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CIRCUIT ANALYSIS USING SPICE AND PSPICE
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15.1 Use PSpice to nd V 3; 4 in the circuit of Fig. 15-23.
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Fig. 15-23 The source le is DC analysis, Fig. 15-23 Vs 2 0 R1 0 1 R2 0 1 DC 36 12 105 V
CIRCUIT ANALYSIS USING SPICE AND PSPICE
[CHAP. 15
R3 R4 R5 R6 .DC .PRINT .END
1 2 3 4 Vs DC
2 3 4 0 105 V(1)
74 16.4 103.2 28.7 105 V(3, 4)
The output le contains the following: DC TRANSFER CURVES Vs V(1) 1:050E 02 1:139E 01 Therefore, V 3; 4 73:07 V. V(3, 4) 7:307E 01
Write the source le for the circuit of Fig. 15-24 and nd I in R4 .
Fig. 15-24
The source le is DC analysis, VS Is R1 R2 R3 R4 .DC .PRINT .END Fig. 15-24 2 0 0 3 0 1 1 2 1 3 3 0 Vs 200 DC I(R4) DC DC 27 47 4 23 200 200 V 20 A
The output le contains the following results: DC TRANSFER CURVE Vs I(R4) 2:000E 02 1:123E 01 Current I R4 11:23 A ows from node 3 to node 0 according to the order of nodes in the data statement for R4.
Find the three loop currents in the circuit of Fig. 15-25 using PSpice and compare your solution with the analytical approach.
CHAP. 15]
CIRCUIT ANALYSIS USING SPICE AND PSPICE
Fig. 15-25
The source le is DC analysis, V1 V2 R1 R2 R3 R4 R5 .DC .PRINT .END Fig. 15-25 2 0 0 4 0 1 1 2 1 3 3 0 3 4 V1 25 DC I(R1) DC DC 2 5 10 4 2 25 I(R3) 25 50
1 I(R5)
The output le includes the following results: DC TRANSFER CURVES V1 I(R1) 2:500E 01 1:306E 00 I(R3) 3:172E 00 I(R5) 1:045E 01
The analytical solution requires solving three simultaneous equations.
Using PSpice, nd the value of Vs in Fig. 15-4 such that the voltage source does not supply any power.
We sweep Vs from 1 to 10 V. The source and output les are DC sweep in the circuit of Fig. 15-4 R1 0 1 500 R2 1 2 3k R3 2 3 1k R4 0 3 1.5 k Vs 3 1 DC 4V Is 0 2 DC 3 mA .DC Vs 1 10 1 .PRINT DC I(Vs) .PROBE .PLOT DC I(Vs) .END The output le contains the following results: DC TRANSFER CURVES Vs I(Vs) 1:000E 00 7:500E 04 2:000E 00 2:188E 12 3:000E 00 7:500E 04
CIRCUIT ANALYSIS USING SPICE AND PSPICE
[CHAP. 15
4:000E 00 5:000E 00 6:000E 00 7:000E 00 8:000E 00 9:000E 00 1:000E 01
1:500E 03 2:250E 03 3:000E 03 3:750E 03 4:500E 03 5:250E 03 6:000E 03
The current in Vs is zero for Vs 2 V.
Perform a dc analysis on the circuit of Fig. 15-26 and nd its Thevenin equivalent as seen from terminal AB.
Fig. 15-26 We include a .TF statement in the following netlist: Thevenin Vs R1 Is .TF .END equivalent of Fig. 15-26 1 0 DC 3 1 2 10 0 2 DC 1 V(2) Is
The output le includes the following results: NODE (1) VOLTAGE 3.0000 NODE (2) VOLTAGE 13.000
VOLTAGE SOURCE CURRENTS NAME CURRENT Vs 1:000E 00 TOTAL POWER DISSIPATION 3:00E 00 WATTS
SMALL-SIGNAL CHARACTERISTICS V 2 =Is 1:000E 01 INPUT RESISTANCE AT Is 1:000E 01 OUTPUT RESISTANCE AT V 2 1:000E 01 The Thevenin equivalent is VTh V2 13 V, RTh 10
Perform an ac analysis on the circuit of Fig. 15-27(a). Find the complex magnitude of V2 for f varying from 100 Hz to 10 kHz in 10 steps.
We add to the netlist an .AC statement to sweep the frequency and obtain V(2) by any of the commands .PRINT, .PLOT, or .PROBE. The source le is AC analysis of Fig. Vs 1 R1 1 R2 2 15-27(a). 0 AC 2 1k 0 2k 10 0
CHAP. 15]
CIRCUIT ANALYSIS USING SPICE AND PSPICE
C .AC .PRINT .PLOT .PROBE .END
2 0 LIN AC AC
1 uF 10 Vm(2) Vm(2) Vm(2)
10000 Vp(2) Vp(2) Vp(2)
Fig. 15-27 The output le contains the following results: AC ANALYSIS FREQ 1:000E 02 1:200E 03 2:300E 03 3:400E 03 4:500E 03 VM(2) 6:149E 00 1:301E 00 6:883E 01 4:670E 01 3:532E 01 VP(2) 2:273E 01 7:875E 01 8:407E 01 8:598E 01 8:696E 01
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