Pspice statements and dot commands used in this circuit




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http://encon.fke.utm.my/SEE1023-05

A Simple note
1. Suggested format for our Spice input file as follows:
Title Statement

Circuit Description

Analysis Request

Output Request


2. The first line must be a title and the last line must be .END. The sequence of the remaining lines is arbitrary.
3. For a circuit that we do know the operating frequency, use that frequency in the Analysis Request. For the cicuit that we don’t know the operating frequency, use w = 1 rad/s or f = 0.15916Hz in the Analysis Request. Convert XL and XC into their Henry and Farad, respectively using the chosen operating frequency.
4. <+node> and <-node> for passive elements are only important when we want to know their currents. We must follow Passive Sign Convention (PSC) to write the Output Request, i.e. the current is considered positive when it enters the positive <+node>.
5. We have to place a voltage source with 0V in the branch in order to sense the current for CCCS (F device) and CCVS (H Device).
6. For the Output Request, if we want to know a voltage we can write the voltage between two nodes. If we want to know a current we can write the current through element according to PSC.
7. Please be careful when to put unit for capacitor; in PSpice F means femto(10-15). It would be better if we let the capacitors without unit. Also, always keep in mind that PSpice is case insensitive.
8. We can use *……… in a new line to insert comment to improve readability.
9. Summary of PSpice statements and Dot Command used in the circuits below as follows:

PSPICE STATEMENTS AND DOT COMMANDS USED IN THIS CIRCUIT

[...] means an optional item; <...> means a required item.







PSpice Statement

Example

Resistor

R[name]<+node><-node><value>

R1 1 2 100

Inductor

L[name]<+node><-node><value>

L2 2 3 200m

Capacitor

C[name]<+node><-node><value>

C5 5 3 400u

AC Current Source

I[name]<+node><-node>AC<mag><phase>

I1 0 1 AC 3A 0

AC Voltage Source

V[name]<+node><-node>AC<mag><phase>

V1 2y 0 AC 18V 30

VCCS (G Device)

V[name]<+node><-node><+Ctrl node> <-Ctrl node><gain>

G1 1 2 1 0 0.2

CCCS (F Device)

V[name]<+node><-node><CTRL device name><gain>

F1 3 0 Vx 2










.AC Command

.AC LIN

.AC LIN 1 1Hz 1Hz

.PRINT AC

.PRINT AC<Output variable list>

.PRINT AC IM(L2) IP(L2)

AC CIRCUIT EXAMPLE 1

I1 0 1 AC 3A 0

R1 0 1 1


C1 0 1 1

R2 1 1x 2

L1 1x 2 2

G1 1 2 1 0 0.2

C2 1 2 0.5

C3 0 2 1


L2 2 2x 2

R3 2x 2y 2

V1 2y 0 AC 18V 30

.AC LIN 1 0.15916Hz 0.15916Hz

.PRINT AC VM(1) VP(1) VR(1) VI(1)

.PRINT AC VM(2) VP(2) VR(2) VI(2)

.PRINT AC IM(L2) IP(L2) IR(L2) II(L2)

.PRINT nopage nobias

.END


AC CIRCUIT EXAMPLE 2

V1 1 0 AC 20V 30

R1 1 2 10

C1 2 2x 0.1

VX 2x 0 0

L1 2 3 1H

F1 3 0 Vx 2

L2 3 0 0.5H

*Vx is required for F1

.AC LIN 1 0.6366HZ 0.6366HZ

.PRINT AC IM(C1) IP(C1) IR(C1) II(C1)

.PRINT AC VM(2) VP(2) VR(2) VI(2)

.OPT nopage nobias

.END



10.3 From Sadiku

Determine Vo and Io in the circuit below using PSpice


This circuit gives the operating frequency. So, use that operating frequency in the .AC Command.

AC CIRCUIT (10.3 Sadiku)

V1 1y 0 AC 16 0

R1 1y 1x 4

C1 1x 1 {1/12}

R2 1 0 4


I1 0 1 AC 2 90

L1 1 1p 2

R3 1p 0 4

.AC LIN 1 0.6366Hz 0.6366Hz

.PRINT AC Vm(1,0) Vp(1,0)

.PRINT AC Im(L1) Ip(L1)

.OPT nopage nobias

.END


10.15 From Sadiku

Solve for the current I and the voltage Vo in the circuit using PSpice.


This circuit did not give the operating frequency. So, use w = 1 rad/s or f = 0.15916Hz as its operating frequency in the .AC Command and to convert XL and XC into their L and C, respectively.
AC CIRCUIT (10.15 Sadiku)

V1 1x 0 AC 20 -90

R1 1x 1 4

C1 1y 0 0.5

L1 1 2 1

I1 1 2 AC 5 0

F1 0 2 Vx 2

R2 0 2 4


Vx 1 1y AC 0 0

*Vx is required for F1

.AC LIN 1 0.15916Hz 0.15916Hz

.PRINT AC Im(C1) Ip(C1)

.PRINT AC Vm(1,2) Vp(1,2)

.OPT nopage nobias

.END

10.19 From Sadiku

Obtain Vo and Io using PSpice


This circuit did not give the operating frequency. So, use w = 1 rad/s or f = 0.15916Hz as its operating frequency in the .AC Command and to convert XL and XC into their L and C, respectively.


AC CIRCUIT (10.19 Sadiku)

*My name:

V1 1 2 AC 12 0

R1 1 0 2

C1 2 0 {1/4}

L1 3 1 2

G1 0 3 1 0 0.2

.AC LIN 1 0.15916Hz 0.15916Hz

.PRINT AC Vm(1) Vp(1)

.PRINT AC Im(L1) Ip(L1)

.OPT nopage nobias

.END

10.62 From Sadiku

Using Thevenin’s theorem, find Vo in the circuit below.


This circuit gives the operating frequency. So, use that operating frequency in the .AC Command.
AC CIRCUIT (10.62 Sadiku)

*MY Name: Nik Din Muhamad

*TO FIND VTH

V1 1x 0 AC 12 0

R1 1x 1 4

C1 1 0 {1/4}

L1 1 2 2

C2 2 0 {1/8}

F1 1 2 V1 -3

*no need Vx for F1

.AC LIN 1 0.15916Hz 0.15916Hz

.PRINT AC Vm(2) Vp(2)

.OPT nopage nobias

.END






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