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ANALIZA CIRCUITELOR ELECTRICE CU
PSPICE. LUCRARI DE LABORATOR/
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IONESCU ANISOARA,
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u1 = f ( u2 ) → u~1 = α dq u~2
u1 = f (i2 ) → u~1 = Rdq ~
i1
29
RlH.!H(#%$(M>Z!$6/'6BNJ<$ >5&/5$
df
~
φ = f (i ) → Φ = Ldq i~ cu Ldq =
q
di
RlH.!H(#%$(M>Z!$6/'6BNJ<$L*'Œ&"
~
~
i = f (φ ) → i = Γdq φ
si
~
u~ = φ
df
cu Γdq =
q
dφ
R+>!$(5$F-V2!1&('(>?!$<A'6†J,$‰2$F-V#Œ&$n
df
q = f ( u) → q~ = Cdq u~ cu Cdq =
q
du
R+>!$(5$F-V2!1&('(>?!$<A'6†J,$P-o">#:$(B
u = f (q ) → u~ = Sdq q~ cu
S dq =
si
q~ = Cdq u~
df
q
dq
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E.5$,&L*v#6>@/!':5KO$($(Z':#:$.#6 8
( Rdq , Gdq , Ldq , Γdq , Cdq , S dq , α dq , β dq )
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30
U 1 = 2 jI 1 + jI 2
#,
H(!H.#:$(',
si
>5&E('6D
U 2 = jI 1 + 2 jI 2
M5;!'%’I=$(j>Z?-1‘-o#,-ADK„@>&F4D#,#J<$j"E(!1l>& Ž 7
I1 =
U1 j
+ U2
j3 3
2
>5&
*6&$(>@46#6!$."
0o#(Ž ; /;@&('%DB[
si
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>5&.546#<#,'6
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U2 j
+ U
3 3 1
j
2
B-/Š-V>"#,C>5&.546#<#,'6 E(!25$4D#<',Z':!t$.!&/#,',!ˆE($,1&L>#:A>5&(#6&.'
e ( t ) = 2 cos 2t
π
i ( t ) = 2 sin( 2t + )
s
4
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J<$ $F-/#:&$(‰@'…>5Z',! !&(BCH.!H(#%$(>5&E('62 8 mY& $(!2546#6#,'<#%$P*1#<!&/BZ>@&(4#:#6':M*2&$(>42#6!$.5"‰'<>@?-A2!
!&(B3H.!H(#%$(-A&$/[ U = 2 jI + jI , U = 2 jI + jI ?>#…>!#%v>5&.#:,&('ˆ5>5x(#’(?'<@$‘s-1DQ>@'†E/5;!$6
1
1
2
2
2
1
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M !Hn-V51’.BŠ>BM5’.5K &$>Z#/>&(#:F"5;!?$(5$ km -/5A#<M>5& 8 R VW 8 0V#F?&(BŠ-1&o-o#,Z':MXD$F-V#:&$(
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31
V5 = 0
V1 = j
j
1
2
2 1
V 2 +
− V1 − V 3 = I 4 + U 2
3
3
3j
3j 3
U2 = −V3
V 2 − V 4 = 3I 3
I3 =
V1 − V 3
1− j
V
V
1
1 2
j
V3
+ + − 1 − 2 = 1 + j + U1
1 − j 3 3j 1 − j
3
3
U1 = V1 − V 2
1
V 4 = −1 − j − I 4
2
#,>ZBX&$P-/#2-1D@K j5>5&.546#6#(>5&Q$(>&$I.-/>5&2?'% V1 ,..., V5 , U1 , U 2 , I 3 , I 4 .
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Vj
k ∈j
Yk −
k ∈i , j
ViYk =
0/#(Z>&(i42#<#,',Š-1&E(',#%KU$25/ 8
k ∈j
I sk
32
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BP%1#RB0% #&+~&2L-75.3-u‡ _ M0 … ŠŠ…M†:‰
4SŠ4ƒ=‰ Š‰&6
i(t) = ioff + iampl*sin[(2*3.14*fc*t)+mod*sin(2*3.14*fm*t)]
43
A
:`\%&L4%53M%
%,
%LL
%\'
L45
6
Li
s 7
=LLwB0+
tu\'&%,+#%1
R5.3-<'#+-+-7
C4%5=<6 #7
R5.3-uBP2,#,#%
B0%, #RB0%"‡
C{!
A
0M… ŠŠ…M†:‰ 4SŠ ˆ< SŠ
%LL
?
†
ˆ%6
ioff + iampl*exp[-(t-td)*df]*sin[2*3.14*freq*(t-td) +
+ 3.14*Phase/180] ,
td < t < TSTOP
:`\%&L4%53M%
s 7
'7L%,%,+-
T ?- :`_E
D
? g?
? g?
? g?
s 7
=LLwB0+
tu\'&%,+#%1
R5.3-
_%,2';(~&2+ 9w)*%
F5+-8 +-%1)*7
F)*
]_^`2'
C C4=
t : TY ? W g Y*\t
C4t Ž W YAt Ž g ??T
C4t Ž D :xW YAt Ž g ??T D ?
CDRvuG ::]ŒT ? Dv‹G :] I T 2t T 2tŒW:RB8WRB8W:SBqr?SBCT@??$RB7K
CPY W ‹Ž g ??$WJt Ž T :Ci‘2I g ??$W g ??$W T g r5A ]
K
J
vu%,+-+M
*B#7
t
t
C)
C)
+ +-
%,
%LL
%\'
Li7
+-
L
D *B0
s 7
' 7L%,%,+-
T ?M :`_E
D
T ? :`8E=D
*†$‚ 2… =ƒ€ˆŠˆƒ% Rˆ=ƒ$‰

44
vu%,+-+M
*B#7
t
t
C)
B d
B
7
F6=7
J<name> <d> <g> <s> <model> [<area>
>]
<d> <g> <s> = nodurile corespunzatoare drenei, grilei si sursei
<model>
= numele modelului
<area>
= factor de arie
Exemple:
JIN 100 1 0 JFAST
J13 22 14 23 JNOM
+ 2.0
K***
- bobine cuplate
Forma generala:
K<name> L<name> < L<name> >* <coupling>
K<name> < L<name> >* <coupling> <model> [<size>]
L<name>* = numele bobinelor cuplate
M
<coupling> = factor de cuplaj 0 ≤ K =
∠1
L1 + L2
<model> = numele modelului. Se poate defini bobina, neliniara, cu miez
feromagnetic, cu histerezis.
<size> = factor de scala pentru aria transversala a bobinei neliniare.
Exemple:
KTUNED L3OUT L4IN .8
KXFR1 LPRIM LSEC .99
KXFR2 L1 L2 L3 L4 .98 KPOT_3C8
L*** bobina
Forma generala:
L<name> <+node> <-node> [model] <value> [IC=<initial>]
<+node>, <-node> = nodurile pozitiv si negativ
45
[model] = numele modelului
<value> = valoare in H
%,%,3-%19!".7
%,%,3-%<'+&4#;5#7+#%1Jyy%,uI{B0=L4&`B0*Bw+M
5
9
IC=<
~;%,%==5= t=‘ B0CB'5%L%5:BC4D(DSK
g
Exemple:
LLOAD 15 0 20mH
L2
1 2 .2e-6
LCHOKE 3 42 LMOD .03
LSENSE 5 12 2uH IC=2mA
M ***- tranzistor MOSFET.
Forma generala:
M<name> <d> <g> <s> <sub> <mdl> [L=<value>] [W=<value>]
+[AD=<value>] [AS=<value>] [PD=<value>] [PS=<value>]
+[NRD=<value>] +[NRS=<value>] [NRG=<value>]
57*B'# )*+M=[NRB=<value>]
7%i$$%&%
,
/ <d>, <g> <s>, <sub>= nodurile
sursei i substratului
6 1 !" #6&(,#,#%
G !Q,# %,6u/0%3-%,
5,#,#%
t }t=:J!Q7%%&
=%L-# )*7
&
7%S/0%SBw# PB0%
D D;:J!"'76+7%%% 53-%,#%&%
‘  ‘ B
 : ‘  ‹ ‘  !"#&
<'+7+-=5%,.&+-=
% Li#`)*%%_%i
B#0B0%i$7%%S/0%SB#yRB+7+#,#% g
Exemple:
M1 14 2 13 0 PNOM L=25u W=12u
M13 15 3 0 0 PSTRONG
M2A 0 2 100 100 PWEAK L=33u w=12u AD=288p AS=288p PD=60u PS=60u
+ NRD=14 NRS=24 NRG=10
Q*** - tranzistor bipolar
Q<
< name>
> <c>
> <b>
> <e>
> [<
<subs>
>] <model>
> [<
<area>
>]
<c> <b> <e> = nodurile corespunzatoare colectorului si bazei
<subs>= nod al substratului ( optional )
<model> = numele modelului
<area> = factor de arie
Exemple:
46
Q1 14 2 13 PNPNOM
Q13 15 3 0 1 NPNSTRONG 1.5
Q7 VC 5 12 [SUB] LATPNP
R*** - rezistor
Forma generala:
R<name> <+node> <-node> [<model>] <value>
<+node>, <-node> = nodurile pozitiv si negativ
<model> = numele modelului
<value> = valoarea rezistentei in Ω
Exemple:
RLOAD 15 0 2k
R2 1 2 2.4e4
R*** N1 N 2 <VALUE> <MNAME> <L= LENGTH> <W = WIDTH>
S*** - intrerupator controlat in tensiune
Forma generala:
S<name> <+node> <-node> <+control>
<-control>
#7%<'$)*%,+M%,.
/P% +-%,.;<model>
=%,+74# '+-#,#%
<+node>,
<-node>
=
5 + 1 5$+719!"$#%&('$)*%,+-%,./0% +-%1.2&(+-SBP%1#%%
F
8
command Uc
<model> = numele modelului
Exemple:
S12 13 17 2 0 SMOD
SRESET 5 0 15 3 RELAY
T*** - linie de transmisie
Forma general :
T<name><A+><A-><B+><B->Z∅
∅=<value>[TD=<val> | F=<val>[NL=<val>]]
47
$#%&('4+M%
Tb/0%`W
<A+>, <A-> =
!Q%,\'3-=575+-%-Bw+M%5
∅
A!"~ +29)*%&<+7RB6%iB0%%
"!QLi75.3-
‘ G ! 1#%,‹5+&%5 ‹%,%&%i#u,# %,6uL4%Z)*%5
G_!‘ G<N 0/ %
*Bw+MC,# %,6
λ λ
(#
uB0\,#,#%=L-5. 3-< I.7<' L4%,%,+-b! ? g WRr[K
]_^`2'
T1 1 2 3 4 Z0=220 TD=115ns
T2 1 2 3 4 Z0=50 F=5MEG NL=0.5
V*** - sursa independent de tensiune
Forma general :
V<name> <+node> <-node> [[DC] <value>] [AC <mag> [<phase>]]
+ [ <transient> ]
]_^:'%53-%%&CB# +FB0%,%7=5#25&
=uB#PB0=%,'+-==5#7+
I***.
Exemple:
VBIAS 13 0 2.3mV
VAC 2 3 AC .001
VACPHS 2 3 AC .001 90
VPULSE 1 0
+ PULSE(-1mV 1mV 2ns 2ns 2ns 50ns 100ns)
V3 26 77 DC .002 AC 1
+ SIN(.002 .002 1.5MEG)
W*** -
…i‚ ˆ
€
+ =ƒ
‚ 4M†5w‚"ƒ€
ˆ‚
Forma general :
W<name> <+node>
<-node> <vname> <model>
F 8 9!"$#%&('$)*%,+-%,./0% +-%1.2&(~ + #'+M4#,#%
. !" #6&CB#0B0% (+-SBP%1#=57%5#7+'7)*%,+M(%16=
5\
1 !" #6&(,#,#% g
Exemple:
W12 13 17 VC WMOD
WRESET 5 0 VRESET RELAY
X*** - subcircuit
48
Forma generala:
X<name> [<node>]* <sname> [PARAMS: <<par>=<val>*>]
<node>* = nodurile la care este conectat subcircuitul
<sname> = numele subcircuitului. Acesta este definit asfel:
SUBCKT <sname> [<node>]*
..........
ENDS <sname>
Examples:
X12 100 101 200 201 DIFFAMP
XBUFF 13 15 UNITAMP
3. Liniile de comand
‚F†54… $†!w‚ƒ€ˆ‚_†54 ˆ
‚†4… )
Forma general :
.AC [LIN][OCT][DEC] <points> <start> <end>
B0(.%)*=L-5. 3-
&%,%0*'=5+-.uB0#x'
5 g
[LIN], [OCT],
[DEC]
'%, +4B79!"#4#<'#5+-CI&+-+M'+#J.% +-<G#C-‘\*'
5+M.('+#J.% -+ E Ž( /0% '=5<'+#x.7%+- C
C
]Ž KF~;57CB0=L45+&#)*
%1)*<t Ž
B+-4+i9!QLi75.3-=%,%13-%
$9!QL-5.3-=L4%,
Example:
.AC LIN 101 10Hz 200Hz
.AC OCT 10 1KHz 16KHz
.AC DEC 20 1MEG 100MEG
‚F†54… $†!w‚ƒ€ˆ‚ƒ
‚4…i‚S€S€
Forma general :
.DC [LIN] <varname> <start> <end> <incr> [<nest>]
.DC [OCT][DEC] <varname> <start> <end> <points> [<nest>]
.DC <varname> LIST
#<value>*
=7[<nest>]
+-#,#%%, '+<+-RB0%,# CB0#;5#7+
<varname>
=
5#% .7<.%&)*
[LIN],
+M[OCT],
4#,#% [DEC]
!".%&3-%=%,%7**'=5+-.CB0#x'=5=(.&%&%
B +-4+i $9!".$%&
%,%,3-%u/0%L4%,('+#J.7
+-$4#1#%
'%, +4B79!"#4#<'#5+-('+#J.7%=7+-#,#% '+&4#;57CB0
L5=%1)* ‹Ž
%,5w9!"'*B#=%,576+M<'+#x.7=7+-#,#%
*B+&9!Q,+7+-_%,'+54#% '7+4#B0<.%&)*
G#C : !Q%1)* ‹Ž '+&4#;x%iB+-=<.7% .,#Q<.$%%
+-$4#1#% .
Exemple:
49
.DC VIN -.25 .25 .05
.DC LIN I2 5mA -2mA 0.1mA
.DC VCE 0v 10v .5v
+ IB 0mA 1mA 50uA
.DC RES RMOD(R) 0.9 1.1 .001
.DC DEC NPN QFAST(IS)
+ 1e-18 1e-14 5
.DC TEMP LIST 0 20 27 50 80
.DC PARAM RS -1 1 0.1
.DC SRCNAM VSTART VSTOP VINCR < SRC2 START2 STOP2 INCR2 >
‡Š0… € S4 ŠM…0…iˆ
€4i€…Sˆ<…-‚ *†ˆ!,q‡
END F- 6=7
, . END
.ENDS -
‡Š0… €
4 5 ‚Sˆ…RˆSˆŠ-…i‚…„…iˆu†
€S‚S€… €Fƒ… ƒ€… F6=7
.ENDS [<name>]
<name> = numele subcircuitului
Exemplu:
.ENDS 741
.FOUR
-Analiza Fourier
Forma general :
.FOUR <freq> <output
L-5. 3-
var>*
5$\'$+-%L-#+-
<freq>
=
#+' #+_.qN\!O.%&y% ' +4# 572B0J */+-x&%1)* # % V5*B+-x%1)*
B0<'+-=L5(# 6% '5
9J =+-&
55#+-=5#2%1)*<~x%1 +7 )*%,+-7%,# g
Exemplu:
.FOUR 10KHz v(5) v(6,7)
.IC - Conditii initiale pentru regimul tranzitoriu
Forma generala:
.IC < <vnode> = <value> >*
%,%13-%
<vnode> = <value> - se atribuie nodurilor < vnode> valorile
<value>
Exemplu:
.IC V(2)=3.4 V(102)=0
.INC - Includerea unui fisier
Forma general :
.INC <name>
L%i/0%#,#%
<name> = numele
, inclusiv calea
Exemple:
50
.INC SETUP.CIR
.INC C:\\PSLIB\\VCO.CIR
.LIB - Utilizarea unei biblioteci de modele si subcircuite
Forma
general
! #:.LIB
[<name>]
%,y 7%%i %,5,#RB0%,. 5& ‹5
' %,5%,'
g
%1'RB0*/w+M* B0 #+-%&%1)*)* y%,y%+-5
NOM.LIB
Exemple:
.LIB
.LIB OPNOM.LIB
.LIB C:\\PSLIB\\QNOM.LIB
.MC -Analiza Monte Carlo
Forma general :
.MC <#runs> [DC][AC][TRAN] <opvar> <func> <option>*
<#runs> = numar de simulari Monte Carlo
[DC] [AC] [TRAN] = tipul de analiz
=%*/0%,7(' +4#;57CB0=L45=%Z)* A 2+-=Ž(&
<opvar>
L-#59!Q=L-#marimile
53-%='%&5+-(.7%{7%,6%&8=%*/0%,7('+#2=y3M%,
2B0%,# <.7=55+-7%iB+-%5 g t 5*B+-<'+L4% B
H AOtJ
] ] I&.,#KP[RtuG_G ] ] &I .,#K
$'+&9!Q%,.PB0='3-%,# %5uB0('$+FB0+-
AOtJ5A C-‘\[ C :]
: [LIST], .PLOT, .PROBE, ALL,
FIRST <n> , EVERY <n> , RUNS <n> .
Exemple:
.MC 10 TRAN V(5) YMAX
.MC 50 DC IC(Q7) MIN LIST
.MC 20 AC VP(13,5) RISE_EDGE(1.0) LIST
OUTPUT ALL
.WCASE -Wort Case Analysis
Forma general :
.WCASE <analysis> <opvar> <func> <option>*
Exemple:
.WCASE DC V(4,5) YMAX
.WCASE TRAN V(1) FALL EDGE(3.5v) VARY BOTH BY RELTOL DEVICES RL
.MODEL - Model.
Forma generala: .MODEL <name> <type> [<param>=<value> [<tol>]]*
<name> = numele modelului
<type> = tipul dispozitivului
<name>
C***
L***
R***
D***
Q***
Q***
<type>
CAP
IND
RES
D
NPN
PNP
Tip
condensator
bobina
rezistor
dioda
tr.bipolar NPN
tr.bipolar PNP
51
Q***
J***
J***
M***
M***
B***
K***
S***
W***
N***
O***
LPNP
NJF
PJF
NMOS
PMOS
GASFET
CORE
VSWITCH
ISWITCH
DINPUT
DOUTPUT
tr.lateral PNP
JFET canal N
JFET canal P
MOSFET canal N
MOSFET canal P
GaAsFet canal N
miez neliniar
~ + #'+M85+&&+~&J+MRB0%,#
~ + #'+M85+&&+~&25# +
dispozitiv digital de intrare
dispozitiv digital de iesire
/0%
<param>
parametri la valorile <value
.+#=5#x<value>
+-&3M[< tol>]
+-1"-~&setarea
J' 5anumitor
+-
vu4)*=%iB+-<'+7%_5
9+-.(+M%,'#7%
%iB'$)*%,+-%1.C/0%#
** pentru rezistor
Nume
R
TC 1
TC 2
TCE
R***
5L%5%Parametru
+S 2'%L%57
'+&4#x7)*%iB+-3-
5L%5%+%,%8
+-2'7+#7
5L%5%+'+7+-%5=
+-2'7+#7
5L%5% +S^'3M%
+-2'7+#7
Unitate
Val. predefinita
1.0
_
C-1
0
0.0
0
0.0
C-2
0
/0 -
0
C
0.0
<valoare> = RNOM* R * [ 1+ TC1( T - TNOM) + TC2( T - TNOM)2]
sau
TCE ( T - TNOM )
<valoare> = !"
RNOM
.7*<R*
)*%i1B+-.01
3-%(+M\'7+#7=‘xE A
RNOM
F‘xE A !"+-\'+# <%, g D+M
L%SBw'5%L4%&5+-(~&2&%,%==5\ g E DF C E(‘<:
** pentru condensator C***
Nume
C
VC1
VC2
TC1
Parametru
coeficient de amplificare pentru
capacitate
coeficient liniar de tensiune
coeficient patratic de tensiune
coeficient liniar de temperature
52
Unitate
-
Val.predefinita
1.0
V-1
V-2
0 -1
C
0.0
0.0
0.0
coeficient patratic de temperature
TC2
0
C-2
0.0
<valoare> = CNOM.C.(1+VC1.ϑ
ϑ3-+VC
ϑ 2).[1+TC1(T-TNOM)+TC2(TTNOM)2]
%% 2.ϑ
CNOM = valoarea capacit
la temperatura TNOM
~
TNOM = valoarea nominal a temperaturii. Se poate specifica n linia de comand
.OPTIONS
ϑ= tensiunea pe condensator
T = temperatura
** pentru bobina L***
Nume
L
5L4%&5%+Parametru
=\'&%L4%&5<'+#
%,#53-+-
5L%5%+%,%8=5#7+
5L%5%+'+7+-%5==5#7+
5L%5% +S&%,%8 (+-2'7+#7
5L%5% +'++M%5
(+-2'7+#7
IL1
IL2
TC1
TC2
Unitate
-
Val.predefinita
1.0
A-1
A-2
0 -1
C
0 -2
C
0.0
0.0
0.0
0.0
2
2
<valoare> = LNOM*L*(1+IL
).[1+TC
(T-TNOM)+TC
2(T-TNOM) ]
%,#5+M1.i+IL
3-%<2.i+-\
'+# 1=
F‘JE A
LNOM
F‘xE A =!"valoarea
+-\'+# <%,
%`!Q5#7+#' %,xy$yy%,
T = temperatura
**
'+#B‚ ˆ
€
‘‹#
D 76+&%%
b)*%iB+- 3-$
b)*%iB+- 3-
L4L +-RB0%,#==5+7 '+&4#B+-7
$ +-RB0%,#==5+7 '+&4#B+-7
L4L -x8
1-x8
N N<'+#B‚ ˆ
€
‘‹#
-x8
x
- _
5 =ƒ
‚ 4M†w‚>ˆ‚0…i€S‚Sˆ (‡ 5 =ƒ
‚ 4M†w‚ƒ€ˆ‚ vC%1+-+-
s
s g '7L%,%,+-
T g?
T] T g?
s
? g?
vC%1+-+-
s g '7L%,%,+-
T g?
T] T] Y
? g?
Ω
Ω
D 76+&%
b)*%iB+- 3-$
b)*%iB+- 3-
L4L 5 # +=5+& ' +4#B+-7$
5 # +=5+& ' +4#B+-7
L4L 53
Ω
Ω
t
t
N N<'+#F… F†
‘‹#
‡
‡
\
‹
 =
D76+&%
5 #7+uB0+#73-%
b)*%-Bw+M3-=%5
5L4%&5%+=6%iB0%
+-%,2';<+`)*%,+
5'5%,+-+- 53M%,#%%
'+-+-%&,#% 53M%,#%%
5L4%&5%+#=77
%
5+-%1.
^:'+<+-2'+&#=5#7+#,#%
CB0+# 3-%&
5L4%&5%+b)*+L4%&5|
^:'+<)* 6$+SL%5|
5L%5%+'+&4#;L42#=5'5%,+M3-%%
~x'7%1)*7
%15+-
+MRB0%,#=uBw+&'# 7
5#7+#<+-RB0%,# CB+7'# vC%1+-+-
t
s g '7L%,%,+-
T g ?$] T
? g?
T g?
? g?
? g?
T g?
? gr
T g TT
Y g?
Ω
:
s
s
?
t
?
s
T
gr
T g ?$∞] Y
DF76+7%% $8,+-_%iB'$)*%,+M%,.uBw# +')* +-3-% ~;+-%1#J~&Jy%,y% L% g
Exemple:
.MODEL RMAX RES (R=1.5 TC=.02 TC2=.005)
.MODEL QDRIV NPN (IS=1e-7 BF=30)
.MODEL DLOAD D (IS=1e-9 DEV 5% LOT 10%)
‰ F
€S‚†5 [„… ˆ*† ƒ
‚ )ˆ
*ˆ‚„wˆ…… ˆ
*†[„…… 4 ‹ˆ ‚
F6=7
.NODESET
- Setarea valorilor potentialelor nodurilor
†
‚
w‚
ƒ€ˆ‚ ƒ
‚ …i‚S€R€
Fˆ‚
€
.NODESET < <node> = <value> >*
<node> = <value> - nodurilor <node> li se atribuie valorile <value>.*
Exemplu:
.NODESET V(2)=3.4 V(3)=-1V
.NOISE - Analiza zgomotului
Forma general :
.NOISE <opvar> <name> [<ival>]
5*B+-%Z)*9B02'+-
pentru
zgomotul ; a
#,+-%%Z)*(# 6% ~<opvar>
2'7#==5#2variabila
%1)*<~;
5# care
+S1+-se
4analizeaza
+-%,. t Ž
g
9!"#
+M4#,#%%,'+<+-RB0%,#CB0#;5#7+5
5RB+-%,+&#%uBw# PB0=<)* 6$+
%,.19! L-5.3-<'+#;5uB0(+-%1'*B+-uB#'%16+M85+7%,y# 3-%=L4%57%
B#0BP=b)*+# +-$+-5%15#%1+#,#% g
54
,€S‚Rƒ% P†5…iƒ Sˆ=Š4€S‚Rƒ„… ‚F†ˆ
.OP
Forma general :
.OP
w‚ƒ€ˆ‚ 8
ƒ
‚
tinuu
.OPTIONS - Optiuni
Forma general :
.OPTIONS [<fopt>*] [<vopt>=<value>*]
3
<fopt>,<vopt> = op iuni care nu cer sau cer atribuirea unor valori <value>
Acestea_sunt:
B0(+M%,'7*BP5<+-%,2'%=55#
, %,%5=+-=%,B#y5%15#%1+-C/0% '76+&%%5*B+-7
. / %iB+-)*=%,%%=%,;L%i/0%=<y%,y&%+-5%&
. ‡
&%iB+-)*
+-=
%1+7
- +-%1'*/+-<+-y1# #7%$
- >- B#'7%,
&%iB+-
- - B#'7%,
&%iB+-<'76+&%&{,#1#%
-, B#'7%,(' %,7
#+M6+M
-,b‡
+M%,'7*/w+M(.&%=%L7%,+-_'3-%,# %
=‡ -. ! .,# B05 %,2y<.7=7%%yRB0,# +-('+#25# + g s
(
( J
‡\!
(#.
(
!
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A!
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.PARAM < <name>=<value> >* <<name>={expr}>*
<name>=<value> - parametrului <name> i se atribuie valoarea<value>.
<name>={expr} - parametrului <name> i se atribuie expresia {expr}.
Referirea la valoarea unui parametru global se face utilizand acoladele { }
Exemple:
.PARAM pi=3.14159265
.PARAM RSHEET=120, VCC=5V
.PLOT - Trasarea unui grafic
Forma general :
.PLOT [DC][AC][NOISE][TRAN] [ [<opvar>*] [(<lo>,<hi>)] ]*
[DC],[AC],[NOISE],[TRAN] - tipul de analiza pentru care se cere trasarea
graficului
*/0%,
<opvar> = numele variabilei de i e pentru care se traseaz graficul. Num rul maxim al
acestor variabile este de 8.
(<lo>,<hi>) = punct de origine al axelor pentru graficul respectiv.
Exemple:
.PLOT DC V(3) V(2,3) V(R1) I(VIN)
.PLOT AC VM(2) VP(2) VG(2)
.PLOT TRAN V(3) V(2,3) (0,5V) ID(M2) I(VCC) (-50mA,50mA)
.PRINT - Tiparirea rezultatelor
Forma general :
.PRINT [DC][AC][NOISE][TRAN]
[<opvar>*]
56
7 ) #,+M+-[DC],[AC],[NOISE],[TRAN]
+M%,'#=%1)*<'+&4#;57CB0=5<+-%,'7%,7
'.w9!".7%y%&
57<#4)*h=L4% +-%1'%1+-
Exemple:
.PRINT DC V(3) V(2,3) V(R1) IB(Q13)
.PRINT AC VM(2) VP(2) VG(5) II(7)
.PRINT NOISE INOISE ONOISE DB(INOISE)
.PROBE -Postprocesorul grafic PROBE
Forma general :
.PROBE[/CSDF]
.PROBE[/CSDF]
'.w9!"[<opvar>*]
.7%y%&('+#257uBP=7*B+-=J%1)*='%5
DSbE (] <'6%,+M(.%Z) #%1)*7(# 8^:' *B0%%<.7%y%==%*/0%,7 g s 7%y%&
%&*/P%1<'`B0%,y%&CB# +P
{'+#x%1\#=5#7+5+-%1##B0#J+& )*%1+-7%,#
102"‚ R%ˆ $(6 !"'+-3-%&,# #,#% 102"&'‚ R%ˆ $ "(w‚ S%ˆ $36{!"+-SBP%1#<~+7(# /0% 102"‚†:‰ %ˆ $(6{!"+-RB0%,# ('
+# %,' F
02"‚†$‰Q%ˆ $(6{!"'+- +-%,# #,#%`^2 ##%+ 6
08"{‚F†$‰Q%ˆ $36{!"+MRB0%,#<~+7<#7%b^6/0% ;<##%&6+2#,+-%1'
08"{‚F†$‰Q%ˆ $36{!Q5# +&# '%12&6+&#%,'& 08"{‚F†$‰Q%ˆ $36!Q5# +# '7%,xy<^2<##%+2#,+-%,'
6"' +-
L% #=%&54#%+<+-%,'#wŽ ? =?&]8?&R? ? =? C4?&G8?& ?s
^ J'+L4%7%57
%1Jy4 =? b? :
I (K
- D/G/S
(J)
- D/G/S/B (M)
- C/B/E/S (Q)
- pentru regimul sinusoidal:
M - amplitudinea
DB - amplitudinea in dB
P - faza
~&+-w)*%==#'
G–
' 4+M(7
'+-=%,6 %,7
'+#;%1)*b)*+#1#%w
- ‡ )* 6$+#=%,+7
- - ‡ )*$6+&#
%&*BP%1
0 - ‡ 6 )*+# %, +7<~;
0 - - ‡ 6+{)* 6$+#=%*/0%,<~;
R#53M%%<#+-%&%1)*y%uB#+P
57
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t t=‘\Ii^K
tubŽ(t=‘\Ii^K
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55+Mx^R*~x %%
7%,.+-=,#%`^J~x'4+5#;yRB05%iB0
%,+-$=,#%`^J~;%,+M4.1#L4%1%,+'+#;yRB05%iB0
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6%1%,2# '43M%% ==,#%`^
6^%,\#'43-%&%7=
1#%`^
Exemple:
.PROBE
.PROBE v(2) I(R2) VBE(Q13) VDB(5)
.PROBE/CSDF
G5&*/P% ) #,+-+<B0 0#\#+-%%Z) 99%,# 75?t 5?M{75;^'*B B0% g Ž: u
*Bw+Mx#
L6+{B'5%L%5<'+#;+-<#7%5
%*/0%,7<~\D : DCŽ ] g
5† 4iƒ€4i€ 4 ˆ‚ … 4 … ) … [„4… 4 .SENS F6=7
,‡
‡
w‚ƒ€ˆ‚ ƒ
‚ …i‚S€R€
"+ )R†%$ '.qN !".7%y%&
%*/0%,7<'+#;5uB0
$*/+-=%1)*
B0 )*%1+-%,.%,+M3-%~x'$4+5#x+-3-% '7+%&%5%15#%1+#,#% g
Exemplu:
.SENS V(9) V(4,3) I(VCC)
.STEP -
‚F†54… $†!w‚
{†
0…
58
Forme generale:
.STEP [LIN] <varname> <start> <end> <incr>
.STEP [OCT][DEC] <varname> <start> <end> <points>
.STEP <varname> LIST <value>*
[LIN], [OCT],[DEC], LIST - se variaza variabila <varname> liniar, pe
B0#x~x'#5+M .,#qN‹Bw'5%L4%&5+-(~&2&%iB+-
octave, peB+-decade
4+i $9!".7%=(~&5'#+{/0%SBPL 90/P%1+S&(.7%y%&
<incr> = incrementul
<points> = num rul de puncte
Exemple:
.STEP VIN -.25 .25 .05
.STEP LIN I2 5mA -2mA 0.1mA
.STEP RES RMOD(R) 0.9 1.1 .001
.STEP TEMP LIST 0 20 27 50 80
.STEP PARAM X 1 5 0.1
.SUBCKT - Definirea unui subcircuit
Formele generale:
.SUBCKT <name> [<node>*] [PARAMS: <par>[=<val>]* ]
<name> = numele subcircuitului
<node>* = nodurile extreme ale subcircuitului. Niciunul nu poate fi zero.
PARAMS: <par> [ = <val>]* - setarea unor parametri
Exemple:
.SUBCKT OPAMP 1 2 101 102
.SUBCKT FILTER IN OUT PARAMS: CENTER, WIDTH=10KHz
.TEMP - Temperatura
Forma generala:
.TEMP <value>*
/
<value>* = valorile temperaturii pentru care se dore te analiza circuitului
Exemple:
.TEMP 125
.TEMP 0 27 125

‹†54iƒ€4i€4S€R‚Sˆ…SŠ4€S‚Rƒ„……Sˆ *†$‚ Šˆ
w‚ƒ€ˆ‚ ƒ
‚ …i‚S€R€
F6=7
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† %$&"… 3 ƒ%$
'.w9!".7%y%&
=%*/0%,7
%,'RB759!"#
+M4#,#%
5#% . <' )*%, +-(.%y%=
%1+
Exemple:
.TF V(5) VIN
59
.TF I(VDRIV) ICNTRL
.TRAN - Analiza de regim tranzitoriu
Forma general :
.TRAN[/OP] <pstep>
[<noprint>
'*<ftime>
B#<+-%,2'J
'+#257u[<ceiling>]]
BP=L%i/0)*(7[SKIPBP]
) #,+-+&#
<pstep>
=
Li+M%,6
!"+-%,2'#L%, '+#;5uB0
*/+-=%Z)*(+7 )*%,+M%&
'7%,+i !"+-%1\'#%1%,+-% ' +4#;57CB0=7*/+-
L%i/07(7) #,+-+M
5%%,$ !
limita superioara a pasului de timp; daca nu se specifica se considera
ceiling=ftime/50 pentru un circuit care contine elemente dinamice sau ceiling=pstep pentru un
circuit rezistiv.
5 %,3-%%&8%,%1+-%=L%,%,+-=%,;%,%%=%,.0B08&6+M
[SKIPBP] = folosirea
=5%,75#%,+{B0#25&5#+-=L4 `B0%,2B0+-7%=%,;%,%&
=5 g CŽ
Exemple:
.TRAN 1nS 100nS
.TRAN/OP 1nS 100nS 20nS SKIPBP
.TRAN 1nS 100nS 0nS .1nS
.WATCH -
€‚… $†ˆ*†1ˆ [ €4 P†5wˆ%42 ‚Š ‰ † ‚S€R‰Qˆ
…iƒ
!w‚<4…i‰
F€
4F†$‚F†54… [ˆ…Rƒ…2ƒ€… €4i€…
F6=7
:
.WATCH [DC] [AC] [TRAN] <<varname> <l, h>>*
[DC], [AC], [TRAN] = tipul analizei
#(.%y%%SI' 9=<+7%.%y%bK_/0%&%,6%1+-
<varname> <l, h> =
B0=%,L%&u/0%SBw# '%&
Exemple:
.WATCH
.WATCH
DC V(1) (1,5) I (R5)
TRAN VC(2) (-0.8, 0.8)
.b€R‚*$…i‰Qˆ*†4…i‚…… 4 .WIDTH
- =7
F6
?
IL(L) (0.1mA, 0.6mA)
w‚"Š-…0…iˆ
€4Rˆb…iˆ0… ˆ
- ")S5† 4 $
.
1 !O##_\5 2L4%Z^`+_'+# %,%%&\ %, L4%i/0%#_2%*/0%,7 g DF+-2L4%~ +
]_^`2'
g C‹ Ebv !
/0%TYY g s 7<' L4%,%,+-=*Bw+M5? g
5?
60
CAP 3. LUCRARI DE LABORATOR
LUCRAREA 1
5 , - . :`uBPuB0%,2#)*=5#xD;:DCŽ
]#+-7
5%15#%1+-CB0%SB0CB0=^:'%&5(7) #,+-+M
] ! D s\V ] ! D v V
]S! T W:s2VC7B Y‹! T t\V
V‹#S! V
! Ω
Ω
!Y
V‹&%!Y
V
Ω
Ω
%& g T g T g
B,
KQ]R! T s\V
y KQ
] S!9Ws\V
5K $ !9W V
!9W:s2V ] !9WFN C V
C"! ! T t2V[ !"
! T
!" ! T V"R!9W ΩV
Ω
Ω
]
Ω
%& g T g W g
(' 7. .!-
RD g T g T g : $yRB04. 5O5%,5#%,+#
# 7y#5& L4$4+-# 6%u%,ŒB#0B0 +MRB0%,# B0#
B05+-%1#%xL4$4+- #%J %, Bw# PB0 5# + g Ž(%,75#%,+#x%1% 7 5%J B0,# +-%O#%5 g : ,#+-%
+-4%,+M
5&
# :DCŽ ] B0<.%&L4%5=5#xy% +# '#+-7%${#+-%&%1)*x+-=%,;L%iB0%#N g $#+ g
DR g T g W g C4 5) #F# #%5%,75#%,+ %,% 5# B#0B0J5$6 +-2BP ' +- 5 +-6%1+#6+7%5%
B0%iB+-2#,#%9 5#3-%%995%,75#%,+&#,#%QB0 B0Π#)* '+# #%,+- .&%" '76+7%$ 5$\ g DF+# +-4%,^`%iB+- +- /0% #%&5%,+-+- B0,#+-%&%(5%,5#%,+#,#%(%, L%#7 T g W g '+&4#
5) # %&‹K7:y K /0%5K B0‹+M4%,‹ +M4#5 %,.+ =+-RB0%,#‹5%,75#%,+&#,#%F%,+7=y$4&bt
/0% g
b)*%-Bw+M+-‹5 %1.+-
~+7by4C5%,75#%,+&#,#%'*B0%,.%1)*+
R
'+-CL%S+-6%,+-C# 'C5# # 4)* g ‹ 5=5*B+S5%15AB#%10+
5$+-%,6 B#0B0;5+M*S'+# 5&5#,#,#%
B0\'&%5
R AB 0
~&+7At B0&
%
# 5# + T t 5 ~ L%#7AW g T g b) #1+-
% g W g T g
+MRB0%,#<v %,;57CB0=+-6%,
R AB 0 =
:`=+-6%,
+-$4#5 %1.+<+-RB0%,#
%, +<y<Bt g
DF+#;+-6%1<7)*%iB+-+-%5%,.+-b*),+bBP<'*B0%,.%Z)*)*
5%,75#%,+&# g
61
U g
1
b) #,+-=5#+-%&%
I1 + I3 = I2
2I1 − 2I 3 = 0
2I = 2I + U
1
Ž # 3
I2 = 1A
‹%,x#1+-%,
5#+-%% ) #,+-<vu!+? 7*B'5+-%,.
R AB 0 =
DF+#;
+-4%,
U AB0
U
= 0Ω
1A
BPuB05%1#25#+-%%=5%,75#%,+#,#%w
I5 = 0
I2 = I4
I1 + I 3 = I 2
I1 + I 2 + 2 I 4 = 2
I 1 + I 2 = 2 + 2 I 1 − U AB 0
b) #,+-
I3 = 1
I1 = −
I2 = I4
I 4 = 1 + I1
I4 =
I1 + I 4 + 2 I 4 = 2
U AB 0 = 2 I 1 + 2 I 4
1
A
4
3
A
4
2 6
U AB 0 = − + = 1V
4 4
b) #,+-=7+-#5%,. +S (+-SBP%1#=(y4(t
K MC ;5*B+5)
U
−E =0
B0% AB0 3
0B %_50%,⋅7I5#= %,0+#x9%,L%,%1+-+-x2B0,#+-%&%_I-7%5
K0V
I ∈R
62
− 2 I 4 = 2 I 1 − U AB 0
y`K+CM;5*B+5)
U
7) AB0
#,+M
− E4 ≠ 0
0⋅I ≠ 0
B0%5%,5#%,+# #;7CB0,# +-%V
5K MC ;5*B+5)
I=
U AB0
R4
B0%O'+&4#
# %5 g
R4 ≠ 0
5%,75#%,+# 7 B01#+-%
:
<.#+-%%Z)*(#+-7
%1RB+4#5+-%,#%
a) Descrierea elementelor de circuit
-
resistor
R<name> <+node> <-node> [<model>] <value>
R<name> = numele rezistorului
<+node> , <-node> = nodurile intre care se
conecteaza
<model> = numele modelului (optional)
<value> = valoarea rezistentei, in Ω
!"$#%"&'()")*'
I<name> <+node> <-node> [[DC] <value>] [AC <mag> [<phase>]]
[ <transient specifications> ]
<+node>, <-node> = nodurile pozitiv si negativ
[[DC] <value>]= semnal de curent continuu de tip Is =
<value> , in A.
-
!"$#%"&'( de tensiune
V<name> <+node> <-node> [[DC] <value>] [AC <mag> [<phase>]]
[ <transient specifications> ]
<+node>, <-node> = nodurile pozitiv si negativ
[[DC] <value>] = semnal de tensiune continua de tip E =
<value> , in V.
- & comandate liniar
- sursa de tensiune comandata in tensiune (VCVS)
63
E<name> <+node> <-node> <+control> <-control>
<gain>
<+node> ,<-node> = nodurile pozitiv si negativ ale sursei
<+contro>,<-control> = nodurile pozitiv si negativ ale
portii de comanda
<gain>
= E/Uc
- sursa de tensiune comandata in curent (CCVS)
H<name> <+node> <-node> <vname> <gain>
<+node>, <-node> = nodurile pozitiv si negativ
<vname>
= sursa de tensiune al carei curent
reprezinta marimea de comanda
<gain>
= E/Ic
- sursa de curent comandata in tensiune (VCCS)
G<name> <+node> <-node> <+control> <-control>
<gain>
<+node> ,<-node>
= nodurile pozitiv si negativ ale
sursei
<+contro>,<-control> = nodurile pozitiv si negativ ale
portii de comanda
<gain>
= Is/Uc
-
sursa de curent comandata in curent (CCCS)
F<name> <+node> <-node> <vname> <gain>
<+node>, <-node> = nodurile pozitiv si negative ale sursei
<vname> = sursa de tensiune al carei curent reprezinta
marimea de comanda
<gain>
= Is/Ic
b) Descrierea tipului de analiza de curent continuu
.OP
- realizeaza determinarea punctului de functionare in current continuu si tipareste valorile
potentialelor nodurilor si ale curentilor prin sursele de tensiune in fisierul *.out.
.DC [LIN] <varname> <start> <end> <incr> [<nest>]
.DC [OCT][DEC] <varname> <start> <end> <points> [<nest>]
.DC <varname> LIST <values [<nest>]
<varname> = numele sursei a carei valoare se variaza;
[LIN],[OCT],[DEC] = variatie de tip liniar, pe octave, sau decadica a valorii sursei;
LIST = variatie dupa o lista de valori <value>;
<start>,<end> = valorile de inceput si sfsrsit ale parametrului considerat;
<incr> = increment;
64
<points> = numarul de puncte in intervalul considerat;
<nest> = cea de-a doua sursa a carei valoare se variaza (optional) ;
LUCRAREA 2
CIRCUITE NELINIARE DE CURENT CONTINUU
1. PROBLEME
1.1.) Sa se simuleze cu PSPICE urmatorul circuit :
E1= 1 .. 10 V, r1= 1 kΩ, r2= 2 kΩ
Sa se reprezinte grafic caracteristica de transfer v(2) = f(V(1)) si sa se indice utilitatea acestui
circuit.
1.2.) Sa se simuleze cu PSPICE urmatorul circuit:
r1= 3.0 Ω, E1=10.0 V
a) Sa se aprecieze numarul de solutii al circuitului prin metoda dreptei de sarcina.
b) Sa se rezolve circuitul plecand de la urmatoarele valori pentr V(2)
(se foloseste comanda .NODESET) : 0.0V ; 3.0V ; 7.0V ;
65
2. INTERPRETAREA REZULTATELOR
La problema 1.1. se obtine cu SPICE urmatoarea caracteristica de transfer: Se constata ca
pentru valori mai mari sau egale cu 6V ale tensiunii de intrare V(1), tensiunea de iesire V(2) este
egala cu 4V. Rezulta ca circuitul este un stabilizator de tensiune.
La punctul a) al problemei 1.2 se traseaza dreapta de sarcina cu ecuatia E = R i + u
respectiv: 10 = 3 i + u .
Dreapta se intersecteaza cu axele in punctele A de coordonate (0, 3,33) si B de coordonate (10, 0).
Se constata ca dreapta de sarcina se intersecteaza cu caracteristica rezistorului neliniar in
punctele corespunzatoare valorilor U1 = 11765
,
V , U 2 = 3,2174 V si U 3 = 4,6102 V .
Rezulta ca circuitul are trei solutii, deci un numar impar de solutii in conformitate cu teorema de
existenta de la paragraful 2.4.1.2. din curs.
Programul SPICE rezolva circuitul utilizand metoda Newton-Raphson. In cazul in care
circuitul are mai multe solutii, prin aceasta metoda se determina numai una dintre ele si anume
aceea care este “mai apropiata” de aproximatia initiala.
In problema, aproximatia initiala se introduce cu instructiunea NODESET. Atribuind
acestuia valorile 0V, 3V si 7V se obtin succesiv cele trei solutii.
Totodata se verifica faptul ca daca aproximatia initiala nu este specificata, programul
SPICE ii atribuie acesteia valoarea nula V (0) = 0 si in acest caz se determina prima solutie.
3. INSTRUCTIUNI SPICE
Se utilizeaza urmatoarele linii pentru descrierea circuitului:
66
1) Descrierea unei diode
D<name> <+node> <-node> <model>[<area>]
<+node> ,<-node> = nodurile pozitiv si negativ
<model>
= numele modelului
<area>
= numarul de diode montate in paralel (daca acest parametru nu
este precizat se considera area =1);
2) Descrierea unui model
.MODEL <name> <type> [<param>=<value> ]
<name>= numele modelului ( de exemplu cel declarat in linia de descriere a
diodei);
<type>= tipul modelului (pentru dioda este D);
<param> = <value> - reprezinta setarea unor parametrii ai modelului.
Principalii parametri ai diodei sunt:
Parametru
Nume
IS
curent
de
saturatie
Unitate
de masura
Valoare
predefinita
A
RS
rezistenta
serie
1.0E-14
Ω
CJ0
BV
capacitaea tensiunea
jonctiunii inversa de
la tensiune strapungere
nula
F
V
0
0
∝
IBV
curentul la
tensiunea
BV
A
1.0E-3
3) Descrierea surselor comandate neliniar:
-
comanda de tip polinom:
E<name> <+node> <-node> POLY(<value>) < <+control> <-control> >*
+
< <coeff> >*
F<name> <+node> <-node> POLY(<value>) < <vname> >* < <coeff> >*
G<name> <+node> <-node> POLY(<value>) < <+control> <-control> >*
+
< <coeff> >*
H<name> <+node> <-node> POLY(<value>)< <vname> >* < <coeff> >*
<value> = dimensiunea polinomului (1;2;3)
< coeff> = coeficientii polinomului
Exemplu:
- polinom de dimensiune 1
e = a0 + a1uc + a2uc2 + ...
- comanda de tip o expresie analitica:
E<name> <+node> <-node> VALUE=[<exp>]
67
G<name> <+node> <-node> VALUE=[<exp>]
<exp> - o expresie analitica;
-
comanda de tip tabel (intre puncte se considera extrapolarea liniara):
E<name> <+node> <-node> TABLE [<exp>] < (inval), (outval) >*
G<name> <+node> <-node> TABLE [<exp>]= < (inval), (outval) >*
Cu acest ultim tip de surse comandate se modeleaza rezistoarele
neliniare cu caracteristici liniare pe portiuni (PWL) .
LUCRAREA 3
CIRCUITE CU EXCITATIE SINUSOIDALA CARE FUNCTIONEAZA LA
SEMNALE MICI SAU LA SEMNALE MARI
1. PROBLEME
1.1.) Sa se simuleze cu PSPICE urmatorul circuit:
Rb = 1MΩ ; Rc = 4kΩ; E1 = 5V; Vcc = 9V; e2 = E2 sin (ωt);
TB: β= 200 ; Cje = 2pF; Cjc = 2pF; Vje = 0.6V; Vjc = 0.6V;
a) Sa se reprezinte grafic caracteristica de transfer in curent continuu Ve = f (Vi), unde
Vi ∈[0,5E 1 ] , iar Ve este tensiunea pe rezistorul Rc.
b) Sa se faca analiza de curent alternativ la semnal mic a circuitului pentru E2=1.0 V si respectiv
E2= 20.0 V. Sa se reprezinte grafic Ve=f (frecv) , frecventa variind decadic intre 0.1 Hz si 100Mhz.
Sa se discute valabilitatea rezultatelor obtinute prin analiza de semnal mic pentru cele doua valori
ale lui E2, timand seama de caracteristica determinate la punctul a).
2. INTERPRETAREA REZULTATELOR
La punctul a) al problemei, din analiza caracteristicii de transfer in curent continuu
Ve = f (Vi ) prezentata in figura 1, rezulta ca aceasta prezinta doua portiuni liniare distincte:
68
1 0 V
5 V
0 V
0 V
5 V
1 0 V
1 5 V
2 0 V
2 5 V
V 1
v (5 )-v (4 )
Fig. 1
prima portiune pentru Vi ≤ 11,25V si a doua portiune pentru Vi > 11,25V . Acest rezultat se
datoreaza faptului ca tranzistorul are caracteristici neliniare de intrare si de iesire atunci cand
functioneaza la semnal mare. Prima portiune liniara a carcateristicii u e = f (ui ) corespunde
conditiilor de functionare la semnal mic.
La punctul b) al problemei in urma analizei in curent alternativ a circuitului rezulta Ve = f ( frecv )
pentru E 2 = 10
. V , fig. 2 si E 2 = 20.0 V , fig. 3.
1 .0 V
0 .5 V
0V
1 0 0 m H z 1 .0 H z
100H z
10K H z
V (5 )-V (4 )
F re q u e n c y
Fig. 2
69
1 .0 M H z
100M H z
2 0 V
1 0 V
0 V
1 0 0 m H z
1 .0 H z
1 0 0 H z
1 0 K H z
1 .0 M H z
1 0 0 M H z
V (5 )-V (4 )
F re q u e n c y
Fig. 3
Se constata ca alura celor doua curbe este identica, curba din figura 3 reproducand la o alta scara
curba din figura 2.
Luand in considerare caracteristica determinata la punctul a) rezulta ca rezultatul obtinut la
punctul b) pentru E 2 = 20.0 V nu este valabil, deoarece pentru Vi > 11,25 V circuitul functioneaza la
semnal mare, deci modelul liniar la semnal mic nu este valabil.
3. INSTRUCTIUNI SPICE
Se vor utiliza urmatoarele linii:
a) descrierea tranzistorului bipolar:
Q<name> <c> <b> <e> [<subs>] <model>
<c> = nodul corespunzator colectorului
<b> = nodul corespunzator bazei
<e> = nodul corespunzator emitorului
<subs> = nodul corespunzator substratului (optional)
<model> = numele modelului
b) descrierea modelului tranzistorului bipolar :
.MODEL <name> <type> [<param>=<value> ]
<name>= numele modelului (declarat in linia de descriere a tranzistorului)
<type>=tipul modelului (pentru tranzistor bipolar este PNP sau NPN):
<param> = <value> - reprezinta setarea unor parametrii ai modelului.
Principalii parametri ai tranzistorului bipolar (modelul Gummel-Poon) sunt:
Parametru
Nume
IS
curent de
BF
amplificarea
RB
rezistenta
70
RE
rezistenta
RC
rezistenta
Unitate
de masura
Valoare
predefinita
Parametru
Nume
Unitate
de masura
Valoare
predefinita
saturatie
A
1.0E-16
in curent β
-
bazei
Ω
100
emitorului
Ω
0
colectorului
Ω
0
0
Vje
tensiune de
deschidere
BE
V
Vjc
tensiune de
deschidere
BC
V
Cje
capacitatea BE
(jonctiune
nepolarizata)
F
Cjc
capacitatea BC
(jonctiune
nepolarizata)
F
0.75
0.75
0
0
c) descrierea analizei in curent alternativ a circuitului echivalent de semnal mic:
.AC [LIN][OCT][DEC] <points> <start> <end>
[LIN][OCT][DEC] = variatia liniara, pe octave, sau decadica a frecventei;
<points> = numarul de puncte in intervalul considerat;
<start>, <end> = capetele intervalului de frecventa considerat;
Pentru sursele independente forma generala este:
V<name> <+node> <-node> AC <mag> [<phase>]
I<name> <+node> <-node> AC <mag> [<phase>]
<+node>, <-node> = nodurile pozitiv si negativ
AC <mag> [<phase>] = semnal de curent alternativ de tipul
x = <mag> * sin (2 πf t + phase), unde x este tensiune sau
curent si f este frecventa specificata in linia de comanda.AC
71
LUCRAREA 4
CIRCUITE LINIARE DE CURENT ALTERNATIV
1. PROBLEME
1.1.) Sa se simuleze cu PSPICE urmatorul circuit:
Se dau: r1=10Ω; r2=1Ω; r3=2Ω; c2=1.5mF; L1=50mH; L2=25mH; L3=50mH; M=25mH; e(t)=30
sin(2πft + π/2); is(t)=2 sin(2πft); f ∈ (10Hz,200Hz)
Se cere sa se calculeze curentul prin r2 (modulul, argumentul, partea reala si partea imaginara)
pentru f=50Hz si sa se verifice concordanta acestor rezultate.
1.2.) Sa se traseze cu PSPICE caracteristicile de amplitudine si de faza ale amplificarii in tensiune
Au= Ve/e pentru urmatoarele filtre in banda de frecvente 1Hz-1MHz. Sa se indice tipul fiecarui
filtru.
Fig.a)
r1 =1kΩ; c1 = 8nF; c2 =11.5nF
L1 = 3mH;L2 =12mH;
e(t)=1 sin (ωt)
72
Fig. b)
r1 =1,5 kΩ; c1 = 10nF; c2 = 8uF
L1 = 2mH;L2 =12mH;
e(t)=1 sin (ωt)
Fig.c)
r1 =1,5 kΩ; c1 = 10nF; c2 = 8uF
L1 = 2mH; c3=1nF
e(t)=1 sin (ωt) ; Rs=1.5kΩ
Fig. d)
r1 =1,5 kΩ; c1 = 0,1uF; c2 = 40uF
L1 = 10mH; L2 =20mH; c3=1uF
e(t)=1 sin (ωt) ; L3=4mH
2. INTERPRETAREA REZULTATELOR
La problema 1.1. chestiunile care se vor urmari sunt:
- modul de descriere a elementelor de circuit, bobine, bobine cuplate si condensatoare, in circuitele
de curent alternativ;
- modul de obtinere si tiparire a marimilor electrice, curenti si tensiuni, la o frecventa precizata;
La problema 1.2., pentru fiecare punct, se va trasa caracteristica de amplificare si se vor obtine
urmatoarele tipuri de filtre: trece jos, trece sus, trece banda si opreste banda.
3. INSTRUCTIUNI SPICE
Se vor utiliza urmatoarele linii:
a) descrierea bobinelor:
L<name> <+node> <-node> [<model>]<value> [IC=<initial>]
<+node> <-node>= nodurile intre care se conecteaza
model = numele modelului (optional)
73
value = valoarea inductivitatii (in H)
IC=<initial>= conditia initiala (pentru curent) (se foloseste pentru .TRAN cu
optiunea SKIPBP)
b) descrierea cuplajului
K<name> L<name> < L<name> >* <coupling>
L<name> < L<name> >* = bobinele intre care exista cuplaj
<coupling> = factorul de cuplaj ( K ≤ M / L1 L2 )
c) descrierea condensatorului
C<name> <+node> <-node> [<model>]<value>[IC=<initial>]
<+node> ,<-node> = nodurile pozitiv si negativ
<value>
= valoarea capacitatii in F
IC=<initial>
= tensiunea initiala, in V, pe condensator
(se foloseste pentru .TRAN cu optiunea SKIPBP)
d) descrierea analizei in curent alternativ a circuitului echivalent de semnal mic:
.AC [LIN][OCT][DEC] <points> <start> <end>
[LIN][OCT][DEC] = variatia liniara, pe octave, sau decadica a frecventei;
<points> = numarul de puncte in intervalul considerat;
<start>, <end> = capetele intervalului de frecventa considerat;
Pentru sursele independente forma generala este:
V<name> <+node> <-node> AC <mag> [<phase>]
I<name> <+node> <-node> AC <mag> [<phase>]
<+node>, <-node> = nodurile pozitiv si negativ
AC <mag> [<phase>] = semnal de curent alternativ de tipul
x = <mag> * sin (2 πf t + phase), unde x este tensiune sau
curent si f este frecventa specificata in linia de comanda .AC
d) Comanda .PRINT pentru analiza de tip .AC
.PRINT AC VM(1) VP(5) IR(r1) VI(5,4)
VM(1) = amplitudinea potentialului nodului 1 (modulul numarului complex);
VP(5) = faza potentialului nodului 5 (argumentul numarului complex);
IR(r1) = partea reala a curentului prin r1;
VI(5,4) = partea imaginara a tensiunii intre nodurile 5 si 4.
74
LUCRAREA 5
SUBCIRCUITE. ANALIZA IN REGIM TRANZITORIU
1. PROBLEME
1.) Sa se determine caracteristica amplificarii A= U2 / U1 pentru urmatorul circuit cu
amplificatoare operationale:
a) E1=10-3sin(2πft) (V), f∈(0.1Hz , 10MHz)
Pentru amplificatorul operational se va utiliza urmatorul model, care va constitui un
subcircuit:
ri = 1MΩ; re=0.1mΩ; Ce=0.1pF
b) sa se simuleze in regim tranzitoriu acelasi circuit cu f=500Hz si E1 = 1mV , E1 = 10V ; sa se
vizualizeze tensiunea de iesire sis a se explice rezultatele obtinute.
2.) Sa se simuleze in regim tranzitoriu urmatorul circuit :
75
Rb = 1MΩ ; Rc = 4kΩ; E1 = 5V; Vcc = 9V; e2 = E2 sin (ωt); f=10kHz;
TB: β= 200 ; Cje = 2pF; Cjc = 2pF; Vje = 0.6V; Vjc = 0.6V;
Sa se faca analiza pentru E2=1.0 V si respectiv E2= 20.0 V. Sa se reprezinte grafic Ve.Sa se
compare formele de unda obtinute pentru cele doua valori ale lui E2 .Sa se discute valabilitatea
rezultatelor obtinute la lucrarea L3 pentru acelasi circuit.
2. INTERPRETAREA REZULTATELOR
La problema 1. chestiunile care se vor urmari sunt:
- Modul de utilizare a subcircuitelor;
- Analiza in curent alternativ (.AC) excitand circuitul in gama de frecvente 0,1 Hz - 10 Mhz;
- Se va verifica prin analiza in regim tranzitoriu ca pentru amplitudinea de 1 mV la intrare se
obtine acelasi rezultat ca la analiza in curent alternativ deci circuitul functioneaza la semnale mici.
Daca tensiunea de intrare are amplitudinea de 10V forma de unda la iesire nu este sinusoidala deci
circuitul functioneaza la semnal mare si rezultatele obtinute cu analiza de semnal mic (AC) nu sunt
valabile.
La problema 2. simuland circuitul in regim tranzitoriu se obtin pentru E2=1.0 V figura 1 si pentru
E2= 20.0 V figura 2.
10V
5V
0V
1 0 0 us
150us
V (5 )-V (4 )
200us
2 5 0 us
V (2 )
T im e
Fig. 1
76
3 0 0 us
350us
400us
40V
0V
-4 0 V
100us
150us
V (5 )-V (4 )
200us
250us
300us
350 us
400us
V (2 )
T im e
Fig. 2
Din figura 2 rezulta ca forma de unda a tensiunii Ve pe rezistorul de sarcina Re, care la o
anumita scara reprezinta curentul de colector, este distorsionata. Acest lucru se produce deoarece
pentru circuitul neliniar cu tranzistor se poate folosi modelul liniar in curent alternativ numai
pentru functionarea la semnale mici. E2=20.0 V corespunde functionarii la semnale mari.
3. INSTRUCTIUNI SPICE
Se folosesc urmatoarele linii:
a) Semnalul sinusoidal amortizat este descris astfel:
Pentru sursele independente forma generala este:
V<name> <+node> <-node> SIN (voff vampl freq td df phase)
I<name> <+node> <-node> SIN (ioff iampl freq td df phase)
I= ioff + iampl*expi-(t-td)*df*sin(2*3.14*freq*(t-td)+3.14*Phase/180) unde td < t < TSTOP
Marime
ioff
iampl
freq
td
df
Phase
Semnificatie
Valoare
predefinita
1/TSTOP
0.0
0.0
0.0
Valoare de offset
Amplitudinea
Frecventa
Timp de intarziere
Factor de amortizare
Faza
Unitate de
masura
A
A
Hz
s
s-1
grade
b) comanda de analiza in regim tranzitoriu este:
.TRAN <pstep> <ftime> [<noprint> [<ceiling>]][SKIPBP]
<pstep> =pasul de timp pentru tiparirea rezultatelor;
<ftime> = timpul final;
<noprint> = timpul de la care se incepe tiparirea (vizualizarea) rezultatelor;
77
<ceiling> =limita superioara a pasului de timp; daca nu se specifica se considera
ceiling=ftime/50 pentru un circuit care contine elemente dinamice sau ceiling=pstep pentru un
circuit rezistiv.
SKIPBP = utilizarea conditiilor initiale impuse (U=<value> pentru condensator
si I=<value> pentru bobina, care sunt specificate in liniile care definesc elementele respective).
c) definirea unui subcircuit
.SUBCKT <name> [<nodes>*] [PARAMS: <par>[=<val>]* ]
..... (descrierea elementelor)
.ENDS
Utilizarea subcircuitului :
X<name> [<nodes>]* <sname> [PARAMS: <<par>=<val>*>]
<name> = numele modelului (definit in linia .SUBCKT)
<sname> = numele elementului modelat cu subcircuitul respectiv;
<nodes> = nodurile prin care subcircuitul se conecteaza cu circuitul
78
LUCRAREA 6
REGIMUL TRANZITORIU AL CIRCUITELOR NELINIARE
1. PROBLEME
1.) Pentru circuitul din figura sa se faca analiza Fourier a tensiunii de iesire pentru E2 =1 mV si
E2 =10 V incazul in care frecventa sursei e2 este f=1 kHz.
Rb = 20 kΩ ;C=100 uF, Rc = 100 Ω; E1 = 5V; Vcc = 9V; e2 = E2 sin (ωt);
TB: β= 200 ; Cje = 200 pF; Cjc = 200 pF; Vje = 0.6V; Vjc = 0.6V;
2.) Sa se simuleze convertorul dc - dc :
L=20mH ; C=1.25uF; R=400Ω ; E=100V;
Comutator: - frecventa de comutatie f=10kHz ;
- durata de inchidere reprezinta 50% din perioada de comutatie;
- rezistente : Rinchis = 1Ω ; Rdeschis = 1e+06Ω ;
D - dioda avand modelul predefinit de PSPICE;
Sa se vizualizeze tensiunea de comanda a comutatorului(pe un interval de 5 perioade de comutatie)
si tensiunea la iesirea convertorului(atat pe intreaga durata a regimului tranzitoriu cat si 3 perioade
din regimul permanent).
Sa se traseze spectrul de frecvente pentru tensiunea de comanda a comutatorului si pentru
tensiunea la iesirea convertorului.
2. INTERPRETAREA REZULTATELOR
79
La problema 1., realizand analiza Fourier a tensiunii de iesire rezulta fig. 1 pentru E2=1.0 mV si
fig. 2 pentru E2=10.0 V.
Din analiza celor doua figuri rezulta ca tensiunea de iesire in cazul E2=10.0 V, care are forma de
unda distorsionata, contine o gama larga de armonici superioare.
5.0V
2.5V
0V
0Hz
5KHz
V(5)- V(4)
10KHz
15KHz
20KHz
25KHz
30KHz
Frequency
Fig. 1
1.0V
0.5V
0V
0Hz
10KHz
20KHz
30KHz
V(5,4)
Frequency
Fig. 2
La problema 2. chestiunile care se vor urmari sunt:
- modul de utilizare a comutatorului;
- descrierea semnalelor de tip puls pentru surse in regim tranzitoriu;
-trasarea spectrului de frecvente al unei marimi periodice
80
40KHz
50KHz
1100mV
800mV
400mV
0V
4.5ms
V(4)
4.6ms
4.8ms
5.0ms
Time
Fig.3
In fig. 3 se prezinta semnalul generat cu functia pulse, iar in fig. 4 se prezinta semnalul la iesirea
convertorului. Spectrele tensiunii de comanda si pentru tensiunea de iesire sunt prezentate in fig. 5
- 7. Figura 7 contine un detaliu al spectrului tensiunii de iesire care permite evidentierea
armonicelor frecventei de comanda.
300V
200V
100V
0V
0s
2.0ms
4.0ms
V(3)
Time
Fig.4
81
5.0ms
800mV
400mV
0V
0Hz
100KHz
200KHz
300KHz
400KHz
V(4)
Frequency
Fig. 5
Se observa ca spectrul tensiunii de comanda este discret (contine numai componentele armonice
ale frecventei de comutatie si componenta de current continuu) deoarece aceasta marime este
periodica. Tensiunea la iesirea convertorului nefiind periodica (deoarece contine si componente
tranzitorii care nu s-au amortizat) spectrul acesteia nu este discret.
200V
100V
0V
0Hz
10KHz
20KHz
30KHz
40KHz
V(3)
Frequency
Fig.6
Pentru trasarea spectrului tensiunii de iesire s-au considerat numai ultimile 5 perioade din raspuns
(o forma de unda aproapiata de o functie periodica).
82
1.0V
0.5V
0V
9KHz
25KHz
50KHz
75KHz
100KHz
V(3)
Frequency
Fig.7
3. INSTRUCTIUNI SPICE
Se vor utiliza urmatoarele linii:
a) Semnalul de tip puls pentru surse in regim tranzitoriu este descris de:
PULSE(i1 i2 td trise tfall pw per)
pentru sursa de curent
PULSE(v1 v2 td trise tfall pw per) pentru sursa de tensiune
Marime
i1
i2
td
trise
tfall
pw
per
Semnificatie
Valoare
predefinita
0.0
TSTEP
TSTEP
TSTOP
TSTOP
Valoare initiala
Valoare de varf
Timp de intarziere
Timp de crestere
Timp de descrestere
Durata impulsului
Perioada
83
Unitate de
masura
A
A
s
s
s
s
s
b) comutatorul comandat in tensiune
S<name> <+node> <-node> <+control> <-control> <model>
<+node> <-node> = nodurile intre care se conecteaza;
<+control> <-control> = nodurile tensiunii de comanda;
<model> = numele modelului;
.MODEL <model> VSWITCH ( RON=... , ROFF=... , VON= ... , VOFF= ... )
Valori predefinite :
RON=1Ω ; ROFF=1MΩ (rezistenta echivalenta a comutatorului in pozitiile inchis
si deschis);
VON=1V; VOFF=0V (valorile tensiunii de comanda la care comutatorul se
inchide si se deschide) ;
d) Analiza Fourier a unui semnal este descrisa de linia:
.FOUR <freq> <output var>*
<freq> = frecventa considerata fundamentala ;
<output var> = semnalele pentru care se face analiza Fourier;
84
LUCRAREA 7
COMPORTAREA CALITATIVA A CIRCUITELOR DE ORDINUL II
1. PROBLEME
1.) Sa se simuleze circuitul cu conditiile initiale si valorile din tabelul 1 :
L=1H ; C=1F;
Caz
1
2
3
4
5
R [Ω]
3
-3
1
-1
1e-9
Conditii initiale (uc(0),il(0))
(1,1) (-1,1) (1,-1) (-1,-1)
(1,1) (-1,1) (1,-1) (-1,-1)
(1,1) (-1,1) (1,-1) (-1,-1)
(1,1) (-1,1) (1,-1) (-1,-1)
(1,1) (-1,1) (1,-1) (-1,-1)
Tabelul 1
Sa se indice in cazurile 1 .. 5 tipul comportarii calitative a circuitului in jurul punctului de
echilibru.
2. a) Sa se simuleze circuitul din figura pentru i (0) = 6A. Sa se vizualizeze i(t) si curba i = f(V(1))
si sa se explice rezultatul obtinut.
b) Intre nodurile 1 si 0 se conecteaza un condensator cu C= 1mF avand conditia initiala Uc0= 8V.
Sa se vizualizeze traiectoria in planul fazelor (iL – uC ). Sa se compare rezultatele obtinute pentru
eroarea relative RELTOL cu valoarea predefinita( 1e-05) si valoarea impusa de 1e-07.
2. INTERPRETAREA REZULTATELOR
85
1. Circuitul din problema 1. este un circuit liniar, dinamic, de ordinul doi, cu excitatie nula, cu
ecuatiile:
uC + Ri + U L = 0
iC = iL = i
Tinand cont de ecuatiile de functionare ale bobinei
si condensatorului
U L = Li L
i C = CU C
rezulta ecuatiile de stare ale circuitului:
UC
iL
0
=
−
1
L
1 C UC R − iL
L
Pentru a studia comportarea calitativa a unui circuit liniar cu excitatia nula se determina valorile
proprii s1 si s2 ale matricei A a ecuatiei x = Ax .
Valorile proprii s1 si s2 sunt solutiile
ecuatiei: 1
0−s
C
det =0
1
R
−
− −s
L
L
respectiv
s2 +
R
1
s+
=0
L
L ⋅C
care pentru L=1H si C=1F devine
s 2 + Rs + 1 = 0
Punctele de echilibru se determina rezolvand
sistemul
de
ecuatii Ax=0 adica
In cazul problemei ∆ = !
0
1
−1 − R
a11
a 21
a12 x1Q
= 0
a 22 x 2 Q
"
#$
si ∆ = 1 deci ∆ ≠ 0 si sistemul admite numai solutia banala si
originea este singurul punct de echilibru adica x1Q = x 2Q = 0 .
Evolutia circuitului plecand de la o stare initiala(uC(0) si iL(0)) poate fi reprezentata printr-o curba
numita traiectorie de faza in planul de coordonate uC si iL numit planul fazelor.
Evolutia circuitului corespunzatoare mai multor stari initiale poate fi reprezentataprintr-o
multime de traiectorii in planul fazelor. Aceste traiectorii formeazaun portret de faza.
Considerand perechile de valori U C 0 si I l 0 din tabelul 1 se determinatraiectoriile
corespunzatoare starilor initiale date, care formeaza portrete de faza in cele 4 cazuri considerate. In
functie de valorile rezistentei R din tabelul 1 rezulta 4 cazuri.
Cazul1. R = 3 Ω si valorile proprii sunt:
s1 =
−3 + 5
2
s2 =
deci s2 < s1 < 0 si originea este un nod stabil.
86
−3 − 5
2
Cazul 2. R = −3 Ω si valorile proprii sunt:
s1 =
3+ 5
2
s2 =
3− 5
2
deci s2 > s1 > 0 si originea este un nod instabil.
Cazul 3. R = 1 Ω si valorile proprii sunt:
s1 = −
1
3
+j
2
2
s2 = −
1
3
−j
2
2
1
2
deci constanta de atenuare α = − si pentru α < 0 , punctul de echilibru este un focar stabil si
corespunde raspunsului periodic amortizat.
Cazul 4. R = −1 Ω si valorile proprii sunt:
s1 =
deci constanta de atenuare α =
1
2
1
3
+j
2
2
s2 =
1
3
−j
2
2
si in acest caz α > 0 , punctul de echilibru este un focar instabil.
87
Solutia x(t) a ecuatiei de stare x = Ax , are doua componente: uC(t) si iL(t).
Cazul 5 R 0 si valorile proprii sunt s1, 2 = ± j si originea este centru
2. Pentru circuitul dinamic neliniar de la problema 2. punctul a), parcursul dinamic este:
unde punctele Q1(1,-1) si Q2(-1,1) sunt puncte de impas iar originea 0 este un nod instabil.
Simuland circuitul cu PSPICE se constata ca traiectoria evolueaza pe curba plecand din
punctul corespunzator conditiei initiale iL(0) si se opreste in punctul de impas intalnit Q1(1,-1).
La punctul b) al problemei 2. se constata ca leg nd la bornele circuitului un condensator C,
circuitul se comporta ca un oscilator de relaxare.
In cazul circuitului real nu este necesara introducerea in circuit a condensatorului C, rolul
acestuia fiind indeplinit de capacitatea parazita dintre bornele circuitului.
Cel de-al doilea punct al problemei 2. are ca scop ilustrarea preciziei analizei efectuata cu
PSPICE in functie de diferite valori date pentru eroarea relativa RELTOL.
88
RELTOL=1e-05
RELTOL=1e-07
3. INSTRUCTIUNI SPICE
.OPTIONS [<fopt>*] [<vopt>=<value>*]
- setarea unor optiuni:
Flag Options
ACCT summary & accounting
EXPAND show subcircuit expansion
LIBRARY list lines from library files
LIST output summary
NODE output netlist
NOECHO suppress listing
NOMOD suppress model parameter listing
NOPAGE suppress banners
OPTS output option values
Value Options
ABSTOL best accuracy of currents
CHGTOL best accuracy of charges
CPTIME CPU time allowed
DEFAD MOSFET default AD
DEFAS MOSFET default AS
DEFL MOSFET default L
89
DEFW MOSFET default W
GMIN minimal conductance, any branch
ITL1 iteration number - DC & bias point blind limit
ITL2 iteration number - DC & bias point guess limit
ITL4 iteration number - transient per-point limit
ITL5 iteration number - transient total, all points
LIMPTS maximal number of points for print/plot
NUMDGT number of digits in output file
PIVREL relative magnitude for matrix pivot
PIVTOL absolute magnitude for matrix pivot
RELTOL relative accuracy of voltages and currents
TNOM default temperature
TRTOL transient accuracy adjustment
VNTOL best accuracy of voltages
WIDTH printing width in output file
90
LUCRAREA 8
PROPRIETATI CALITATIVE IN CIRCUITELE ELECTRICE
NELINIARE IN REGIM TRANZITORIU
1. PROBLEME
Sa se simuleze circuitul cu urmatorii parametri:
- dioda D : IS=8.3e-15A; RS=9.6Ω; TT=4e-06s (timp de tranzit) ; CJO=300pF ; M=0.4
(gradient); VJ=0.75V (potential de jonctiune); R = 15Ω ; L = 10mH; e(t)= E0sin (ωt) ;
f=100kHz.
Se vor utiliza un pas de afisare de T/2000=5.e-09 s si o valoare RELTOL=1e-6.
Sa se reprezinte grafic dependenta i(t) = f ( e(t)) si sa se interpreteze rezultatele in urmatoarele
situatii:
a) E0 = 0.5 V . Se vor simula 40 perioade , dintre care se vor afisa ultimele 4.
b) E0 = 1.5 V . Se vor simula 40 perioade, dintre care se vor afisa ultimele 4.
c) E0 = 6.5 V . Se vor simula 60 de perioade, dintre care se vor afisa ultimele 20.
2. INTERPRETAREA REZULTATELOR
In urma simularii se determina raspunsul I(L1) pentru cele trei cazuri:
a) pentru E0=0,5 V rezulta un raspuns periodic de frecventa excitatiei.
91
400uA
400uA
0A
0A
-400uA
360us
I(l1)
370us
380us
390us
400us
-400uA
-500mV
I(l1)
Time
0V
500mV
V(1)
Fig. 1. E0=0,5 V
Comportarea circuitului in acest caz este o comportare obisnuita deoarece pentru amplitudini
relativ mici ale excitatiei comportarea diodei poate fi aproximata prin modelul de semnal mic.
b) pentru E0=1,5 V se observa raspunsul subarmonic(periodic de frecventa ω /2).
1.0mA
1.0mA
0.5mA
0.5mA
0A
0A
-0.5mA
360us
I(l1)
370us
380us
390us
-0.5mA
-2.0V
I(l1)
400us
Time
-1.0V
0V
1.0V
2.0V
V(1)
Fig. 2. E0=1,5 V
c) pentru E0=6,5 V se observa comportarea haotica a circuitului(raspuns neperiodic la excitatie
periodica).
92
4.0mA
4.0mA
2.0mA
2.0mA
0A
0A
-2.0mA
400us
I(l1)
450us
500us
550us
-2.0mA
-8.0V
I(l1)
600us
-4.0V
0V
V(1)
Time
Fig. 3. E0=6,5 V
93
4.0V
8.0V
CAP 4. SOLUTIILE PROBLEMELOR
LUCRAREA 1
Problema 1
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10/5.04
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DGEO
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jY7BbL9 & k
LUCRARAEA 2
Problema 1
*Circ cu dioda Zener cdz1.cir
R1 1 2 1k
R2 2 0 2k
D 0 2 Dioda_Zener
*Modelul diodei
*Bv=4 tens de strapungere inversa
.MODEL Dioda_Zener D(Is=10f Rs=0
+
Bv=4 Ibv=10m Cjo=0)
*Tensiunea de polarizare
V1 1 0 1
*Analiza in curent continuu
.DC LIN V1 -10 15 .1
.PROBE
.END
****
Diode MODEL PARAMETERS
Dioda_Zener
IS 10.000000E-15
BV 4
IBV .01
97
' 2 222
,
-%
;V & CGDJ%
4.0V
2.0V
0V
-2.0V
-10V
V(2)
-5V
0V
5V
10V
15V
V(1)
Problema 2
a. pct
*circuit cu rezistenta neliniara cnl1.cir
R1 1 2 3
V1 1 0 10
G1 2 0 TABLE {V(2)}=
+(0,0), (2,5), (4,0.5), (8,9)
.NODESET V(2)=0.0
.OP
.END
****
SMALL SIGNAL BIAS SOLUTION
NODE VOLTAGE
NODE VOLTAGE
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE
( 1) 10.0000 ( 2) 1.1765
VOLTAGE SOURCE CURRENTS
NAME
V1
CURRENT
-2.941E+00
TOTAL POWER DISSIPATION 2.94E+01 WATTS
**** VOLTAGE-CONTROLLED CURRENT SOURCES
NAME
I-SOURCE
G1
2.941E+00
98
NODE VOLTAGE
b. pct
*circuit cu rezistenta neliniara cnl1.cir
R1 1 2 3
V1 1 0 10
G1 2 0 TABLE {V(2)}=
+(0,0), (2,5), (4,0.5), (8,9)
.NODESET V(2)=3.0
.OP
.END
****
SMALL SIGNAL BIAS SOLUTION
NODE VOLTAGE
NODE VOLTAGE
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE
NODE VOLTAGE
( 1) 10.0000 ( 2) 3.2174
VOLTAGE SOURCE CURRENTS
NAME
V1
CURRENT
-2.261E+00
TOTAL POWER DISSIPATION 2.26E+01 WATTS
**** VOLTAGE-CONTROLLED CURRENT SOURCES
NAME
G1
I-SOURCE
c.
pct
2.261E+00
*circuit cu rezistenta neliniara cnl1.cir
R1 1 2 3
V1 1 0 10
G1 2 0 TABLE {V(2)}=
+(0,0), (2,5), (4,0.5), (8,9)
.NODESET V(2)=7.0
.OP
.END
****
SMALL SIGNAL BIAS SOLUTION
TEMPERATURE = 27.000 DEG C
99
NODE VOLTAGE
NODE VOLTAGE
NODE VOLTAGE
NODE VOLTAGE
( 1) 10.0000 ( 2) 4.6102
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
-1.797E+00
TOTAL POWER DISSIPATION 1.80E+01 WATTS
**** VOLTAGE-CONTROLLED CURRENT SOURCES
NAME
G1
I-SOURCE 1.797E+00
d. pct
*circuit cu rezistenta neliniara cnl1.cir
R1 1 2 3
V1 1 0 10
G1 2 0 TABLE {V(2)}=
+(0,0), (2,5), (4,0.5), (8,9)
*.NODESET V(2)=7.0
.OP
.END
****
SMALL SIGNAL BIAS SOLUTION
NODE VOLTAGE
NODE VOLTAGE
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE
( 1) 10.0000 ( 2) 1.1765
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
-2.941E+00
TOTAL POWER DISSIPATION 2.94E+01 WATTS
**** VOLTAGE-CONTROLLED CURRENT SOURCES
NAME
G1
I-SOURCE 2.941E+00
100
NODE VOLTAGE
LUCRAREA 3
Punctul a.
**** CIRCUIT DESCRIPTION
*etb31.cir circ cu tranz bipolare
*.options reltol=1e-6 itl5=0 numdgt=8
V1 2 0 DC 5
V2 5 0 DC 9
R1 2 3 1MEG
R2 5 4 4K
Q1 4 3 0 TB
.MODEL TB NPN(Is=10.f Bf=200 Cje=2p
+
Cjc=2p Vje=.6 Vjc=.6)
.DC LIN V1 0 25 .05
*.PRINT DC V(R2)
.PROBE
.END
****
BJT MODEL PARAMETERS
TB
NPN
IS 10.000000E-15
BF 200
NF 1
BR 1
NR 1
CJE 2.000000E-12
VJE .6
CJC 2.000000E-12
VJC .6
CN 2.42
D .87
JOB CONCLUDED
101
10V
5V
0V
0V
5V
10V
15V
v(5)-v(4)
V1
Punctul b
**** CIRCUIT DESCRIPTION
*etb32.cir circ cu tranz bipolare
V1 2 1 DC 5
V10
+ DC 0
+ AC 1 0
V2 5 0 DC 9
R1 2 3 1MEG
R2 5 4 4K
Q1 4 3 0 TB
.MODEL TB NPN(Is=10.f Bf=200 Cje=2p
+
Cjc=2p Vje=.6 Vjc=.6)
*.DC LIN V1 0 25 .05
.AC DEC 100 .1 100MEG
.OP
.PROBE
.END
****
BJT MODEL PARAMETERS
TB
NPN
IS 10.000000E-15
BF 200
NF 1
BR 1
NR 1
CJE 2.000000E-12
VJE .6
CJC 2.000000E-12
VJC .6
CN 2.42
102
20V
25V
D
****
.87
SMALL SIGNAL BIAS SOLUTION
NODE VOLTAGE
NODE VOLTAGE
( 1) 0.0000 ( 2) 5.0000 ( 3)
( 5) 9.0000
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE
.6515 ( 4) 5.5212
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
V
V2
-4.349E-06
-4.349E-06
-8.697E-04
TOTAL POWER DISSIPATION 7.85E-03 WATTS
**** BIPOLAR JUNCTION TRANSISTORS
NAME
Q1
MODEL
TB
IB
4.35E-06
IC
8.70E-04
VBE
6.51E-01
VBC
-4.87E+00
VCE
5.52E+00
BETADC
2.00E+02
GM
3.36E-02
RPI
5.95E+03
RX
0.00E+00
RO
1.00E+12
CBE
3.49E-12
CBC
9.64E-13
CJS
0.00E+00
BETAAC
2.00E+02
CBX/CBX2 0.00E+00
FT/FT2
1.20E+09
103
NODE VOLTAGE
800mV
400mV
0V
100mHz 1.0Hz
V(5)- V(4)
100Hz
10KHz
Frequency
**** CIRCUIT DESCRIPTION
*etb32.cir circ cu tranz bipolare
V1 2 1 DC 5
V10
+ DC 0
+ AC 20 0
V2 5 0 DC 9
R1 2 3 1MEG
R2 5 4 4K
Q1 4 3 0 TB
.MODEL TB NPN(Is=10.f Bf=200 Cje=2p
+
Cjc=2p Vje=.6 Vjc=.6)
*.DC LIN V1 0 25 .05
.AC DEC 100 .1 100MEG
.OP
.PROBE
.END
****
BJT MODEL PARAMETERS
TB
NPN
IS 10.000000E-15
BF 200
NF 1
BR 1
NR 1
CJE 2.000000E-12
VJE .6
CJC 2.000000E-12
VJC .6
CN 2.42
D .87
104
1.0MHz
100MHz
**** SMALL SIGNAL BIAS SOLUTION
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
( 1) 0.0000 ( 2) 5.0000 ( 3)
( 5) 9.0000
.6515 ( 4) 5.5212
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
V
V2
-4.349E-06
-4.349E-06
-8.697E-04
TOTAL POWER DISSIPATION 7.85E-03 WATTS
**** BIPOLAR JUNCTION TRANSISTORS
NAME
Q1
MODEL
TB
IB
4.35E-06
IC
8.70E-04
VBE
6.51E-01
VBC
-4.87E+00
VCE
5.52E+00
BETADC
2.00E+02
GM
3.36E-02
RPI
5.95E+03
RX
0.00E+00
RO
1.00E+12
CBE
3.49E-12
CBC
9.64E-13
CJS
0.00E+00
BETAAC
2.00E+02
CBX/CBX2 0.00E+00
FT/FT2
1.20E+09
105
20V
10V
0V
100mHz 1.0Hz
V(5)- V(4)
100Hz
10KHz
1.0MHz
100MHz
Frequency
LUCRARAEA 4
Problema 1
**** CIRCUIT DESCRIPTION
*circ lin c.a. CLA1.CIR
R1 2 3 10
R2 3 5 1
R3 3 4 2
L1 1 2 50m
L2 6 7 25m
L3 0 4 50m
K1 l1 l3 0.5
K2 l2 l3 0.707
C2 5 6 1.5m
V_SIN 7 0
+DC 0
+AC 30 90
I_SIN 0 1
+DC 0
+AC 2 0
.AC LIN 1 50 50
.PRINT AC IM(R2) IP(R2) IR(R2) II(R2)
.PROBE
.END
**** SMALL SIGNAL BIAS SOLUTION
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
( 1) 0.0000 ( 2) 0.0000 ( 3) 0.0000 ( 4) 0.0000
( 5) 0.0000 ( 6) 0.0000 ( 7) 0.0000
VOLTAGE SOURCE CURRENTS
106
NAME
V_SIN
CURRENT
0.000E+00
TOTAL POWER DISSIPATION 0.00E+00 WATTS
**** AC ANALYSIS
TEMPERATURE = 27.000 DEG C
FREQ
IM(R2) IP(R2) IR(R2) II(R2)
5.000E+01 1.138E-01 -6.592E+01 4.645E-02 -1.039E-01
Problema 2a
**** CIRCUIT DESCRIPTION
*CIRC LIN DE CA FIL1
R1 1 2 1K
L1 2 3 3M
L2 3 4 12M
C1 3 0 8N
C2 4 0 11.5N
V1 1 0
+DC 0
+AC 1 0
+SIN(0 1 50 0 0 0)
.AC LIN 100 1 1MEG
.PROBE
.END
**** SMALL SIGNAL BIAS SOLUTION
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
( 1) 0.0000 ( 2) 0.0000 ( 3) 0.0000 ( 4) 0.0000
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
0.000E+00
TOTAL POWER DISSIPATION 0.00E+00 WATTS
107
1.0V
0.5V
0V
1.0Hz
V(4)
100Hz
10KHz
1.0MHz
Frequency
Problema 2b
**** CIRCUIT DESCRIPTION
*CIRC LIN DE CA FIL2
R1 1 2 1.5K
L1 3 0 2M
L2 4 0 12M
C1 2 3 10N
C2 3 4 8U
V1 1 0
+DC 0
+AC 1 0
+SIN(0 1 50 0 0 0)
.AC LIN 100 1 2MEG
.PROBE
.END
****
SMALL SIGNAL BIAS SOLUTION
NODE VOLTAGE
NODE VOLTAGE
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE
( 1) 0.0000 ( 2) 0.0000 ( 3) 0.0000 ( 4) 0.0000
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
0.000E+00
TOTAL POWER DISSIPATION 0.00E+00 WATTS
108
NODE VOLTAGE
1.0V
0.5V
0V
1.0Hz
V(4)
100Hz
10KHz
1.0MHz
100MHz
Frequency
Problema 2c
**** CIRCUIT DESCRIPTION
*CIRC LIN DE CA FIL3
R1 1 2 1.5K
RS 4 0 1.5K
L1 2 3 2M
C1 2 0 10N
C2 3 4 8U
C3 4 0 1N
V1 1 0
+DC 0
+AC 1 0
+SIN(0 1 50 0 0 0)
.AC DEC 100 1 2MEG
.PROBE
.END
****
SMALL SIGNAL BIAS SOLUTION
NODE VOLTAGE
NODE VOLTAGE
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE
( 1) 0.0000 ( 2) 0.0000 ( 3) 0.0000 ( 4) 0.0000
VOLTAGE SOURCE CURRENTS
NAME
V1
CURRENT
0.000E+00
TOTAL POWER DISSIPATION 0.00E+00 WATTS
109
NODE VOLTAGE
500mV
250mV
0V
1.0Hz
V(4)
100Hz
10KHz
1.0MHz
100MHz
Frequency
Problema 2d
**** CIRCUIT DESCRIPTION
*CIRC LIN DE CA FIL4
R1 1 2 1.5K
L1 5 0 10M
L2 2 3 20M
L3 4 0 4M
C1 2 5 .1U
C2 2 3 40U
C3 3 4 1U
V1 1 0
+DC 0
+AC 1 0
+SIN(0 1 50 0 0 0)
.AC DEC 100 1 2MEG
.PROBE
.END
**** SMALL SIGNAL BIAS SOLUTION
NODE VOLTAGE
NODE VOLTAGE
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE
( 1) 0.0000 ( 2) 0.0000 ( 3) 0.0000 ( 4) 0.0000
( 5) 0.0000
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
0.000E+00
110
NODE VOLTAGE
TOTAL POWER DISSIPATION 0.00E+00 WATTS
1.0V
0.5V
0V
1.0Hz
V(3)
100Hz
10KHz
Frequency
LUCRAREA 5
Problema 1
****
CIRCUIT DESCRIPTION
*est51.cir CIR cu AO/ SUBcirc
V1 1 0
+ DC 0
+ AC 1m 0
+ SIN(0 1m 500 0 0 0)
*
R1 1 2 1
C1 1 2 1
R2 2 3 1
C2 3 6 1
R3 2 4 1
R4 4 5 1
R5 5 6 1
C5 5 6 1
R6 1 7 1
C6 7 5 1
*
X1 [0 2 4] AO
X2 [0 5 6] AO
111
1.0MHz
100MHz
*
.SUBCKT AO 2 1 4
Ri 1 2 1MEG
Re 3 4 .1m
Ce 4 2 .1p
E 3 2 TABLE { V(2)-V(1)} = (-.0015,-15)(0,0)(.0015,15)
.ENDS
*
.AC DEC 100 .1 10MEG
*.tran 1n 4.m
.probe
.options reltol=1e-6 itl5=0 numdgt=8
.end
****
SMALL SIGNAL BIAS SOLUTION
NODE VOLTAGE
NODE VOLTAGE
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE
NODE VOLTAGE
( 1) 0.00000000 ( 2) 0.00000000 ( 3) 0.00000000 ( 4) 0.00000000
( 5) 0.00000000 ( 6) 0.00000000 ( 7) 0.00000000 ( X1.3) 0.00000000
( X2.3) 0.00000000
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
0.000E+00
TOTAL POWER DISSIPATION 0.00E+00 WATTS
1.0mV
0.5mV
0V
100mHz
1.0Hz
V(6)
100Hz
10KHz
Frequency
112
1.0MHz
1.0
0.5
0
100mHz
1.0Hz
V(6)/ V(1)
100Hz
10KHz
1.0MHz
Frequency
*est51.cir CIR cu AO/ SUBcirc
V1 1 0
+ DC 0
+ AC 1m 0
+ SIN(0 1m 500 0 0 0)
*
R1 1 2 1
C1 1 2 1
R2 2 3 1
C2 3 6 1
R3 2 4 1
R4 4 5 1
R5 5 6 1
C5 5 6 1
R6 1 7 1
C6 7 5 1
*
X1 [0 2 4] AO
X2 [0 5 6] AO
*
.SUBCKT AO 2 1 4
Ri 1 2 1MEG
Re 3 4 .1m
Ce 4 2 .1p
E 3 2 TABLE { V(2)-V(1)} = (-.0015,-15)(0,0)(.0015,15)
.ENDS
*
*.AC DEC 100 .1 10MEG
.tran 1n 4.m
.probe
.options reltol=1e-6 itl5=0 numdgt=8
.end
**** INITIAL TRANSIENT SOLUTION
TEMPERATURE = 27.000 DEG C
113
NODE VOLTAGE
NODE VOLTAGE
NODE VOLTAGE
NODE VOLTAGE
( 1) 0.00000000 ( 2) 0.00000000 ( 3) 0.00000000 ( 4) 0.00000000
( 5) 0.00000000 ( 6) 0.00000000 ( 7) 0.00000000 ( X1.3) 0.00000000
( X2.3) 0.00000000
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
0.000E+00
TOTAL POWER DISSIPATION 0.00E+00 WATTS
1.0mV
0V
-1.0mV
0s
1.0ms
2.0ms
V(6)
Time
****
CIRCUIT DESCRIPTION
*est51.cir CIR cu AO/ SUBcirc
V1 1 0
+ DC 0
+ AC 1m 0
+ SIN(0 10 500 0 0 0)
*
R1 1 2 1
C1 1 2 1
R2 2 3 1
C2 3 6 1
R3 2 4 1
R4 4 5 1
R5 5 6 1
C5 5 6 1
R6 1 7 1
C6 7 5 1
*
X1 [0 2 4] AO
114
3.0ms
4.0ms
X2 [0 5 6] AO
*
.SUBCKT AO 2 1 4
Ri 1 2 1MEG
Re 3 4 .1m
Ce 4 2 .1p
E 3 2 TABLE { V(2)-V(1)} = (-.0015,-15)(0,0)(.0015,15)
.ENDS
*
*.AC DEC 100 .1 10MEG
.tran 1n 4.m
.probe
.options reltol=1e-6 itl5=0 numdgt=8
.end
****
INITIAL TRANSIENT SOLUTION
NODE VOLTAGE
NODE VOLTAGE
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE
NODE VOLTAGE
( 1) 0.00000000 ( 2) 0.00000000 ( 3) 0.00000000 ( 4) 0.00000000
( 5) 0.00000000 ( 6) 0.00000000 ( 7) 0.00000000 ( X1.3) 0.00000000
( X2.3) 0.00000000
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
0.000E+00
TOTAL POWER DISSIPATION 0.00E+00 WATTS
10mV
0V
-10mV
0s
1.0ms
2.0ms
V(6)
Time
Problema 2
****
CIRCUIT DESCRIPTION
115
3.0ms
4.0ms
*etb52.cir circ cu tranz bipolare
.options reltol=1e-6
V1 1 0 DC 5
C3 2 3 100u
*V3 3 0 AC 1m SIN(0 1m 1K 0 0)
V3 3 0 AC 10 SIN(0 10 1K 0 0)
V2 5 0 DC 9
R1 1 2 20K
R2 5 4 100
Q1 4 2 0 TB
.MODEL TB NPN(IS=10.F BF=200 CJE=200p
+
CJC=200p)
*.DC LIN V1 0 25 .05
*.AC DEC 100 .1 100MEG
.TRAN .1u 10m
.OP
.PROBE
.END
****
BJT MODEL PARAMETERS
TB
NPN
IS 10.000000E-15
BF 200
NF 1
BR 1
NR 1
CJE 200.000000E-12
CJC 200.000000E-12
CN 2.42
D .87
****
SMALL SIGNAL BIAS SOLUTION
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE
NODE VOLTAGE
NODE VOLTAGE
( 1) 5.0000 ( 2)
( 5) 9.0000
.7521 ( 3) 0.0000 ( 4) 4.7521
VOLTAGE SOURCE CURRENTS
NAME
V1
CURRENT
-2.124E-04
116
NODE VOLTAGE
V3
V2
0.000E+00
-4.248E-02
TOTAL POWER DISSIPATION 3.83E-01 WATTS
****
OPERATING POINT INFORMATION
TEMPERATURE = 27.000 DEG C
**** BIPOLAR JUNCTION TRANSISTORS
NAME
Q1
MODEL
TB
IB
2.12E-04
IC
4.25E-02
VBE
7.52E-01
VBC
-4.00E+00
VCE
4.75E+00
BETADC
2.00E+02
GM
1.64E+00
RPI
1.22E+02
RX
0.00E+00
RO
1.00E+12
CBE
3.35E-10
CBC
1.09E-10
CJS
0.00E+00
BETAAC
2.00E+02
CBX/CBX2 0.00E+00
FT/FT2
5.89E+08
****
INITIAL TRANSIENT SOLUTION
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE
NODE VOLTAGE
NODE VOLTAGE
( 1) 5.0000 ( 2)
.7521 ( 3) 0.0000 ( 4) 4.7521
( 5) 9.0000
VOLTAGE SOURCE CURRENTS
NAME
V1
V3
V2
CURRENT
-2.124E-04
0.000E+00
-4.248E-02
TOTAL POWER DISSIPATION 3.83E-01 WATTS
117
NODE VOLTAGE
4.50V
4.25V
4.00V
0s
2ms
V(5)- V(4)
4ms
6ms
8ms
10ms
Time
10V
0V
-10V
0s
2ms
V(5)- V(4)
4ms
6ms
Time
LUCRAREA 6
Problema 1
**** CIRCUIT DESCRIPTION
*etb61.cir circ cu tranz bipolare
.options reltol=1e-6
V1 1 0 DC 5
118
8ms
10ms
C3 2 3 100u
V3 3 0 AC 1m SIN(0 1m 1K 0 0)
*V3 3 0 AC 10 SIN(0 10 1K 0 0)
V2 5 0 DC 9
R1 1 2 20K
R2 5 4 100
Q1 4 2 0 TB
.MODEL TB NPN(IS=10.F BF=200 CJE=200p
+
CJC=200p)
*.TRAN .1u 10m
.TRAN .1u 10m 5m
.FOUR 1K 10 V(5,4)
*.OP
.PROBE
.END
****
BJT MODEL PARAMETERS
TB
NPN
IS 10.000000E-15
BF 200
NF 1
BR 1
NR 1
CJE 200.000000E-12
CJC 200.000000E-12
CN 2.42
D .87
****
INITIAL TRANSIENT SOLUTION
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE
NODE VOLTAGE
NODE VOLTAGE
( 1) 5.0000 ( 2)
( 5) 9.0000
.7521 ( 3) 0.0000 ( 4) 4.7521
NODE VOLTAGE
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
V3
V2
-2.124E-04
0.000E+00
-4.248E-02
TOTAL POWER DISSIPATION 3.83E-01 WATTS
****
FOURIER ANALYSIS
TEMPERATURE = 27.000 DEG C
119
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(5,4)
DC COMPONENT = 4.247607E+00
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE
NORMALIZED
NO
(HZ) COMPONENT COMPONENT (DEG)
PHASE (DEG)
1
2
3
4
5
6
7
8
9
10
1.000E+03
2.000E+03
3.000E+03
4.000E+03
5.000E+03
6.000E+03
7.000E+03
8.000E+03
9.000E+03
1.000E+04
1.625E-01
1.589E-03
5.077E-04
1.470E-04
3.713E-04
1.359E-04
2.854E-04
1.450E-04
2.411E-04
1.767E-04
1.000E+00
9.777E-03
3.125E-03
9.045E-04
2.285E-03
8.364E-04
1.756E-03
8.922E-04
1.484E-03
1.088E-03
7.960E-01
-8.396E+01
-5.091E+01
1.237E+02
7.320E+01
-1.204E+02
-1.695E+02
-4.044E+00
-4.427E+01
1.118E+02
0.000E+00
-8.476E+01
-5.171E+01
1.229E+02
7.240E+01
-1.212E+02
-1.703E+02
-4.840E+00
-4.506E+01
1.110E+02
TOTAL HARMONIC DISTORTION = 1.092516E+00 PERCENT
5.0V
2.5V
0V
0Hz
5KHz
V(5)- V(4)
10KHz
15KHz
Frequency
120
20KHz
25KHz
30KHz
1.0V
0.5V
0V
0Hz
2KHz
V(5)- V(4)
4KHz
6KHz
Frequency
****
CIRCUIT DESCRIPTION
*etb61.cir circ cu tranz bipolare
.options reltol=1e-6
V1 1 0 DC 5
C3 2 3 100u
*V3 3 0 AC 1m SIN(0 1m 1K 0 0)
V3 3 0 AC 10 SIN(0 10 1K 0 0)
V2 5 0 DC 9
R1 1 2 20K
R2 5 4 100
Q1 4 2 0 TB
.MODEL TB NPN(IS=10.F BF=200 CJE=200p
+
CJC=200p)
*.TRAN .1u 10m
.TRAN .1u 10m 5m
.FOUR 1K 10 V(5,4)
*.OP
.PROBE
.END
****
BJT MODEL PARAMETERS
TB
NPN
IS 10.000000E-15
BF 200
NF 1
BR 1
NR 1
CJE 200.000000E-12
CJC 200.000000E-12
121
8KHz
10KHz
CN 2.42
D .87
****
INITIAL TRANSIENT SOLUTION
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE
NODE VOLTAGE
NODE VOLTAGE
( 1) 5.0000 ( 2)
( 5) 9.0000
.7521 ( 3) 0.0000 ( 4) 4.7521
NODE VOLTAGE
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
V3
V2
-2.124E-04
0.000E+00
-4.248E-02
TOTAL POWER DISSIPATION 3.83E-01 WATTS
****
FOURIER ANALYSIS
TEMPERATURE = 27.000 DEG C
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(5,4)
DC COMPONENT = 2.677038E-01
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE
NORMALIZED
NO
(HZ) COMPONENT COMPONENT (DEG)
PHASE (DEG)
1
2
3
4
5
6
7
8
9
10
1.000E+03
2.000E+03
3.000E+03
4.000E+03
5.000E+03
6.000E+03
7.000E+03
8.000E+03
9.000E+03
1.000E+04
5.345E-01
5.316E-01
5.269E-01
5.203E-01
5.120E-01
5.019E-01
4.903E-01
4.771E-01
4.625E-01
4.465E-01
1.000E+00
9.947E-01
9.858E-01
9.736E-01
9.580E-01
9.392E-01
9.173E-01
8.927E-01
8.653E-01
8.355E-01
1.600E-01
-8.959E+01
-1.794E+02
9.083E+01
1.041E+00
-8.875E+01
-1.785E+02
9.168E+01
1.897E+00
-8.788E+01
0.000E+00
-8.975E+01
-1.795E+02
9.067E+01
8.813E-01
-8.891E+01
-1.787E+02
9.152E+01
1.737E+00
-8.804E+01
TOTAL HARMONIC DISTORTION = 2.791763E+02 PERCENT
122
1.0V
0.5V
0V
0Hz
10KHz
20KHz
30KHz
40KHz
50KHz
8KHz
10KHz
V(5,4)
Frequency
1.0V
0.5V
0V
0Hz
2KHz
4KHz
6KHz
V(5,4)
Frequency
Problema 2
**** CIRCUIT DESCRIPTION
*etR61.cir CIR cu SWITCH
options reltol=5.e-7 itl5=0 numdgt=8
V1 1 0
+ DC 100
*
V4 4 0 PULSE (0 1 50.u 1.p 1.p .05m .1m )
*
L1 1 2 20m
C1 3 0 1.25u
R1 3 0 400
*
d1 2 3 dioda
.model dioda d
123
*
S1 2 0 4 0 S1
*
.MODEL S1 VSWITCH ( RON=1, ROFF=1MEG,
+ VON=1, VOFF=0 )
*
.TRAN 10u 5m
*.TRAN 10u 5m 4.5m
*.FOUR 10.K V(3),V(4)
.PROBE
.END
****
Diode MODEL PARAMETERS
dioda
IS 10.000000E-15
****
Voltage Controlled Switch MODEL PARAMETERS
S1
RON
ROFF
VON
VOFF
****
1
1.000000E+06
1
0
INITIAL TRANSIENT SOLUTION
NODE VOLTAGE
NODE VOLTAGE
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE
( 1) 100.00000000
( 2) 100.00000000
( 3) 99.20229970
( 4) 0.00000000
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
V4
-2.481E-01
0.000E+00
TOTAL POWER DISSIPATION 2.48E+01 WATTS
124
NODE VOLTAGE
1100mV
800mV
400mV
0V
4.5ms
V(4)
4.6ms
4.8ms
5.0ms
Time
300V
200V
100V
0V
0s
2.0ms
4.0ms
5.0ms
V(3)
Time
800mV
400mV
0V
0Hz
100KHz
200KHz
V(4)
Frequency
125
300KHz
400KHz
200V
100V
0V
0Hz
10KHz
20KHz
30KHz
40KHz
V(3)
Frequency
1.0V
0.5V
0V
9KHz
25KHz
50KHz
V(3)
Frequency
126
75KHz
100KHz
LUCRAREA 7
Problema 1
****
CIRCUIT DESCRIPTION
*circ in reg tranz cetr1.cir
.options reltol=1e-6 itl5=0 numdgt=8 NOMOD NOPAGE NOECHO
R1 0 1 3
L1 1 2 1 IC=-.10E+01
C1 2 0 1 IC=.10E+01
.TRAN 10M 15 SKIPBP
.PRINT TRAN V(2),I(L1)
.PROBE
.END
****
TIME
TRANSIENT ANALYSIS
V(2)
TEMPERATURE = 27.000 DEG C
I(L1)
1.0000000E-02 9.9011134E-01 -9.8027822E-01
2.0000000E-02 9.8041165E-01 -9.6102829E-01
3.0000000E-02 9.7091674E-01 -9.4228926E-01
4.0000000E-02 9.6166173E-01 -9.2414761E-01
5.0000000E-02 9.5251726E-01 -9.0628115E-01
6.0000000E-02 9.4349691E-01 -8.8872298E-01
7.0000000E-02 9.3484698E-01 -8.7208478E-01
8.0000000E-02 9.2619704E-01 -8.5544659E-01
9.0000000E-02 9.1754711E-01 -8.3880839E-01
1.0000000E-01 9.0942287E-01 -8.2347094E-01
…………………………………………………………….
1.4960000E+01 2.3808301E-03 -9.0939618E-04
1.4970000E+01 2.3716028E-03 -9.0587170E-04
1.4980000E+01 2.3623756E-03 -9.0234722E-04
1.4990000E+01 2.3531484E-03 -8.9882274E-04
1.5000000E+01 2.3439212E-03 -8.9529826E-04
127
1.0
0
-1.0
0s
5s
I(L1)
10s
15s
V(2)
Time
1.0A
0A
-1.0A
0V
0.2V
0.4V
0.6V
I(L1)
V(2)
****
CIRCUIT DESCRIPTION
*circ in reg tranz cetr1.cir
.options reltol=1e-6 itl5=0 numdgt=8 NOMOD NOPAGE NOECHO
R1 0 1 3
L1 1 2 1 IC=.10E+01
C1 2 0 1 IC=.10E+01
.TRAN 10M 15 SKIPBP
.PROBE
.END
128
0.8V
1.0V
1.0A
0A
-1.0A
0s
5s
I(L1)
10s
15s
1.0V
1.5V
V(2)
Time
1.0A
0A
-1.0A
0V
0.5V
I(L1)
V(2)
****
CIRCUIT DESCRIPTION
*circ in reg tranz cetr1.cir
.options reltol=1e-6 itl5=0 numdgt=8 NOMOD NOPAGE NOECHO
R1 0 1 3
L1 1 2 1 IC=.10E+01
C1 2 0 1 IC=-.10E+01
129
.TRAN 10M 15 SKIPBP
.PROBE
.END
1.0A
0A
-1.0A
0s
5s
I(L1)
10s
15s
V(2)
Time
1.0A
0A
-1.0A
-1.0V
-0.8V
I(L1)
-0.6V
-0.4V
V(2)
****
CIRCUIT DESCRIPTION
*circ in reg tranz cetr1.cir
.options reltol=1e-6 itl5=0 numdgt=8 NOMOD NOPAGE NOECHO
R1 0 1 -3
L1 1 2 1 IC=.10E+01
C1 2 0 1 IC=.10E+01
130
-0.2V
0V
.TRAN 1M 2 SKIPBP
.PROBE
.END
5.0A
2.5A
0A
0s
I(L1)
0.2s
V(2)
0.4s
0.6s
0.8s
1.0s
Time
5.0A
2.5A
0A
0V
0.5V
1.0V
1.5V
I(L1)
V(2)
131
2.0V
2.5V
3.0V
20KA
15KA
10KA
5KA
0A
0V
2.0KV
4.0KV
6.0KV
8.0KV
I(L1)
V(2)
****
CIRCUIT DESCRIPTION
*circ in reg tranz cetr1.cir
.options reltol=1e-6 itl5=0 numdgt=8 NOMOD NOPAGE NOECHO
R1 0 1 3
L1 1 2 1 IC=.10E+01
C1 2 0 -1 IC=.10E+01
.TRAN 1M 15 SKIPBP
.PROBE
.END
1.0A
0A
-1.0A
0s
I(L1)
0.5s
V(2)
1.0s
Time
132
1.5s
2.0s
1.0A
0A
-1.0A
0V
0.5V
1.0V
1.5V
2.0V
I(L1)
V(2)
****
CIRCUIT DESCRIPTION
*circ in reg tranz cetr1.cir
.options reltol=1e-6 itl5=0 numdgt=8 NOMOD NOPAGE NOECHO
R1 0 1 1
L1 1 2 1 IC=.10E+01
C1 2 0 1 IC=.10E+01
.TRAN 1M 15 SKIPBP
.PROBE
.END
2.0A
0A
-2.0A
0s
I(L1)
0.2s
V(2)
0.4s
0.6s
Time
133
0.8s
1.0s
1.0A
0A
-1.0A
-0.5V
I(L1)
0V
0.5V
1.0V
1.5V
V(2)
****
CIRCUIT DESCRIPTION
*circ in reg tranz cetr1.cir
.options reltol=1e-6 itl5=0 numdgt=8 NOMOD NOPAGE NOECHO
R1 0 1 -1
L1 1 2 1 IC=-.10E+01
C1 2 0 1 IC=-.10E+01
.TRAN .1M 15 SKIPBP
.PROBE
.END
200A
100A
0A
-100A
-200A
0s
I(L1)
2s
V(2)
4s
6s
Time
134
8s
10s
5.0A
2.5A
0A
-2.0V
I(L1)
-1.0V
0V
1.0V
2.0V
V(2)
150A
100A
50A
0A
-50A
-75V
-50V
0V
I(L1)
V(2)
**** CIRCUIT DESCRIPTION
*circ in reg tranz cetr1.cir
.options reltol=1e-6 itl5=0 numdgt=8 NOMOD NOPAGE NOECHO
R1 0 1 1n
L1 1 2 1 IC=.10E+01
C1 2 0 1 IC=.10E+01
.TRAN .1M 15 SKIPBP
.PROBE
135
50V
.END
2.0A
0A
-2.0A
0s
I(L1)
2s
V(2)
4s
6s
8s
10s
Time
2.0A
0A
-2.0A
-1.5V
-1.0V
I(L1)
-0.5V
0V
V(2)
Problema 2
**** CIRCUIT DESCRIPTION
*circ nelin de ord 2 cnd2.cir
L1 0 1 1 IC=6
G1 1 0 TABLE {V(1)}=
+ (-10,-8),(-1,1),(1,-1),(10,8)
.TRAN .1M 7 SKIPBP
.PROBE
.END
136
0.5V
1.0V
1.5V
2.0A
0A
-2.0A
-4.0V
I(L1)
-2.0V
0V
2.0V
4.0V
6.0V
2.0V
4.0V
6.0V
V(1)
*circ nelin de ord 2 cnd2.cir
L1 0 1 1 IC=6
G1 1 0 TABLE {V(1)}=
+ (-10,-8),(-1,1),(1,-1),(10,8)
C1 1 0 1M IC=8
.TRAN .1M 7 SKIPBP
.PROBE
.END
2.0A
0A
-2.0A
-4.0V
I(L1)
-2.0V
0V
V1(C1)
137
*circ nelin de ord 2 cnd2.cir
L1 0 1 1 IC=6
G1 1 0 TABLE {V(1)}=
+ (-10,-8),(-1,1),(1,-1),(10,8)
C1 1 0 1M IC=8
.TRAN .1M 7 SKIPBP
.PROBE
.OPTIONS RELTOL=1e-7
.END
2.0A
0A
-2.0A
-4.0V
I(L1)
-2.0V
0V
2.0V
V1(C1)
138
4.0V
6.0V
LUCRAREA 8
Problema 1a
****
CIRCUIT DESCRIPTION
*el81.cir circ el nelin reg tran
.options reltol=1e-6 itl5=0 numdgt=8
r1 1 2 15
l1 2 3 10m
V1 1 0 sin(0 0.5 1.e+5 0 0 0)
d 3 0 dioda
.model dioda d(Is=8.3e-15 Rs=9.6
+
Cjo=300p Vj=.75
+
M=.4 Tt=4.e-06)
.tran 5n 400u 360u
*.tran 2u 200u 100u 1u
.probe
.end
****
****
Diode MODEL PARAMETERS
dioda
IS 8.300000E-15
RS 9.6
TT 4.000000E-06
CJO 300.000000E-12
VJ .75
M .4
INITIAL TRANSIENT SOLUTION
NODE VOLTAGE
NODE VOLTAGE
TEMPERATURE = 27.000 DEG C
NODE VOLTAGE
( 1) 0.00000000 ( 2) 0.00000000 ( 3) 0.00000000
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
0.000E+00
139
NODE VOLTAGE
400uA
0A
-400uA
360us
I(l1)
370us
380us
390us
400us
Time
400uA
0A
-400uA
-500mV
I(l1)
0V
500mV
V(1)
140
Problema 1b v=1.5V
1.0mA
0.5mA
0A
-0.5mA
360us
I(l1)
370us
380us
390us
400us
Time
1.0mA
0.5mA
0A
-0.5mA
-2.0V
I(l1)
-1.0V
0V
1.0V
V(1)
141
2.0V
Problema 1c v=6.5V
4.0mA
2.0mA
0A
-2.0mA
400us
I(l1)
450us
500us
550us
600us
Time
4.0mA
2.0mA
0A
-2.0mA
-8.0V
I(l1)
-4.0V
0V
4.0V
V(1)
142
8.0V
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