Vp = 0,7 + Vk Vcc

1
VC C
4k
Vth =
15 · 20k
= 12
5k + 20k
Rth =
5k · 20k
= 4k
5k + 20k
5k
10k
Q1
PN P
PUT
20k
2 0k
C
1uF
50
Rk
Vcc = 4k · ie + 0,7 + Rth ib + Vth
VCC
4k
ie
Q1
PNP
ib
Vth
Vcc = 4k (β + 1) · ib + 0,7 + Rth ib + Vth
Rth
ic
C
1uF
ib =
15 − 12 − 0,7
= 2,846µA
4k · 201 + 4k
ic = βib = 0,569mA
T
Vp
Vc
Vc =
ic
t + Vv
C
Vv
t2
t1
Vp-Vv
Vc (t = t1 ) = Vp
VRK
t1 =
10 · 1µf
(Vp − Vv ) C
=
= 17,574ms
ic
0,569mA
Figura 1. Gráficas de la pregunta 1
t2 = Rk C · ln
Vp = 0,7 + Vk
Vk =
Vcc · 20k
= 10
10k + 20k
Vp = 10,7
Vp
Vv
= 50 · 1µf · ln
10,7
0,7
T = t1 + t2 = 17,71ms
= 0,136ms
2
VF = 100 sin(wt) f = 60Hz
Pregunta 2.
R1 = 1k R2 = 2k
π
3π
α1 =
α2 =
4
2
x1 = arcsin
VF
p
0
−30
100
R3 = 1k
E = 30
= −0,3 = −17,45◦
2p
x2 = π − x1 = 3,446 = 197,45◦
Vg
a2
a1
x3 = 2π + x1 = 5,97 = 342,54◦
π
≤ wt ≤ x2
4
100 sin (wt) + 30
VR1 = VR3 =
2
VR2 = 0
VF+E
x3
x2
x1
3π
≤ wt ≤ x3
2
2 (100 sin (wt) + 30)
VR2 =
3
100 sin (wt) + 30
VR3 =
3
VR1 = 0
VR1
a1
x2
VR2
a2
x3
"
Vdc (R3 ) =
Vdc (R3 ) =
VR3
a2
a1
1
2π
1
2π
´x2 (100 sin(wt)+30)
π
4
2
dwt +
´x3
3π
2
#
(100 sin(wt)+30)
dwt
3
h x
50 − cos(wt) π2 + 15 x2 − π4 · · ·
4
x3
x
· · · + 100
− cos(wt) 3π3 + 10 x3 −
3
x2
2
Vdc (R3 ) =
Figura 2. Gráficas de la pregunta 2
1
2π
[83,055 + 39,912 − 31,8 + 12,66]
Vdc (R3 ) = 16,5245
3π
2
i
3
T1
R1
Vf
VF
π
0
L1
π
√
VF = 220 2 sin(wt) f = 60Hz
R1 = 1Ω L1 = 2,65mH
π
3π
α=
y α=
3
2
Vg
2
1
XL = wL = 2πf L = 2π · 60 · 2,65mH = 1
q
Z = R2 + XL2 = 1,4142
XL
φ = arctan
= 0,7853 = 45◦
R
IRL
2
1
β2
β1
α=
π
3
α=
y
π
2
φ=
y
π
=⇒ β1 = 3,9122 = 224,15◦
4
φ=
β2 = π + β
VRL
β1
β2
√
2
=
Vrms
(220
2)
2
(220
2)
2
···
β2
√
2
Vrms
=
(220
´β1
2
sin (wt)dwt +
π
3
2π
IT1
2
"
2π
√
2
=
Vrms
β1
=⇒ β2 = 6,9965 = 400,87◦
2
1
1
π
=⇒ β = 3,855 = 220,87◦
4
2)
2π
h
3π
β1 +β2 − π
3− 2
2
Vrms = 187,96
1
β1
β2-2π
Figura 3. Gráficas de la pregunta 3
2
#
2
sin (wt)dwt
3π
2
···
3π
sin(2 π
3 )+sin(2 2 )−sin(2β1 )−sin(2β2 )
+
4
2
[2,574 − 0,2807]
√ q 2,2933
Vrms = 220 2
2π
VAK1
´β2