Respuesta de Frecuencia(Electrnica II)

Joshua Mndez

CI. 24712677

Rs= 17 ohms

R1= 10 Kohms

RE= 10 ohms

RL= 1 Kohms

R2= 70 Kohms

Rc = 7kohms

c1= c2= 100 microfaradios

CE =10 microfaradios

La resistencia en paralelo a

CE es de 10 ohms

hfe=100

Cbe= 7pf

Cce= 1pf

Cbc= 1pf

Solucin

Asumo 20v=Vcc

En frecuencias bajas

=

+

=2

1+2(Divisor de voltaje)

20(70)

10+70= 17.5

=>Rb=R1//R2=10k//70k=8.75k

=>Ica=17.50.7 100

8.75100(10)= 172.31

=>hie=26

=

100(26)

172.31= 15.1

=>hie=15.1

Asi Rest=Rb// (hie+BRe)

=8.75k (15.1+100(10))

=8.75k//1.015k

Rest=909.5

As Rc1=Rs+Rest

=17+909.5

=926.5

=>1

211=

1

2 926.5 100

=>Fc1=1.72h

=//+

+ //1

=17//8,75

100+ 10 //10=5

=> =1

2=

1

2 5 10=1,013khertz

2 = + 2 = 7 + 1 = 8

=> 2=1

222=

1

2 8 100=0,199khertz

Luego

=> = max , 1, 2=max(1.013,1.72,0.199)Hertz

=1013Hertz

=1,013khertz

As: F1=1,013khertz

En Altas Frecuencias

C1,C2 y Ce en corto

La capacitancia de Miller es:

= 1 ; = = 15

=//

+=

100(1//7)

15,1+1= - 86,2

: = 1 + 86,2 1 = 87,2

Luego

= 0////Rb +hgeRe

= 17//8,75//15 +100

=1k

7

=> =1

2=

1

2(+)

=> =1

2(1)(82,2+7)= 1,78

= + //RL

= 10+1k//7k

= 885

=> =1

2 885 1)= 179,93

=> = min , = min 1,783, 179,93

= ,