Appendix 8. Training Exercises to Laboratory Works

Transcription

1 Appendix 8 Training Exercises to Laboratory Works BGU 214

2 2 Introduction Simulation run time is based on the time required to establish the steady state of process under study. The proposed exercises can start with 115 periods (23mS) Maximum step size : 23uS for 5Hz.. Figures report should be submitted only 23 last period. Figures report should be kept only major grid. When printed on a black white printer, the currents and voltages waves in the figures of the report should be named in the appropriate manner.

3 3 Experiment No.1 1. Construct schemes in SPICE: separately for Thyristors, separately for TRIAC's and separately for MOSFET; 2. Spend simulations (Thyristor and TRIAC circuits )at different values of parameters RC ignition circuit and compare results (6 8 variants ); 3. Compare gate voltage (V G, V G2 and V G3, V G4 ) at pulse excitation and excitation from RC ignition circuit; 4. Spend simulations (TRIAC) at different values of parameters Lo, Ro and compare results; 5. Simulation results presents the report on template form for experiment No.1

4 FREQ = 5 VAMPL = 34 VOFF = V1 Ro1 T A 2 FREQ = 5 VAMPL = 34 VOFF = V5 G Ro1,Ro2, Ro3, Ro4 [17] Ohm D1 Ro5 T5 A5 2 R_12 {Rc} C1 {Cc} G5 Lo 9mH X_Lo << Ro; X_Lo ~ Ro: X_Lo >> Ro pls4 4 Vp5 TD = {ign} FREQ = 5 VAMPL = 34 VOFF = PARAMETERS: Rc = Cc = 1u ig_deg = 6 ign = {ig_deg/36*2m} Vpls = 1 Tw = 3m V2 Rc = [, 2.5k, 5k, 7.5k, ] ig_deg = [, 6, 9, 12,18] degree X1 2N6397 Ro2 T2 A2 2 X2 2N6397 G2 D2 pls2 B M1 IRF83 Vp2 TD = {ign} PW = 5u PER = 2m FREQ = 5 VAMPL = 34 VOFF = FREQ = 5 VAMPL = 34 VOFF = V3 T4 V4 TRIAC 2N6348 Vdrm=6v Idrm=1u Ih=6ma dvdt=1e6 Ton=1.5u Igt=12ma Vgt=.9v Vtm=1.3v Itm=11A Ro3 T3 A3 Ro4 1 V2 = {Vpls} PER = 2m 2 X3 2N6348 X4 2N6348 V_pls = 1,3,5,7,9,11 V by Tw=5mS Tw = 5uS, 2mS, 5mS, 8mS; 1mS by V_pls=1V A4 G3 D32 G4 D31 D41 D42 B3 R_13 {Rc} C3 {Cc} Vp_41 Vp_42 TD = {ign} PW = 5u PER = 2m PW = 5u PER = 2m TD = {ign1m}

5 Experiment No. 2

6 1. Sinusoidal Voltage Source 6 Ls source leakage negligible Ls V_E 1n D1 Vph Vout L1 {Lo} Vr VOFF = FREQ = 5 Vin1 R1 {Rload} 1.Construct schemаtic; 2. Spend simulations at different values of parameters Lo, Rload : a) Lo =5nH, Rload=1 Ω; b) Lo=5mH, Rload= 5Ω. 3. Submit current I(R1) and voltage V_E waves in one plot 2. Sinusoidal Voltage Source Ls source leakage negligible Ls D1 V_E Vph Vout L1 1n {Lo} Vr VOFF = FREQ = 5 AC = Vin D2 R1 {Rload} 1.Construct schemаtic; 2.Spend simulations at different values of parameters Lo, Rload : Lo =5nH, Rload=1 Ω; Lo=5mH, Rload= 5Ω. 3.Submit current and voltage waves : V_E, I(R1); I(D1), I(D2). 4. Compare the simulation results for different variants of parameters and evaluate the influence of parameters. 5. Measure and explain the changes diodes D1,D2 conduction angles ( )

7 7 3. Sinusoidal Voltage Source Ls source leakage negligible V_E Ls Vph D4 1nH Vout VOFF = FREQ = 5 AC = V2 C1 {Co} R1 {Rload} 1.Construct schemаtic; 2.Spend simulations at different values of parameters Lo, Rload, Co: Co =5nF, Rload=1 Ω; Co=1uF, Rload= 1Ω. 3.Submit current and voltage waves : V_E, I(R1); V_E, I(D4); V_E, I(C1); I(D4), I(C1). 4. Compare the simulation results for different variants of parameters and evaluate the influence of parameters. 5.Measure and explain the changes diode conduction angle ( ). 4. Sinusoidal Voltage Source with midpoint VOFF = FREQ = 5 PHASE = E11 V8 mid Ls1 1mH L1 ph11 Vr D1 R6 VOFF = FREQ = 5 PHASE = V9 E12 {Lo} Ls2 ph12 D11 {Rload} 1mH 1.Construct schemаtic; 2.Spend simulations at different values of parameters Lo, Rload: Lo =5nH, Rload=1 Ω; Lo=1mH, Rload= 1Ω.

8 3.Submit current and voltage waves : V_E11, V_E12, I(R6); V_E11, I(D1); I(D1), I(D11), 8 4. Compare the simulation results for different variants of parameters and evaluate the influence of parameters. 5.Measure and explain the changes diode ignition angle ( ), conduction angle ( ), commutation angle( ). 5. Sinusoidal Voltage Source VOFF = FREQ = 5 AC = E1 V3 {Ls} Ls Vph D4 D5 C2 {Co} C3 {Co} Vr Vr R3 {Rload} 1.Construct schemаtic; 2.Spend simulations at different values of parameters Ls,Co, Rload: Ls =1nH, Co=1uF, Rload= Ω, 1 Ω, 1 Ω; Ls=1uH, Co=1uF, Rload= Ω, 1 Ω, 1 Ω; 3.Submit current and voltage waves : V_E1, (Vr) (Vr), V_E1, I(R3), V_E1, I(D4), V_E1, I(D5), I(D4), I(C2). 4.Compare the simulation results for different variants of parameters and evaluate the influence of parameters. 5.Measure and explain the changes diode ignition angle ( ), conduction angle ( ), commutation angle( ).

9 9 6. Sinusoidal Voltage Source Ls source leakage negligible E6 Ls ph6 D1 out6 D2 L1 {Lo} r6 VOFF = FREQ = 5 AC = V1 1nH R7 {Rload} D4 D3 r7 1.Construct schemаtic; 2.Spend simulations at values of parameters Co, Rload: Lo =1nH,, Rload=1 Ω; 3.Submit current and voltage waves : V_E6, I(R7), V_E6, I(D1), V_E6, I(Ls). 4. Compare the simulation results for different variants of parameters and evaluate the influence of parameters. 5.Measure and explain the changes diode D1 ignition angle ( ), conduction angle ( ), commutation angle( ). 7. Secondary winding of 3phase transformer FREQ = 5 VOFF = PHASE = rod "a" Va Las 1u pha Da r Ro {Rload} FREQ = 5 VOFF = PHASE = 12 Vb rod "b" Lbs 1u phb Db FREQ = 5 VOFF = PHASE = 24 rod "c" Vc Lcs 1u phc Dc *Each phase contains a single coil, which sits on one of the rod of the 3phase transformer

10 1 1.Construct schemаtic; 2.Spend simulations at Rload=1 Ω. 3.Submit current and voltage waves : V_Ea, V_(Ea, Eb); V_Ea, V_pha; V_Ea, V_Eb, V_Ec ; V_Ea,I(Ro); I(Da), I(Db), I(Dc), V_Ea, I(Da), Rod current I(Las) and average I(Las). 4.Measure diode s Da ignition angle ( ), conduction angle ( ), commutation angle( ). 8. FREQ = 5 PHASE = Va1 Ea La 1u pha1 phb1 rod "a" rod "b" FREQ = 5 PHASE = Va2 Laa 1u ph1 Ro {Rload} Da r FREQ = 5 PHASE = 12 Vb1 Eb Lb 1u FREQ = 5 PHASE = 12 Vb2 Lbb 1u ph2 Db phc1 FREQ = 5 PHASE = 24 Vc1 Ec Lc 1u rod "c" FREQ = 5 PHASE = 24 Vc2 Lcc 1u ph3 Dc **Every phase contains two identical coils sitting at one rod of the 3ph transformer 1.Construct schemаtic; 2.Spend simulations at Rload=1 Ω. 3.Submit current and voltage waves : V_Ea, V_(ph1, ph2); /single coil EMF and transformer line voltage/ V_Ea, V_pha1; /single coil EMF and voltage/ V_pha1, (V_ph3) V(pha1) ; / coil voltage at rod a and coil voltage at rod c components of phase3 / V_Ea,I(La); I(La), I(Laa) Rod current I(La) I(Laa) and average I(La) I(Laa) ; 4.Compare and explain the results of exp.7 and exp.8.

11 11 9. Secondary winding of 3phase transformer FREQ = 5 VOFF = PHASE = Va rod "a" A La {Ls} D1 D2 L1 {Lo} D3 FREQ = 5 VOFF = PHASE = 12 Vb rod "b" B Lb {Ls} Ro {load} FREQ = 5 VOFF = PHASE = 24 V19 rod "c" C Lc {Ls} D6 D5 D4 1.Construct schemаtic; 2.Spend simulations at different values of parameters Ls, Rload: Ls =1nH, Rload=1 Ω; Ls=1uH, Rload=1 Ω; 3.Submit current and voltage waves : V_A, V(A,B); V_A, I(La); V_A, I(Ro); V_A, I(D1); Measure and explain the changes diode D1 ignition angle ( ), conduction angle ( ), commutation angle( ). I(La), I(Lb); Rod a current I(La) and average I(La); Show the sequence of phase operation during one period 4. Compare the simulation results for different variants of parameters and evaluate the influence of parameters.

12 Experiment No.3

13 in in2 X12 2N4444 X22 2N4444 Vin 15Vdc V2 = 2 TD = s1 V1 Vdc a X1 2N4444 L1 4mH V_Ia r1 R1 7 PARAMETERS: freq = 2 Tp = {1/freq} Tw = 5u V2 = 2 TD = {Tp/2} c1 4u C1 s2 V2 Vdc b X2 2N4444 V_Ib Vin2 15Vdc V2 = 2 TD = s12 V12 Vdc a2 L12 4mH V_Ia2 D1 r12 R12 7 V22 V2 = 2 TD = {Tp/2} c12 4u C12 s22 Vdc b2 V_Ib2 D2 V2 = 2 TD = {Tp/2} s4 V4 X4 2N4444 V2 = 2 TD = s3 V3 X3 2N4444 V2 = 2 TD = {Tp/2} s42 V42 X42 2N4444 D4 V2 = 2 TD = X32 2N4444 s32 V32 D3 Thyristor: 2N4444 Vdrm=6v Vrrm=6v Ih=6ma Vtm=1.v Itm=5 dvdt=5e7 Igt=7ma Vgt=.75v Ton=1u Toff=15u Idrm=1u Fig. 3.A1.3. Schematic of bridge Voltage Inverter (without/with clamping diodes) Vin; R1; C1; L1; Vin2; R12; C12; L12; freq = var; I thyristor ;V thyristor ; V ab ; I R1 ; V ab2 ; I R12 =?? Summary

14 R3 K K1 K_Linear COUPLING = 1 L1 = L12 L2 = L11 L3 = L2 tr1 O1 L2 R2 O2 {Rload} 8mH tr2 R32 K K12 K_Linear COUPLING = 1 L1 = L12_2 L2 = L11_2 L3 = L2_2 tr12 R22 O12 1 O22 L2_2 8mH tr22 R4 5mH L11 5mH L12 R5 L11_2 5mH 5mH L12_2 R42 R52 Lin 1 1 L mH a C1 b 5mH a2 C12 b2 2Vdc Rin 1u Vin D1 Vdc X1 2N4444 V_Ia s1 Vs1 V2 = 5 TD = PW = 5u 6u V2 = 5 TD = {Tp/2} PW = 5u X2 2N4444 s2 Vs2 Vdc V_Ib D2 2Vdc R53 1u V5 D12 Vdc S1 V_Ia2 s12 V42 V2 = 5 TD = PW = {Tw2} 6u V2 = 5 TD = {Tp/2} PW = {Tw2} s22 V52 S2 Vdc V_Ib2 D22 PARAMETERS: freq = 1 Tp = {1/freq} Tw = 5u Tw2 = {.497*Tp} Rload = 1 Fig. 3.A1.4. Schematic of Current Inverter (without/with clamping diodes) Vin; Rload; C1; Vin2; C12; freq = var; I thyristor (I(V_la));V thyristor ; V o1,o2 ; I R1 ; V ab2 ; V o12,o22 =?? Summary

15 Experiment No.4

16 Controlled rectifier simulation model Appendix Va2 a2 FREQ = 5 VAMPL = 3 VOFF = PHASE = Ra2 1m La2 {Lph} D1 D2 D3 Lo2 {Lo} Oo Vb2 b2 FREQ = 5 VAMPL = 3 VOFF = PHASE = 12 Rc2 Vc2 c2 1m FREQ = 5 VAMPL = 3 VOFF = PHASE = 24 Rb2 1m Lb2 {Lph} Lc2 D6 {Lph} D5 D4 Oo Ro2 1 Vo2 215mv PARAMETERS: Tp = 2m Tig = {Kig*Tp/3} Kig = Lph = Lo = a) Kig=,.1,.3; Lph=; Lo= b) Kig=; Lph=.1mH,.2mH; Lo= c) Kig=; Lph=; Lo =, 5mH, 15mH d) Kig=.3; Lph=.75mH; Lo= e) Kig=.3; Lph=; Lo=3m 2N645: V_DRM,_RRM =8V; I_T(rms) =32A; I_G =5mA Va a FREQ = 5 VAMPL = 3 VOFF = PHASE = Vb b FREQ = 5 VAMPL = 3 VOFF = PHASE = 12 Ra 1u Rb 1u La {Lph} Lb {Lph} G1 R1 4 R6 G6 Waveforms Study: Vph, Iph, Ith, Vpulse, Io X1 2N645 K1 2N645 X6 G2 R5 R2 4 G5 X2 G3 2N645 K2 R3 2N645 X5 R4 4 G4 Lo {Lo} X3 2N645 K3 X4 2N645 O O Ro 1 2n Vo Vc c FREQ = 5 VAMPL = 3 VOFF = PHASE = 24 Rc 1u Lc {Lph} 4 K4_6 4 4 G1 G2 G3 G4 G5 G6 V3 V5 V4 V6 V7 V8 PW = 1u PER = 2m TD = {Tp/12Tig} PW = 1u PER = 2m TD = {5*Tp/12Tig} PW = 1u PER = 2m TD = {3*Tp/4Tig} PW = 1u PER = 2m TD = {Tp/4Tig} PW = 1u PER = 2m TD = {11*Tp/12Tig} PW = 1u PER = 2m TD = {7*Tp/12Tig} PW = 1u PER = 2m TD = {Tp/12TigTp/6} K1 V9 V1 PW = 1u PER = 2m K2 TD = {5*Tp/12TigTp/6} PW = 1u PER = 2m V11 K3 TD = {3*Tp/4TigTp/6} PW = 1u PER = 2m V12 V13 PW = 1u PER = 2m K4_6 K4_6 TD = {11*Tp/12TigTp/6} TD = {Tp/4TigTp/6} PW = 1u PER = 2m K4_6 V14 TD = {7*Tp/12TigTp/6}

17 Experiment No.5 1. Introduction in chopper techniques. IN 24Vdc V1 TD = {To} PW = {Tp*DU} Vp3 S2 Ro1 {load} PARAMETERS: f req = 1 Tp = {1/f req} To = load = 1 Du =.35 Hz 1/S 1/S Ohm construct presented schematic in SPICE; carry out simulations (transient) for different values (68) of parameter Du ( 1); draw the diagram of dependence of average value of loud current from parameter Du

18 2. chopper with external commutation source General circuit IN Rint.5 25 V1 X1 2N645 R G1 K12 Ro {load} PARAMETERS: f req = 1 Hz Tp = {1/f req} S To = S load = 1 Ohm Du =.1 ( 1) Pw_S = 5u S X2 2N645 R2 2 G2 R3 2 G1 G2 S1 C1 A TD = {To} TR = 2u TF = 2u PW = 1u Vp1 K12 Vp2 TD = {ToTp*Du} TR = 2u TF = 2u PW = 1u Vp3 V2 = 4 TD = {To} TR = 2u TF = 2u PW = {Pw_S} 3Vdc V2 D1 2u R4 1 Thyristor's Stopper S1 S1.subckt 2N645 Vdrm=8v Vrrm=8v Ih=6ma Vtm=1.7v Itm=32 dvdt=5e6 Igt=5ma Vgt=.7v Ton=1u Toff=15u Idrm=1u construct schematic in SPICE; carry out simulations (transient) for different values of frequencies (2 8Hz) find the range of frequencies of working ability of the scheme at the specified parameters; draw the current waveform in loud, thyristors, capacitor for one value of frequency; explain currents forms. Device's principle of action: The pulse on a gate G1 opens the first thyristor X1. The current in load Ro starts to flow. In the same time the switch S1 opens and a source V2 charges the capacitor C1. At the necessary moment the pulse on a gate G2 opens the thyristor X2 and the positive pole A of a capacitor C1 is put to the cathode of the thyristor X1 and the negative pole is put to the its anode. The current through thyristor X1 stops and the Thiristor X1 is closed. The Thyristor X2 will be closed when capacitor C1 will be discharged.

19 3. RCchopper G1 R1 2 K1 IN PARAMETERS: f req = 1 Tp = {1/f req} To = load = 1 Du =.35 Hz S S Ohm ( 1) Rint.5 25 V1 G2 X1 2N645 2N645 X2 K2 C1 22u R2 1 Ro {load} G1 G2 TD = {To} TR = 2u TF = 2u PW = 1u Vp1 Vp2 TD = {ToTp*Du} TR = 2u TF = 2u PW = 1u K1 construct schematic in SPICE; execute simulations (transient) for different values of frequencies (2 8Hz) find the range of frequencies of working ability of the scheme at the specified parameters; draw the current waveform in loud, thyristors, capacitor for one value of frequency; explain currents forms. K2

20 4. LC chopper G1 R1 2 K1 IN PARAMETERS: f req = 1 Tp = {1/f req} To = load = 1 Du =.35 Hz S S Ohm ( 1) Rint.5 25 V1 G2 D1 R2 X1 2N645 2N645 X2 2 L1 K2 C1 5u Ro {load} 1mH G1 G2 TD = {Tp} TR = 2u TF = 2u PW = 1u Vp1 Vp2 TD = {ToTp*Du} TR = 2u TF = 2u PW = 1u K1 K2 construct schematic in SPICE; carry out simulations (transient) for different values of frequencies (2 8Hz) find the range of frequencies of working ability of the scheme at the specified parameters; draw the current waveform in loud, thyristors, capacitor for one value of frequency; explain currents forms.

21 Experiment No.6

22

23

24

25 Experiment No. 7 3

26 4 Exercise 1 6step inverter Running time ~ >152 periodes S1 S1 D1 S2 S2 D2 S3 S3 D3 L1 R1 55Vdc Vin {Lo} L2 {Lo} L3 {Lo} {Ro} R2 {Ro} R3 {Ro} S6 S6 D6 S5 S5 X_L=2*PI*f*L=6.28*5*.1=31.4 Ohm D5 S4 S4 D4 PARAMETERS: f req = 5 DU =.495 Tp = {1/f req} Tw = {DU*Tp} TD = {Tp/3} Ro = 15 Lo = 5u (3,... 3n) Ohm (1n,....1m) H S1 S2 S3 S6 S5 S4 V2 = 5 TD = V1 V2 = 5 TD = {TD} V2 V2 = 5 TD = {2*TD} V3 V2 = 5 TD = {Tp/2} V4 V2 = 5 TD = {Tp/2TD} V5 V2 = 5 TD = {Tp/22*TD} V6 Result voltages and currents of switch and load in the scheme without diodes: (Z~3 Ohm) 1. for active load : 2. for activeinductive load; 3. for mainly inductive load. The same in the scheme with diodes

27 5 PWM forming Exercise 2 PWM are created at comparison of a signal of carrier frequency with a signal of reference frequency Carrier sawtooth voltage generator V2 = 1 TD = TR = {.49*Tp_sw} TF = {.49*Tp_sw} PW = 1n PER = {Tp_sw} V4 A1 Reference frequency generator R14 A2.3*sin(2*PI*(f req_env )*time) R15 Comparator E9 E1 A3 IF(V(A2)>V(A1),1,) R16 PARAMETERS: f req_env = 25 f req_switch = 1 Tp_sw = {1/f req_switch} PI = freq_env output frequency (reference); freq_switch internal switching frequency (carrier frecuency) V2 = 1 TD = TR = {.49*Tp_sw} TF = {.49*Tp_sw} PW = 1n PER = {Tp_sw} V2 B1 R8 B2.7*sin(2*PI*(f req_env )*time) R9 E5 E6 B3 IF(V(B2)>V(B1),1,) R1 Study process of pulses formation on the specified schemes; Compare spectra of output signals; Draw spectra curves V2 = 1 TD = TR = {.49*Tp_sw} TF = {.49*Tp_sw} PW = 1n PER = {Tp_sw} V3 C1 R11 C2 1.2*sin(2*PI*(f req_env )*time) R12 E7 E8 C3 IF(V(C2)>V(C1),1,) R13 For calculation of spectra use SPICE option : Simulation Setting\Output File Options\ Center frequency freq_env; Number of harmonics 45; Output variables V(A3), V(B3), V(D3) V2 = 1 TD = TR = {.49*Tp_sw} TF = {.49*Tp_sw} PW = 1n PER = {Tp_sw} V5 D1 R17 D2 1.2*sin(2*PI*(3*f req_env )*time) R18 E11 E12 D3 IF(V(D2)>V(D1),1,) R19

28 6 PWM 3phase inveter D15 D16 D17 Exercise 3 Running time ~ >152 periodes of envelope curve X_L=2*PI*f*L=6.28*5*.1H=3.14 Ohm Vin 55Vdc A3 A32 S3 D2 S1 B3 S4 C3 S7 D3 D4 D9 D14 B32 D19 S9 D13 C32 D18 S8 U2 V2 W2 D1 Load L1 R1 {Lo} {Ra} V L2 R2 {Lo} {Ra} L3 R3 {Lo} {Ra} IGBT simple model PARAMETERS: f req = 5 envelope frequency Tp = {1/f req} TD = {Tp/3} f req_car = 1carrier frequency DU_car =.495 Tw_car = {DU_car*Tp_car} Tp_car = {1/f req_car} Lo = 1nH 1mH Ra = 3.14( 3.14n) V_du =.8 ( Pi= ) Pulse Generator Result voltages and currents of switch and load 1. for active load (Z~3 Ohm); 2. for activeinductive load (Z~3 Ohm) Carrier high frequency (saw) R16 Envelope low frequency (SIN) E11 A22 R17 Envelope low frequency (SIN) E9 A2 R15 Comparator A3 E1 A3 Comparator A32 E12 A32 R18 V_du*sin(2*PI*f req*time) IF(V(A2)>V(S),3,) V_du*sin(2*PI*f req*timepi) IF(V(A22)>V(S),3,) V2 = 1 TD = TR = {.49*Tp_car} TF = {.49*Tp_car} PW = 1n PER = {Tp_car} Vcar S R14 R21 E15 B22 V_du*sin(2*PI*f req*(timetd)pi) R22 E16 E13 B2 R2 V_du*sin(2*PI*f req*(timetd)) B3 E14 B3 IF(V(B2)>V(S),3,) IF(V(B22)>V(S),3,) B32 B32 R23 C2 R25 C3 C3 R26 E19 C22 R27 E2 E17 E18 C32 C32 R28 V_du*sin(2*PI*f req*(time2*td)) IF(V(C2)>V(S),3,) V_du*sin(2*PI*f req*(time2*td)pi) IF(V(C22)>V(S),3,)