EENG 2910: Circuit Design and Analysis Using PSpiceClass 5a: AC Circuit Analysis 5b: Op-Amp Circuit Analysis

Oluwayomi Adamo

Department of Electrical Engineering

College of Engineering, University of North Texas

EENG 2910, Class 5 2

5a. New Parts Needed for AC Analysis AC Sources

VAC and IAC Both are form “source.olb” The ACMAG and ACPHASE properties can be configured in the Property Editor. For AC sources, you can also set DC value. But for AC analysis, you can set DC

value to zero. The frequency f of the AC source is configured in Simulation Profile Editor. The

frequency needs to be in the Hz unit. The simulation type for AC analysis is AC Sweep/Noise.

Printer Parts IPRINT is used to print current into the output file. VPRINT1 (with one lead) and VPRINT2 (with two lead) are used to print voltage

into the output file. Both IPRINT and VPRINT are from “special.olb” In Property Editor you can configure to select which property to print. For

example, REAL, IMAG, MAG, PHASE. Put “y” for a property, if you want to print that property.

If one of the fields does not appear in the property editor for some of the printers, add it by pressing the “New Column …” button in the property editor.

AAetfA j )2cos(

EENG 2910, Class 5 3

New Command Needed in PSpice AD To plot magnitude

V( ) or VM( ) and I() or IM( ) for peak value in linear scale VDB( ) and IDB in dB scale

To plot phase VP( ) and IP( ) for phase

To plot real part and imaginary part VR( ), VI( ), IR( ), II( )

EENG 2910, Class 5 4

Coupled Inductors

The symbol for mutual coupling is K. The mutual coupling part is the “K-Linear” from “analog.olb”. K-Linear can couple up to eight inductors.

The mutual coupling K couples two or more inductors with a coupling coefficient k. The coupling coefficient k:

PSpice assumes a dot on the first node of each inductor. Read the book, page 145, for more details.

21

,10LL

Mkk

M is mutual inductance

EENG 2910, Class 5 5

Example 5.2

9

V1ACMAG = 1VACPHASE = 0

R3

8VV

0

V

8

6

L3

50uH1 2

3

7

C310uF

L2

50uH1 2

C210uF

5R2

1

R1

2

L1

50uH1 21

C110uF

V3ACMAG = 1VACPHASE = 0

2

V2ACMAG = 1VACPHASE = 0

4

Figure 5.2.1

Figure 5.2.2

Frequency

100Hz 10KHz 100KHzVP(3) VP(6) VP(9)

-200d

-100d

0dVM(3)/VM(1) VM(6)/VM(1) VM(9)/VM(1)

0

1.0

2.0

3.0

SEL>>

EENG 2910, Class 5 6

Example 5.4

Time

0s 25ms 50ms-I(Ra) -I(Rb) -I(Rc) -I(Vx)

-40A

0A

40AAVG(V(1)*I(Ra))+AVG(V(2)*I(Rb))+AVG(V(3)*I(Rc))

0W

2.5KW

5.0KW

SEL>>

C1

150uF

12

Van

FREQ = 60

VAMPL = 169.7VOFF = 0

PHASE = 0

R3

10

b

0

2

6

5Ry

1

Rz

1

Load

I

3

Ra

0.5

n

4

cRc0.5

C

Vx

0Vdc

10

9

Transmission Line

L1

120mH

1 2

R1

5

R2

10

Vbn

FREQ = 60

VAMPL = 169.7VOFF = 0

PHASE = -120

Generator

I

I

11

A 71 a

B

Rx

1

Rb0.5

Vcn

FREQ = 60

VAMPL = 169.7VOFF = 0

PHASE = -240

8

I

The AC source is VSIN

Both of these vertical lines are dashed lines, not wires!

Figure 5.4.1

Figure 5.4.2

EENG 2910, Class 5 7

5b. Op Amps Circuit Example 10.2

Time

0s 2.0ms 4.0msV(OUT)

-4.0V

0V

4.0V

SEL>>

V(V1:+)-2.0V

0V

2.0V

Figure 10.2.1

(Your final drawing of circuit schematics)

Figure 10.2.2

Question: Please explain the relation between V(V1:+) and V(out).

EENG 2910, Class 5 8

Example 10.4

Frequency

10KHz 100MHz10Hzabs(VP(7))

0d

50d

100d

SEL>>

VM(7)0V

0.5V

1.0V

Figure 10.4.1

(Your final drawing of circuit schematics)

Figure 10.4.2

EENG 2910, Class 5 9

Assignment 5

Repeat and reproduce the results for Examples 5.2, and 5.4.

Repeat and reproduce the results for Examples 10.2, and 10.4,