Boylestad 11th Edition Halaman 362 - 364 No. 11 & 17
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11. For the network of Fig. 5.151 : a. Determine Zi and Zo . b. Find Av . c. Repeat parts (a) and (b) with ro = 20 k
12 V
220 k
2.2 k
Io Vo Ξ² = 60 ro = 40 k
Vi
Zo Iin
Zin DC analysist:
12 V
220 k
2.2 k
Ξ² = 60 ro = 40 k
-Vcc + VRB + VBE = 0 -Vcc + IBRB + VBE = 0
IB =
ππΆπΆ β ππ΅πΈ π π΅
=
12β0.7 220π
= 51.36 Β΅A
IE = (Ξ² + 1) IB = (60+1) x 51.36 x 10-6 = 3.13318 mA
AC analysist with re parameter:
ib
b
c
ic
+
+
io
vin
Ξ² ib
Ξ² re
220 k
ro
vo
2.2 k
-
-
e
e
Zin
re =
Zo
26 ππ πΌπΈ
26 π₯ 10β3
= 3.13318π₯10β3 = 8.3 β¦
a. Zi = 220k // (Ξ²re) =
220k(60x8.3) 220k+60x8.3
= 496.77 β¦
Because the ro β₯ 10Rc , therefore Zo = Rc = 2.2 k β¦ b. Av =
ππ ππ
βπππ₯π π
βπππ₯π π
= ππ π₯ π½ππ = ππ π₯ π½ππ =
βπ½πππ₯π π ππ π₯ π½ππ
=
βπ π ππ
=
β2.2π 8.3
= -265.06
220k(60x8.3)
c. Zi = 220k // (Ξ²re) = 220k+60x8.3 = 496.77 β¦ Zo = Rc // ro = Av =
ππ ππ
=
2.2π π₯ 20π 2.2π+20π
βπππ₯(π π//ππ) ππ π₯ π½ππ
=
= 1.98 kβ¦ βπππ₯(π π//ππ) ππ π₯ π½ππ
17. For the network of Fig. 5.156 : a. Determine re . b. Calculate VB and VC . c. Determine Zi and Av = Vo>Vi.
=
βπ½πππ₯(π π//ππ) ππ π₯ π½ππ
=
β(π π//ππ) ππ
=
β1.98π 8.3
= -238.79
Vcc = 20 V
4.7 k
220 k
VC
VB
Vi
Ξ² = 180 gos = 30 Β΅S
Vo Cc
Cc
Zin 56 k
2.2 k
DC analysist :
Vcc = 20 V
4.7 k
Ξ² = 180 gos = 30 Β΅S RTH
2.2 k VTH
CE
56π
VTH = 56π+220π x 20 = 4.06 V RTH = 56k // 220k =
56π π₯ 220π 56π+220π
= 44637.68 β¦
βV = 0 -VTH + VRTH + VBE + VRE = 0 -VTH + IBRTH + VBE + IERE = 0 -VTH + IBRTH + VBE + (Ξ²+1)IBRE = 0 IB = π
πππ» β ππ΅πΈ ππ»
4.06β0.7
+ (π½+1)π πΈ
= 44637.68 +(180+1)x2200 = 7.59 Β΅A
IE = (Ξ² + 1) IB = 181 x 7.59 x 10-6 = 1.373 mA AC analysist with re parameter :
ib
b
c
ic
+
+
io
vin
Ξ² re
RTH
Ξ² ib
ro
vo
4.7 k
-
-
e
e
Zin
Zo
a. re =
26 ππ πΌπΈ
=
26 π₯ 10β3 1.373 π₯ 10β3
= 18.93 β¦
b. VB = VBE + VRE = VBE + IERE = 0.7 + 1.373x10-3x2.2k = 3.72 V VC = Vcc β VRC = Vcc β Ξ²IBRC = 20 β 180x7.59x10-6x4.7k = 13.58 V c. Zi = RTH || (Ξ²re) =
44637.68 x (180x18.93) 44637.68+180x18.93 1
ro = π = ππ
1 30 π₯ 10β6
= 3165.74 β¦
= 33.3 k β¦
Zo = ro || RC = 33.3k || 4.7k =
33.3π π₯ 4.7π
Av = = = = = =
33.3π+4.7π
= 4119.19369 β¦
ππ ππ βππ(π π//ππ) ππ π₯ π½ππ βππ(π π//ππ) ππ π₯ π½ππ βπ½ππ(π π//ππ) ππ π₯ π½ππ β(π π//ππ) ππ β4119.19369 18.93
= -217.6
12 V
220 k
2.2 k
Io Vo Ξ² = 60 ro = 40 k
Vi
Zo Iin
Zin DC analysist:
12 V
220 k
2.2 k
Ξ² = 60 ro = 40 k
-Vcc + VRB + VBE = 0 -Vcc + IBRB + VBE = 0
IB =
ππΆπΆ β ππ΅πΈ π π΅
=
12β0.7 220π
= 51.36 Β΅A
IE = (Ξ² + 1) IB = (60+1) x 51.36 x 10-6 = 3.13318 mA
AC analysist with re parameter:
ib
b
c
ic
+
+
io
vin
Ξ² ib
Ξ² re
220 k
ro
vo
2.2 k
-
-
e
e
Zin
re =
Zo
26 ππ πΌπΈ
26 π₯ 10β3
= 3.13318π₯10β3 = 8.3 β¦
a. Zi = 220k // (Ξ²re) =
220k(60x8.3) 220k+60x8.3
= 496.77 β¦
Because the ro β₯ 10Rc , therefore Zo = Rc = 2.2 k β¦ b. Av =
ππ ππ
βπππ₯π π
βπππ₯π π
= ππ π₯ π½ππ = ππ π₯ π½ππ =
βπ½πππ₯π π ππ π₯ π½ππ
=
βπ π ππ
=
β2.2π 8.3
= -265.06
220k(60x8.3)
c. Zi = 220k // (Ξ²re) = 220k+60x8.3 = 496.77 β¦ Zo = Rc // ro = Av =
ππ ππ
=
2.2π π₯ 20π 2.2π+20π
βπππ₯(π π//ππ) ππ π₯ π½ππ
=
= 1.98 kβ¦ βπππ₯(π π//ππ) ππ π₯ π½ππ
17. For the network of Fig. 5.156 : a. Determine re . b. Calculate VB and VC . c. Determine Zi and Av = Vo>Vi.
=
βπ½πππ₯(π π//ππ) ππ π₯ π½ππ
=
β(π π//ππ) ππ
=
β1.98π 8.3
= -238.79
Vcc = 20 V
4.7 k
220 k
VC
VB
Vi
Ξ² = 180 gos = 30 Β΅S
Vo Cc
Cc
Zin 56 k
2.2 k
DC analysist :
Vcc = 20 V
4.7 k
Ξ² = 180 gos = 30 Β΅S RTH
2.2 k VTH
CE
56π
VTH = 56π+220π x 20 = 4.06 V RTH = 56k // 220k =
56π π₯ 220π 56π+220π
= 44637.68 β¦
βV = 0 -VTH + VRTH + VBE + VRE = 0 -VTH + IBRTH + VBE + IERE = 0 -VTH + IBRTH + VBE + (Ξ²+1)IBRE = 0 IB = π
πππ» β ππ΅πΈ ππ»
4.06β0.7
+ (π½+1)π πΈ
= 44637.68 +(180+1)x2200 = 7.59 Β΅A
IE = (Ξ² + 1) IB = 181 x 7.59 x 10-6 = 1.373 mA AC analysist with re parameter :
ib
b
c
ic
+
+
io
vin
Ξ² re
RTH
Ξ² ib
ro
vo
4.7 k
-
-
e
e
Zin
Zo
a. re =
26 ππ πΌπΈ
=
26 π₯ 10β3 1.373 π₯ 10β3
= 18.93 β¦
b. VB = VBE + VRE = VBE + IERE = 0.7 + 1.373x10-3x2.2k = 3.72 V VC = Vcc β VRC = Vcc β Ξ²IBRC = 20 β 180x7.59x10-6x4.7k = 13.58 V c. Zi = RTH || (Ξ²re) =
44637.68 x (180x18.93) 44637.68+180x18.93 1
ro = π = ππ
1 30 π₯ 10β6
= 3165.74 β¦
= 33.3 k β¦
Zo = ro || RC = 33.3k || 4.7k =
33.3π π₯ 4.7π
Av = = = = = =
33.3π+4.7π
= 4119.19369 β¦
ππ ππ βππ(π π//ππ) ππ π₯ π½ππ βππ(π π//ππ) ππ π₯ π½ππ βπ½ππ(π π//ππ) ππ π₯ π½ππ β(π π//ππ) ππ β4119.19369 18.93
= -217.6
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