1 AC Analysis ω=const, E(ω)=E I(ω)=I, U(ω)=U, α(ω)=α, etc PSPICE Simulation updated 2014.05.27
1
AC Analysis
ω=const, E(ω)=E
I(ω)=I,
U(ω)=U,
α(ω)=α, etc
PSPICE Simulation
updated 2014.05.27
PSpice AC Simulation -Content
1. Single-frequency Analysis
2. Parameter Sweep (ω = ω*= const)
3. Frequency Sweep (ω = var)
4. Frequency Sweep with Variable Parameter
5. Impedance Frequency Characteristic
6. Resonant Frequency
7. Amplitude Response
8. Bode Plot, cut-off Frequency
9. Bandwidth, Half-power Frequencies
10. Parameter Sweep (ω = var) – Performance Analysis
11. Commonly used Goal Functions
12. Coupled Coils
13. Ideal Transformer
14. 3-phase Circuit
Single-frequency Analysis
p pico 10-12
n nano 10-9
u micro 10-6
m mili 10-3
k kilo 103
meg mega 106
e.g. 50u=50 10-6
Netlist
Statements
xx
Rxx X Y Value (in [Ω] )
Vxx X Y AC Value1 Value2
Ixx X Y AC Value1 Value2
Passive Element
Description
.PRINT AC Vy(X) Vy(X,Y) Iy(Vxx)
.END
Title Statement
Description of Outputs
resistor
Cxx X Y Value (in [F] )
Lxx X Y Value (in [H] )
capacitor
inductor
or
Value1 = rms, in [A] or [V]
Value2 = phase, in [deg]
Value3 = frequency, in [Hz] .AC LIN 1 Value3 Value3
Source Description
single frequency analysis, f =ω/(2π) = const
Control
Statements
y=M (magnitude)
y=P (phase)
y=R (Re part)
y=I (Im part)
Calculation of the DC bias point is always the first step of a PSpice run !!! – see Problem AC2.0
4
Examplary Circuit
V5)0exp(5)( o jjUR
10)90exp(10)( o jjjU L
5)90exp(5)( o jjjUC
I( jω)=0.5 A exp180°= –1
(passive s.c.)
0
2 1 3
)( jI
)( jE
LjX R
CjX
m6420 LLX L
u31810/1 CCXC 10R
09.72/10V)45314sin(10 o Etef = 50
AC2 RLC
Circuit.cir
Draw Wire Get New Part Create Circuit Diagram
using
Get New Part Browser and
Draw Wire
To enable PRINT add Printpoints:
VPRINT1: node voltage
VPRINT2: voltage between two nodes
IPRINT: current
Examplary Circuit – Schematics
https://www.youtube.com/watch?v=65j8sH8MwXc
Set values of Circuit Parts
To enable PRINT (magnitude and phase) set:
AC=1, MAG=1, PHASE=1
Examplary Circuit – Schematics
AC Sweep in Analysis Setup
set Start=End Frequency
Examplary Circuit – Schematics
N1 N4 N2
N3
Save File, then Simulate
8
0
N4 N2 N1
e
L R
C
N3
Examplary Circuit – Schematics
Schematics assigns nodes automatically
The assignment can be found in the Netlist
(Analysis: Examine Netlist, Output File)
It is possible to assign another label
by double clicking on the node
9
when Simulation is completed:
Netlist and Results may be viewed in
Output File
AC2 RLC Circuit
VIN 3 0 AC 7.09 45
R1 1 2 10
L1 2 3 64m
C1 0 1 318u
.AC LIN 1 50 50
.PRINT AC V(3,2) VP(3,2) V(2,1) V(1,0) V(2,0) I(VIN) IP(VIN)
.END
0
2 1 3
e
L R
C
9
0
N4 N2 N1
e
L R
C
N3
Examplary Circuit – Schematics
10)90exp(10)( o jjjU L
AC ANALYSIS TEMPERATURE = 27.000 DEG C
******************************************************************************
FREQ VM($N_0002) VP($N_0002)
5.000E+01 4.994E+00 -9.028E+01
******************************************************************************
FREQ VM($N_0004) VP($N_0004)
5.000E+01 7.059E+00 -4.530E+01
******************************************************************************
FREQ VM($N_0001,$N_0004) VP($N_0001,$N_0004)
5.000E+01 1.003E+01 8.973E+01
******************************************************************************
FREQ VM($N_0004,$N_0002) VP($N_0004,$N_0002)
5.000E+01 4.989E+00 -2.750E-01
******************************************************************************
FREQ IM(V_PRINT10) IP(V_PRINT10)
5.000E+01 4.989E-01 -2.750E-01
V5)0exp(5)( o jjUR
5)90exp(5)( o jjjUC
V)45exp(06.7)(N4 jjU
A5.0)( jI
10
0
N4 N2 N1
e
L R
C
N3
Examplary Circuit – Schematics
PSPICE Problems\TR\RLCall.sch
AC2.0 Find indication of the voltmeter, if C=1μF, R=2kΩ and . V1000sin10 tu
Ru Vu
C
C0
3
1
2
ERROR -- Node 2 is floating
ERROR -- Node 1 is floating
Ru Vu
C
C0
3
1
2
R
FREQ V(2,1) I(VIN)
1.592E+02 5.015E+00 2.507E-03
AC2.0 corrected
VIN 3 0 AC 7.09 0
R 2 1 2k
C2 2 3 1u
C1 0 1 1u
Rinf 0 1 10MEG
.AC LIN 1 159.24 159.24
.PRINT AC V(2,1) I(VIN)
.END
AC2.0
VIN 3 0 AC 7.09 0
R 2 1 2k
C2 2 3 1u
C1 0 1 1u
.AC LIN 1 159.24 159.24
.PRINT AC V(2,1) I(VIN)
.END
Calculation of the DC bias point is always the first step of a PSpice run !!!
Some large resistance,
eg. 10MEG, must be
connected in parallel
with one the capacitors
AC2.0.cir
AC2.0
corrected.cir
In DC analysis capacitor is an o.c.
Nodes with no DC path to ground aren't allowed in
PSpice !!!
AC2.14’’ Find the phase shift between the source voltage u and its current i, if R=2 kΩ, XL=2kΩ,
|XC |=1kΩ, u=10√2sin(10000t)V, time in seconds.
)()(111
)( j
LC
eYXX
jR
jY
12
45
11arctan
LC
iuXX
R
u
i
FREQ IP(Vin)
1.592E+03 -1.350E+02
AC2.14''
Vin 1a 0 AC 10 0
Rt 1a 1 0.001
R 1 0 2k
L 1 0 0.2
C 1 0 0.1u
.AC LIN 1 1592 1592
.PRINT AC IP(Vin)
.END
45180135i
450 iu
Source satisfies passive
sign convention, then:
PSPICE Problems\AC\AC2.14''.cir
Parameter Sweep (ω = ω*= const)
It is possible to visually verify, in PROBE, the Circuit response Y(ω*)=y(Z) :
Output Variable or function built of Variables, using Analog Operators and Functions,
by performing a sweep of a parameter Z (R, C or L)
Design parameter sweep in the Netlist
Z=var
Y(ω*)=y(Z) X(jω*)
E(jω*) or J(jω*)
.MODEL ZMOD ZZZ(Z=1)
.STEP DEC ZZZ ZMOD(Z) Start Stop Increments/Dec
Z n m ZMOD 1
-------------------------------------------------------------------
.AC LIN 1 F1 F2=F1
n m
Z= C or L or R
ZZZ=CAP or IND or RES
After pressing run Simulation, from Available Selections select All
Parameter Sweep (ω = ω*= const), cont.
Frequency sweep will appear, then select Performance Analysis.
Parameter Sweep (ω = ω*= const), cont.
Frequency sweep will appear, then select Performance Analysis. Frequency will be
replaced by ZMOD (Z); Z=C or L or R
Parameter Sweep (ω = ω*= const), cont.
Press Add Trace, then complete Trace Expression:
single Simulation Output Variable or function built of Output Variables,
using Analog Operators and Functions
Parameter Sweep (ω = ω*= const), cont.
or
AC2 PFC C Sweep
Vin 1 0 AC 200
Rl 1 2 0.1 ………………. line resistance
.MODEL CMOD CAP(C=1) …............................. CAP MODEL coding
C 2 0 CMOD 1 ……………………………............ CAP MODEL invoking
.STEP DEC CAP CMOD(C) 0.1u 500u 1000 ……. CAP SWEEP specification
* Start Stop Points/Decade
L 2 3 32m
R 3 0 10
.AC LIN 1 50 50
.PROBE
.END
18
Sizing a capacitor for the PFC
mH32;10 LR
7.0,kW2,Hz50,V200 pfPfU
For properly sized Capacitor:
φ = αu – αi = 0
I=Imin, Pl=Pl,min
Lj
R
Cj
1
)( jI
)( jU
)( jI L)( jIC
Rl
• )( jUin
0
1
2
• 3
see AC2.40
)( jI )( jU
)( jIC
)( jI L
Lj
R
Cj
1
)( jI
)( jU
)( jI L)( jIC
Rline
•
)( jU in
0
1a
1
2
• 3
AC2 PFC C Sweep
Vin 1 0 AC 200
Rt 1 2 0.1
.MODEL CMOD CAP(C=1)
C 2 0 CMOD 1
.STEP DEC CAP CMOD(C) 0.1u 500u 1000
L 2 3 32m
R 3 0 10
.AC LIN 1 50 50
.PROBE
.END
AC2 PFC C Sweep.cir
mH32;10 LR
1.0;Hz50,V200 linein RfU
P=Rline I(f)2=F(C)
C
Performance
Analysis
Add
Trace
Toggle
Cursor
Cursor
Min
Sizing a capacitor for the PFC
20
Sizing a capacitor for the PFC
Lj
R
Cj
1
)( jI
)( jU
)( jI L)( jIC
Rl
• )( jUin
0
1
2
• 3
AC2 PFC C Sweep fi=0
Vin 1 0 AC 200
Rt 1 2 0.1
.MODEL CMOD CAP(C=1)
C 2 0 CMOD 1
.STEP DEC CAP CMOD(C) 0.1u 500u 1000
L 2 3 32m
R 3 0 10
.AC LIN 1 50 50
.PROBE
.END
φ(C)=αu−αi(C)
AC2 PFC C Sweep
fi=0.cir
mH32;10 LR
7.0,kW2,Hz50,V200 pfPfU
see AC2.40
C
Sizing a capacitor for the PFC: Schematics
1. Create Circuit Diagram using Get New Part Browser and Draw Wire
2. Set values of Circuit Parts, for C1 set {Cvar}
3. Get PARAM Part and set NAME1=Cvar, VALUE1=1u (arbitrary)
22
4. Analysis Setup
• AC Sweep:Total Pts=1; Start=End Frequency=50
• Parametric: Global; Name=Cvar; Decade; Start=0.1u End=500u, Pts/Dec=1000
5. Save File 6. Simulate,
when completed select All
7. Analysis: Examine Netlist
Sizing a capacitor for the PFC: Schematics
23
8. Add Plot to Window, Add Traces: P(V1(C1)) – P(I(Rl)) and I(Rl), then, for
• φ=P(V1(C1)) – P(I(Rl)): Cursor Search and „search forward level (0)”;
• I=I(Rl): Cursor Min
φ(C)=αu−αi(C)
I(C)
C
PSPICE Problems\PSpice Schematics\PFC Capacitor.sch
Sizing a capacitor for the PFC: Schematics
24
AC4.6c’ For the RC circuit sketch the gain K(ω)=U2(ω)/U1(ω) in terms of C (frequency and
resistance are fixed). Size C, such that the cut-off frequency fc=1kHz for R=1592Ω.
AC5 Filter RC var C
VIN 1 0 AC 1 0
R 2 1 1592
.MODEL CMOD CAP(C=1)
.STEP DEC CAP CMOD(C) 0.01u 99u 100
C 2 0 CMOD 1
.AC LIN 1 1k 1k
.PROBE
.END
100nF2
1
2
1
RfC
RCf
c
c
PSPICE
Problems\AC\AC5
Filters\AC5 Filter
RC var C.cir
K(1kHz)
1)(1 jU Cj
1)(2 jU
R 2
0
1
25
1)(1 jU
Cj
1)(2 jU
R
2
0
1
AC4.6d’ For the CR circuit sketch the gain K(ω)=U2(ω)/U1(ω) in terms of C (frequency and
resistance are fixed). Size C, such that the cut-off frequency fc=1kHz for R=1592Ω.
100nF2
1
2
1
RfC
RCf
c
c
K(1kHz)
AC5 Filter CR var C
VIN 1 0 AC 1 0
R 2 0 1592
.MODEL CMOD CAP(C=1)
.STEP DEC CAP CMOD(C) 1n 9.9u 100
C 2 1 CMOD 1
.AC LIN 1 1k 1k
.PROBE
.END
PSPICE
Problems\AC\AC5
Filters\AC5 Filter
CR var C.cir
Frequency Sweep (ω = var)
p pico 10-12
n nano 10-9
u micro 10-6
m mili 10-3
k kilo 103
meg mega 106
e.g. 50u=50 10-6
Netlist
Statements
xx
Rxx X Y Value (in [Ω] )
Passive Element
Description
.PRINT AC Vy(X) Vy(X,Y) Iy(Vxx)
or
.PLOT AC Vy(X) Vy(X,Y) Iy(Vxx)
.END
Title Statement
Description of Outputs
resistor
Cxx X Y Value (in [F] )
Lxx X Y Value (in [H] )
capacitor
inductor
Vxx X Y AC Value1 Value2
Ixx X Y AC Value1 Value2
or Value1 = rms, in [A] or [V]
Value2 = phase, in [deg]
N=number of frequencies
ND=number of frequencies in decade
F1,F2 = frequencies, in [Hz]
.AC LIN N F1 F2
or
.AC DEC ND F1 F2
Source Description
Control
Statements
y=M (magnitude)
y=P (phase)
y=R (Re part)
y=I (Im part) .PROBE
Remember: Calculation of the dc bias point is always the first step of a Pspice run !!!
27
RLC Circuit
.AC LIN N F1 F2
fi= F1+(F2−F1)(i −1)/(N−1); i=1,...,N
.AC DEC ND F1 F2
fi= F1·10 (i−1)/ND; i=1,...,N=0.1·F2/F1+1
5 intervals/decade
101
C
L
RQ
10 frequencies
AC4 Filter
RLC.cir AC4 Filter
RLCdec.cir
0
1 3 2
1V 10μF 0.1H
•
10Ω Hz1602
1
LCf r
28
RLC Circuit
0
1 3 2
1V 10μF 0.1H
•
Filter RLC PROBE
VIN 1 0 AC 1 0
R 2 0 10
L 1 3 0.1
C 3 2 10u
.AC DEC 500 10 10k
.PROBE
.END
PSPICE
Problems\AC\AC4
Filter RLCdec
PROBE.cir 10Ω
I(f)
Draw Wire Get New Part Create Circuit Diagram
using
Get New Part Browser and
Draw Wire
RLC Circuit: Schematics
Set values of Circuit Parts
https://www.youtube.com/watch?v=daJ3CSidBKI
AC Sweep in Analysis Setup
set Pts/Decade, Start & End Frequency
RLC Circuit: Schematics
N3
N1 N2
Save File, then Simulate
Schematics assigns nodes automatically
Assignment can be found in the Netlist
(Analysis: Examine Netlist or Output File)
0
N3 N1 N2
1V 10μF 0.1H
•
10Ω
RLC Circuit: Schematics
when Simulation is completed:
Netlist may be viewed in Output File
Frequency sweep displayed in PROBE
after adding Trace
0
1 3 2
1V 10μF 0.1H
•
0
N3 N1 N2
1V 10μF 0.1H
•
Filter RLC PROBE
VIN 1 0 AC 1 0
R 2 0 10
L 1 3 0.1
C 3 2 10u
.AC DEC 500 10 10k
.PROBE
.END
PSPICE
Problems\AC\RLC
sweep.sch
10Ω
10Ω
RLC Circuit: Schematics
33
0
N3 N1 N2
1V 10μF 0.1H
10Ω
•
** Analysis setup **
.ac DEC 500 10 10k
* From [PSPICE NETLIST] section of pspiceev.ini:
.lib "nom.lib„
.INC "RLC sweep.net"
**** INCLUDING "RLC sweep.net" ****
* Schematics Netlist *
C_C1 $N_0001 $N_0002 10uF
L_L1 $N_0003 $N_0001 0.1H
V_V1 $N_0003 0 DC 0V AC 1V 0
R_R2 $N_0002 0 10
I(f)
RLC Circuit: Schematics
Frequency Sweep with Variable Parameter
It is possible to visually verify, in PROBE, the Circuit frequency response Y(ω):
Output Variable or function built of Variables, using Analog Operators and Functions,
for few values of the Circuit parameter Z (R, C or L)
Design parameter sweep and frequency sweep in the Netlist
Z=var
Y(ω) X(jω)
E(jω) or J(jω)
.MODEL ZMOD ZZZ(Z=1)
.STEP LIN ZZZ ZMOD(Z) Start Stop Increment; or
.STEP DEC ZZZ ZMOD(Z) Start Stop Incr/Dec
Z n m ZMOD 1
--------------------------------------------------------------
.AC LIN N F1 F2; or .AC DEC ND F1 F2
n m
Z= C or L or R
ZZZ=CAP or IND or RES
After pressing run „Simulation”, from „Available Selections” select few values: for
LIN: Start, Stop and selectected values designated by the Increment,
set Increment such that these values are available or
DEC: Start, Stop and selectected values for the set Incr/Dec=1
Frequency Sweep with Variable Parameter, cont.
Frequency Sweep with Variable Parameter, cont.
Press Add Trace, then complete Trace Expression:
single Simulation Output Variable or function built of Output Variables,
using Analog Operators and Functions or
37
UR for: R=10: Q=10; R=20: Q=5; R=50: Q=2; R=100: Q=1
R=10
R=20
R=50
R=100
AC4 Filter
RLC,
Rvar.cir
Filter RLC
VIN 1 0 AC 1 0
.MODEL RMOD RES(R=1)
.STEP RES RMOD(R) 10 100 10
R 2 0 RMOD 1
L 1 3 0.1
C 3 2 10u
.AC LIN 100 40 400
.PROBE
.END
0
1 3 2
1V 10μF 0.1H
R
•
RLC Circuit, UR : variable R
Filter R LCpar loaded, var R
VIN 1 0 AC 1 0
.MODEL RMOD RES(R=1)
.STEP DEC RES RMOD(R) 400 40k 1
R 1 2 1k
L 2 0 0.1
C 2 0 10u
Rl 2 0 RMOD 1
.AC DEC 1000 10 1000
.PROBE
.END
U1 (jω)
R
U2(jω) Rl
AC5.1c’ Find the gain K(ω)=U2(ω)/U1(ω) of the given band-pass filter: R=1kΩ, C=10µF,
L=100mH, loaded by the Rl=400, 4kΩ, 40kΩ resistance, at cut-off frequencies.
Filter R LCpar loaded,
var R.cir
Rl =40k Kmax=0.98
Rl =4k Kmax=0.8
Rl =0.4k
Kmax=0.29
l
l
RR
RK
max
39
Filter R LCpar C sweep
Vin 1 0 AC 1 0
R 1 2 100
L 2 0 0.1
.MODEL CMOD CAP(C=1)
.STEP DEC CAP CMOD(C) 0.1u 10u 1
C 2 0 CMOD 1
.AC DEC 100 1 100k
.PROBE
.END
AC5.1c For the sketched tank filter study the bandwidth as a function of the capacitance.
Parameter (capacitance) sweep: C=0.1, 1, 10 µF
Filter R LCpar
C sweep.cir
10µF 1µF 0.1µF
L
C
GQ
1
quality factor
Q Q
Q
parallel RLC circuit:
0
2 1
U1 (jω)=1 U2 (jω)
K(f)
f
AC5.1c For the sketched tank filter study the bandwidth as a function of the capacitance.
Schematics: Parametric Analysis Setup
AC5.1c For the sketched tank filter study the bandwidth as a function of the capacitance.
PSPICE Problems\PSpice Schematics\Tank Filter.sch
K(f)
f
10µF 1µF 0.1µF
Schematics: Parametric Performance Analysis Setup
AC5.1c For the sketched tank filter study the bandwidth as a function of the capacitance.
AC5.1c For the sketched tank filter study the bandwidth as a function of the capacitance.
Performance Analysis
Add Trace, Goal Function: Bandwidth(V$N_0001),3)
CRf
2
1
C μF 0.1 1 10
Δf kHz 15.9 1.59 0.159
AC5.1c For the sketched tank filter study the bandwidth as a function of the capacitance.
PSPICE Problems\PSpice Schematics\Tank Filter Bandwidth.sch
Δf(C)
C
45
Filter RLC bandwidth, L=var
VIN 1 0 AC 1 0
R 2 0 100
.MODEL CMOD CAP(C=1)
.STEP DEC CAP CMOD(C) 0.1u 10u 1
C 3 2 CMOD 1
L 1 3 0.1
.AC LIN 1000 40 400
.PROBE
.END
0
1 3 2
1V 10μF 0.1H
R=100Ω
• L
R
Q
rlu
Hz1592
L
Rf
AC5.1k Find bandwidth of the RLC band-pass filter, in Hz, if C=10µF; L=0.1H; R=100Ω.
C=0.1µF:
fr=1592Hz
C=10µF:
fr=159Hz
C=1µF:
fr=505Hz
PSPICE
Problems\AC\AC4
Filter RLC
bandwidth, L=var.cir
AC5.1k Find bandwidth of the RLC band-pass filter, in Hz, if C=10µF; L=0.1H; R=var.
Schematics: Parametric Analysis Setup
47
K(f)
Δf(R)
f
R
L
Rf
2
AC5.1k Find bandwidth of the RLC band-pass filter, in Hz, if C=10µF; L=0.1H; R=var.
1kΩ
100Ω
316Ω
PSPICE Problems\PSpice
Schematics\CLRvarfvar.sch
R Ω 100 316 1000
Δf Hz 159 502 1590
Practical Coil – Impedance
48
mAC4 Practical
Coil
CL=0.1 pF :
ωr=107 rad/sec, fr=1.6 MHz
L=0.1 H, RL=10 Ω
fr=1.6 MHz
LLR
LC
tR
1 0 2
3 •
•
AC4 Practical Coil
VIN 3 0 AC 1 0
Rt 3 1 10
L 2 0 100m
CL 1 0 0.1p
RL 1 2 10
.AC DEC 1000 0.01 10000MEG
.PROBE
.END
LC
C RC
49
AC4 Practical Capacitor
VIN 1 0 AC 1 0
L 1 2 0.01u
C 2 0 1u
Rl 2 0 5MEG
.AC DEC 1000 0.001 1000MEG
.PROBE
.END
LC=10 nH :
ωr=107 rad/sec, fr=1.6 MHz
C=1 μF, RC=5 MΩ
AC4 Practical
Capacitor
Practical Capacitor - Impedance
RC=5MΩ
fr=1.6 MHz
AC3.1b For the circuit shown: C=1nF; L=0.1mH; R=1kΩ, find 1) the resonant frequency ωr ;
2) impedance Z(ωr) ; 3) rms current I(ωr), for the rms supply voltage of 1V.
50
AC3.1b cont
VIN 1 0 AC 1 0
L 1 2 0.1m
C 2 0 1n
R 2 0 1k
.AC DEC 100 10k 1MEG
.PROBE
.END
αi(f)
477kHz
current leads voltage ≡ capacitive character
L
CR
0
i 2 1
51
AC3.1b cont
VIN 1 0 AC 1 0
L 1 2 0.1m
C 2 0 1n
R 2 0 1k
.AC DEC 100 10k 1MEG
.PROBE
.END
AC3.1b cont imp.cir
minimum not exactly at the resonant frequency
477kHz
αi(f)
Z(f)
L
CR
0
i 2 1
AC3.1b For the circuit shown: C=1nF; L=0.1mH; R=1kΩ, find 1) the resonant frequency ωr ;
2) impedance Z(ωr) ; 3) rms current I(ωr), for the rms supply voltage of 1V.
52
AC3.1b For the circuit shown: C=1nF; L=0.1mH; R=1kΩ, find 1) the resonant frequency ωr ;
2) impedance Z(ωr) ; 3) rms current I(ωr), for the rms supply voltage of 1V.
L
CR
0
i 2 1 AC3.1b cont rms current
VIN 1 0 AC 1 0
L 1 2 0.1m
C 2 0 1n
R 2 0 1k
.AC DEC 1000 10k 1MEG
.PROBE
.END
αi(f)
I(f)
PSPICE
Problems\AC\AC3.1
b cont rms
current.cir
53
180/radiansin ii
degreesin)( fileading power factor
capacitive character
lagging power factor
inductive character
PSPICE
Problems\AC\AC4.7
Filter RLC pf'.cir
]180/)(cos[)]/)((cos[arctan)( fRfXfpf i
)( jI
)( jU)/(1 Cj
R
Lj
AC4.7 Filter RLC pf'
VIN 1 0 AC 10 0
R 2 0 10
L 1 3 0.1
C 3 2 10u
.AC LIN 400 1 400
.PROBE
.END 0
1 3
2
AC3.7a The series RLC circuit, R=10Ω, L=100mH, C=10μF is connected to 10V rms source of
adjustable frequency. Determine the source frequency for unity power factor.
1)( rfpf
54
)( jI
)( jU)/(1 Cj
R
Lj
AC3.7b The series RLC circuit, R=10Ω, L=100mH, C=10μF is connected to 10V rms source of
adjustable frequency. Determine the maximum real power supplied by the source, in W.
PSPICE
Problems\AC\AC4.7
Filter RLC P,Q.cir
AC4.7 Filter RLC pf'
VIN 1 0 AC 10 0
R 2 0 10
L 1 3 0.1
C 3 2 10u
.AC LIN 400 1 400
.PROBE
.END
W10)()(22
2
maxmax
R
UR
R
URIfPfP r
UfIRfICLR
UfP )()(
)/1()( 2
22
2
0
1 3
2
55
Multiple-resonance - Examplary Circuit
)( jI
)( jU
1LjX
CjXR
0
1 2 1LjX
Hz2250rad/s210 2
4
21
212
rr f
CLL
LLHz1590rad/s10
11
4
2
1 rr fCL
U(ω)/I(ω)=Z(ω))
AC4 Multiple-resonance
I 0 1 AC 1 0
R 1 0 1k
L1 1 2 10m
C 2 0 1u
L2 2 0 10m
.AC LIN 1000 10 10k
.PROBE
.END
AC4 Multiple-
resonance.cir
7 1
2
3 4
5
6
Frequency Response of RC Circuit, LIN
56
)(1 jUCj
1 )(2 jU
R
2
0
1 f 0 fc=159.2 ∞
U2(f) 1 0 2/1
AC5 Filter RC
VIN 1 0 AC 1 0
R 2 1 1k
C 2 0 1u
.AC LIN 100 10 2k
.PLOT AC VM(2)
.PROBE
.END
AC5 Filter RC.cir
Hz2.1592
1
2
1μF1,1
RCTfCkR c
fTU
UK
2;
)(1
1
)(
)()(
21
2
K(f)=U2(f)
U1(f)=1
57
Frequency Response of RC Circuit, DEC
)(1 jUCj
1 )(1 jU
R
2
0
1
AC5 Filter RC.dec
VIN 1 0 AC 1 0
R 2 1 1k
C 2 0 1u
.AC DEC 10 10 10k
.PLOT AC VM(2)
.PROBE
.END
f 0 fc=159.2 ∞
U2(f) 1 0 2/1
AC5 Filter RCdec.cir fT
K
2;)(1
1)(
2
Hz2.159μF1,1 cfCkR
K(f)=U2(f)
U1(f)=1
Bode Plot of RC Circuit, cut-off frequency
58
)(1 jUCj
1
)(2 jUR 2
0
1
AC5 Filter RC Bode
VIN 1 0 AC 1 0
R 2 1 1592
C 2 0 1u
.AC DEC 10 10 10k
.PLOT AC VM(2)
.PROBE
.END Hz100
2
1
2
1μF1,1592
RCTfCR c
20dB/Dec
−3dB
AC5 Filter RC
Bode.cir
1)(2
2
1
2dB
1)(log20)2(1log20
)(
)(log20)(
fUfUfT
fU
fUfK
AC4.10d Filter RC.dec
VIN 1 0 AC 1 0
R 2 1 1k
C 2 0 1u
Rout 2 0 1k
.AC DEC 10 10 10k
.PROBE
.END
Hz3181
2
1
2
RCTf cc
2
dB )(1log202/1log20)( TK
–6dB
μF1,k1 CR
2/,1
2/1)( RCT
TjjK
2)(1
2/1)(
TK
1 2
)(1 jU Cj
1)(2 jUR
R
0
1 2
or
Logaritmic (dB) scale
AC4.10d’ For the given RC circuit, find a) the cut-off frequency and b) the dB gain at 1KHz
frequency: KdB(1kHz).
AC4.10d Filter RC.dec
VIN 1 0 AC 1 0
R 2 1 1k
C 2 0 1u
Rout 2 0 1k
.AC DEC 10 10 10k
.PROBE
.END
2)(1
2/1)(
TK
Hz3181
2
1
2
RCTf cc
353.02
2/1)( cfK
AC4.10d’ For the given RC circuit, find a) the cut-off frequency and b) the dB gain at 1KHz
frequency: KdB(1kHz).
)(1 jU Cj
1)(2 jUR
R
0
1 2
Linear scale
PSPICE
Problems\AC\AC4.10d
Filter RC dec.cir
5.0)0( max KK
0inI
)(1 U
+
lR0inI
)()( 2 UUout
)(inU
R
C
)(1 U
lR
)()( 2 UUout )(inUR
C
k=1
AC4.10e’ For the given RC circuit, find the cut-off frequency. C=1µF, R=Rl=1kΩ.
The op-amps low output impedance prevents the filter’s cut-off frequency point
from being affected by changes in the impedance of the load
62
)(1 U
lR
)(2 U)(inUR
C
)(inU
0
2 1a 1
AC4.10e’ For the given RC circuit, find the cut-off frequency. C=1µF, R=Rl=1kΩ.
AC4.10e Filter CR oa
VIN 1 0 AC 1 0
R 1a 0 1k
C 1 1a 1u
E 2 0 1a 0 1
Rout 2 0 1k
.AC DEC 10 10 10k
.PROBE
.END
Hz1592
1
2
1
RCTfc
1)( max KK
The load resistance does not affect neither, the maximum gain and the cut-off
frequency
AC5.1a Find bandwidth of the RLC band-stop filter, in Hz, if C=10µF; L=0.1H; R=100Ω.
63
Filter RLC B-s bandwidth
VIN 1 0 AC 1 0
R 1 2 100
L 0 3 0.1
C 3 2 10u
.AC LIN 1000 40 400
.PROBE
.END
L
R
Q
rlu
AC4 Filter RLC B-s
bandwidth.cir
1 2
0
• 3 U1(jω)=1 U2(jω)
Hz1592
L
Rf
K(f)=U2(f)
Linear scale
64
Filter RLC B-s bandwidth
VIN 1 0 AC 1 0
R 1 2 100
L 0 3 0.1
C 3 2 10u
.AC LIN 1000 40 400
.PROBE
.END
L
R
Q
rlu
AC4 Filter RLC B-s
bandwidth db.cir
Hz1592
L
Rf
AC5.1a Find bandwidth of the RLC band-stop filter, in Hz, if C=10µF; L=0.1H; R=100Ω.
KdB(f)=20logU2(f)
dB scale
1 2
0
• 3 U1(jω)=1 U2(jω)
65
Filter RLC bandwidth
VIN 1 0 AC 1 0
R 2 0 100
L 1 3 0.1
C 3 2 10u
.AC LIN 1000 40 400
.PROBE
.END
L
R
Q
rlu
AC4 Filter RLC
bandwidth GF.cir
AC5.1k Find bandwidth of the RLC band-pass filter, in Hz, if C=10µF; L=0.1H; R=100Ω.
Hz1592
L
Rf
0
1 3 2
1V 10μF 0.1H
R=100Ω
•
dB scale
Parameter Sweep (ω = var) – Performance Analysis It is possible to visually verify, in PROBE, the Goal Function as function of the Circuit
parameter: g=h(Z); Z=R, C or L, by performing a parameter sweep for a frequency sweep
Design parameter sweep and frequency sweep in the Netlist
.MODEL ZMOD ZZZ(Z=1)
.STEP DEC ZZZ ZMOD(Z) Start Stop Increments/Dec
Z n m ZMOD 1
------------------------------------------------------------------
.AC LIN N F1 F2; or .AC DEC ND F1 F2
Z= C or L or R
ZZZ=CAP or IND or RES
Z=var
design specification (optional)
X(jω)
E(jω) or J(jω)
n m
g=h(Z)
After pressing run Simulation, from Available Selections select All
Parameter Sweep – Performance Analysis, cont.
Frequency sweep will appear, then select Performance Analysis.
Parameter Sweep – Performance Analysis, cont.
Frequency sweep will appear, then select Performance Analysis. Frequency will be
replaced by ZMOD (Z); Z=C or L or R
Parameter Sweep – Performance Analysis, cont.
Press Add Trace, then select Goal Function and complete Trace Expression, using
the selected Simulation Output Variable and the design specification, optionally
Parameter Sweep (ω = ω*= const), cont.
or
Commonly used Goal Functions
XatNthY(Y,value,n)
Determines the X position(s), usualy time or frequency, at which the
trace crosses a given threshold the n-th time – if n=1, then the 1st time
YatX(X,value)
Determines the Y position, at which the trace argument, usualy time or
frequency, is equal the set value
ZPBW(Y,3); Z=L(ow), H(igh), B(and)
Determines the 3dB bandwidth of the ZPass Filter, where Y=K=Vout
is the voltage gain (Vin=1)
Bandwidth(Y,3), same as BPBW
CenterFreq(Y,3)
Determines the Center frequency of the 3dB Pass Band
72
1)(1 jU
Cj
1 )(2 jU
R
2
0
1
AC4.6d’’ For the CR circuit sketch the cut-off frequency in terms of R (capacitance is fixed:
C=100nF). Size R, such that the cut-off frequency fc=1kHz .
fc(R)=XatNthY(V(2),0.7071,1)
R
7071.02
1
Kc f
RCf
fc=1kHz→R=1.592kΩ
AC5 Filter CR var R’
VIN 1 0 AC 1 0
R 2 0 RMOD 1
.MODEL RMOD RES(R=1)
.STEP DEC RES RMOD(R) 0.1k 99k 100
C 2 1 100n
.AC DEC 100 10 20k
.PROBE
.END
PSPICE
Problems\AC\AC5
Filters\AC5 Filter CR
var R'.cir
AC4.6d’’ For the CR circuit sketch the cut-off frequency in terms of R (capacitance is fixed:
C=100nF). Size R, such that the cut-off frequency fc=1kHz .
Schematics:
Parametric Analysis Setup
AC4.6d’’ For the CR circuit sketch the cut-off frequency in terms of R (capacitance is fixed:
C=100nF). Size R, such that the cut-off frequency fc=1kHz .
Performance Analysis
Add Trace, Select Goal Function: fc(R)=XatNthY(V($N_0001),0.7071,1)
AC4.6d’’ For the CR circuit sketch the cut-off frequency in terms of R (capacitance is fixed:
C=100nF). Size R, such that the cut-off frequency fc=1kHz .
Search Cursor
search forward level (1k)
PSPICE Problems\PSpice Schematics\CRvarfvar.sch
fc(R)=XatNthY(V($N_0001),0.7071,1)
R
76
0
1 3 2
1V 10μF 0.1H
R=100Ω
• L
R
Q
rlu
Hz1592
L
Rf
AC5.1k Find bandwidth of the RLC band-pass filter, in Hz, if C=10µF; L=0.1H; R=100Ω.
PSPICE
Problems\AC\AC4
Filter RLC
bandwidth, C=var.cir
Filter RLC bandwidth, C=var
VIN 1 0 AC 1 0
R 2 0 100
.MODEL CMOD CAP(C=1)
.STEP DEC CAP CMOD(C) 0.1u 10u 100
C 3 2 CMOD 1
L 1 3 0.1
.AC DEC 1000 10 10k
.PROBE
.END
Δf=BPBW(V(2),3)
C
77
0
1 3 2
1V 10μF 0.1H
R=100Ω
• L
R
Q
rlu
Hz1592
L
Rf
AC5.1k Find bandwidth of the RLC band-pass filter, in Hz, if C=10µF; L=0.1H; R=100Ω.
PSPICE
Problems\AC\AC4
Filter RLC
bandwidth, var L.cir
Filter RLC bandwidth,var L
VIN 1 0 AC 1 0
R 2 0 100
.MODEL LMOD IND(L=1)
.STEP DEC IND LMOD(L) 0.001 0.99 100
C 3 2 10u
L 1 3 LMOD 1
.AC DEC 1000 10 100k
.PROBE
.END
L
Δf=BPBW(V(2),3)
AC5.1c’ Find the gain K(ω)=U2(ω)/U1(ω) of the given band-pass filter: R=1kΩ, C=10µF,
L=100mH, loaded by the Rl=4kΩ resistance, at cut-off frequencies.
Filter R LCpar
loaded.cir
V/V2/8.0)(2
2/max
l
l
RR
RK
U1 (jω)
R
U2(jω) Rl
L1=10 mH, L2=2.5 mH, M=3 mH
The coupling coefficient k (k must be larger than 0 but smaller than 1):
The DOTS are always associated with the first node of the inductances
(i.e. node no. 1 of L1 and no. 2 of L2).
PSpice will assign the positive node voltage to the first node of the inductance (i.e. the
DOT) and use the passive sign convention for the current direction.
6.0/ 21 LLMk
0 0
)( jZ t
)(1 jI Mj )(2 jI
)( jEo 2Lj
)(2 jU
)( jZ l
1Lj
)(1 jU
1 3 2
L1 1 0 10m
L2 2 0 2.5m
k L1 L2 0.6
Coupled Coils
Coupled Coils
Vo 3 0 AC 10 0
Rt 3 1 0.01
Rl 2 0 500
L1 1 0 10m
L2 2 0 2.5m
k L1 L2 0.6
.AC LIN 1 100 100
.PRINT AC V(1) V(2), I(Vo)
.END
Remarks:
SPICE always needs a reference node numbered 0. One can never leave any node floating. For
that reason we have connected the two coils together to node 0 (or put a large resistor between
it). As there is no current flowing through this section, the circuit will not change by connecting
these nodes.
SPICE does not allow to connect a voltage source in parallel with an inductor, as it thinks that
the voltage source is short circuited. You can place a small resistor (ex. 0.001 Ohm) in series
with the voltage source to overcome this problem.
Coupled Coils Examplary Circuit
L1=10mH, L2=2.5 mH, M=3 mH, Rt =0.01 Ω, Rl =500 Ω,
eo =14.1sin(2π100t) V
Coupled
Coils.cir
tR
)(1 jI Mj )(2 jI
)( jEo 2Lj
)(2 jU
lR
1Lj
)(1 jU
1 3 2
0 0 •
Coupled Coils
Vo 3 0 AC 10 0
Rt 3 1 0.01
Rl 2 0 500
L1 1 0 10m
L2 2 0 2.5m
k L1 L2 0.6
.AC LIN 1 100 100
.PRINT AC V(1) VP(1), I(Vo) IP(Vo)
.END
Coupled Coils – Basic Transformer Examplary Circuit
L1=10mH, L2=2.5 mH, M=3 mH,
Rt =0.01 Ω, Rl =500 Ω, eo =14.1sin(2π100t) V
Coupled Coils
f=100Hz.cir
2
22
1
1
11
)()(
)()(
LjjZ
MLj
jI
jUjZ
l
28.657.1500
55.328.6
4/28.6500
10362828.6)(
622
1 jj
jj
jjZ
28.6592.1
10
)(
)()(
1
11
I
UZ
90)90(0)()()( 111 IU
FREQ V(1) VP(1) I(Vo) IP(Vo)
1.000E+02 1.000E+01 9.119E-02 1.592E+00 9.016E+01
tR
)(1 jI Mj )(2 jI
)( jEo 2Lj
)(2 jU
lR
1Lj
)(1 jU
1 3 2
0 0 •
82
Coupled Coils
Vo 3 0 AC 10 0
Rt 3 1 0.01
Rl 2 0 500
L1 1 0 10m
L2 2 0 2.5m
k L1 L2 0.6
.AC DEC 1 1 10MEG
.PROBE
.END
Coupled Coils
Gain.cir
213m
fc
3.0)()(
)()(500
1
2
11
2
nL
Lk
MML
M
U
UKMRl
)( 1 MLj )(1 jU
)(2 jU
)(1 jI 0)(2 jI
)( 2 MLj
Mj
V/V213.02
3.0)( cK
Coupled Coils – Basic Transformer Examplary Circuit, f=var
)( jEo
)(1 jItR )(2 jI
)(1 jU lR)(2 jU
● ●
)(2 jnI )(1 jnU
0cV
0
1 2 3 4
Ideal Transformer Model using dependent sources
Ideal Transformer
Vo 3 0 AC 100 0
Rt 3 1 0.001
F 0 1 Vc 0.1
E 4 0 1 0 0.1
Vc 2 4 0
Rl 2 0 500
.AC LIN 1 50 50
.PRINT AC V(1) VP(1)
V(2) I(Vo) IP(Vo)
.END
Assume: Rt=0.001, Rl=500, n=0.1
FREQ V(1) VP(1) V(2) I(Vo) IP(Vo)
5.000E+01 1.000E+02 0.000E+00 1.000E+01 2.000E-03 1.800E+02
k50500001.0
500)()(
221n
jZjZ l
k50002.0
100)(1 Z 0)180180(0)(1
passive sign convention
Ideal Transformer
Contr S'ces
Ideal Transformer
Practial Transformer
Vo 3 0 AC 100 0
.........................
C1 1 0 100p
C2 2 0 100p
C12 1 2 100p
Rl 2 0 500
.AC DEC 10 1 100k
.PROBE
.END
Practical (Iron-Core) Transformer Examplary Circuit with Parasitic Capacitances
Rt =10 Ω, Rl =500 Ω, n=0.1, eo =141sin(2π50t) V
L1=L2=0.1m, Lm=100m
R1=R2=10 Ω
C1=C2=C12=100pF
)(1 jU lR)(2 jU
● ●
)(2 jnI )(1 jnU
0cV
0
1b 2
)( jEo
tR3 2c 1R
2R
• • • • 1
1a
2b 2a
• • 0
1Lj 2Lj
mLj
● ●
C12
85
Practical (Iron-Core) Transformer Examplary Circuit, Frequency Response
Caution: A PSpice model may have very limited range of usefulness – be sure that the parameters
of the model closely match the simulated conditions !!!
Effect of Lm Effect of L1, L2
and C1, C2, C12
fenomenon of
resonance appears
Practical Trafo
HFGain.cir
)(3 jJ
)(1 jJ
)(2 jJ
)(3 jI
)(2 jI
)(3 jZ
)(1 jZ
)(1 jI
)(2 jZ
A
B
C
)(1 jE
)(2 jE
)(3 jE
1(A)
2(B)
3(C)
0(N)
AC7.4
V1 1 0 AC 100 0
V2 2 0 AC 100 -120
V3 3 0 AC 100 120
R1 1 2 100
R2 2 3 100
R3 1 3 100
.AC LIN 1 50 50
.PRINT AC I(V1) IP(V1)
+ I(V2) IP(V2) I(V3) IP(V3)
.END
FREQ I(V1) IP(V1) I(V2) IP(V2) I(V3) IP(V3)
5.000E+01 3.000E+00 1.800E+02 3.000E+00 6.000E+01 3.000E+00 -6.000E+01
AC7.4.cir
AC7.4 For the balanced wye-delta system: E=100V, Z(jω)=100, find rms value of a line
current.
)(3 jI
)(2 jI
)(3 jZ
)(1 jZ
)(1 jI
)(2 jZ
A
B
C
)(1 jE
)(2 jE
)(3 jE
1(A)
2(B)
3(C)
0
N
AC7.11a
V1 1 0 AC 100 0
V2 2 0 AC 100 -120
V3 3 0 AC 100 120
R1 1 N 0.00001
R2 2 N 10
R3 1 N 10
.AC LIN 1 50 50
.PRINT AC I(V2) IP(V2)
.END
AC7.11a.cir
FREQ I(V2) IP(V2)
5.000E+01 1.732E+01 3.000E+01
)210exp(310A. j
30+180=210
1,N
3
2
0
)(3 jE
)(1 jE
)(2 jV)(2 jE
AC7.11a Given the three-phase 3-wire balanced system in Y-Y configuration (α1=0°, α2=−120°,
α3=120°): E=100V, Z(jω)=10Ω. The 1st phase is shorted. What is the 2nd phase current?
)(3 jI
)(2 jI
)(3 jZ
)(1 jZ
)(1 jI
)(2 jZ
A
B
C
)(1 jE
)(2 jE
)(3 jE
1(A)
2(B)
3(C)
0
N
)(23 jU
)(2 jV
)(3 jV
AC7.11b
V1 1 0 AC 100 0
V2 2 0 AC 100 -120
V3 3 0 AC 100 120
R1 1 N 1MEG
R2 2 N 10
R3 1 N 10
.AC LIN 1 50 50
.PRINT AC I(V2) IP(V2)
.END
AC7.11b.cir
)270exp(35A. j
90+180=270
1
3
2
0
)(3 jE
)(1 jE)(23 jU
)(2 jE
N
)(2 jV
FREQ I(V2) IP(V2)
5.000E+01 8.660E+00 9.000E+01
AC7.11b Given the three-phase 3-wire balanced system in Y-Y configuration (α1=0°, α2=−120°,
α3=120°): E=100V, Z(jω)=10Ω. The 1st phase is opened. What is the 2nd phase current?
89
© copyright J.Rutkowski
any suggestions and remarks to
Silesian University of Technology – Gliwice, 2014