[PDF] EXERCICES Exercice 1 Sachant que R1 = 2.5kΩ et R2 = 45kΩ

View - Download EXERCICES Exercice 1 Sachant que R1 = 2.5kΩ et R2 = 45kΩ

Previous PDF

Next PDF

E.K.Boukas, 2002Mecatronique

EXERCICES

Exercice1SachantqueR1=2:5k

etR2=45k ,trouvezlegaindel'ampli-opsuivant: ViVoR 1R 2

Exercice2SachantqueR1=2:5k

etR2=25k ,trouvezlegaindel'ampli-opsuivant: +ViVoR 1R 2

Exercice3SachantqueR1=1k

,R2=20k etR3=100k ,trouvezlegaindel'ampli-opsui- vant: +ViV oR 1R2 R 3

Exercice4SachantqueR1=1k

,R2=1k ,R3=12k ,R4=15k ,V1=3VetV2=1:5V, trouvezlasortieVodel'ampli-opsuivant: +Vi2V i1 V oR 1 R 2R 3 R 4 page1of12

E.K.Boukas, 2002Mecatronique

6 Vi V o R 1R4R 3R2 6Vi V oR 1 R 4R 3R2 iidel'amplicateuren fonctiondesresistances. 6 6iii oR1 R 4R 3R2 toutengardantlam^emechargeRL=1k page2of12

E.K.Boukas, 2002Mecatronique

RLP...........................

+15V 15VV oVi+15V

Exercice9SachantqueR1=10k

,R2=20k etR=20k ,calculezlatensiondesortieVodans lesdeuxcassuivants: cas1:Vi1=+5VetVi2=+1V. cas2:Vi1=+12VetVi2=+3V.

15V+15V

Vi1V i2 V oR 1R 2R i 1i 2i 6 seur.

Exercice12DeterminezlegainVo

Videl'amplicateuroperationnelsuivant:

+An

15V+15V

ViVoR 1R i 1i6 page3of12

E.K.Boukas, 2002Mecatronique

ViVo10k

10F

ViVo20k

100k
Vn

1Solutions

Solutiondel'exercice1:

Lecircuitestunsimpleampli-opinverseur.

Legainducircuitest:

gain=R2

R1=452:5=18

Solutiondel'exercice2:

de1.

Legaindel'ampli-opinverseurest:

gainpartiel=R2

R1=252:5=10

Legainglobalest:

gainglobal=(gaindusuiveur)(gainpartiel) gainglobal=(1)(10)=10 page4of12

E.K.Boukas, 2002Mecatronique

Solutiondel'exercice3:

gain=1+R2

R1=1+201=21

Solutiondel'exercice4:

V o=R3

R1(Vi2Vi1)=R3R2(Vi2Vi1)=121(1:5V3V)=18V

Solutiondel'exercice5:

6 Vi V o R 1R4R 3R2 i 1 i 2i3i 3xy V +=V V x=Vi i

1=i2+i3(1a)

V yVx=VyVi=R2i3=)Vy=R2i3+Vi(1b) V oVy=R3i1=)Vo=R1i3+Vy(1c) V y=R4i2=)i2=Vy

R4(1d)

V x=Vi=R1i3=)i3=Vi

R1(1e)

page5of12

E.K.Boukas, 2002Mecatronique

i

1=i2+i3=Vy

R4+ViR1

i

1=R2i3+Vi

R4+ViR1=R2(Vi

R1)+Vi

R4+ViR1=ViR1+R2R1R4+1R1

(2a) V oVy=R3i1 V o(R2i3+Vi)=R3i1 V o(R2(Vi

R1)+Vi)=R3

V iR1+R2R1R4+1R1 V o=ViR2

R1+1+R3R1+R2R1R4+1R1

(3a)

Enn,l'equationchercheeest:

V o

Vi=R2R1+1+R3R1+R2R1R4+1R1

etlegainest: gain=R2 siR1!1=)gain=1+R3 R4 siR2!0=)gain=1+R3(R1+R4) R1R4

Solutiondel'exercice6:

6Vi V oR 1 R 4R 3R2x y -i1 i 2i 3 i 4 V +=V page6of12

E.K.Boukas, 2002Mecatronique

V x=0 i

3=i2+i4(4a)

i 1=Vi

R1(4b)

i 2=Vy

R2(4c)

V oVy=R3i3(4d) V y=R4i4(4e) i

2+i1=0=)i2=i1

V x=Vi=R1i3=)i3=Vi

R1(4g)

Enutilisant4b,4cet4e,onobtient:

V y=R4i4=R2i2=R2i1=R2Vi

R1(5a)

V o=Vy+R3i3 V o=Vy+R3(i2+i4) V o=Vy+R3(Vi

R1+VyR4)

V o=R2Vi R1+R3

ViR1R2ViR1R4

V o=ViR2

R1+R3R1+R2R3R1R4

(6a)

Enn,l'equationchercheeest:

V o=ViR2R4+R3R4+R2R3 R1R4 etlegainest: gain=R2R4+R3R4+R2R3 R1R4 siR4!1=)gain=R2+R3 R1 siR3!0=)gain=R2 R1 page7of12

E.K.Boukas, 2002Mecatronique

Solutiondel'exercice7:

6 6iii oR1 R 4R 3R2 ?i i i zx V +=V

CommeV+=0,alorsonendeduitque:V+=V=0.

Deplus,puisqueV=Vx,alorsonaaussiVx=0.

i i=io+i V (potentielsxetz) i iR2+iR4=0(7a) i

1R2=iR4(7b)

i

1R2=(iiio)R4(7c)

i i[R2+R4]=ioR4(7d) i i=ioR4 R2+R4 (7e)

Finalement,onobtient:

i o ii=1+R2R4

Solutiondel'exercice8:

RLP...........................

+15V 15VV oVi+15V -i1=0 page8of12

E.K.Boukas, 2002Mecatronique

V o=Vi V +=Vi V =Vo9 ;V o=A(V+V) V o=A(ViVo)=)Vo(1+A)=AVi V o

Vi=A1+AetavecAtresgrand,onobtient=)Vo=Vi

V o=Vi=+5V

Leschemaequivalentducircuitest:

Vi=5VR

1 R 2+15V i 1=0 ?i i=15V

R1+R2=15V10k

=1:5mA R 2=Vi i=5V1:5mA=3:33k R 1=10k

R2=10k

3:33k =6:66k

Vi=1:53VR

1 R

LR2+15V

i 16=0? ?i i 2

LecourantiatraversR1est:

i=15V

R1+R2RLR2+RL=(15V)(6:66k

)+(3:33k )(1k )(3:33k )+(1k )2mA

LatensionViest:

i=i1+i2=2mA i=Vi

R2+ViRL=Vi1R2+1RL

=2mA V i=2mA 1

R2+1RL=2mA

1 (3:33k )+1(1k )1:53V potentiometreetdel'impedancedecharge. page9of12

E.K.Boukas, 2002Mecatronique

Solutiondel'exercice9:

cas1:Latensiondesortiedel'ampli-opest: V o=R

R1Vi1RR2Vi2

V o=20k 10k (5V)20k 20k (1V) V o=11V cas2:Latensiondesortiedel'ampli-opest: V o=R

R1Vi1RR2Vi2

V o=20k 10k (12V)20k 20k (3V) V o=27V ensaturationcarleresultatserafausse.

Solutiondel'exercice10:

del'ampli-op.

L'equationdel'ampli-opinverseurest:

V o=R2 R1Vi etsongainest: A=R2

R1=VoVi=5V0:1V=50

SionchoisitR1=1k

alorsR2=50(1k )=50k

Eneet:

V i gainAVo

0V500V

0:1V505V

page10of12

E.K.Boukas, 2002Mecatronique

Solutiondel'exercice11:

Legaindel'ampli-opinverseurestdenipar:

A=1+R2

R1

50=1+R2

R1

SionchoisitR1=1k

alorsona: R

2=(501)(R1)=(49)(1k

)=49k V i gainAVo

0V500V

0:1V505V

Solutiondel'exercice12:

ona: V iVn

R1+VoVnR=0

Commelegaindel'ampli-opestA,alors:

Vquotesdbs_dbs10.pdfusesText_16

[PDF] exercices corrigés analyse de la valeur

[PDF] exercices corrigés analyse du bilan

[PDF] exercices corrigés bascules rs jk d pdf

[PDF] exercices corrigés capteur de température pdf

[PDF] exercices corrigés champs et forces 1ère s pdf

[PDF] exercices corrigés changement de variable intégrale

[PDF] exercices corrigés chimie des solutions

[PDF] exercices corrigés chimie minérale pdf

[PDF] exercices corrigés chromatographie sur couche mince pdf

[PDF] exercices corrigés commandes linux

[PDF] exercices corrigés comptabilité de gestion

[PDF] exercices corrigés d auto induction

[PDF] exercices corrigés d'algorithmique

[PDF] exercices corrigés damplificateur opérationnel

[PDF] exercices corrigés d'architecture des ordinateurs pdf