Transcript Pspice Slide - Electrical Engineering

EENG 2910:
Circuit Design and Analysis Using PSpice
Class 5a: AC Circuit Analysis
5b: Op-Amp Circuit Analysis
Oluwayomi Adamo
Department of Electrical Engineering
College of Engineering, University of North Texas
5a. New Parts Needed for AC Analysis
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AC Sources
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A cos(2 f t   )  Ae j  A
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
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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.
EENG 2910, Class 5
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New Command Needed in PSpice AD
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To plot magnitude
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To plot phase
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V( ) or VM( ) and I() or IM( ) for peak value in linear scale
VDB( ) and IDB in dB scale
VP( ) and IP( ) for phase
To plot real part and imaginary part
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VR( ), VI( ), IR( ), II( )
EENG 2910, Class 5
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Coupled Inductors
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The symbol for mutual coupling is K.
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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:
0  k  1,
k
M
M is mutual
inductance
L1L2


PSpice assumes a dot on the first node of each inductor.
Read the book, page 145, for more details.
EENG 2910, Class 5
4
Example 5.2
R1
1
L1
R2
21
2
4
2
50uH
3
V1
L2
R3
51
50uH
1
Figure 5.2.1
C1
10uF
L3
81
2
50uH
8
6
V
V2
ACMAG = 1V
ACPHASE = 0
7
2
9
V
V
V3
ACMAG = 1V
ACPHASE = 0
C2
10uF
ACMAG = 1V
ACPHASE = 0
C3
10uF
0
3.0
2.0
1.0
SEL>>
0
VM(3)/VM(1)
VM(6)/VM(1)
VM(9)/VM(1)
0d
-100d
Figure 5.2.2
-200d
100Hz
VP(3)
EENG 2910, Class 5
10KHz
VP(6)
VP(9)
Frequency
100KHz
5
Generator
Example 5.4
Transmission Line
Rx
1
a
Ra
4
Load
R1
A
10
1
0.5
150u F
5
I
Van
C1
7
VAMPL = 16 9.7
VOF F = 0
FRE Q = 60
PHA SE = 0
The AC source is VSIN
Vbn
0.5
2 Rb
n
Ry
b
5
R2
B
8
1
VAMPL = 16 9.7
VOF F = 0
FRE Q = 60
PHA SE = -1 20
L1
11 1
10
I
2
120m H
Vcn
Figure 5.4.1
VAMPL = 16 9.7
VOF F = 0
FRE Q = 60
PHA SE = -2 40
5.0KW
Rz
3
Rc
0.5
c
6
1
I
2.5KW
0
SEL>>
0W
AVG(V(1)*I(Ra))+AVG(V(2)*I(Rb))+AVG(V(3)*I(Rc))
40A
R3
C
9
12
10
Vx
0Vdc
I
Both of these vertical lines are
dashed lines, not wires!
0A
-40A
0s
-I(Ra)
-I(Rb)
25ms
-I(Rc)
Time
50ms
-I(Vx)
Figure 5.4.2
EENG 2910, Class 5
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5b. Op Amps Circuit
Example 10.2
2.0V
0V
(Your final drawing of
circuit schematics)
Figure 10.2.1
-2.0V
V(V1:+)
4.0V
0V
SEL>>
-4.0V
0s
2.0ms
4.0ms
V(OUT)
Time
Figure 10.2.2
Question: Please explain the relation between V(V1:+) and V(out).
EENG 2910, Class 5
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Example 10.4
1.0V
0.5V
(Your final drawing of
circuit schematics)
Figure 10.4.1
0V
VM(7)
100d
50d
SEL>>
0d
10Hz
10KHz
100MHz
abs(VP(7))
Frequency
Figure 10.4.2
EENG 2910, Class 5
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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,
EENG 2910, Class 5
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