how to generate barcode in ssrs report VOLTAGE SOURCE CURRENTS NAME CURRENT Vs 1.200E 02 TOTAL POWER DISSIPATION 1.44E 01 WATTS in Software

Generator Quick Response Code in Software VOLTAGE SOURCE CURRENTS NAME CURRENT Vs 1.200E 02 TOTAL POWER DISSIPATION 1.44E 01 WATTS

VOLTAGE SOURCE CURRENTS NAME CURRENT Vs 1.200E 02 TOTAL POWER DISSIPATION 1.44E 01 WATTS
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SMALL-SIGNAL CHARACTERISTICS V(3)/Vs 2:000E + 00 INPUT RESISTANCE AT Vs 1.000E + 03 OUTPUT RESISTANCE AT V(3) 0.000E + 00
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.SUBCKT Statement A subcircuit is de ned by a set of statements beginning with .SUBCKT hnamei hexternal terminalsi
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and terminating with an .ENDS statement. Within a netlist we refer to a subcircuit by Xaa hnamei hnodesi
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Hence, the .SUBCKT statement can assign a name to the model of an op amp for repeated use.
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EXAMPLE 15.12 Given the circuit of Fig. 15-12(a), nd Is , If , V2 , and V6 for Vs varying from 0.5 to 2 V in 0.5-V steps. Assume a practical op amp [Fig. 15-12(b)], with Rin 100 k
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, Cin 10 pF, Rout 10 k
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, and an open loop gain of 105 . The source le employs the subcircuit named OPAMP of Fig. 15-12(b) whose description begins with .SUBCKT and ends with .ENDS. The X1 and X2 statements describe the two op amps by referring to the OPAMP subcircuit. Note the correspondence of node connections in the X1 and X2 statements with that of the external terminals speci ed in the .SUBCKT statement. The source le is Op amp circuit of Fig. 15-12 .SUBCKT OPAMP Rin 1 2 Cin 1 2 Rout 3 5 using .SUBCKT 1 2 3 4 10 E5 10 pF 10 k
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CIRCUIT ANALYSIS USING SPICE AND PSPICE
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[CHAP. 15
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Eout .ENDS Vs Rs R1 R2 R3 R4 Rf X1 X2 .DC .PRINT .TF .END
5 4 1 0 1 2 2 3 3 4 4 5 5 6 6 2 0 3 0 5 Vs DC V(6)
10 E5
4 0 6 0 0.5 V(2) Vs
DC .5 1k 5k 9k 1.2 k 6k 40 k OPAMP OPAMP 2 0.5 V(6) I(Vs)
I(R1)
I(Rf)
Fig. 15-12
The output le is DC TRANSFER CURVES Vs V(2) 5:000E 01 5:000E 01 1:000E 00 1:000E 00 1:500E 00 1:500E 00 2:000E 00 2:000E 00 V(6) 4:500E 00 9:000E 00 1:350E 01 1:800E 01 I(Vs) 3:372E 09 6:745E 09 1:012E 08 1:349E 08 I(R1) 1:000E 04 2:000E 04 3:000E 04 4:000E 04 I(Rf) 9:999E 0 2:000E 0 3:000E 0 4:000E 0
CHAP. 15]
CIRCUIT ANALYSIS USING SPICE AND PSPICE
NODE (1) (5)
VOLTAGE NODE .5000 (2) 13:00E 06 (6)
VOLTAGE .5000 4.4998
NODE (3) (X1.5)
VOLTAGE NODE VOLTAGE 9:400E 06 (4) :9000 9:3996 (X2.5) 12:9990
VOLTAGE SOURCE CURRENTS NAME CURRENT Vs 3:372E 09 TOTAL POWER DISSIPATION 1:69E 09 WATTS
SMALL-SIGNAL CHARACTERISTICS V 6 =Vs 9:000E 00 INPUT RESISTANCE AT Vs 1:483E 08 OUTPUT RESISTANCE AT V 6 7:357E 02 There is no voltage drop across Rs . Therefore, V 2 Vs and the overall gain is V 6 =Vs V 2 =Vs 9. current drawn by R1 is provided through the feedback resistor Rf .
AC STEADY STATE AND FREQUENCY RESPONSE
Independent AC Sources Independent ac sources are described by a statement with the following syntax: hnamei hnodesi AC hmagnitudei hphase in degreesi
Voltage sources begin with V and current sources with I. The convention for direction is the same as that for dc sources.
EXAMPLE 15.13 Write data statements for the sources shown in Fig. 15-13.
Fig. 15-13 AC Source Voltage Current hnamei Vs Is hnodesi 2 1 3 4 htypei AC AC hmagnitudei 14 2.3 hphasei 45 105
.AC Statement The .AC command sweeps the frequency of all ac sources in the circuit through a desired range or sets it at a desired value. The syntax is :AC hsweep typei hnumber of pointsi hstarting fi hending fi
For the ac steady state, hsweep typei is LIN. In order to have a single frequency, the starting and ending frequencies are set to the desired value and the number of points is set to one. .PRINT AC and .PLOT AC Statements The .PRINT AC statement prints the magnitude and phase of the steady-state output. The syntax is :PRINT AC hmagnitudesi hphasesi
CIRCUIT ANALYSIS USING SPICE AND PSPICE
[CHAP. 15
The magnitudes and phases of voltages are Vm(variable) and Vp(variable), respectively, and the magnitudes and phases of currents are Im(variable) and Ip(variable), respectively. The syntax for .PLOT AC is similar to that for .PRINT AC.
EXAMPLE 15.14 In the series RLC circuit of Fig. 15-14(a) vary the frequency of the source from 40 to 60 kHz in 200 steps. Find the magnitude and phase of current I using .PLOT and .PROBE. The source le is AC analysis of series RLC, Fig. 15-14 Vs 1 0 AC 1 0 R 1 2 32 L 2 3 2m C 3 0 5n .AC LIN 200 40 k 60 k .PLOT AC Im(Vs) Ip(Vs) .PROBE Vm(1, 2) Vm(2,3) Vm(3) Im(Vs) .END
Ip(Vs)
The graph of the frequency response, plotted by Probe, is shown in Fig. 15-14(b).
Fig. 15-14
CHAP. 15]
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