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CIRCUIT ANALYSIS USING SPICE AND PSPICE
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[CHAP. 15
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.PRINT prints the value of variables. :PRINT
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The syntax is htypei houtput variablesi
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htypei is DC, AC, or TRAN (transient). .PLOT line-prints variables. The syntax is :PLOT htypei houtput variablesi
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.PROBE generates a data le *.DAT which can be plotted in post-analysis by evoking the Probe program. The syntax is :PROBE houtput variablesi
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EXAMPLE 15.8 Find the value of Vs in the circuit in Fig. 15-8 such that the power dissipated in the 1-k
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resistor is zero. Use the .DC command to sweep Vs from 1 to 6 V in steps of 1 V and use .PRINT to show I Vs , V(1,2), and V(2).
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Fig. 15-8
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The source le is DC sweep, Fig. 15-8 Vs 1 0 DC 1V Is 0 2 DC 1 mA R1 1 2 1k R2 0 2 2k .DC Vs 1 6 1 .PRINT DC I(Vs) V(1,2) V(2) .END The results in the output le are DC TRANSFER CURVES Vs I(Vs) 1:000E 00 3:333E 04 2:000E 00 1:333E 12 3:000E 00 3:333E 04 4:000E 00 6:667E 04 5:000E 00 1:000E 03 6:000E 00 1:333E 03 The answer is Vs 2 V. EXAMPLE 15.9 Write the source le for the circuit in Fig. 15-9(a) using commands .DC, .PLOT, and .PROBE to nd the I-V characteristic equation for I varying from 0 to 2 A at the terminal AB. First, we connect a dc current source Iadd at terminal AB, sweep its value from 0 to 2 A using the .DC command, and plot V versus I. Since the circuit is linear, two points are necessary and su cient. However, for clarity of the plot, ten steps are included in the source le as follows: V(1,2) 3:333E 01 1:333E 09 3:333E 01 6:667E 01 1:000E 00 1:333E 00 V(2) 1:333E 00 2:000E 00 2:667E 00 3:333E 00 4:000E 00 4:667E 00
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CHAP. 15]
CIRCUIT ANALYSIS USING SPICE AND PSPICE
Fig. 15-9
Terminal Characteristic, Fig. 15-9 Iadd 0 5 DC Is 0 4 DC Vs 3 2 DC R1 0 1 1 R2 1 2 2 R3 3 4 3 R4 4 5 2 .DC Iadd 0 2 .PLOT DC V(5) .PROBE .END The output is shown in Fig. 15-9(b).
0 0.6 A 5V
The I-V equation is V 8I 8:6.
CIRCUIT ANALYSIS USING SPICE AND PSPICE
[CHAP. 15
15.6 THEVENIN EQUIVALENT .TF Statement The .TF command provides the transfer function from an input variable to an output variable and produces the resistances seen by the two sources. It can thus generate the Thevenin equivalent of a resistive circuit. The syntax is :TF houtput variablei hinput variablei
EXAMPLE 15.10 15-10.
Use the command .TF to nd the Thevenin equivalent of the circuit seen at terminal AB in Fig.
Fig. 15-10 The node numbers and element names are shown on Fig. 15-10. Transfer Function in Fig. 15-10 Vs 1 0 DC 12 E1 4 0 2 0 10 R1 1 2 1k R2 2 0 2k R3 2 3 1k R4 3 4 200 .TF V(3) Vs .END The output le contains the following results: NODE (1) VOLTAGE 12.0000 NODE (2) VOLTAGE 2:0000 NODE (3) VOLTAGE 17:0000 NODE (4) VOLTAGE 20:000 The source le is
VOLTAGE SOURCE CURRENTS NAME CURRENT Vs 1:400E 02 TOTAL POWER DISSIPATION 1:68E 01 WATTS SMALL-SIGNAL CHARACTERISTICS V(3)/Vs 1:417E 00 INPUT RESISTANCE AT Vs 8:571E 02 OUTPUT RESISTANCE AT V(3) 6:944E 01 Therefore, VTh 1:417 12 17 V and RTh 69:44
OP AMP CIRCUITS
Operational ampli ers may be modeled by high input impedance and high gain voltage-controlled voltage sources. The model may then be used within a net list repeatedly.
EXAMPLE 15.11 Find the transfer function V3 =Vs in the ideal op amp circuit of Fig. 15-11(a). The op amp is replaced by a voltage-dependent voltage source with a gain of 106 [see Fig. 15-11(b)]. The source le is
CHAP. 15]
CIRCUIT ANALYSIS USING SPICE AND PSPICE
Fig. 15-11 Inverting Vs E1 R1 R2 .TF .END op amp circuit, Fig. 15-11 1 0 DC 12 3 0 0 2 1E6 1 2 1k 2 3 2k V(3) Vs
The transfer function is written in the output le: NODE (1) VOLTAGE 12.0000 NODE (2) VOLTAGE 24.00E 06 NODE (3) VOLTAGE 24.0000
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