PSPICE Lecture - AC Circuits
PSPICE – AC Circuits
Reference: (see course web site)
• PSPICE Assignment #2
Topics to be presented:
• AC circuits – graphing sinusoidal waveforms using
Transient Analysis
• AC circuits – finding phasor values using AC Sweep
• Varying frequency with AC Sweep
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PSPICE Lecture - AC Circuits
2
AC Analysis in PSPICE
There are two common ways to analyze AC circuits in PSPICE:
1) Transient Analysis
• Used for graphing sinusoidal waveforms
• Time is varied, to time is on the x-axis of all graphs
• Use the sources VSIN and ISIN
• Angles referenced to sin( ), not cos( )
2) AC Sweep
• Used for calculating phasor voltages, phasor currents, etc (use the same
starting and ending frequency for a single analysis)
• Can also be used to vary frequency, so frequency (in Hz) is on the x-axis
of all graphs
• Use the sources VAC and IAC
• Angles are relative. If you use angles referenced to sin( ), then angles are
referenced to sin( ). If you use angles referenced to cos( ), then angles
are referenced to cos( ).
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PSPICE Lecture - AC Circuits
Example - Analyze the circuit below by hand.
Find the phasors V R , V C , and I. Also find v R (t), v C (t), and i(t).
+
VR
-
250
+
+
50sin(2000t + 30o) V
VC
1uF
-
-
Solution:
w  2000  rad/s, so f 
Phasor circuit:
I
ZC 
+
VR
I
+
+
-

wC
-
250
5030o V
-j
1uF
VC
V
Z Total

-j
2  fC

w
2
 1000 Hz
-j
2  (1000 )( 1 E  6 )
50  30 
 - j159
 0.169  62.5  A
250 - j159
V R  I  R  0.169  62.5   250   42.2  62.5  V
V C  I  Z C  0.169  62.5  - j159   26.8   27 . 5  V
i(t)  0.169sin(1
000t  62.5  ) A
v R (t)  42.2sin(10
00t  62.5  ) V
v C (t)  26.8sin(10
00t  27.5  ) V
PSPICE Lecture - AC Circuits
4
Example (continued) Use PSPICE to graph v(t) and vC(t) for 3 periods.
Solution: Since we want to graph the waveforms, used a transient analysis and
use part VSIN for the source.
1) Create a New Project and draw the circuit in PSPICE. Since a transient
analysis will be performed, use part VSIN for the voltage source. Add node
labels as well.
Notes: Since v(t) = 50sin(2000t + 30), use the following values:
• VOFF – This is the DC value. Use 0V.
• VAMPL – This is the peak value of the sinusoid. Use 50V.
• FREQ – This is the frequency in Hz. Use 1000Hz.
(continued on next slide)
PSPICE Lecture - AC Circuits
Notes: (continued)
• AC – This is the phasor voltage. We will use
VAC instead for phasors. Delete this
property.
• PHASE – This is the phase angle in degrees.
It does not show by default. Double-click on
the part and select Edit Part. The Parts
Editor window to the right should appear.
Select the property named Phase. Select
Display and change it to Name and Value.
Close the Parts Editor. Enter 30 degrees for
the phase.
• The source below now has the correct values.
• The units are not required, but are
recommended.
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PSPICE Lecture - AC Circuits
2) Select PSPICE – New Simulation Profile
• Analysis Type: Select Time Domain (Transient)
• Run to time: Since T = 1/f = 1/1000 = 1ms, enter 3ms for 3 periods
• Select OK.
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PSPICE Lecture - AC Circuits
7
3) Select PSPICE – Run
4) The graphing window should appear.
• Select Trace – Add Trace
• Select V(B) for the source voltage and V(C) for the capacitor voltage
• Note that the voltages are a little choppy. This can be fixed by using
more points for the analysis. By default PSPICE uses about 100 points so
the time increment is about 3ms/100 = 30us. If we change this to 3us,
PSPICE will use 1000 points instead of 100 points. See next slide.
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PSPICE Lecture - AC Circuits
Blank value changed to
3us to use 3ms/3us = 1000
points for smoother graphs
PSPICE Lecture - AC Circuits
• Select Plot – Label – Text to add labels to each waveform
• Select Window – Copy to Clipboard so that you can easily paste the
graph into a Word document without the black background.
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PSPICE Lecture - AC Circuits
10
Example (continued) Use PSPICE to find the phasors I , V R , and V C
Solution: Since we want to find phasors, use an AC Sweep analysis and use
part VAC for the source.
1) Create a New Project and draw the circuit in PSPICE. Since an AC Sweep
will be performed, use part VAC for the voltage source. Add node labels as
well.
• Note that PSPICE only shows the values of two properties. Doubleclick on the part to open the parts editor. Select each property and
change Display from Value Only to Name and Value. Also display
Name and Value for the property named ACPHASE.
PSPICE Lecture - AC Circuits
Notes: Since v(t) = 50sin(2000t + 30), use the following values:
• ACMAG – This is the phasor magnitude. Use 50V.
• ACPHASE – This is the phasor phase in degrees. Use 30degrees.
• DC – This is the DC value. Use 0V or delete this property.
• Units are not required, but are recommended.
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PSPICE Lecture - AC Circuits
2) Add voltage and current printers
• An AC sweep analysis only calculates DC node voltages (typically all 0) by default.
In order to find phasor voltages and currents, we must add printers.
• Voltage Printers - must be placed in parallel with the desired component(s)
• Current Printers – must be placed in series with the desired component(s)
• Polarity – Note the (-) sign on each printer and rotate/flip if necessary for the correct
voltage polarity or current direction.
• For both printers set the following properties:
Use part VPRINT2 for
• AC: Yes
the voltage printer and
part IPRINT for the
• MAG: Yes
current printer. Both are
• PHASE: Yes
in the Special library.
PSPICE Lecture - AC Circuits
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3) Select PSPICE – New Simulation Profile
• Analysis Type: Select AC Sweep/Noise
• AC Sweep Type: Use the same Start Frequency and End Frequency
(1000Hz in this case) for a single analysis. Even though the frequency
isn’t actually being varied, a value must still be entered for Points/Decade
• Select OK.
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PSPICE Lecture - AC Circuits
4) Select PSPICE – Run
5) View Results
• The values from the voltage printers should now appear in the .OUT file.
• Select PSPICE – View Output File
• Note that the voltage printers express the voltages using node voltages
• Add text to the Output file to clearly identify all results
• A portion of the Output file is shown below:
so V C  26.85   27 . 52  V
so V R  42.18  62.48  V
All results agree
with the hand
calculations!
so I  0.1687  62.48  A
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PSPICE Lecture - AC Circuits
Example: Use PSPICE to graph total circuit impedance as frequency varies to
show that |Z| is minimum at resonance. The circuit below is a series resonant
circuit.
As frequency increases XL becomes
more positive and XC becomes more
negative. At the resonant frequency
XL and XC cancel leaving a resistive
circuit and minimum total impedance
(equal to R).
Z Total  R  j X L  j X C
Z Total 
R  X L  X C 
2
Z Total  R  j2  fL 
2
-j
2  fC
Z Total  100  j2  f(0.1) 
-j
2  f(1E - 6)
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PSPICE Lecture - AC Circuits
The resonant frequency is:
wo 
fo 
wo
2
1
LC

1

 3162.3 rad/s
(0.1)(1E - 6)
3162.3
2
 503.3 Hz
Total impedance was calculated below as
frequency varies from 100 to 10,000 Hz.
Note that |Z| reaches a minimum at f = 500
Hz (close to the resonant frequency)
We will now graph |Z| vs f in PSPICE to
verify this result.
PSPICE Lecture - AC Circuits
1) Create a new project and draw the circuit in PSPICE
• Use part VAC for the source since an AC Sweep analysis will be
performed.
• Use any non-zero value for ACMAG and ACPHASE
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PSPICE Lecture - AC Circuits
2) Create a new Simulation Profile
• Let frequency vary from 100Hz to 10000Hz
• Frequency is typically shown on a log scale.
• Select 50 points/decade (for a total of 100 points in this case)
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PSPICE Lecture - AC Circuits
19
3) Analyze the circuit - Select PSPICE – Run
4) Graph the results (PSPICE’s graphing window should open automatically)
• Select Trace – Add Trace and enter V(A)/I(R1) for the total impedance
• Use a cursor to mark the minimum point on the curve
• Add text
• The results are correct!