dispositivos electronicos

Problema N1
Determinar V e Id








V = 12 – 0.7 – 0.3
V = 11 V
Hallando Id
12 = 0.7 + 0.3 + 5.6K(Id)
Id = 1.96 mA
Problema N2
Determinar I, V1, V2 y V







-10 + 4.7K(I) + 0.7 + 2.2K(I) – 5 = 0
6.9K(I) = 14.3
I = 2.07 mA
Hallando V1 Hallando V2 Hallando V

V1 = 4.7K(2.07)m V2 = 2.2K(2.07)m V = 2.2K(2.07)m - 5
V1 = 9.729 V V2 = 4.554 V V = -0.44 VProblema N3
Determinar V

V = 10 – 0.7
V = 9.3 v
Problema N4
Para la red de diodo zener. Determinar Vrl, Vr, Iz, Pz. Vz = 10v



Hallando si el diodo trabaja
Vrl de ser mayor o igualque Vz

Vrl = RLx16/(RL + R)
Vrl = 1.2Kx16/(1K + 1.2K)
Vrl = 8.72v

Pero el voltaje del zener es 10v, por lo tanto el diodo está en el estado de apagado
Vrl = 8.72v

-16 + Vr + Vrl = 0
Vr =16 – 8.72
Vr = 7.28v

Iz = 0 A
Pz = 0 W
Problema N5
Determinar Zi, Zo, Vi, Vo. Vbe tiende a cero




β = 90 = Ic/Ib









Aplicando Thevenin







Rbb =8.2Kx(56k)/(56 + 8.2)k
Rbb = 7.15KΩ
Vbb = 8.2kx(22)/(56 + 8.2)k
Vbb = 2.8v
-2.8 + 7.15K(Ib) + Vbe + 1.5K(Ic) = 0
7.15K(Ib) + 1.5K(90Ib) = 2.8
Ib = 19.6µA
Ic=1.7mA
Analizando:

Hie = 25mVxhfe/Ic Vi= hie(Ib)
hie = 25mVx90/1.7mA Vi = 0.025v
hie = 1.3KΩ
Zi = 7.15K(1.3K)/(7.15 + 1.3)K Vo = -hfe(Ib)Rc
Zi = 1.1KΩ Vo = -11.99v
Zo = 6.8KΩ

Problema N6
Determinar Zi, Zo, Vi, Vo. Vbetiede a cero
β = 210 = Ic/Ib
Aplicando Thevenin






Rbb = 10Kx(90K)/100K
Rbb = 9KΩ
Vbb = 10K(16)/100k
Vbb = 1.6v
1.6 = 9k(Ib) + Vbe + 0.68k(Ic)
1.6 = 9k(Ib) + 0.68k(210Ib)
Ib =10.5µA Entonces Ic = 2.2mA
Hie = 25mV(210)/2.2mA
hie = 2.3KΩ
Analizando:

Zi = 9k(2.3k)/11.3k
Zi = 1.83KΩ
Zo = 2.2KΩ
Vi = hie(Ib)
Vi = 2.3kx(10.5)µ
Vi = 0.024
Vo = -hfe.Ib.Rc
Vo =-210(10.5)µx2.2k
Vo = -4.85

Problema N7
Determinar Zi, Zo, Vi, Vo
Vbe tiende a cero

Β = 120 = Ic/Ib




20 = 470k(Ib) + Vbe + 0.56k(Ic)
20 = 470k(Ib) + 0.56k(120Ib)
Ib = 37.2µA
Ic = 4.4mA...