platforma_proiect pac

7/29/2019 Platforma_proiect PAC

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Continutul proiectului:

Cap . 1 : Tema Proiectului

Cap . 2 : Cerintele proiectului

2.1: Analiza de curent continuu:

a) Caracteristica de iesire a tranzistoarelor

b) Functia de transfer

c) Punctul static de functionare al amplificatorului

2.2: Analiza de curent alternativ :

a) Folosind analiza de curent alternativ sa se determine spectrul de frecvente al

amplificatorului

b) Sa se determine rezistententele de intrare/iesire ale amplificatorului

2.3: Analiza in domeniul timp

a) La intrarea amplificatorului se va aplica un semnal sinusoidal. Frecventa semnalului

de intrare va fi aleasa atat din interiorul, cat si din afara spectrului de frecvente. Cum

se modifica aplitudinea semnalului de iesire in functie de frecventa?

b) Analiza parametrica

c) Analiza cu temperatura 2.4: Analiza Fourier

2.5: Cablajul in Layout al circuitului

2.6: Concluzii

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2.7: Bibliografie

Cap.1: TEMA PROIECTULUI

Sa se aleaga individual o schema electrica a unui amplificator cu tranzistoare

(sau amplificatoare operationale), pentru care se va prezenta modul initial de

functionare (denumire amplificator, modul de functionare, unde poate fi

intrebuintat etc). Pe baza schemei alese se vor realiza in OrCAD

(Pspice/Capture/Layout) analizele prezentate la punctul B. Rezultatele analizelor

vor fi comparate cu datele de catalog ale amplificatorului ales.

Obs: Schema electriva va contine cel putin doua tranzistoare.

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1.1.: Amplificator audio 150W

Bazat pe doua trazistoare de putere Darlington,TIP142 si TIP147 ,acest

circuit poate reda o putere de 150W RMS pe o sarcina de impedanta 4 ohmi.

TIP 147 si 142 sunt tranzistori Darlington pereche,complementary, care pot

face fata la un current de 5Asi o tensiune de 100V ,acestia fiind faimosi pentru

fidelitatea lor. In aceasta shema mai avem si doi tranzistori de tip BC 558, Q5 si Q6,

conectati in etajul de preamplificare, iar TIP 142 ,TIP 147 impreuna cu TIP42

(Q1,Q2,Q3) formeaza etajul final de amplificare cu iesirea pe difuzor.Acest circuit

este proiectat atat de bine incat poate fi realizat pe o placa de test sau chiar prin

metoda liprii pin cu pin.Acest circuit poate fi alimentat cu o sursa simetrica de

tensiune de +/-45V la un current de 5A.

2.2: Schema electrica in Capture :

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Cap . 2 : Cerintele proiectului

2.1: Analiza de curent continuu :

a) Caracteristica de iesire a tranzistoarelor:

1) TIP142:

program_TIP142

.OPTIONS NOPAGE NOECHO

VBE 1 0 DC 12V

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VX 3 0 DC 15V

Q1 3 1 4 TIP142

Q2 3 4 0 TIP142A .6286

D1 3 1 Dmodel

R1 1 4 8.000E3

R2 4 3 40

.MODEL Dmodel D

+ IS=4.3887E-12

+ N=3.4748

+ RS=1.0000E-3

+ CJO=1.0000E-12

+ M=.3333

+ VJ=.75

+ ISR=100.00E-12+ BV=120

+ IBV=1.00E-6

+ TT=5.0000E-9

.MODEL TIP142 NPN

+ IS=1.605E-12 BF=745.1 NF=1 VAF=100

+ IKF=.3592 ISE=3.554E-9 NE=1.998 BR=.1018

+ NR=1 VAR=100 IKR=2.814 ISC=3.257E-12

+ NC=2.370 RB=0.159612 NK=.4489

+ RE=0 RC=0.106336 EG=1.110

+ CJE=2.000E-12 VJE=.75 MJE=.33 TF=23.00E-9

+ XTF=1 VTF=10 ITF=10.00E-3 CJC=2.000E-12

+ VJC=.75 MJC=.33 XCJC=.9 FC=.5

+ TR=1.50E-6

.MODEL TIP142A NPN

+ IS=1.605E-12 BF=745.1 NF=1 VAF=100

+ IKF=.3592 ISE=3.554E-9 NE=1.998 BR=.1018

+ NR=1 VAR=100 IKR=2.814 ISC=3.257E-12

+ NC=2.370 RB=0.159612 NK=.4489

+ RE=0 RC=0.106336 EG=1.110

+ CJE=2.000E-12 VJE=.75 MJE=.33 TF=23.00E-9

+ XTF=1 VTF=10 ITF=10.00E-3 CJC=0

+ VJC=.75 MJC=.33 XCJC=.9 FC=.5

+ TR=1.50E-6

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.DC VBE 1 0 0.2 VX 0 -4 3

.PLOT DC I(VX)

.OP

.PROBE

.END

Rezultate:

**** BJT MODEL PARAMETERS

X_Q1.Q1model X_Q1.Q2model

NPN NPN

IS 1.605000E-12 1.605000E-12

BF 745.1 745.1

NF 1 1

VAF 100 100

IKF .3592 .3592

ISE 3.554000E-09 3.554000E-09

NE 1.998 1.998

BR .1018 .1018

NR 1 1

VAR 100 100

IKR 2.814 2.814

ISC 3.257000E-12 3.257000E-12

NC 2.37 2.37

NK .4489 .4489

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RB .159612 .159612

RC .106336 .106336

CJE 2.000000E-12 2.000000E-12

CJC 2.000000E-12

XCJC .9 .9

TF 23.000000E-09 23.000000E-09

XTF 1 1

VTF 10 10

ITF .01 .01

TR 1.500000E-06 1.500000E-06

CN 2.42 2.42

D .87 .87

NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE

( 1) 11.6820 ( 2) 11.6820 ( 3) 12.0000 ( 4) 12.0000

( 5) 11.1460

VOLTAGE SOURCE CURRENTS

NAME CURRENT

V1 -2.891E-01

TOTAL POWER DISSIPATION 3.47E+00 WATTS

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CONCLUZII:

Tranzistorul darlington de tip NPN TIP142 are in componenta defapt 2 tranzistori npn

obisnuiti la care coeficientul de amplificare Darlington=1+2 de aceea se explica cresterea

curentului pentru valori mai mari de 1.2V la care tranzistorul incepe sa conduca :

VBE=VB1+VB2~2VBE1.

Castigul in curent este mult mai mare la acest tip de tranzistoare decat la cele obisnuitefata de tensiunea aplicata intre emiter si collector. Dupa temperature de 27C acesta isi pierde din

caracteristici si drept urmare pentru o amplificare eficienta a curentului de iesire acesta se

recomanda a fi montat pe un radiator.

2) TIP147 :

program_TIP147

I1 6 5 5uA

V1 0 5 12V

R3 2 6 0.001

R4 1 0 0.001

R5 3 5 0.001

Q1 3 2 4 TIP147

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Q2 3 4 1 TIP147A 5.131

D1 1 3 Dmodel

R1 2 4 8.000E3

R2 4 3 40

.MODEL Dmodel D

+ IS=4.3887E-12

+ N=3.4748

+ RS=1.0000E-3

+ CJO=1.0000E-12

+ M=.3333

+ VJ=.75

+ ISR=100.00E-12

+ BV=120

+ IBV=1.00E-6

+ TT=5.0000E-9

.MODEL TIP147 PNP

+ IS=350.7E-15 BF=2.997E3 NF=1 VAF=100

+ IKF=.7028 ISE=20.45E-12 NE=1.522 BR=.1001

+ NR=1 VAR=100 IKR=4.029 ISC=21.47E-9

+ NC=1.562 RB=5.07314 NK=.7315

+ RE=0 RC=0.00101053 EG=1.110

+ CJE=2.000E-12 VJE=.75 MJE=.33 TF=15.00E-9

+ XTF=1 VTF=10 ITF=10.00E-3 CJC=2.000E-12

+ VJC=.75 MJC=.33 XCJC=.9 FC=.5

+ TR=8.00E-5

.MODEL TIP147A PNP

+ IS=350.7E-15 BF=2.997E3 NF=1 VAF=100

+ IKF=.7028 ISE=20.45E-12 NE=1.522 BR=.1001

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+ NR=1 VAR=100 IKR=4.029 ISC=21.47E-9

+ NC=1.562 RB=5.07314 NK=.7315

+ RE=0 RC=0.00101053 EG=1.110

+ CJE=2.000E-12 VJE=.75 MJE=.33 TF=15.00E-9

+ XTF=1 VTF=10 ITF=10.00E-3 CJC=0

+ VJC=.75 MJC=.33 XCJC=.9 FC=.5

+ TR=8.00E-5

.DC V1 1V 12V 1V I1 5UA 10UA 2.5UA

.PLOT DC IC(Q1)

.OP

.PROBE

.END

REZULTATE:

BJT MODEL PARAMETERS

TIP147 TIP147A

PNP PNP

IS 350.700000E-15 350.700000E-15

BF 2.997000E+03 2.997000E+03

NF 1 1

VAF 100 100

IKF .7028 .7028

ISE 20.450000E-12 20.450000E-12

NE 1.522 1.522

BR .1001 .1001

NR 1 1

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VAR 100 100

IKR 4.029 4.029

ISC 21.470000E-09 21.470000E-09

NC 1.562 1.562

NK .7315 .7315

RB 5.07314 5.07314

RC 1.010530E-03 1.010530E-03

CJE 2.000000E-12 2.000000E-12

CJC 2.000000E-12

XCJC .9 .9

TF 15.000000E-09 15.000000E-09

XTF 1 1

VTF 10 10

ITF .01 .01

TR 80.000000E-06 80.000000E-06

CN 2.2 2.2

D .52 .52

SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C


( 1) -3.4651 ( 2) -4.4147 ( 3) -8.5349 ( 4) -4.3745

( 5) -12.0000 ( 6) -4.4147


NAME CURRENT

V1 -3.465E+03


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3) TIP41:

program_TIP41

.OPTIONS NOPAGE NOECHO

V1 5 0 DC 12V

I1 0 4 5UA

R1 2 4 0.001

R2 3 0 0.001

R3 1 5 0.001

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Q1 1 2 3 TIP41

.MODEL TIP41 NPN(Is=457.5f Xti=3 Eg=1.11 Vaf=50 Bf=156.7 Ise=1.346p Ne=1.34

+ Ikf=3.296 Nk=.5961 Xtb=2.2 Br=7.639 Isc=604.1f Nc=2.168

+ Ikr=8.131m Rc=91.29m Cjc=278.7p Mjc=.385 Vjc=.75 Fc=.5 Cje=433p

+ Mje=.5 Vje=.75 Tr=1.412u Tf=37.34n Itf=35.68 Xtf=1.163 Vtf=10

+ Rb=.1)

.DC V1 0 12 1 I1 5UA 25UA 5UA

.PLOT DC IC(Q1)

.OP

.PROBE

.END

REZULTATE:


TIP41

NPN

IS 457.500000E-15

BF 156.7

NF 1

VAF 50

IKF 3.296

ISE 1.346000E-12

NE 1.34

BR 7.639

NR 1

IKR 8.131000E-03

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ISC 604.100000E-15

NC 2.168

NK .5961

RB .1

RC .09129

CJE 433.000000E-12

MJE .5

CJC 278.700000E-12

MJC .385

TF 37.340000E-09

XTF 1.163

VTF 10

ITF 35.68

TR 1.412000E-06

XTB 2.2

CN 2.42

D .87

SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C


( 1) 12.0000 ( 2) .5142 ( 3) 247.7E-09 ( 4) .5142

( 5) 12.0000


NAME CURRENT

V1 -2.427E-04

TOTAL POWER DISSIPATION 2.91E-03 WATTS

Concluzii:

Tranzistorul TIP41 are rolul de adaptare de impedanta pentru intrarea in tranzistorul Darlington

pnp TIP147 de aceea amplificarea in curent a acestui tranzistorului este usor mai mica decat al

tranzsitorului TIP147 ce are rolul de etaj final de amplificare.

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4) BC558 :

PROGRAM BC558

V1 5 0 DC 12V

I1 4 0 5UA

R1 4 2 0.001

R2 3 0 0.001

R3 1 5 0.001

Q1 3 2 1 BC558

.MODEL BC558 PNP(Is=1.02f Xti=3 Eg=1.11 Vaf=73.79 Bf=174.7 Ise=10.73f

+ Ne=1.644 Ikf=.1023 Nk=.5033 Xtb=1.5 Br=4.432 Isc=14.74f

+ Nc=1.296 Ikr=2.237 Rc=1.039 Cjc=9.81p Mjc=.332 Vjc=.4865 Fc=.5

+ Cje=30p Mje=.3333 Vje=.5 Tr=10n Tf=830.3p Itf=.8981 Xtf=10.32

+ Vtf=10)

.DC V1 0V 12V 1V I1 5UA 25UA 5UA

.PLOT DC I(R2)

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.PROBE

.END

REZULTATE:


BC558

PNP

IS 1.020000E-15

BF 174.7

NF 1

VAF 73.79

IKF .1023

ISE 10.730000E-15

NE 1.644

BR 4.432

NR 1

IKR 2.237

ISC 14.740000E-15

NC 1.296

NK .5033

RC 1.039

CJE 30.000000E-12

VJE .5

MJE .3333

CJC 9.810000E-12

VJC .4865

MJC .332

TF 830.300000E-12

XTF 10.32

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VTF 10

ITF .8981

TR 10.000000E-09

XTB 1.5

CN 2.2

D .52


( 1) 12.0000 ( 2) 11.2900 ( 3) 961.0E-09 ( 4) 11.2900

( 5) 12.0000


NAME CURRENT

V1 -9.660E-04

TOTAL POWER DISSIPATION 1.16E-02 WATTS

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Concluzii:

Dupa cum se poate observa cele 2 tranzistoare BC558 in configuratia emitor comun pe intrarea

circuitului ridica curentul ce intra in baza acestora usor spre 50mA spre a putea fi mai departe

introdus in etajul final de amplificare cu transistor Darlington.Deasemenea acestea au si rolul de

adaptare de impedanta, facand parte din circuitul de preamplificare , pentru a putea folosi tranzistorii

de etaj final la potential maxim.

b)Functia de Transfer:

program_PAC

V1 1 0 DC 45V

V2 0 2 DC 45V

VIN 15 0 SIN(0 1V 500HZ)

R1 2 9 1.5K

R2 12 13 220

R3 6 8 3.3K

R4 10 12 22K

R5 8 3 .3K

R6 1 11 22K

R7 16 0 27K

R8 5 10 0.33

R9 10 19 0.33

R10 18 2 33

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C1 8 10 10UF

C2 16 15 10UF

C3 13 0 100UF

D1 7 6 1N4007

D2 17 7 1N4007

Q1 1 2 21 TIP142

Q2 1 21 5 TIP142A .6286

D3 3 1 Dmodel

R11 2 4 8.000E3

R12 4 3 40

Q4 2 17 20 TIP147

Q5 2 17 19 TIP147A 5.131

D4 17 19 Dmodel1

R13 2 20 8.000E3

R14 20 19 40

.MODEL Dmodel1 D( IS=4.3887E-12 N=3.4748 RS=1.0000E-3 CJO=1.0000E-12 M=.3333 VJ=.75

+ ISR=100.00E-12 BV=120 IBV=1.00E-6 TT=5.0000E-9)

.MODEL TIP147 PNP( IS=350.7E-15 BF=2.997E3 NF=1 VAF=100 IKF=.7028 ISE=20.45E-12

+ NE=1.522 BR=.1001 NR=1 VAR=100 IKR=4.029 ISC=21.47E-9

+ NC=1.562 RB=5.07314 NK=.7315 RE=0 RC=0.00101053 EG=1.110 CJE=2.000E-12 VJE=.75

+ MJE=.33 TF=15.00E-9 XTF=1 VTF=10 ITF=10.00E-3 CJC=2.000E-12

+ VJC=.75 MJC=.33 XCJC=.9 FC=.5 TR=8.00E-5)

.MODEL TIP147A PNP( IS=350.7E-15 BF=2.997E3 NF=1 VAF=100 IKF=.7028 ISE=20.45E-12

NE=1.522 BR=.1001 NR=1

+ VAR=100 IKR=4.029 ISC=21.47E-9

+ NC=1.562 RB=5.07314 NK=.7315 RE=0 RC=0.00101053 EG=1.110 CJE=2.000E-12

+ VJE=.75 MJE=.33 TF=15.00E-9 XTF=1 VTF=10 ITF=10.00E-3 CJC=0


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Q3 17 9 18 TIP41

Q6 2 12 11 BC558

Q7 9 16 11 BC558

.MODEL 1N4007 D (IS=14.11E-9 N=1.984 RS=33.89E-3 IKF=94.81 XTI=3

+ EG=1.110 CJO=51.17E-12 M=.2762 VJ=.3905 FC=.5 ISR=100.0E-12

+ NR=2 BV=100.1 IBV=10 TT=4.761E-6)

.MODEL Dmodel D ( IS=4.3887E-12 N=3.4748 RS=1.0000E-3 CJO=1.0000E-12 M=.3333 VJ=.75

+ ISR=100.00E-12 BV=120 IBV=1.00E-6 TT=5.0000E-9)

.MODEL TIP142 NPN ( IS=1.605E-12 BF=745.1 NF=1 VAF=100 IKF=.3592 ISE=3.554E-9 NE=1.998

+ BR=.1018 NR=1 VAR=100 IKR=2.814 ISC=3.257E-12




.MODEL TIP142A NPN ( IS=1.605E-12 BF=745.1 NF=1 VAF=100 IKF=.3592 ISE=3.554E-9


+ NC=2.370 RB=0.159612 NK=.4489 RE=0 RC=0.106336 EG=1.110 CJE=2.000E-12 VJE=.75 MJE=.33

TF=23.00E-9

+ XTF=1 VTF=10 ITF=10.00E-3 CJC=0


.MODEL TIP41 NPN(Is=457.5f Xti=3 Eg=1.11 Vaf=50 Bf=156.7 Ise=1.346p Ne=1.34 Ikf=3.296

Nk=.5961 Xtb=2.2 Br=7.639 Isc=604.1f Nc=2.168

+ Ikr=8.131m Rc=91.29m Cjc=278.7p Mjc=.385 Vjc=.75 Fc=.5 Cje=433p Mje=.5 Vje=.75 Tr=1.412u

Tf=37.34n Itf=35.68 Xtf=1.163 Vtf=10

+ Rb=.1)

.MODEL BC558 PNP (Is=1.02f Xti=3 Eg=1.11 Vaf=52.31 Bf=306.5 Ise=10.27f Ne=1.764

+ Ikf=91.85m Nk=.5351 Xtb=1.5 Br=6.48 Isc=1.472f

+ Nc=1.294 Ikr=.5584 Rc=1.086 Cjc=9.81p Mjc=.332 Vjc=.4865 Fc=.5 Cje=30p Mje=.3333

+ Vje=.5 Tr=10n Tf=611.6p Itf=1.373 Xtf=26.05 Vtf=10)

.TF V(10) VIN

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.PROBE

.END

REZULTATE:

SMALL-SIGNAL CHARACTERISTICS

V(10)/VIN = 0.000E+00

INPUT RESISTANCE AT VIN = 1.000E+20

OUTPUT RESISTANCE AT V(10) = 8.008E+03

C) Punctul static de functionare:

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Punctul static de functionare are loc pentru tensiunea de intrare de 36,28mV si curentul de

1,344A la care avem tensiunea de iesire de -1,658V si curentul de 52,59mA.

2.2: Analiza de curent alternativ :

a) in spectrul de frecvente:

curent_alternativ

.AC DEC 5000 30KHZ 1000GHZ

V1 1 0 DC 45V

V2 0 2 DC 45V

VIN 15 0 AC 10V

R1 2 9 1.5K

R2 12 13 220

R3 6 8 3.3K

R4 10 12 22K

R5 8 3 .3K

R6 1 11 22K

R7 16 0 27K

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R8 5 10 0.33

R9 10 19 0.33

R10 18 2 33

C1 8 10 10UF

C2 16 15 10UF

C3 13 0 100UF

D1 7 6 1N4007

D2 17 7 1N4007

Q1 1 2 21 TIP142

Q2 1 21 5 TIP142A .6286

D3 3 1 Dmodel

R11 2 4 8.000E3

R12 4 3 40

Q4 2 17 20 TIP147

Q5 2 17 19 TIP147A 5.131

D4 17 19 Dmodel1

R13 2 20 8.000E3

R14 20 19 40


+ ISR=100.00E-12 BV=120 IBV=1.00E-6 TT=5.0000E-9)






.MODEL TIP147A PNP( IS=350.7E-15 BF=2.997E3 NF=1 VAF=100 IKF=.7028 ISE=20.45E-12 NE=1.522

BR=.1001 NR=1

+ VAR=100 IKR=4.029 ISC=21.47E-9

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Q3 17 9 18 TIP41

Q6 2 12 11 BC558

Q7 9 16 11 BC558



+ NR=2 BV=100.1 IBV=10 TT=4.761E-6)


+ ISR=100.00E-12 BV=120 IBV=1.00E-6 TT=5.0000E-9)


+ BR=.1018 NR=1 VAR=100 IKR=2.814 ISC=3.257E-12







TF=23.00E-9

+ XTF=1 VTF=10 ITF=10.00E-3 CJC=0


.MODEL TIP41 NPN(Is=457.5f Xti=3 Eg=1.11 Vaf=50 Bf=156.7 Ise=1.346p Ne=1.34 Ikf=3.296 Nk=.5961

Xtb=2.2 Br=7.639 Isc=604.1f Nc=2.168

+ Ikr=8.131m Rc=91.29m Cjc=278.7p Mjc=.385 Vjc=.75 Fc=.5 Cje=433p Mje=.5 Vje=.75 Tr=1.412u

Tf=37.34n Itf=35.68 Xtf=1.163 Vtf=10

+ Rb=.1)



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.PLOT AC VM(10) VP(10)

.PROBE

.END

b)Sa se determine rezistententele de intrare/iesire ale amplificatorului:

INPUT RESISTANCE AT V2 = 2.211E+04

OUTPUT RESISTANCE AT V(10) = 8.008E+03

2.3: Analiza in domeniul timp:

a) La intrarea amplificatorului se va aplica un semnal sinusoidal. Frecventa

semnalului de intrare va fi aleasa atat din interiorul, cat si din afara spectrului de

frecvente. Cum se modifica aplitudinea semnalului de iesire in functie de frecventa.

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PROGRAM:

DOMENIU TIMP

.TRAN 1NS 10MS

V1 1 0 DC 45V

V2 0 2 DC -45V

VIN 15 0 SIN(0 0.5V 1KHZ)

R1 2 9 1.5K

R2 12 13 220

R3 6 8 3.3K

R4 10 12 22K

R5 8 3 .3K

R6 1 11 22K

R7 16 0 27K

R8 5 10 0.33

R9 10 19 0.33

R10 18 2 33

R15 10 0 8

C1 8 10 10UF

C2 16 15 10UF

C3 13 0 100UF

D1 7 6 1N4007

D2 17 7 1N4007

Q1 1 2 21 TIP142

Q2 1 21 5 TIP142A .6286

D3 3 1 Dmodel

R11 2 4 8.000E3

R12 4 3 40

Q4 2 17 20 TIP147

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Q5 2 17 19 TIP147A 5.131

D4 17 19 Dmodel1

R13 2 20 8.000E3

R14 20 19 40


+ ISR=100.00E-12 BV=120 IBV=1.00E-6 TT=5.0000E-9)







BR=.1001 NR=1

+ VAR=100 IKR=4.029 ISC=21.47E-9




Q3 17 9 18 TIP41

Q6 2 12 11 BC558

Q7 9 16 11 BC558



+ NR=2 BV=100.1 IBV=10 TT=4.761E-6)


+ ISR=100.00E-12 BV=120 IBV=1.00E-6 TT=5.0000E-9)


+ BR=.1018 NR=1 VAR=100 IKR=2.814 ISC=3.257E-12


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TF=23.00E-9

+ XTF=1 VTF=10 ITF=10.00E-3 CJC=0



Xtb=2.2 Br=7.639 Isc=604.1f Nc=2.168

+ Ikr=8.131m Rc=91.29m Cjc=278.7p Mjc=.385 Vjc=.75 Fc=.5 Cje=433p Mje=.5 Vje=.75 Tr=1.412u Tf=37.34n

Itf=35.68 Xtf=1.163 Vtf=10

+ Rb=.1)





.PLOT TRAN V(R15)

.PROBE

.END

REZULTATE:


( 1) 45.0000 ( 2) 45.0000 ( 3) 45.0000 ( 4) 45.0000

( 5) 43.5280 ( 6) 45.0000 ( 7) 44.2960 ( 8) 45.0000

( 9) 43.6450 ( 10) 42.1100 ( 11) 43.6460 ( 12) 44.2970

( 13) 44.2970 ( 15) 0.0000 ( 16) 42.8700 ( 17) 43.6240

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( 18) 45.0000 ( 19) 42.4280 ( 20) 42.7130 ( 21) 44.3600


NAME CURRENT

V1 -4.298E+00

V2 9.668E-01

VIN 0.000E+00


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B) Analiza Parametrica:

IN TIMP:

DOMENIU TIMP

V1 1 0 DC 45V

V2 0 2 DC -45V


R1 2 9 1.5K

R2 12 13 220

R3 6 8 3.3K

R4 10 12 {R4}

R5 8 3 .3K

R6 1 11 22K

R7 16 0 27K

R8 5 10 0.33

R9 10 19 0.33

R10 18 2 33

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R15 10 0 8

C1 8 10 10UF

C2 16 15 10UF

C3 13 0 100UF

D1 7 6 1N4007

D2 17 7 1N4007

Q1 3 4 21 TIP142

Q2 3 21 5 TIP142A .6286

D3 3 5 Dmodel

R11 2 4 8.000E3

R12 4 3 40

Q4 2 17 20 TIP147

Q5 2 20 19 TIP147A 5.131

D4 17 19 Dmodel1

R13 2 20 8.000E3

R14 20 19 40


+ ISR=100.00E-12 BV=120 IBV=1.00E-6 TT=5.0000E-9)






.MODEL TIP147A PNP( IS=350.7E-15 BF=2.997E3 NF=1 VAF=100 IKF=.7028 ISE=20.45E-12 NE=1.522BR=.1001 NR=1

+ VAR=100 IKR=4.029 ISC=21.47E-9



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Q3 17 9 18 TIP41

Q6 2 12 11 BC558

Q7 9 16 11 BC558



+ NR=2 BV=100.1 IBV=10 TT=4.761E-6)


+ ISR=100.00E-12 BV=120 IBV=1.00E-6 TT=5.0000E-9)


+ BR=.1018 NR=1 VAR=100 IKR=2.814 ISC=3.257E-12







TF=23.00E-9

+ XTF=1 VTF=10 ITF=10.00E-3 CJC=0



Xtb=2.2 Br=7.639 Isc=604.1f Nc=2.168


Itf=35.68 Xtf=1.163 Vtf=10

+ Rb=.1)





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.TRAN 0.01US 1MS

.PARAM R4=10

.STEP PARAM R4 LIST 4 10 1K

.PROBE

.END

REZULTAT:

INITIAL TRANSIENT SOLUTION TEMPERATURE = 27.000 DEG C

CURRENT STEP PARAM R4 = 4


( 1) 45.0000 ( 2) 45.0000 ( 3) 34.5150 ( 4) 34.5670

( 5) 33.6460 ( 6) 34.7430 ( 7) 35.1740 ( 8) 34.5340

( 9) 36.3850 ( 10) 33.6450 ( 11) 42.6280 ( 12) 41.6760

( 13) 41.6760 ( 15) 0.0000 ( 16) 41.7550 ( 17) 35.6050

( 18) 43.2260 ( 19) 34.3700 ( 20) 44.3130 ( 21) 34.1430


NAME CURRENT

V1 -1.078E-04

V2 4.207E+00

VIN 0.000E+00



CURRENT STEP PARAM R4 = 10


( 1) 45.0000 ( 2) 45.0000 ( 3) 31.9760 ( 4) 32.0400

( 5) 31.0980 ( 6) 32.3370 ( 7) 32.7920 ( 8) 32.0060

( 9) 34.0320 ( 10) 31.0980 ( 11) 43.5440 ( 12) 42.6080

( 13) 42.6080 ( 15) 0.0000 ( 16) 42.6700 ( 17) 33.2470

( 18) 43.1700 ( 19) 32.0000 ( 20) 44.2820 ( 21) 31.6080

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NAME CURRENT

V1 -6.620E-05

V2 3.889E+00

VIN 0.000E+00


**** INITIAL TRANSIENT SOLUTION TEMPERATURE = 27.000 DEG C

**** CURRENT STEP PARAM R4 = 1.0000E+03


( 1) 45.0000 ( 2) 45.0000 ( 3) 28.2720 ( 4) 28.3540

( 5) 27.3880 ( 6) 28.8170 ( 7) 29.2930 ( 8) 28.3170

( 9) 30.5190 ( 10) 27.3880 ( 11) 30.7260 ( 12) 44.0500

( 13) 44.0500 ( 15) 0.0000 ( 16) 29.8750 ( 17) 29.7690

( 18) 43.6010 ( 19) 28.5110 ( 20) 44.2380 ( 21) 27.9130


NAME CURRENT

V1 -6.488E-04

V2 3.424E+00

VIN 0.000E+00


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c)Analiza temperaturii:

#

V1 1 0 DC 45V

V2 0 2 DC -45V


R1 2 9 1.5K

R2 12 13 220

R3 6 8 3.3K

R4 10 12 22k

R5 8 3 .3K

R6 1 11 22K

R7 16 0 27K

R8 5 10 0.33

R9 10 19 0.33

R10 18 2 33

R15 10 0 8

C1 8 10 10UF

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C2 16 15 10UF

C3 13 0 100UF

D1 7 6 1N4007

D2 17 7 1N4007

Q1 3 4 21 TIP142

Q2 3 21 5 TIP142A .6286

D3 3 5 Dmodel

R11 2 4 8.000E3

R12 4 3 40

Q4 2 17 20 TIP147

Q5 2 20 19 TIP147A 5.131

D4 17 19 Dmodel1

R13 2 20 8.000E3

R14 20 19 40


+ ISR=100.00E-12 BV=120 IBV=1.00E-6 TT=5.0000E-9)







BR=.1001 NR=1

+ VAR=100 IKR=4.029 ISC=21.47E-9




Q3 17 9 18 TIP41

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Q6 2 12 11 BC558

Q7 9 16 11 BC558



+ NR=2 BV=100.1 IBV=10 TT=4.761E-6)


+ ISR=100.00E-12 BV=120 IBV=1.00E-6 TT=5.0000E-9)


+ BR=.1018 NR=1 VAR=100 IKR=2.814 ISC=3.257E-12







TF=23.00E-9

+ XTF=1 VTF=10 ITF=10.00E-3 CJC=0



Xtb=2.2 Br=7.639 Isc=604.1f Nc=2.168


Itf=35.68 Xtf=1.163 Vtf=10

+ Rb=.1)





.TRAN 0.01US 1MS

.TEMP 0 20 100

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.PROBE

.END

REZULTATUL:



( 1) 45.0000 ( 2) 45.0000 ( 3) 27.9310 ( 4) 28.0150

( 5) 27.0030 ( 6) 28.3200 ( 7) 28.8440 ( 8) 27.9630

( 9) 30.0900 ( 10) 27.0020 ( 11) 30.1100 ( 12) 44.2040

( 13) 44.2040 ( 15) 0.0000 ( 16) 29.2960 ( 17) 29.3680

( 18) 44.4660 ( 19) 28.1150 ( 20) 44.1780 ( 21) 27.5380


NAME CURRENT

V1 -6.768E-04

V2 3.376E+00

VIN 0.000E+00


**** TEMPERATURE-ADJUSTED VALUES TEMPERATURE = 20.000 DEG C


( 1) 45.0000 ( 2) 45.0000 ( 3) 27.9260 ( 4) 28.0100

( 5) 27.0270 ( 6) 28.4210 ( 7) 28.9090 ( 8) 27.9670

( 9) 30.0890 ( 10) 27.0270 ( 11) 30.1110 ( 12) 44.2290

( 13) 44.2290 ( 15) 0.0000 ( 16) 29.3210 ( 17) 29.3970

( 18) 44.4160 ( 19) 28.1400 ( 20) 44.2190 ( 21) 27.5580


NAME CURRENT

V1 -6.768E-04

V2 3.379E+00

VIN 0.000E+00

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**** INITIAL TRANSIENT SOLUTION TEMPERATURE = 100.000 DEG C


( 1) 45.0000 ( 2) 45.0000 ( 3) 27.7310 ( 4) 27.8160

( 5) 27.1320 ( 6) 28.8210 ( 7) 29.1660 ( 8) 27.8220

( 9) 30.0880 ( 10) 27.1310 ( 11) 30.1170 ( 12) 44.3330

( 13) 44.3330 ( 15) 0.0000 ( 16) 29.4260 ( 17) 29.5120

( 18) 44.2080 ( 19) 28.2490 ( 20) 44.3850 ( 21) 27.5340


NAME CURRENT

V1 -6.765E-04

V2 3.392E+00

VIN 0.000E+00


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2.4: Analiza Fourier:

#

V1 0 1 DC 15V

V2 0 2 DC -15V


R1 2 9 1.5K

R2 12 13 220

R3 6 8 3.3K

R4 10 12 22k

R5 8 3 .3K

R6 1 11 22K

R7 16 0 27K

R8 5 10 0.33

R9 10 19 0.33

R10 18 2 33

R15 10 0 8

C1 8 10 10UF

C2 16 15 10UF

C3 13 0 100UF

D1 7 6 1N4007

D2 17 7 1N4007

Q1 3 4 21 TIP142

Q2 3 21 5 TIP142A .6286

D3 3 5 Dmodel

R11 2 4 8.000E3

R12 4 3 40

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Q4 2 17 20 TIP147

Q5 2 20 19 TIP147A 5.131

D4 17 19 Dmodel1

R13 2 20 8.000E3

R14 20 19 40


+ ISR=100.00E-12 BV=120 IBV=1.00E-6 TT=5.0000E-9)







BR=.1001 NR=1

+ VAR=100 IKR=4.029 ISC=21.47E-9




Q3 17 9 18 TIP41

Q6 2 12 11 BC558

Q7 9 16 11 BC558



+ NR=2 BV=100.1 IBV=10 TT=4.761E-6)


+ ISR=100.00E-12 BV=120 IBV=1.00E-6 TT=5.0000E-9)


+ BR=.1018 NR=1 VAR=100 IKR=2.814 ISC=3.257E-12

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TF=23.00E-9

+ XTF=1 VTF=10 ITF=10.00E-3 CJC=0



Xtb=2.2 Br=7.639 Isc=604.1f Nc=2.168


Itf=35.68 Xtf=1.163 Vtf=10

+ Rb=.1)





.TRAN 0.001US 0.002S

.FOUR 5MEGHZ V(10)

.PROBE

.END

REZULTATE:

FOURIER ANALYSIS TEMPERATURE = 27.000 DEG C

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(10)

DC COMPONENT = 0.000000E+00

HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED

NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)

1 5.000E+06 1.000E-30 1.000E+00 0.000E+00 0.000E+00

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2 1.000E+07 1.000E-30 1.000E+00 0.000E+00 0.000E+00

3 1.500E+07 1.000E-30 1.000E+00 0.000E+00 0.000E+00

4 2.000E+07 1.000E-30 1.000E+00 0.000E+00 0.000E+00

5 2.500E+07 1.000E-30 1.000E+00 0.000E+00 0.000E+00

6 3.000E+07 1.000E-30 1.000E+00 0.000E+00 0.000E+00

7 3.500E+07 1.000E-30 1.000E+00 0.000E+00 0.000E+00

8 4.000E+07 1.000E-30 1.000E+00 0.000E+00 0.000E+00

9 4.500E+07 1.000E-30 1.000E+00 0.000E+00 0.000E+00

TOTAL HARMONIC DISTORTION = 2.828427E+02 PERCENT

2.5:

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Cablajul in Layout al circuitului

Cu piese

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Fata cablaj fara piese.

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2.6: Concluzii

In lucrarea de fata au fost prezentate probleme generale ale amplificatoarelor

cu tranzistoare si probleme specifice amplificatoarelor cu tranzistoare. Problemele

generale sunt structura unui etaj sau a unui lant de amplificare, parametrii

principali, caracteristicile grafice, modul de analiza. Probleme specifice sunt clasele

de functionare sau schemele echivalente utilizate pentru analiza amplificatoarelor

cu tranzistoare.

Un dezavantaj este o dublare aproximative a tensiunii baza-emitor. Din moment ce sunt

doua intersectii intre baza si emitor la tranzistoarelei Darlington, echivalentul tensiunii baza-emitor este suma celor doua tensiuni baza-emitor:

VBE = VBE1 + VBE2=~ 2VBE1

Pentru tehnologia pe baza de siliciu, in cazul in care fiecare VBEi este aproximativ 0,65 V,

atunci cnd aparatul functioneaza in regiunea activa sau saturatie, necesarul de tensiune baza-

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emitor pentru o pereche este de 1,3 V.

O alta problema este o reducere a vitezei de comutare, pentru ca primul tranzistor nu poate

inhiba curentul activ de baza de cel de al doilea , ceea ce face aparatul lent. Pentru a atenua acest

lucru, tranzistorul al doilea are deseori un rezistor de cteva sute de ohmi conectat intre baza si

terminale de emitor. Acest rezistor ofera o cale de descarcare de gestiune, impedanta scazuta

pentru acumularea de sarcina in jonctiunea baza-emitor, tranzistorul inchidandu-se astfel mai

repede.

2.7.Bibliografie:

Curs Proiectare Asistata de Calculator prof. D. Farini

Proiect prof. O. Tomescu

www.alltransistors.com

www.alldatasheet.com
http://www.alltransistors.com/http://www.alldatasheet.com/http://www.alltransistors.com/http://www.alldatasheet.com/

platforma_proiect pac

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