EXPERIMENT NO. 1 AIM: Simulate and study V-I characteristics of a Diode using PSPICE windows. CIRCUIT DIAGRAM: 1 VD 1Vdc D1 D1N914 2 Vx 0Vdc PROGRAM: ** Diode Characteristic VD 1 0 DC 1V *DC input voltage is overridden during DC sweep D1 1 2 D1N914 ; Diode with model D1N914 * anode cathode model VX 2 0 DC 0V ; measures the diode current ID * DIode model defines the model parameters .MODEL D1N914 D (IS=3.93E-9 RS=1 BV=100V IBV=5E-6 CJO=1.7PF TT=2NS) .DC VD 0 1V 0.01V ; DC sweep from 0 to 2 V with 0.01 V increment .PLOT DC I(VX) ; Plots the diode current on the output file .PROBE ; Graphical waveform analyzer .END ; End of circuit file RESULT: CONCLUSION: The forward biased silicon diode in an electronic system under dc condition has a drop of 0.7v across it in conduction state at any value of diode current EXPERIMENT NO.2 AIM: Simulate and study V-I characteristics of a NPN-BJT using PSPICE windows. CIRCUIT DIAGRAM: 1 Q1 Q2N2222A VCE 12Vdc IB 1mAdc PROGRAM: **** NPN-BJT Characteristics IB 0 1 DC 1MA ; Base current VCE 2 0 DC 12V ; Collector-emitter voltage Q1 2 1 0 Q2N2222A ; BJT statement .MODEL Q2N2222A NPN (IS=2.105E-16 BF=173 VA=83.3V CJE=29.6PF CJC=19.4PF + TF=489.88PS TR=4.9NS) ; Model parameters .DC VCE 0 10V 0.02V IB 0 1MA 200UA ; DC sweep for VCE and IB .PROBE ; Graphical waveform analyzer .OP .END ; End of circuit file RESULT: CONCLUSION:The collector to emitter voltage will influence the magnitude of collector current as shown in graph. EXPERIMENT NO.3 AIM: Simulate and study frequency response of R-L-C series circuit using PSPICE windows. CIRCUIT DIAGRAM: R1 1 L1 2 2 1 2 3 50uH VIN C1 VAMPL = 10V FREQ = 5KHZ 10UF PROGRAM: ***ExpTransient Response of an RLC-circuit with a sinusoidal input voltage * SIN (VO VA FREQ) ; Simple sinusoidal source VIN 1 0 SIN (0 10V 5KHZ) ; sinusoidal input voltage R1 1 2 2 L1 2 3 50UH C1 3 0 10UF .TRAN 1US 500US ; Transient analysis .PLOT TRAN V(3) V(1) ; Plots on the output file .PROBE ; Graphical waveform analyzer .END ; End of circuit file RESULT: EXPERIMENT NO. 4 AIM: Simulate and study Darlington pair amplifier circuit using PSPICE windows and determine quiescent condition. CIRCUIT DIAGRAM: RB 47k 1 Q1 QM 3 VIN 5Vdc VCC 12Vdc Q2 QM 4 RE 4.7k PROGRAM: **** Darlington Pair VCC 2 0 DC 12V VIN 1 0 DC 5V * BJTs with model QM Q1 2 1 3 QM Q2 2 3 4 QM RB 2 1 47k RE 4 0 4.7K * Model QM for NPN BJTs .MODEL QM NPN (BF=100 BR=1 RB=5 RC=1 RE=0 VJE=0.8 VA=100) * Transfer function analysis to calculate dc gain, input resistance * and output resistance .TF V(4) VIN .END RESULT: **** 10/10/05 22:36:42 ********* PSpice 9.1 (Mar 1999) ******** ID# 0 ******** **** Darlington Pair **** SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C ************************************************************* NODE VOLTAGE ( 1) 5.0000 ( NODE VOLTAGE 2) 12.0000 ( 3) 4.3560 ( NODE VOLTAGE 4) NODE VOLTAGE 3.5909 VOLTAGE SOURCE CURRENTS NAME CURRENT VCC VIN -9.129E-04 1.489E-04 TOTAL POWER DISSIPATION 1.02E-02 WATTS **** SMALL-SIGNAL CHARACTERISTICS V(4)/VIN = 9.851E-01 INPUT RESISTANCE AT VIN = 4.696E+04 OUTPUT RESISTANCE AT V(4) = 6.677E+01 JOB CONCLUDED TOTAL JOB TIME 0.00 CONCLUSION:The collector to emitter voltag will influence the magnitude of collector current as shown in graph. EXPERIMENT NO.5 AIM: Simulate and study Diode Clipper and Clamper circuits using PSPICE windows. CIRCUIT DIAGRAM: (CLIPPERS) R R4 2 1 1 .22K .22K D1 D1N3940 VAC VOFF = 0 VAMPL = 5V FREQ = 30kHz 2 VOFF = 0 VAMPL = 5V FREQ = 30kHz D1 D1N3940 VAC Figure 1 Figure 4 R R 2 1 .22K D1 3 D1N3940 VAC VOFF = 0 VAMPL = 5V FREQ = 30kHz 1 VDC 1Vdc VOFF = 0 VAMPL = 5V FREQ = 30kHz VAC 2 .22K VDC 1Vdc Figure 2 Figure 5 R R 1 VAC VOFF = 0 VAMPL = 5V FREQ = 30kHz 2 .22K 1 D1 3 D1N3940 VDC 1Vdc VOFF = 0 VAMPL = 5V FREQ = 30kHz VAC R 2 .22K D2 D1N3940 D1 D1N3940 VDC2 1Vdc VDC1 1.5Vdc Figure 7 PROGRAM: Prog. 1 .22K D1 3 D1N3940 Figure 6 1 VAC 2 VDC 1Vdc Figure 3 VOFF = 0 VAMPL = 5V FREQ = 30kHz D1 3 D1N3940 *******CLIPPER 1 R 1 2 .22K D1 2 0 D1N3940 VAC 1 0 SIN(0 5V 30KHZ) .MODEL D1N3940 D( + IS = 4E-10 + RS = .105 + N = 1.48 + TT = 8E-7 + CJO = 1.95E-11 + VJ = .4 + M = .38 + EG = 1.36 + XTI = -8 + KF = 0 + AF = 1 + FC = .9 + BV = 600 + IBV = 1E-4 +) .PROBE .TRAN 0US 100US .END Prog. 2 *******CLIPPER 1 R 1 2 .22K D1 2 3 D1N3940 VAC 1 0 SIN(0 5V 30KHZ) VDC 3 0 DC 1V .MODEL D1N3940 D( + IS = 4E-10 + RS = .105 + N = 1.48 + TT = 8E-7 + CJO = 1.95E-11 + VJ = .4 + M = .38 + EG = 1.36 + XTI = -8 + KF = 0 + AF = 1 + FC = .9 + BV = 600 + IBV = 1E-4 +) .PROBE .TRAN 0US 100US .END Prog. 3 *******CLIPPER 3 R 1 2 .22K D1 2 3 D1N3940 VAC 1 0 SIN(0 5V 30KHZ) VDC 0 3 DC 1V .MODEL D1N3940 D( + IS = 4E-10 + RS = .105 + N = 1.48 + TT = 8E-7 + CJO = 1.95E-11 + VJ = .4 + M = .38 + EG = 1.36 + XTI = -8 + KF = 0 + AF = 1 + FC = .9 + BV = 600 + IBV = 1E-4 +) .PROBE .TRAN 0US 100US .END Prog. 4 *******CLIPPER 4 R 1 2 .22K D1 0 2 D1N3940 VAC 1 0 SIN(0 5V 30KHZ) .MODEL D1N3940 D( + IS = 4E-10 + RS = .105 + N = 1.48 + TT = 8E-7 + CJO = 1.95E-11 + VJ = .4 + M = .38 + EG = 1.36 + XTI = -8 + KF = 0 + AF = 1 + FC = .9 + BV = 600 + IBV = 1E-4 +) .PROBE .TRAN 0US 100US .END Prog. 5 *******CLIPPER 5 R 1 2 .22K D1 3 2 D1N3940 VAC 1 0 SIN(0 5V 30KHZ) VDC 0 3 DC 1V .MODEL D1N3940 D( + IS = 4E-10 + RS = .105 + N = 1.48 + TT = 8E-7 + CJO = 1.95E-11 + VJ = .4 + M = .38 + EG = 1.36 + XTI = -8 + KF = 0 + AF = 1 + FC = .9 + BV = 600 + IBV = 1E-4 +) .PROBE .TRAN 0US 100US .END Prog. 6 *******CLIPPER 6 R 1 2 .22K D1 3 2 D1N3940 VAC 1 0 SIN(0 5V 30KHZ) VDC 3 0 DC 1V .MODEL D1N3940 D( + IS = 4E-10 + RS = .105 + N = 1.48 + TT = 8E-7 + CJO = 1.95E-11 + VJ = .4 + M = .38 + EG = 1.36 + XTI = -8 + KF = 0 + AF = 1 + FC = .9 + BV = 600 + IBV = 1E-4 +) .PROBE .TRAN 0US 100US .END Prog. 7 *******CLIPPER 7 R 1 2 .22K D1 3 2 D1N3940 D2 2 4 D1N3940 VAC 1 0 SIN(0 5V 30KHZ) V1_VDC1 4 0 DC 1V V2_VDC2 0 3 DC 1.5V .MODEL D1N3940 D( + IS = 4E-10 + RS = .105 + N = 1.48 + TT = 8E-7 + CJO = 1.95E-11 + VJ = .4 + M = .38 + EG = 1.36 + XTI = -8 + KF = 0 + AF = 1 + FC = .9 + BV = 600 + IBV = 1E-4 +) .PROBE .TRAN 0US 100US .END RESULT: Result of CLIPPER 1 Result of CLIPPER 2 Result of CLIPPER 3 Result of CLIPPER 4 Result of CLIPPER 5 Result of CLIPPER 6 Result of CLIPPER 7 CIRCUIT DIAGRAM: (CLAMPERS) C1 C1 1 2 1 .01uf 2 .01uf VAC VAC D1N3491 VOFF = 0v VAMPL = 5v FREQ = 30khz R1 D1 560k D1N3491 VOFF = 0v VAMPL = 5v FREQ = 30khz Figure 1 C1 2 .01uf 1 D1N3491 VAC D1 VOFF = 0v VAMPL = 5v FREQ = 30khz 2 .01uf D1N3491 VAC R1 560k VDC D1 VOFF = 0v VAMPL = 5v FREQ = 30khz 560k 1Vdc Figure 5 Figure 2 C1 C1 1 VAC R1 VDC 1Vdc VOFF = 0v VAMPL = 5v FREQ = 30khz 560k Figure 4 C1 1 2 .01uf R1 D1 1 VAC D1 R1 560k VDC 1Vdc 2 .01uf D1N3491 VOFF = 0v VAMPL = 5v FREQ = 30khz D1N3491 D1 R1 560k VDC 1Vdc Figure 3 PROGRAM: Prog. 1 ****CLAMPER1 C_C1 1 2 .01uf R_R1 2 0 560k D_D1 2 0 D1N3491 V_VAC 1 0 sin(0 5v 30khz) .model D1N3491 D(Is=68.65f Rs=3.786m Ikf=1.774 N=1 Xti=2 Eg=1.11 Cjo=1.457n + M=.9735 Vj=.75 Fc=.5 Isr=11.02u Nr=2 Tt=6.059u) * Motorola pid=1N3491 case=DO21 * 88-08-24 rmn .probe .tran 0us 100us .end Figure 6 Prog. 2 ****CLAMPER2 C_C1 1 2 .01uf R_R1 2 0 560k D_D1 2 3 D1N3491 V_VDC 3 0 1V V_VAC 1 0 sin(0 5v 30khz) .model D1N3491 D(Is=68.65f Rs=3.786m Ikf=1.774 N=1 Xti=2 Eg=1.11 Cjo=1.457n + M=.9735 Vj=.75 Fc=.5 Isr=11.02u Nr=2 Tt=6.059u) * Motorola pid=1N3491 case=DO21 * 88-08-24 rmn .probe .tran 0us 100us .end Prog. 3 ****CLAMPER3 C_C1 1 2 .01uf R_R1 2 0 560k D_D1 2 3 D1N3491 V_VDC 0 3 1V V_VAC 1 0 sin(0 5v 30khz) .model D1N3491 D(Is=68.65f Rs=3.786m Ikf=1.774 N=1 Xti=2 Eg=1.11 Cjo=1.457n + M=.9735 Vj=.75 Fc=.5 Isr=11.02u Nr=2 Tt=6.059u) * Motorola pid=1N3491 case=DO21 * 88-08-24 rmn .probe .tran 0us 100us .end Prog. 4 ****CLAMPER4 C_C1 1 2 .01uf R_R1 2 0 560k D_D1 0 2 D1N3491 V_VAC 1 0 sin(0 5v 30khz) .model D1N3491 D(Is=68.65f Rs=3.786m Ikf=1.774 N=1 Xti=2 Eg=1.11 Cjo=1.457n + M=.9735 Vj=.75 Fc=.5 Isr=11.02u Nr=2 Tt=6.059u) * Motorola pid=1N3491 case=DO21 * 88-08-24 rmn .probe .tran 0us 100us .end Prog. 5 ****CLAMPER5 C_C1 1 2 .01uf R_R1 2 0 560k D_D1 3 2 D1N3491 V_VDC 3 0 1V V_VAC 1 0 sin(0 5v 30khz) .model D1N3491 D(Is=68.65f Rs=3.786m Ikf=1.774 N=1 Xti=2 Eg=1.11 Cjo=1.457n + M=.9735 Vj=.75 Fc=.5 Isr=11.02u Nr=2 Tt=6.059u) * Motorola pid=1N3491 case=DO21 * 88-08-24 rmn .probe .tran 0us 100us .end Prog. 6 ****CLAMPER6 C_C1 1 2 .01uf R_R1 2 0 560k D_D1 3 2 D1N3491 V_VDC 0 3 1V V_VAC 1 0 sin(0 5v 30khz) .model D1N3491 D(Is=68.65f Rs=3.786m Ikf=1.774 N=1 Xti=2 Eg=1.11 Cjo=1.457n + M=.9735 Vj=.75 Fc=.5 Isr=11.02u Nr=2 Tt=6.059u) * Motorola pid=1N3491 case=DO21 * 88-08-24 rmn .probe .tran 0us 100us .end RESULT: Result of CLAMPER1 Result of CLAMPER2 Result of CLAMPER3 Result of CLAMPER4 Result of CLAMPER5 Result of CLAMPER6 EXPERIMENT NO.6 AIM: Simulate and study transient & frequency response of a BJT amplifier in common-emitter configuration using PSPICE windows. CIRCUIT DIAGRAM: RC 10k R1 47k Rs 1 500 6 10uF C1 2 C2 4 3 10uF Q2N2907 RL 5 VIN VOFF = 0 VAMPL = 10mV FREQ = 1kHz Vcc Q1 20k R2 5k RE 2k PROGRAM: *Transient & Freq. Response of Bipolar Transistor Amplifier .OPTIONS NOPAGE NOECHO * Transient analysis for 0 to 2 ms with 50 æs increment * Print details of transient analysis operating point. .TRAN/OP 50US 2MS * AC analysis from 1 Hz to 10 KHz with a decade increment and * 10 points per decade .AC DEC 10 1HZ 10KHZ * Print the details of AC analysis operating point .OP * Input voltage is 10 mV peak for ac analysis and for transient response * it is 10 mV peak at 1 kHz with zero-offset value VIN 1 0 AC 10MV SIN (0 10MV 1KHZ) VCC 0 7 DC 15V RS 1 2 500 R1 7 3 47K R2 3 0 5K RC 7 4 10K CE 10uF 15Vdc RE 5 0 2K RL 6 0 20K C1 2 3 10UF C2 4 6 10UF CE 5 0 10UF * Transistor Q1 with model QM Q1 4 3 5 0 QM * Model QM for PNP transistors .MODEL QM PNP (IS=2E-16 BF=100 BR=1 RB=5 RC=1 RE=0 TF=0.2NS TR=5NS + CJE=0.4PF VJE=0.8 ME=0.4 CJC=0.5PF VJC=0.8 CCS=1PF VA=100) * Plot the results of transient analysis for voltages at nodes 4, 6 and 1 .PLOT TRAN V(4) V(6) V(1) * Plot the results of ac analysis for the magnitude and phase angle * of output voltage at node 6 .PLOT AC VM(6) VP(6) .PROBE .END RESULT: EXPERIMENT NO.7 AIM: Simulate and study Half-wave & Full-wave Rectifier using PSPICE windows. CIRCUIT DIAGRAM: Rs D1 2 1 3 5 10 mod1 L1 Vin VOFF = 0 VAMPL = 220V FREQ = 50Hz L2 RL 2000uH 20uH 500 Half-wave Rs 2 D2 1 3 10 D1N4009 L2 10uH Vin VOFF = 0 VAMPL = 220V FREQ = 50Hz RL L1 4 5 2000uH L3 1000 10uH D1N4009 D1 Full-wave Rectifier PROGRAM: *HALF WAVE rectifier Vin 2 0 sin(0 220 50 ) RL 5 0 500 RS 2 1 10 L1 1 0 2000 L2 3 0 20 K1 L1 L2 0.99999 D1 3 5 mod1 .model mod1 D (IS=1e-14, n=1) .tran 0.2m 200m .plot tran v(3), v(5) .probe .end *FULL WAVE rectifier Vin 2 0 sin(0 230V 50HZ) RL 5 4 1000 RS 2 1 10 L1 1 0 2000 L2 3 4 10 L3 4 0 10 K1 L1 L2 L3 0.99 D1 0 5 D1N4009 D2 3 5 D1N4009 .model D1N4009 D(Is=544.7E-21 N=1 Rs=.1 Ikf=0 Xti=3 Eg=1.11 Cjo=4p M=.3333 + Vj=.75 Fc=.5 Isr=30.77n Nr=2 Bv=25 Ibv=100u Tt=2.885n) .tran 0.2ms 200ms .probe .end RESULT: Half-wave Rectifier Full-wave Rectifier EXPERIMENT NO.8 AIM: Simulate and study active low-pass, high-pass & band-pass filter using PSPICE windows. CIRCUIT DIAGRAM: RF 10k 100k 3 V+ 8 RG + R1 2 1 Vo V- OUT - UA741 4 10k Vin C1 0.1uF Low-pass Filter R4 10k R2 20k R4 C2 10k R2 8 1n UA741 20k 8 V3 + R3 - 2 R5 1 - UA741 10k Vo V- 2 10k 8 OUT OUT V1 + 10k 4 V1 - 3 1n V- OUT 2 R1 R5 1 V+ C2 20k 4 + 20k V+ 3 V+ R1 R6 UA741 4 100k R3 10k R7 100k Band-pass Filter High-pass Filter 1 Vo PROGRAM: 1. *LOWpass Filter Circuit * Input voltage is 1 V peak for ac analysis or frequency response VIN 1 0 AC 1 RG 2 0 10k RF 2 4 100K C1 3 0 .1UF R1 1 3 10K * Subcircuit call for op-amp OPAMP XA1 2 3 4 0 OPAMP * Subcircuit definition for OPAMP .SUBCKT OPAMP 1 2 7 4 RI 1 2 2.0E6 * Voltage-controlled current source with a gain of 0.1M GB 4 3 1 2 0.1M R2 3 4 10K C2 3 4 1.5619UF * Voltage-controlled voltage source of gain 2E+5 EA 4 5 3 4 2E+5 RO 5 7 75 * End of subcircuit definition .ENDS OPAMP * AC analysis for 10 Hz to 100 MHz with a decade increment and * 10 points per decade .AC DEC 10 10HZ 100MEGHZ * Plot the results of ac analysis .PLOT AC VM(4) VP(4) .PROBE .END 2. *Highpass Filter Circuit * Input voltage is 1 V peak for ac analysis or frequency response VIN 1 0 AC 1 R1 1 2 20K R2 2 4 20K R3 3 0 10K R4 1 5 10K R5 4 5 10K RL 4 0 100K C1 2 4 0.01UF * Subcircuit call for op-amp OPAMP XA1 2 3 4 0 OPAMP * Subcircuit definition for OPAMP .SUBCKT OPAMP 1 2 7 4 RI 1 2 2.0E6 * Voltage-controlled current source with a gain of 0.1M GB 4 3 1 2 0.1M R2 3 4 10K C2 3 4 1.5619UF * Voltage-controlled voltage source of gain 2E+5 EA 4 5 3 4 2E+5 RO 5 7 75 * End of subcircuit definition .ENDS OPAMP * AC analysis for 10 Hz to 100 MHz with a decade increment and * 10 points per decade .AC DEC 10 10HZ 100MEGHZ * Plot the results of ac analysis .PLOT AC VM(5) VP(5) .PROBE .END 3. *Bandpass Filter Circuit * Input voltage is 1 V peak for ac analysis or frequency response VIN 1 0 AC 1 R1 1 2 20K R2 2 4 20K R3 3 0 10K R4 1 5 10K R5 4 5 10K R6 6 7 100K R7 6 0 100k RL 7 0 100K C1 2 4 0.01UF * Subcircuit call for op-amp OPAMP XA1 2 3 4 0 OPAMP XA2 5 6 7 0 OPAMP * Subcircuit definition for OPAMP .SUBCKT OPAMP 1 2 7 4 RI 1 2 2.0E6 * Voltage-controlled current source with a gain of 0.1M GB 4 3 1 2 0.1M R2 3 4 10K C2 3 4 1.5619UF * Voltage-controlled voltage source of gain 2E+5 EA 4 5 3 4 2E+5 RO 5 7 75 * End of subcircuit definition .ENDS OPAMP * AC analysis for 10 Hz to 100 MHz with a decade increment and * 10 points per decade .AC DEC 10 10HZ 100MEGHZ * Plot the results of ac analysis .PLOT AC VM(7) VP(7) .PROBE .END RESULT: 1. 2. 3. EXPERIMENT NO.9 EXPERIMENT NO.9 AIM: Simulate and study Integrator using PSPICE windows. CIRCUIT DIAGRAM: C1 0.1uF R1 Rf 2 4 1MEG 8 2.5k Vin 3 + V+ 1 3 - 4 2 output voltage V- OUT 1 Rx 2.5k Rl 100k PROGRAM: * Integrator Circuit * The input voltage is represented by a piece-wise linear waveform. * To avoid convergence problems due to a rapid change of the input * voltage, the input voltage is assumed to have a finite slope. VIN 1 0 PWL(0 0 1NS -1V 1MS -1V 1.0001MS 1v 2ms 1v 2.0001ms -1V 3MS -1V 3.0001MS 1V 4MS 1V) R1 1 2 2.5K RF 2 4 1MEG RX 3 0 2.5K RL 4 0 100K C1 2 4 0.1UF * Calling subcircuit OPAMP XA1 2 3 4 0 OPAMP * Subcircuit definition for OPAMP .SUBCKT OPAMP 1 2 7 4 RI 1 2 2.0E6 * Voltage-controlled current source with a gain of 0.1M GB 4 3 1 2 0.1M R1 3 4 10K C1 3 4 1.5619UF * Voltage-controlled voltage source with a gain of 2E+5 EA 4 5 3 4 2E+5 RO 5 7 75 * End of subcircuit OPAMP .ENDS * Transient analysis for 0 to 4 ms with 50 us increment .TRAN 50US 4MS * Plot the results of transient analysis for the voltage at node 4 .PLOT TRAN V(4) V(1) .PLOT AC VM(4) VP(4) .PROBE .END RESULT: EXPERIMENT NO.10 AIM: Simulate and study Differentiator using PSPICE windows. CIRCUIT DIAGRAM: C1 2 4 0.4uF 10k 8 100 Rf 3 Vin 3 + V+ R1 1 5 - 4 2 output voltage V- OUT 1 Rx 10k Rl 100k PROGRAM: * Differentiator Circuit * The maximum number of points is changed to 410. The default * value is only 201. .OPTIONS NOPAGE NOECHO LIMPTS=410 * Input voltage is a piece-wise linear waveform for transient analysis. VIN 1 0 PWL(0 0 1MS 1 2MS 0 3MS 1 4MS 0) R1 1 2 100 RF 3 4 10K RX 5 0 10K RL 4 0 100K C1 2 3 0.4UF * Calling op-amp OPAMP XA1 3 5 4 0 OPAMP * Op-amp subcircuit definition .SUBCKT OPAMP 1 2 7 4 RI 1 2 2.0E6 * Voltage-controlled current source with a gain of 0.1M GB 4 3 1 2 0.1M R1 3 4 10K C1 3 4 1.5619UF * Voltage-controlled voltage source with a gain of 2E+5 EA 4 5 3 4 2E+5 RO 5 7 75 * End of subcircuit OPAMP .ENDS OPAMP * Transient analysis for 0 to 4 ms with 50 æs increment .TRAN 10US 4MS * Plot the results of transient analysis 4 .PLOT TRAN V(4) V(1) .PROBE .END RESULT: Prog. 4 ****CLAMPER4 C_C1 1 2 .01uf R_R1 2 0 560k D_D1 0 2 D1N3491 V_VAC 1 0 sin(0 5v 30khz) .model D1N3491 D(Is=68.65f Rs=3.786m Ikf=1.774 N=1 Xti=2 Eg=1.11 Cjo=1.457n + M=.9735 Vj=.75 Fc=.5 Isr=11.02u Nr=2 Tt=6.059u) * Motorola pid=1N3491 case=DO21 * 88-08-24 rmn .probe .tran 0us 100us .end Prog. 5 ****CLAMPER5 C_C1 1 2 .01uf R_R1 2 0 560k D_D1 3 2 D1N3491 V_VDC 3 0 1V V_VAC 1 0 sin(0 5v 30khz) .model D1N3491 D(Is=68.65f Rs=3.786m Ikf=1.774 N=1 Xti=2 Eg=1.11 Cjo=1.457n + M=.9735 Vj=.75 Fc=.5 Isr=11.02u Nr=2 Tt=6.059u) * Motorola pid=1N3491 case=DO21 * 88-08-24 rmn .probe .tran 0us 100us .end Prog. 6 ****CLAMPER6 C_C1 1 2 .01uf R_R1 2 0 560k D_D1 3 2 D1N3491 V_VDC 0 3 1V V_VAC 1 0 sin(0 5v 30khz) .model D1N3491 D(Is=68.65f Rs=3.786m Ikf=1.774 N=1 Xti=2 Eg=1.11 Cjo=1.457n + M=.9735 Vj=.75 Fc=.5 Isr=11.02u Nr=2 Tt=6.059u) * Motorola pid=1N3491 case=DO21 * 88-08-24 rmn .probe .tran 0us 100us .end RL 4 0 100K C1 2 4 0.01UF * Subcircuit call for op-amp OPAMP XA1 2 3 4 0 OPAMP * Subcircuit definition for OPAMP .SUBCKT OPAMP 1 2 7 4 RI 1 2 2.0E6 * Voltage-controlled current source with a gain of 0.1M GB 4 3 1 2 0.1M R2 3 4 10K C2 3 4 1.5619UF * Voltage-controlled voltage source of gain 2E+5 EA 4 5 3 4 2E+5 RO 5 7 75 * End of subcircuit definition .ENDS OPAMP * AC analysis for 10 Hz to 100 MHz with a decade increment and * 10 points per decade .AC DEC 10 10HZ 100MEGHZ * Plot the results of ac analysis .PLOT AC VM(5) VP(5) .PROBE .END 4. *Bandpass Filter Circuit * Input voltage is 1 V peak for ac analysis or frequency response VIN 1 0 AC 1 R1 1 2 20K R2 2 4 20K R3 3 0 10K R4 1 5 10K R5 4 5 10K R6 6 7 100K R7 6 0 100k RL 7 0 100K C1 2 4 0.01UF * Subcircuit call for op-amp OPAMP XA1 2 3 4 0 OPAMP XA2 5 6 7 0 OPAMP * Subcircuit definition for OPAMP .SUBCKT OPAMP 1 2 7 4 RI 1 2 2.0E6 * Voltage-controlled current source with a gain of 0.1M GB 4 3 1 2 0.1M