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EE421N - SOLUCIONARIO EF - 2020 - II

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UNIVERSIDAD NACIONAL DE
INGENIERIA
Facultad de Ingeniería Eléctrica y
Electrónica
TEMA:
“SOLUCION EXAMEN FINAL – 2020 - II”
PROFESOR:
LOPEZ ARAMBURU FERNANDO MAXIMILIANO
ALUMNO
• DELEGADO 1: GOMEZ RODRIGUEZ CHRISTIAN MANUEL
CODIGO
20154079K
CURSO-SECCION
EE421 – “N”
Marzo
2021
SOLUCIONARIO EXAMEN FINAL – 2020 II
PREGUNTA 01:
SOLUCION:
ANALISIS EN DC:
𝑅=
12 − (−10.6)
𝐼0
𝑅=
22.6
… (1)
𝐼0
β„Žπ‘–π‘’3 = β„Žπ‘–π‘’4 =
25π‘šπ‘‰
3𝐼0⁄
200
β„Žπ‘–π‘’1 = β„Žπ‘–π‘’2 =
25π‘šπ‘‰
3𝐼0
ANALISIS EN AC:
DATO: 𝑍𝑖𝑛 = 40𝐾
𝑍𝑖𝑛 = β„Žπ‘–π‘’1 + β„Žπ‘–π‘’2 + β„Žπ‘–π‘’3 + β„Žπ‘–π‘’4
40𝐾 = 2 ∗ (
25π‘šπ‘‰
25π‘šπ‘‰
)
+
3𝐼0⁄
3𝐼0
200
𝐼0 = 83.75 πœ‡π΄ … (2)
De (2) en (1): 𝑅 = 269.85𝐾Ω
5βˆ†π‘–
π‘‰π‘š = −2 (
6
) (20𝐾)
βˆ†π‘–
𝑉𝑖 = − (200) (𝑍𝑖𝑛 )
π‘‰π‘š
= βˆ†π‘‰ = −166.6
𝑉𝑖
π‘†π‘Žπ‘π‘’π‘šπ‘œπ‘  𝑓 ′ : 𝑉𝑖 |
π‘šπ‘Žπ‘₯
= 25π‘šπ‘‰ = 𝑉𝑇
|π‘‰π‘š | = (166)(25) → |π‘‰π‘š | = 4.16 π‘‰π‘œπ‘™π‘‘π‘ 
PROBLEMA 02:
SOLUCION:
ANALISIS EN AC:
En Q2:
𝑉𝑋 = (1.2𝐾 + 6.4𝐾 ∗ (101)) ∗ 𝑖𝑏2
𝑉𝑋 = 647.6𝐾 ∗ 𝑖𝑏2 → 𝑖𝑏2 =
π‘₯ = 100𝑖𝑏2 +
𝑉𝑋
… (1)
647.6𝐾
π‘‰π‘œ
… (2)
1.5𝐾
𝑉𝑋 = 10𝐾. π‘₯ + π‘‰π‘œ
𝑉𝑋 = 10𝐾. (100𝑖𝑏2 +
𝑉𝑋 = 106 . (
𝑉𝑋
) + 7.6π‘‰π‘œ
647.6𝐾
𝑉𝑋 = 1.54𝑉𝑋 + 7.6π‘‰π‘œ
−0.54𝑉𝑋 = 7.6π‘‰π‘œ
π‘‰π‘œ
= 𝐴𝑉2 = −0.071
𝑉𝑋
π‘π‘’π‘ž2 =
𝑉𝑋
𝑖
π‘‰π‘œ
) + π‘‰π‘œ … (3)
1.5𝐾
𝑖 = π‘₯ + 𝑖𝑏2 , π‘Ÿπ‘’π‘’π‘šπ‘π‘™π‘Žπ‘§π‘Žπ‘›π‘‘π‘œ π‘π‘œπ‘› (2)
𝑖 = 101. 𝑖𝑏2 +
π‘‰π‘œ
1.5𝐾
𝑉𝑋
−0.071. 𝑉𝑋
)+
𝑖 = 101. (
647.6𝐾
1.5𝐾
𝑖 = 10.8 ∗ 10−5 . 𝑉𝑋
π‘π‘’π‘ž2 =
𝑉𝑋
= 9.21𝐾Ω
𝑖
En Q1:
72
𝑉𝐺 = 𝑉𝑖 ( ) → 𝑉𝐺 = 0.88𝑉𝑖
82
𝑉𝑆 = (2.5𝑉𝐺𝑆 )(2.26) → 𝑉𝑆 = 5.6𝑉𝐺𝑆
𝑉𝐺𝑆 = 𝑉𝐺 − 𝑉𝑆 = 𝑉𝐺 = 0.88. 𝑉𝑖 − 5.6. 𝑉𝐺𝑆
6.6. 𝑉𝐺𝑆 = 0.88. 𝑉𝑖
𝑉𝑖 = 7.6𝑉𝐺𝑆 →
𝑉𝐺𝑆
= 0.13
𝑉𝑖
𝑉𝑆 = 𝑉𝑋 = 5.6𝑉𝐺𝑆 →
𝑉𝑋
= 5.6
𝑉𝐺𝑆
𝑉0 𝑉π‘₯ 𝑉𝐺𝑆
𝐴𝑉 = ( ) ( ) ( ) = (−0.071)(0.13)(5.6) → 𝐴𝑉 = −0.052
𝑉π‘₯ 𝑉𝐺𝑆 𝑉𝑖
Ahora: 𝑓𝐻 = β‹― ? ?
πΆπœ‡ :
πœπœ‡ = (10𝐾 βˆ₯ 72𝐾). 5𝑝𝐹
πœπœ‡ = 43.9 ∗ 10−9
πœ”πœ‡ =
1
= 22.78 𝑀 π‘Ÿπ‘Žπ‘‘⁄𝑠𝑒𝑔 → π‘“πœ‡ = 3.6 𝑀𝐻𝑧
πœπœ‡
πΆπœ‹ :
π‘‰πœ‹ = 8.87𝐾. 𝑖 + (𝑖 − 2.5 ∗ 10−3 . π‘‰πœ‹ )(2.26𝐾)
π‘‰πœ‹ = 11.13𝐾. 𝑖 − 5.65π‘‰πœ‹
6.65π‘‰πœ‹ = 11.13𝑖 →
π‘‰πœ‹
11.13𝐾
= π‘…π‘’π‘ž−πœ‹ =
𝑖
6.65
π‘…π‘’π‘ž−πœ‹ = 1.67𝐾Ω
πœπœ‹ = (1.67𝐾)(5𝑝𝐹) = 8.35 ∗ 10−9
πœ”πœ‹ =
1
= 119.7𝑀 π‘Ÿπ‘Žπ‘‘⁄𝑠𝑒𝑔 → π‘“πœ‡ = 19 𝑀𝐻𝑧
πœπœ‹
En Q2:
πΆπœ‹ :
Calculemos 𝑍′0 (𝑉𝑖 = 0): (π‘†π‘Žπ‘™π‘–π‘‘π‘Ž 𝑑𝑒𝑙 𝐹𝐸𝑇)
𝑉 = (2.5𝑉𝐺𝑆 )(3𝐾) → 𝑉 = 7.5𝐾. 𝑉𝐺𝑆 → 𝑉𝐺𝑆 =
𝑉
7.5𝐾
𝑉
𝑉
)+𝑖 =
2.5 (
→ 𝑖 = 0 → 𝒁′𝟎 = ∞
7.5𝐾
3𝐾
𝑖𝑏2 =
π‘‰πœ‹
100
= 0.83π‘šπ‘‰πœ‹ → 100. 𝑖𝑏2 = 83.33π‘šπ‘‰πœ‹ =
𝑉
1.2𝐾
1.2𝐾 πœ‹
Por nodos:
𝑉𝑖 − 𝑉𝐸 𝑉𝑖 − 𝑉𝑂
+
=0
1.2𝐾
10𝐾
𝑉𝐸 = 312.74π‘‰πœ‹
𝑉𝑂
𝑉𝑂 − 𝑉𝑖
100
+
=−
𝑉 𝑉𝑖 = 271.38π‘‰πœ‹
1.5𝐾
10𝐾
1.2𝐾 πœ‹ 𝑉 = −73.30𝑉
π‘œ
πœ‹
𝑉𝐸
100
𝑉𝑖 − 𝑉𝐸
=
π‘‰πœ‹ +
6.4𝐾 1.2𝐾
1.2𝐾 }
π‘‰πœ‹
𝑉𝑖 − π‘‰π‘œ
π‘‰πœ‹
𝑉𝑖 − π‘‰π‘œ
𝑖=
+
=
+
1.2𝐾
10𝐾
1.2𝐾
10𝐾
𝑖 = π‘‰πœ‹ [0.0353]
π‘‰πœ‹
= π‘…π‘’π‘ž−πœ‹ = 28.33Ω
𝑖
πœπœ‹ = (28.33)(5𝑝𝐹) = 1.42 ∗ 10−10
πœ”πœ‹ =
1
= 7.06 𝐺 π‘Ÿπ‘Žπ‘‘⁄𝑠𝑒𝑔
πœπœ‹
π‘“πœ‡ = 1.12 𝐺𝐻𝑧
1
344.68
𝑖 = π‘‰πœ‹ . [
+
]
1.2𝐾
10𝐾
πΆπœ‡ :
𝑖=
π‘‰πœ‹
1.2𝐾
π‘‰πœ‡
= 1.2𝐾(664.92)
𝑖
100. π‘‰πœ‹
π‘‰πœ‹
π‘‰πœ‡ = π‘‰πœ‹ + (6.4𝐾 + 1.5𝐾) (
+
)
1.2𝐾
1.2𝐾
π‘‰πœ‡ = π‘‰πœ‹ [1 + (7.9𝐾)(84.16π‘š)]
π‘‰πœ‡ = π‘‰πœ‹ (664.92) → π‘‰πœ‹ =
π‘…π‘’π‘ž−πœ‡ =
πœπœ‡ = (797.9𝐾). 5𝑝𝐹
1
664.92
π‘‰πœ‡
π‘‰πœ‹
π‘ƒπ‘’π‘Ÿπ‘œ: 𝑖 =
→ 𝑖=
1.2𝐾
1.2𝐾(664.92)
π‘‰πœ‡
= 797.9𝐾Ω
𝑖
πœπœ‡ = 3.99 ∗ 10−6
πœ”πœ‡ =
1
= 250.6 𝐾 π‘Ÿπ‘Žπ‘‘⁄𝑠𝑒𝑔
πœπœ‡
π‘“πœ‡ = 39.89 𝐾𝐻𝑧
PROBLEMA 03:
SOLUCION:
Calculemos πΌπ‘œ :
πΌπ‘œ =
0 − (−10)
→ πΌπ‘œ = 5 π‘šπ΄
2𝐾
2. βˆ†π‘– = πΌπ‘œ →
β„Žπ‘–π‘’1 = β„Žπ‘–π‘’2 =
25π‘šπ‘‰
𝐼𝑏1
=
πΌπ‘œ
= βˆ†π‘– = 2.5π‘šπ΄
2
25π‘šπ‘‰
βˆ†π‘–⁄
𝛽
=
25π‘šπ‘‰
2.5π‘šπ΄⁄
100
= 1000
β„Žπ‘–π‘’1 = β„Žπ‘–π‘’2 = 1000Ω
βˆ†π‘–
𝛽
𝑉𝑖 = (2. β„Žπ‘–π‘’1 ) ( ) ∧ π‘‰π‘œ = (2. βˆ†π‘–)(10𝐾)
π‘‰π‘œ
𝑉𝑖
=
(2. βˆ†π‘–)(10𝐾)
(2. β„Žπ‘–π‘’1 ) (βˆ†π‘–⁄𝛽 )
→ 𝐴𝑉 = 1000
𝑍𝑖𝑛 = (2. β„Žπ‘–π‘’ βˆ₯ 5𝐾) = (2𝐾 βˆ₯ 5𝐾)
𝑍𝑖𝑛 = 1.42𝐾
∧ π‘π‘œπ‘’π‘‘ = 10𝐾
PROBLEMA 04:
SOLUCION:
Para: 𝑽 = −πŸπŸ“ 𝑽
𝑉𝐺 = −24.3 𝑉
∧
𝑉𝑆 = 3𝐾. 𝐼𝐷 − 25
𝑉𝐺𝑆 = −24.3 − 3. 𝐼𝐷 + 25
𝑉𝐺𝑆 = 0.7 − 3𝐾. 𝐼𝐷
Aplicando Ecuaciones FET:
𝑉𝐺𝑆 2
𝐼𝐷 = 𝐼𝐷𝑆𝑆 . [1 −
]
𝑉𝑃𝑂
0.7 − 3𝐾. 𝐼𝐷 2
𝐼𝐷 = 10π‘š. [1 −
]
−5
𝐼𝐷 =
10π‘š
𝐼𝐷1 = 1.3π‘šπ΄ → 𝑉𝐺𝑆 = −3.197 √ → 𝑉𝐷𝑆 = 21.102 π‘‰π‘œπ‘™π‘‘
. [5.7 − 3𝐾. 𝐼𝐷 ]2 {
25
𝐼𝐷2 = 2.78π‘šπ΄ → 𝑉𝐺𝑆 = −7.636 𝑋
Para: 𝑽 = −πŸ‘πŸ 𝑽
𝐼𝐷 =
{
10π‘š
. [5.7 − 3𝐾. 𝐼𝐷 ]2
32
𝐼𝐷1 = 1.237π‘šπ΄ → 𝑉𝐺𝑆 = −3.01 √
𝐼𝐷2 = 2.919π‘šπ΄ → 𝑉𝐺𝑆 = −8.056 𝑋
𝑉𝐷𝑆 = 32 − 3 ∗ 103 (1.237π‘š)
𝑉𝐷𝑆 = 28.289 π‘‰π‘œπ‘™π‘‘
βˆ†π‘‰π·π‘† = 7.187
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