Electronica

-ALEJANDRO SARABIA VALENTINEZ

MATERIA: ANALOGICA 2

SEMESTRE: 6

CARRERA: ING. ELECTRONICA

UNIDAD: 1

PROFESOR: ING.CUAUTHEMOC MOLINA AGUILAR



Calcular: Zi, Zo,∆v
ETAPA DEL BJT:
Req=15k∥4.7k=3.57k
Veq=20v(4.7k)(15k+4.7k)=4.77v
Ib(4.77v-0.7)3.57k+2001k=19.9uA
Ic=20019.9uA=3.98mA
Ie=19.9uA+3.98mA=3.99mA
re=26mV3.99mA=6.51Ωhie=2006.51Ω=1.3k
Zi=3.57k∥1.3k=952.97
Δv=-1k∥2.2k6.51Ω=-105.60v



ETAPA DEL FET:
Id=10mA1--x680-4v2
Id=2.7mA
Vgs=-2.7mA680Ω=-1.83v
gm=210mA4v(1--1.83v-4v)=2.7mAΔV=-2.7mA2.4k1k(3.3M)((2.4k+1k)(1+2.7mA)(3.3M)=-1.90v



Calcular: Zi, Zo, ∆T






CIRCUITO HIBRIDO DE LA 2 ETAPA SIN Cs:

Zo=2.4k∥1k=705.8Ω
CIRCUITO HIBRIDO DELA 1 ETAPA CON Cs:




Zi=3.3MΩ
ETAPA 2 DEL FET:

Id=10mA1--x680-4v2
Id=2.7mA
Vgs=-2.7mA680Ω=-1.83v
gm=210mA4v1—1.83v-4v=2.7mAΔV=-2.7mA2.4k1k(3.3M)((2.4k+1k)(1+2.7mA)(3.3M)=-1.90v
Id=10mA1--x680-4v2=2.7mA
Vgs=-2.7mA680Ω=-1.83v

ΔV=-2.7mA2.4k3.3M(10M)((2.4k+3M)(1+2.7mA)(3.3M)=-21.5v

ΔVT=(-1.90V)(-21.5V)=-40.8Calcular: Zi, Zo, ∆v,∆i

Req=15k∥4.7k=3.57k
Veq=20v(4.7k)(15k+4.7k)=4.77v
Ib(4.77v-0.7)3.57k+2001k=19.9uA
Ic=20019.9uA=3.92mA
Ie=19.9uA+3.92mA=3.93mAVce=20v-3.92mA2.2k-3.93mA1k=7.4v
re=26mV3.92mA=6.6Ω
hie=2006.6Ω=1.3k
Zb=1.3k+2001k=201.300k
Zi=3.57k∥1.3k=947.9Ω
Req=15k∥4.7k=3.57k=3.5k
Zo=50Ω∥2.2k=48.8 Ω →ETAPA 2

Zo=2.KΩ∥1k=687.5 Ω →ETAPA 1Δv1=-2.2k∥947.9Ωre=-100.36v→ETAPA 1

Δv2=-2.2k∥50Ωre=-7.40v→ETAPA 2






Δv=-5
Ic=150mA
Rc=300 Ω
Rb=1k
RL=25 Ω

Icmax=10v1k+860=5.37mAIb=5.37mA200=26.85uA
Ie=26.85uA+5.37mA=5.39mA
re=26mv5.39mA=4.82Ω
hie=2004.82 Ω=964
Zt=300∥964=228.7Ω
Zo=300∥25Ω=24.3Ω
Zb=964+1+200680=1.3k
∆v=-20030025(1k)300+251k+1.3k(228.7)=-5.06v