Clase

  • May 2020
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HALLAR U 0 = ? SOLUCION: U 1 = U 1+ − U 1−

U 2 = U 2+ − U 2−

U 1− = 0

U 2− = 0

U 1 = U 1+ U 2 = U 2+ Aplicando ley de corrientes de kirchoff en nodo U − I1 = I 2 + I 5

U1 − U − U − − U 0 U − − 0 = + R1 R2 R5  1 U1 U 0 1 1  ..................(1) + = U −  + + R1 R2 R R R 2 5   1 Por TEOREMA DIVISOR DE TENSION para U + R4 U+ = U 2 .............(2) R3 + R4 (2) en (1) U1 U 0 R4  1 1 1   + U 2 + = + R1 R2 R3 + R4  R1 R2 R5  U0 R4 = R2 R3 + R4 U0 =

R4 R3 + R4

 1 U 1 1   + U 2 − 1 + R1  R1 R2 R5   R2 R  R  + 1 + 2 U 2 − 2 U 1 R5  R1  R1

HALLAR U 0 = ?

U x = U x+ − U x− ..................(1) Aplicando ley de corrientes de kirchoff en nodo I1 + I 2 = 0 U U x− − U x+ + =0 R R 2U x− + U x− − U x+ = 0 − x

3U x− = U x+ ........................( 2 ' )

U x−

1 U x− = U x+ ......................( 2)en(1) 3 1 U x = U x+ − U x+ 3 2 U x = U x+ 3 3 U x+ = U x .........................(3) 2 Aplicando ley de corrientes de kirchoff en nodo

I2 = I3 − x

U x+

+ x

U −U U x+ − U x = 2R 3R − 3U x + 2U y = U x+ (2 + 3) De 2 '

3U x− + 2U y = U x+ (2 + 3) U x+ + 2U y = 5U x+ 2U y = 4U x+ U y = 2U x+ De 3

3 Uy = 2 Ux 2 U y = 3U x Aplicando ley de corrientes de kirchoff en nodo U

I3 = I4 + I5 + x

U −U y 3R

=

Uy

+

U y −U0

12 R 36 R U y U y −U0 U x+ − U y = + 4 12 U 1 1  U x+ + 0 = U y  + + 1 12  4 12  De (3) y (4) U 3  12 + 3 + 1  U x + 0 = 3U x   2 12  12  U0  16 3  =Ux −  12  4 2  10  U 0 = 12U x   4 U 0 = 30U x

y

Diseñar un multibibrador estable capas de suministrar una señal de salida de frecuencia 30KHz con un ciclo útil de a) 70% b)30% +5

R1 10k

4

VCC

8

U1

R

Q DC

5

3

R3

7

230

CV

R2

C2

2

TR

TH

6

D1 LED-BLUE

1

0.1uF

GND

7.5k

555

C1 1.924nF

SOLUCION: a)

0 ≤ D% ≤ 100% f = 30000 Hz D% = 70% tc = ( R1 + R2 ) C ln 2 td = ( R2 ) C ln 2

T = tc + td = ( R1 + R2 ) C ln 2 + ( R2 ) C ln 2 T = ( R1 + 2 R2 ) C ln 2 f =

1 1 = T ( R1 + 2 R2 ) C ln 2 tc D% = * 100% tc + td

( R1 + R2 ) C ln 2 *100% ( R1 + 2 R2 ) C ln 2 0.7( R1 + 2 R2 ) = ( R1 + R2 ) D%

R1 (0.7 − 1) = R2 (1 − 1.4) 0.3R1 = 0.4 R2

0 .3 R1 0 .4 si __ R1 = 10 KΩ R2 =

⇒ R2 = 7.5 KΩ Para el capacitor: 1 1 f = = T ( R1 + 2 R2 ) C ln 2 C= SOLUCION: b)

1 f ( R1 + 2 R2 ) ln 2

C = 1.923 * 10 −9 F t c = ( R1 ) C ln 2

t d = ( R2 ) C ln 2

T = t c + t d = ( R1 ) C ln 2 + ( R2 ) C ln 2 T = ( R1 + R2 ) C ln 2 f =

1 1 = T ( R1 + R2 ) C ln 2 tc D% = *100% tc + td

( R1 ) C ln 2 *100% ( R1 + R2 ) C ln 2 0.3( R1 + R2 ) = R1 D%

0.3R1 = R2 (1 − 0.3) 0 .7 R2 = R1 0 .3 si __ R1 = 10 KΩ

⇒ R2 = 23.3KΩ Para el capacitor: 1 1 f = = T ( R1 + R2 ) C ln 2 C=

1 f ( R1 + R2 ) ln 2

C = 1.444 *10 −9 F

Graficar y hallar la frecuencia de oscilaciones

R2 40k

-15

U1 4 1 5

Uo

R1

2 6

10k

7

3

UA741

C2

R3

+15 0.1uF

R4

C1

20k

0.1uF

20k

Del lado U − Es un amplificador operacional en su configuración de inversor

 R  U −f = 1 + 2 U 0 R1   Del lado U +

1 SC 1 R+ SC U +f = U0 1 R* SC + R + 1 1 SC R+ SC R RSC + 1 U +f = U R 1 0 +R+ RSC + 1 SC R U +f = U SCR + SCR ( SCR + 1) + SCR + 1 0 SC SCR U +f = U 2 ( SCR ) + 3SCR + 1 0 1 ϖ0 = RC 1 f = 2πRC R*

R2 40k

-15

U1 4 1 5

Uo

R1

2 6

10k

7

3

UA741

R5

R6 230

230

C2

R3

+15 0.1uF

R4

C1

20k

0.1uF

20k

D1

D2

LED-BLUE

LED-RED

ientes

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