a diode is determined by the reverse saturation current IS, the emission coefficient N, and the ohmic resistance RS. Reverse FIGURE 7.1 SPICE diode model with reverse-biased condition. FIGURE 7.2 SPICE small-signal diode model. A A K K CDIDVDD1 ID Rs + − A K CDRDVD + − Rs Diode Rectifiers 203 breakdown is modeled by an exponential increase in the reverse diode current and is determined by the reverse breakdown voltage BV, and the current at breakdown voltage IBV. The charge storage effects are modeled by the transit time TT and a nonlinear depletion layer capacitance, which depends on the zero- bias junction capacitance CJO, the junction potential VJ, and grading coefficient M. The temperature of the reverse saturation current is defined by the gap acti- vation energy (or gap energy) EG and the saturation temperature exponent XTI. The most important parameters for power electronics applications are IS, BV, IBV, TT, and CJO. FIGURE 7.3 Static diode model with reverse-based condition. TABLE 7.1 Parameters of Diode Model Name Area Model parameter Unit Default Typical IS * Saturation current A 1E−14 1E−14 RS * Parasitic resistance W 0 10 N Emission coefficient 1 1 TT Transit time sec 0 0.1NS CJO * Zero-bias p-n capacitance F 0 2PF VJ Junction potential V 1 0.6 M Junction grading coefficient 0.5 0.5 EG Activation energy eV 1.11 11.1 XTI IS temperature exponent 3 3 KF Flicker noise coefficient 0 AF Flicker noise exponent 1 FC Forward-bias depletion capacitance coefficient 0.5 BV Reverse breakdown voltage V • 50 IBV * Reverse breakdown current A 1E−10 A K Rs VD + − ID 206 SPICE for Power Electronics and Electric Power, Second Edition At a specified temperature, the leakage current Is remains constant for a given diode. For power diodes, the typical value of Is is 10−15 A. The plots of the diode characteristics can be viewed from the View menu of the PSpice Model Editor as shown in Figure 7.6. These include forward current (Vfwd vs. Ifwd), junction capacitance (Vrev vs. Cj), reverse leakage (Vrev vs. Irev), reverse breakdown (Vz, Iz, Zz), and reverse recovery (Trr, Ifwd, Irev). 7.5 DIODE PARAMETERS allows specifying many parameters, we use only the parameters that significantly affect power converter output. From the data sheet, we have Reverse breakdown voltage, BV = 1200 Reverse breakdown current, IBV = 13 mA Instantaneous voltage, vF = 1 V at iF = 150 A Reverse recovery charge, QRR = 194 µC at IFM = 200 A FIGURE 7.6 PSpice Model Editor. Diode Rectifiers 207 From the data for the V−I forward characteristic of a diode, it is possible to determine the value of n, VT, and Is of the diode [4]. Assuming that n = 1 and VT = 25.8 mV, we can apply Equation 7.1 to find the saturation current IS : FIGURE 7.7 Data sheets for IR diodes of type R18. (Courtesy of International Rectifier.) R18C, R18S, R18CR & R18SR SERIES 1800–1200 VOLTS RANGE 185 AMP AVG STUD MOUNTED DIFFUSED JUNCTION RECTIFIER DIODES VOLTAGE RATINGS MAXIMUM ALLOWABLE RATINGS VOLTAGE CODE (1) VRRH′ VR − (V) VRSH − (V) Max. rep. peak reverse and direct voltage Max. non-rep. peak reverse voltage TJ = 0° to 200°C TJ = −40° to 0°C TJ = 25° to 200°C 50 Hz half cycle sine wave Non-lubricated threads Lubricated threads Non-lubricated threads Lubricated threads 60 Hz half cycle sine wave Initial TJ = 200°C, rated VRPM applied after surge. Initial TJ = 200°C, rated VRPM applied after surge. 50 Hz half cycle sine wave 60 Hz half cycle sine wave Initial TJ = 200°C, no voltage applied after surge. Initial TJ = 200°C, no voltage applied after surge. 18A 16A 14B 12B 1710 1520 1330 1200 1900 1700 1500 1300 1800 1600 1400 1200 PARAMETER SERIES (2) VALUE UNITS NOTES T Tstg IF(AV) Mounting torque Min. Max. N*m (lbf− in) N*m (lbf− in) Min. Max. Min. Max. Min. Max. Storage temperature Tj Junction temperature ALL Average current IF(AV) Max. av. current (3) IF(RMS) Max. RMS current (3) IFSH Max. peak non–rep. surge current I 2 t I 2 t for time tx = I 2√ − t · √ − tx * 0.1 ≤ tx ≤ 10 ms. Max. I 2 t capability I 2√ − t Max. I 2√ − t capability −40 to 200 °C ALL ALL −40 to 200 °C R18C/S R18C/CR 185 180° half sine wave, TC = 140°C 180° half sine wave, TC = 133°C, 18C/S TC = 145°C, R18CR/SR A 200 A ALL 314 ALL 3820 4000 4550 4750 ALL 73 ALL 1040 14.1 (125) 17.0 (150) 12.2 (108) 15.0 (132) R18S/SR 11.3 (100) 14.1 (125) 9.5 (85) 12.5 (110) 67 104 95 A A kA 2 s kA 2√ − s t = 10 ms t = 8.3 ms Initial TJ = 200°C, no voltage applied after surge. t = 10 ms t = 8.3 ms (1) To complete the part number, refer to the Ordering Information table. (2) R18C & R18S series have cathode-to-case polarity. R18CR & R18SR series have anode-to-case polarity. (3) For devices assembled in Europe, max. I F(AV) is 175 A and max. I F(RMS) is 275 A. 208 SPICE for Power Electronics and Electric Power, Second Edition which gives IS = 2.2 E – 15 A. Let us call the diode model name DMOD. The values of TT and CJO are not available from the data sheet. Some versions of SPICE (e.g., PSpice) support device library files. The software PARTS of PSpice can generate SPICE models from the data sheet parameters of transistors and diodes. The transit time τT can be calculated approximated from We shall assume the typical value CJO = 2PF. Thus, the PSpice model statement is .MODEL DMODD (IS=2.22E−15 BV=1200 IBV=13E−3 CJO=2PF TT=1US) FIGURE 7.7 (continued). CHARACTERISTICS PARAMETER SERIES MIN. TYP. MAX. UNITS TEST CONDITIONS VFM Peak forward voltage VF(TO)1 Low-level threshold VF(TO)2 High-level threshold rF1 Low-level resistance rF2 High-level resistance ta Reverse current rise tb Reverse current fail IRM(REC) Reverse current QRR Recovered charge IRM Peak reverse current RthJC Thermal resistance, junction-to-case RthCS Thermal resistance, case-to-sink wt Weight Case style ALL ALL ALL ALL ALL ALL ALL ALL ALL ALL R18C/S R18CR/SR R18C/S R18C/S R18CR/SR R18CR/SR R18C/CR R18S/SR 1.30 18.0 3.5 18 194 13 110(3.5) DO-205AC (DO-30) DO-205AA (DO-8) 1.42 0.703 0.738 1.100 1.080 20 0.200 0.250 0.271 0.221 0.275 0.225 0.10 V V πΩ µs µs µc °C/W °C/W °C/W °C/W g(oz.) JEDEC A mA Initial TJ = 25°C, 50–80 Hz half sine, Ipeak = 628 A.— — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — Tj = 200°C Tj = 175°C, IFM = 200 A, diR/dt = 1.0 A/µs. Av. power = VF(TO) * IF(AV) + rF * (IF(RMS)) 2 Use low level values for IFM ≤ πIF(AV) IFM QRR iRM (REC) ts tb i t diR dt 1/4 IRM (REC) Tj = 175°C. 180° sine wave 120° rectangular wave Mtg. surface smooth, flat and greased. Max. rated VRRM. DC operation I I e I e V nV D s / s D T= − = −× − ( ) ( )/( . ) 1 150 11 25 8 10 3 τ µ µT RR FM C sec= = ≈Q I 194 200 1 Diode Rectifiers 211 7.5.2 TABULAR DATA EXAMPLE 7.1 DESCRIBING THE DIODE CHARACTERISTIC OF TABULAR DATA DC = 220 V and RL = 0.5 Ω. Use PSpice to calculate the diode current and the diode voltage. Use the diode charac- teristics in Table 7.2. FIGURE 7.7 (continued). 6.70 (0.264) Dia. Min. 16.51 (0.650) Max. Ceramic housing 26.97 (1.062) Max. Across flats 1.2.20 UNF-2A 114.3 (4.5) Nom. 41.27 (1.625) Max. 25.16 (0.951) Max. 8.81 (0.347) Max. 16.26 (0.64) Max. + R18C, R18S, R18CR & R18SR SERIES 1800–1200 VOLTS RANGE 212 SPICE for Power Electronics and Electric Power, Second Edition FIGURE 7.7 (continued). TABLE 7.2 Typical I–V Data of a Power Diode ID(A) 0 20 40 100 500 800 1000 1600 2000 3000 3900 VD(V) 0 0.8 0.9 1.0 1.26 1.5 1.7 2.0 2.3 3 3.5 6.70 (0.264) Dia. Min. 16.51 (0.650) Max. Ceramic housing 26.97 (1.062) Max. Across flats 3/8-24UNF-2A 114.3 (4.5) Nom. 41.27 (1.625) Max. 25.16 (0.951) Max. 8.81 (0.347) Max. 16.76 (0.64) Max. + Dimensions in millimeters and (Inches) Diode Rectifiers 213 in Figure 7.9(a). The diode is modeled by an ETABLE from the abm.slb library. The current is related to the voltage in table form as shown in Figure 7.9(b). The listing of the circuit file follows: FIGURE 7.8 Diode circuit. (a) Diode circuit, (b) PSpice circuit. FIGURE 7.9 PSpice schematic for Example 7.1. (a) ETABLE, (b) ETABLE parameters. Example 7.1 Diode circuit VDD 1 0 DC 220V ; DC voltage of 15 V VX 3 2 DC 0V ; measures the diode current ID RL 2 0 0.4 * The diode is represented by a table Ediode 1 3 TABLE { I(VX) } = + (0, 0) (20, 0.8) (40, 0.9) (100, 1.0) (500, 1.26) (800, 1.5) + (1000, 1.7) (1600, 2.0) (2000, 2.3) (3000, 3) (3900, 3.5) .OP ; Prints the details of operating point .END ; End of circuit file +VD VDC RL VDC Ediode + + + − − + − − − 1 1 D1 Vx ID 2 0 0 0.5 0V RL 0.5 2 3 VDD 220 V + − Vx 0 V +− RL 0.4 0 E1 {I(VX)} Etable 1 2 3 IN + O U T + IN + O U T − (a) 216 SPICE for Power Electronics and Electric Power, Second Edition FIGURE 7.11 PSpice schematic for Example 7.2. (a) Schematic, (b) diode model param- eters, (c) transient setup, (d) Fourier setup. 2 Vs 170 V 60 Hz + − V i o 1 4 R 0.5 3 + V o − L 6.5 mH 0 (a) Vx 0 V + − DMOD D1 Diode Rectifiers 217 FIGURE 7.11 (continued). FIGURE 7.12 Plots for Example 7.2. 15 ms 200 V I (Vx) 0 V 0 A 50 A 100 A Load current (24.277 m, 102.101) −200 V 20 ms 30 ms Time Output voltage 40 ms 50 ms SEL>> V (R:2) 218 SPICE for Power Electronics and Electric Power, Second Edition (b) Fourier coefficients and THD will depend slightly on the internal time step TMAX discussed in Subsection 6.9.2. (c) To find the input power factor, we need to find the Fourier series of the input current, which is the same as the current through diode D1. DC input current Iin(DC) = 44.39 A Rms fundamental input current, I1(rms) = 55.07/ = 38.94 A THD of input current THD = 15.48% = 0.1548 Harmonic input current, Ih(rms) = I1(rms) × THD = 38.94 × 0.1548 = 6.028 A Rms input current Is = [I2in(dc) + I2r(rms) + I2h(rms)]1/2 = (44.392 + 38.942 + 6.0282)1/2 = 59.36 A Displacement angle φ1 = −64.38° Displacement factor DF = cos φ1 = cos(−67.38) = 0.3846(lagging) THE FOURIER COMPONENTS OF TRANSIENT RESPONSE V (2) DC COMPONENT = 2.218909E+01 Harmonic No Frequency (Hz) Fourier Component Normalized Component Phase (Deg) Normalized Phase (Deg) 1 6.000E+01 1.377E+02 1.000E+00 1.119E+01 0.000E+00 2 1.200E+02 3.646E+01 2.647E−01 −1.642E+02 −1.754E+02 3 1.800E+02 2.651E+01 1.925E−01 −1.076E+02 −1.188E+02 4 2.400E+02 1.649E+01 1.197E−01 −4.324E+01 −5.442E+01 5 3.000E+02 9.585E+00 6.958E−02 3.968E+01 2.849E+01 6 3.600E+02 8.108E+00 5.887E−02 1.366E+02 1.254E+02 7 4.200E+02 8.486E+00 6.161E−02 −1.431E+02 −1.542E+02 8 4.800E+02 7.607E+00 5.523E−02 −7.090E+01 −8.209E+01 9 5.400E+02 5.897E+00 4.281E−02 6.598E+00 −4.591E+00 TOTAL HARMONIC DISTORTION = 3.720415E+01 PERCENT THE FOURIER COMPONENTS OF TRANSIENT RESPONSE I (D1) DC COMPONENT = 4.438586E+01 Harmonic No Frequency (Hz) Fourier Component Normalized Component Phase (Deg) Normalized Phase(Deg) 1 6.000E+01 5.507E+01 1.000E+00 −6.738E+01 0.000E+00 2 1.200E+02 7.448E+00 1.352E−01 1.107E−02 1.781E+02 3 1.800E+02 3.634E+00 6.598E−02 1.681E+02 2.355E+02 4 2.400E+02 1.606E+00 2.916E−02 −1.317E+02 −6.432E+01 5 3.000E+02 8.217E−01 1.492E−02 −5.059E+01 1.678E+01 6 3.600E+02 5.774E−01 1.048E−02 5.326E+01 1.206E+02 7 4.200E+02 4.422E−01 8.030E−03 1.264E+02 1.938E+02 8 4.800E+02 4.177E−01 7.584E−03 −1.633E+02 −9.597E+01 9 5.400E+02 2.892E−01 5.251E−03 −7.644E+01 −9.061E+00 TOTAL HARMONIC DISTORTION = 1.548379E+01 PERCENT 2 Diode Rectifiers 221 Note the following: (a) The PSpice plots of the instantaneous output voltage V(3, 4) and load current load current contains ripples and has not reached steady-state conditions. (b) The input current, which is the same as the current through the voltage source VY, is equal to I(VY). DC input current Iin(DC) = −2.56 A, which should ideally be zero Rms fundamental input current I1(rms) = 259.5/ = 183.49 A THD of input current THD = 42.26% = 0.4226 Rms harmonic current Ih(rms) = I1(rms) × THD = 183.49 × 0.4226 = 77.54 A D1 2 3 DMOD D3 0 3 DMOD D2 4 0 DMOD D4 4 2 DMOD .MODEL DMOD D(IS=2.22E−15 BV=1200V IBV=13E− 3 CJO=2PF TT=1US) ANALYSIS


TRAN 10US 50MS 33.3333MS 10US ; Transient analysis .FOUR 60HZ 1(VY) ; Fourier analysis of * input current .PROBE ; Graphic POST- post-processor .OPTIONS ABSTOL = 1.0 N RELTOL = .01 BNTOL = 1.0M ITL5=10000 ; * Convergence .END THE FOURIER COMPONENTS OF TRANSIENT RESPONSE I(VY) DC COMPONENT = −2.56451E + 00 Harmonic No Frequency (Hz) Fourier Component Normalized Component Phase (Deg) Normalized Phase(Deg) 1 6.000E + 01 2.595E+02 1.000E+00 −3.224E+00 0.000E+00 2 1.200E + 02 7.374E−01 2.842E−03 1.410E+02 1.442E+02 3 1.800E + 02 8.517E+01 3.282E−01 4.468E+00 7.693E+00 4 2.400E + 02 5.856E−01 2.257E−03 1.199E+02 1.232E+02 5 3.000E + 02 5.118E+01 1.972E−01 3.216E+00 6.440E+00 6 3.600E + 02 5.526E−01 2.130E−03 1.111E+02 1.143E+02 7 4.200E + 02 3.658E+01 1.410E−01 2.868E+00 6.092E+00 8 4.800E + 02 5.406E−01 2.083E−03 1.065E+02 1.097E+02 9 5.400E + 02 2.846E+01 1.097E−01 2.822E+00 6.047E+00 TOTAL HARMONIC DISTORTION = 4.225668E + 01 PERCENT 2 222 SPICE for Power Electronics and Electric Power, Second Edition Rms input current Is = (I2in(DC) + I21(rms) + I2h(rms))1/2 = (2.562+183.492+77.542)1/2=199.22 A Displacement angle φ1 = −3.22 Displacement factor DF = cos φ1 = cos(−3.22) = 0.998 (lagging) Thus, the input power factor is Assuming that Iin(DC) = 0, Equation 7.3 gives the power factor as Note: The load current is continuous. The input power factor (0.919) is much higher compared to that (0.3802) of the half-wave rectifier. EXAMPLE 7.4 FINDING THE PERFORMANCE OF A SINGLE-PHASE BRIDGE RECTIFIER sinusoidal input voltage has a peak of 169.7 V, 60 Hz. The load inductance L is 10 mH, and the load resistance R is 40 Ω. The filter inductance Le is 30.83 mH, and filter capacitance Ce is 326 µF. Use PSpice (a) to plot the instantaneous output FIGURE 7.15 Plots for Example 7.3. 212.5 A 200.0 A 187.5 A 200 V 150 V 100 V SEL>> 0 V 32 ms 35 ms V (3,4) I (VX) AVG (V(3,4)) 40 ms Time 45 ms 50 ms Output voltage Output current Average output voltage PF rms= = × = I Is 1 1 183 49 199 22 0 998 0 9192( ) cos . . . .φ (lagging) PF / (lagging)= + × =1 1 0 4226 1 2 0 9981 0 9193 2( . ) . . Diode Rectifiers 223 voltage vo and the load current io, (b) to calculate the Fourier coefficients of the output voltage, (c) to calculate the Fourier coefficients of the input current and input power factor, and (d) plot the instantaneous output voltage for Ce = 1 µF, 100 µF, and 326 µF. SOLUTION e variable CVAL and the setup for parametric sweep is shown in Figure 7.17(b). Peak voltage Vm = 169.7 V, and f = 60 Hz. The listing of the circuit file is as follows: FIGURE 7.16 Single-phase bridge rectifier with load filter. Example 7.4 Single-phase bridge rectifier with RL load SOURCE
VS 1 0 SIN (0 169.7V 60Hz) CIRCUIT

LE 3 7 30.83MH .PARAM CVAL = 326UF CE 7 4 326UF R 7 5 40 L 5 6 10MH VX 6 4 DC 0V ; Voltage source to measure the output current VY 1 2 DC 0V ; Voltage source to measure the input current D1 2 3 DMOD D3 0 3 DMOD D2 4 0 DMOD D4 4 2 DMOD .STEP PARAM LIST 1UF 100UF 3264F .MODEL DMOD D(IS=2.22E−15 BV=1200V IBV=13E−3 TT=1US) 30.83 mH 0 V + + − − 1 2 4 3 7 L R 326 µF Le 0 V 6 5 10 mH 40 Ω 0 Vyis vs CeVo io D3 Vx D1 D2D4 226 SPICE for Power Electronics and Electric Power, Second Edition .FOUR 60HZ I(VY) ; Fourier analysis of input current we get THD of input current THD = 42.16% = 0.4216 Displacement angle φ1= −3.492° Displacement factor DF = cos φ1 = cos(−3.492) = 0.9981(lagging) Neglecting the DC input current Iin(DC) = −2.229 A, which is small relative to the fundamental component, we can find power factor from Equation 7.3 as (d) ripple on the output voltage is reduced. RECTIFIER a balanced three-phase balanced supply whose per-phase voltage has a peak of 169.7 V, 60 Hz. The load inductance L is 6.5 mH, and the load resistance R is 0.5 Ω. Use PSpice (a) to plot the instantaneous output voltage vo and line (phase) current ia, (b) to plot the rms and average currents of diode D1, (c) to plot the average output power, and (d) to calculate the Fourier coefficients of the input current and the input power factor. FOURIER COMPONENTS OF TRANSIENT RESPONSE I(VY) DC COMPONENT = −2.229026 E+00 Harmonic No Frequency (Hz) Fourier Component Normalized Component Phase (Deg) Normalized Phase(Deg) 1 6.000E+01 2.555E+02 1.000E+00 −3.492E+00 0.000E+00 2 1.200E+02 1.146E+00 4.486E−03 1.192E+02 1.227E+02 3 1.800E+02 8.372E+01 3.277E−01 3.125E+00 6.617E+00 4 2.400E+02 1.092E+00 4.273E−03 1.044E+02 1.079E+02 5 3.000E+02 5.024E+01 1.967E−01 1.036E+00 4.528E+00 6 3.600E+02 1.082E+00 4.237E−03 9.819E+01 1.017E+02 7 4.200E+02 3.586E+01 1.404E−01 −1.011E−01 3.391E+00 8 4.800E+02 1.070E+00 4.187E−03 9.446E+01 9.795E+01 9 5.400E+02 2.788E+01 1.091E−01 −9.085E−01 2.583E+00 TOTAL HARMONIC DISTORTION = 4. 216000E+01 PERCENT PF (lagging)= + × =1 1 0 4216 0 9981 0 9197 2 1 2( . ) . . / Diode Rectifiers 227 SOLUTION Peak voltage per phase Vm = 169.7 V, and f = 60 Hz. The listing of the circuit file is as follows: FIGURE 7.19 Effects of filter capacitances for Example 7.4. Example 7.5 Three-phase bridge rectifier SOURCE
Van 8 0 SIN (0 169.7V 60HZ) Vbn 2 0 SIN (0 169.7V 60Hz 0 0 120DEG) Vcn 3 0 SIN (0 169.7V 60Hz 0 0 240DEG) Ce = 1 µF 326 µF 100 µF 150 V 100 V 50 V 0 V 32 ms 35 ms V (Le:2, D4:1) 40 ms Time 45 ms 50 ms 8 Van Vy ib ia io ic ce von 0 1 0 V Vcn Vbn D1 D3 D4 D6 D5 R L 6.5 mH 0 VVx 0.5 Ω D2 + − 3 5 2 4 7 6+ + − − + + 228 SPICE for Power Electronics and Electric Power, Second Edition CIRCUIT

CE 4 5 1UF ; Small capacitance to aid convergence R 4 6 0.5 L 6 7 6.5MH VX 7 5 DC 0V ; Voltage source to measure the output current VY 8 1 DC 0V ; Voltage source to measure the input current D1 1 4 DMOD D3 2 4 DMOD D5 3 4 DMOD D2 5 3 DMOD D6 5 2 DMOD D4 5 1 DMOD .MODELDMODD (IS=2.2 2E−15 BV=1200V IBV=13E− 3 CJO=2PF TT=1US) ANALYSIS


.TRAN 10US 33.3333MS 0 10US ; Transient analysis .FOUR 60Hz 1(VY) ; Fourier analysis of line current .PROBE ; Graphics post-processor .OPTIONS ABSTOL = 1.0N RENTOL = 1.0M VNTOL = 1.0M ITL5=10000 ; Convergence * . END FIGURE 7.21 Three-phase bridge rectifier schematic for Example 7.5. ib V+ 7 DMOD D6 ia −DMOD D1 DMOD D5 Ce 1 uF Vbn 170 V 60 Hz +− 6 DMOD D4 8 + Vy 0 V + DMOD D3 Vx 0 V + − 3 0 ic 2 Van + − 4 R 0.5 L 6.5 mH 1 n Vcn+− io DMOD D2 I 5 V o V− − Diode Rectifiers 231 Plot the worst-case line current ia if the resistances, the inductances and capacitances change by ±20%. SOLUTION measure the line current. The setup for worst-case analysis is shown in Figure 7.26(b). FIGURE 7.24 Instantaneous load power for Example 7.5. FIGURE 7.25 Three-phase bridge rectifier with source inductance. 170 KW 165 KW 160 KW 155 KW 150 KW 0 s AVG (V(4,5)*I (VX)) 20 ms Average load power (49, 062 M, 157.081 K) Time 40 ms 50 ms 1 Van Vbn Vy = φV L1 L2 L3 ia ib io ic vo n 0 4 2 0.5 mH 0.5 mH 0.5 mH Vcn D1 D3 D4 D6 D5 R L 6.5 mH 0 VVx 0.5 Ω D2 + − 6 8 5 4 10 9 232 SPICE for Power Electronics and Electric Power, Second Edition Peak phase voltages Vm = 169.7 V, and f = 60 Hz. The listing of the circuit file is as follows: FIGURE 7.26 Three-phase bridge rectifier with line inductances for Example 7.6. (a) Sche- matic, (b) worst-case setup Example 7.6 Three-phase bridge rectifier with source inductances SOURCE
Van 1 0 SIN (0 169.7V 60HZ) CIRCUIT

L1 11 4 0.5MH Vbn 2 0 SIN (0 169.7V 60HZ 0 0 120DEG) L2 2 5 0.5MH Vcn 3 0 SIN (0 169.7V 60HZ 0 0 240DEG) Vy 1 11 DC L3 3 6 0.5MH R 7 9 0.5 L 9 10 6.5MH VX 10 8 DC 0V ; Voltage source to measure the output current D1 4 7 DMOD D3 5 7 DMOD Van Vbn 0 V 2 0.5 mH 20% L 6.5 mH 20% L1 7 + − L2 ib ia 5 9 1 8 6 4 ic 3 L3 V o Vx Ce 0 V 1 uF 20% 0.5 mH 20% 0.5 mH 20% 11 1 Vy Vcn 170 V 60 Hz D4 D6 D1 D3 D5 R 0.5 20% D2 DMOD DMOD DMOD DMOD DMOD DMOD 0 n (a) Diode Rectifiers 233 The PSpice plots of the instantaneous currents through diode D1, I(D1), through diode D3 I(D3), and through diode D5, I(D5), and the line voltages V(1,3) and V(2,3) are shown in Figure 7.27. Because of the source inductances, a commutation interval exists. During this interval, the current through the incoming diode rises and that through the outgoing diode falls. The sum of these currents must equal the load current. Note: Because of the line inductances, there is a transition time for switching the line currents from one diode to another diode as shown in Figure 7.27. This causes a drop in the output voltage because of the commutation of the currents [1]. D5 6 7 DMOD D2 8 6 DMOD D6 8 5 DMOD D4 8 4 DMOD .MODEL DMOD D(IS=2.22E−15 BV=1200V IBV=13E- 3 CJO=2PF TT=1US) ANALYSIS


.TRAN 10US 50MS 33.3333MS 10US ; Transient analysis .PROBE ; Graphics post-processor . OPTIONS ABSTOL = 0.0NRELTOL = 0.01VNTOL = 1.0M ITL5 = 10000 ; Convergence .END FIGURE 7.27 Plots of line voltages and diode currents for Example 7.6 I (D1) I (D3) V (L2:1, L3:1) V (Vy:+, L2:1) I (D5) 32 ms 35 ms 40 ms Time 45 ms 50 ms 400 V Line voltage vab Line voltage Vbc Diode currents ID:5 ID:1 ID:3 0.6 KA 0 V 0 A SELL>> −400 V 236 SPICE for Power Electronics and Electric Power, Second Edition 7.7.3 EXPERIMENT DR.3 THREE-PHASE BRIDGE RECTIFIER FIGURE 7.30 Single-phase bridge rectifier. Objective To study the operation and characteristics of a three-phase bridge rectifier under various load conditions. Applications A three-phase bridge rectifier is used an an input stage in variable-speed AC motor drives, etc. Textbook Apparatus Same as Experiment DR.1, except that six diodes are required. Warning Experimental procedure Set up the circuit as shown in Figure 7.31 and follow the steps for Experiment DR.1. Report Repeat the steps of Experiment DR.1. FIGURE 7.31 Three-phase bridge rectifier. SW 120 V ac 60 Hz Ac A Ac V DcV R L 20 mH A Dc + – is D1 D2 D3 D4 vo io 10 Ω • • • • • • • • • • • • • • • SW SW n120 V, 60 Hz per phase Dc R L 20 mH A Dc + – iavan vcn D1 D2 vo io 10 Ω • • • • • • • • • • • • • • • • • D4 D6 D5D3 SW vbn A V V Diode Rectifiers 239 7.8 (a) Design an output LC filter for the three-phase bridge rectifier of Prob- lem 7.7. The harmonic content of the load current should be less than 5% of the value without the filter. (b) Use SPICE to verify your design in part (a). 7.9 (a) Design an output C filter for the three-phase bridge rectifier of Problem 7.7. The harmonic content of the load voltage should be less than 5% of the value without the filter. (b) Use SPICE to verify your design in part (a). 7.10 Repeat Example 7.2 with an antiparallel (or freewheeling) diode connected across 7.11 Complete the following table for the single-phase half-wave rectifier in Figure Rectifier % THD of Input Current, THDi Input Power Factor, PFi % THD of Output Voltage, THDvo % THD of Load Current, THDio Half-wave Half-wave with a freewheeling diode Rectifier % THD of Input Current, THDi Input Power Factor, PFi % THD of Output Voltage, THDvo % THD of Load Current, THDio Single-phase half- wave rectifier Single-phase bridge rectifier Three-phase bridge rectifier 240 7.12 TABLE P7.12(A) Filter Capacitance Ce = 1 µF % THD of Input Current, THDi Input Power Factor, PFi % THD of Output Voltage, THDvo % THD of Load Current, THDio Single-phase half- wave rectifier Single-phase bridge rectifier Three-phase bridge rectifier TABLE P7.12(B) Filter Capacitance Ce = 300 µF % THD of Input Current, THDi Input Power Factor, PFi % THD of Output Voltage, THDvo % THD of Load Current, THDio Single-phase half- wave rectifier Single-phase bridge rectifier Three-phase bridge rectifier TABLE P7.12(C) Filter Capacitance Ce = 600 µF % THD of Input Current, THDi Input Power Factor, PFi % THD of Output Voltage, THDvo % THD of Load Current, THDio Single-phase half- wave rectifier Single-phase bridge rectifier Three-phase bridge rectifier