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Uploaded by Leonardo Estrada

Seminario PC4-20-1-NEW

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PREGUNTA 1
Datos:
𝐼𝐷𝑆𝑆 = 6π‘šπ΄
𝛽 = 200
πΆπœ‡ = πΆπœ‹ = 15𝑝𝐹
𝑉𝑃𝑂 = −2𝑉
πœ‡ = 400
πΆπœ‡ = πΆπœ‹ = 10𝑝𝐹
Solución.
ANÁLISIS EN DC.
Q1:
Q2:
𝑉𝑃𝑂 = −2𝑉
𝑉𝐺𝑆 = −0.5𝐼𝐷
𝐼𝐷 = 6π‘š [1 −
𝑉𝐺𝑆 2
]
𝑉𝑃𝑂
Resolviendo:
𝐼𝐷 = 2.67 π‘šπ΄
π‘”π‘š = 2π‘šπ΄/𝑉
π‘Ÿπ·π‘† = 200𝐾Ω
ANÁLISIS EN AC:
LOW FREQ.:
Zero [FET]:
1
1
=
𝑅𝑆 𝐢𝑆 10 × 10−6 × 0.5 × 103
𝒓𝒂𝒅
= 𝟐𝟎𝟎
⇒ π’‡π‘ΆπŸ = πŸ‘πŸ. πŸ–π‘―π’›
𝒔
πœ”π‘‚1 =
πŽπ‘ΆπŸ
Polo [FET]
πœ”π‘ƒ1 =
πŽπ‘·πŸ = πŸ’πŸŽπŸŽ
𝑔𝑆 + π‘”π‘š
𝐢𝑆
𝒓𝒂𝒅
⇒
𝒔
π’‡π‘·πŸ = πŸ”πŸ‘. πŸ”π‘―π’›
Zero [BJT]:
1
1
=
𝑅𝐸 𝐢𝐸 (1.5π‘˜)(47πœ‡)
𝒓𝒂𝒅
= πŸπŸ’. πŸπŸ–
⇒ π’‡π‘ΆπŸ = 𝟐. πŸπŸ“π‘―π’›
𝒔
πœ”π‘‚2 =
πŽπ‘ΆπŸ
Polo [BJT]:
πœ”π‘ƒ2 =
1
1
2.7 + 1.18
[( 200 ) ||1.5] 47πœ‡
= 19.4Ω||1.5π‘˜Ω ≅ 19.4 Ω
π‘…π‘ž∗ 𝐢𝐸
∗
𝑅𝐸𝑄
=
πŽπ‘·πŸ = πŸπŸŽπŸ”πŸ‘. πŸ–
𝒓𝒂𝒅
𝒔
⇒
π’‡π‘·πŸ = πŸπŸ•πŸ– 𝑯𝒛
MEDIUM FREQ.:
2.7
2.7 + 1.18
𝑉𝑂 = −𝛽𝑖𝑏 𝑅𝑂
𝑖𝑏 = −π‘”π‘š 𝑉𝐺𝑆
𝑅𝑂 = 1.5||3.3 π‘˜Ω
𝑽𝑢
= 𝑨𝑽𝑢 = πŸ‘πŸ—πŸ•. πŸ“
π‘½π’Š
HIGH FREQ.:
𝐢𝑇1 = πΆπœ‹ + πΆπœ‡ (1 + π‘”π‘š 𝑅𝑂 𝐹𝐸𝑇 )
𝐢𝑇2 = πΆπœ‹ + πΆπœ‡ (1 − 𝐴𝑉 )
𝑅𝑂 𝐹𝐸𝑇 = 0.8 π‘˜Ω
𝐢𝑇2 = 10𝑝 + 10𝑝 (1 − 𝐴𝑉 )
𝛽𝑅𝑂
200(1,43)
𝐴𝑉 = −
=−
= −348.7
β„Žπ‘–π‘’
0.82
𝐢𝑇2 = 3507 𝑝𝐹
1
πœ”π»2 =
(3507.8 × 10−12 )(0.8𝐾)
𝒓𝒂𝒅
πŽπ‘―πŸ = 𝟎. πŸ‘πŸ“πŸ• 𝑴
𝒔
π’‡π‘―πŸ = πŸ“πŸ”. πŸ– π’Œπ‘―π’›
𝐢𝑇1 = 10𝑝𝐹 + 10𝑝𝐹[1 + 2(0.8)]
πœ”π»1
𝐢𝑇1 = 36.42 𝑝𝐹
1
1
=
∗ = (36.48
−12
𝐢𝑇1𝑅𝐸𝑄
× 10 )(0.2 × 103 )
πœ”π»1 =
109 1000
𝒓𝒂𝒅
=
× 106 = πŸπŸ‘πŸ•. πŸ’ 𝑴
6.83 6.83
𝒔
π’‡π‘―πŸ = 𝟐𝟏. πŸ– 𝑴𝑯𝒛
DIAGRAMA DE BODE:
PROBLEMA 2
Datos:
𝛽1 = 𝛽2 = 150
Calcular:
𝛽3 = 𝛽4 = 300
𝛽5 = 𝛽6 = 30
𝐢π‘₯ = 30𝑝𝐹
Av:…………….. Tq’:…………
Solución:
β„Žπ‘–π‘’1 = β„Žπ‘–π‘’2 = 0.93𝐾Ω
𝐼𝐢3 + 𝐼𝐢4 = 4.2π‘šπ΄
80𝐾𝐼𝐢3 = 0.60.5𝐾𝐼𝐢4
𝐼_𝐢3 = 0.033π‘šπ΄
𝐼_𝐢4 = 4.16π‘šπ΄
β„Žπ‘–π‘’3 = 233𝐾
β„Žπ‘–π‘’4 = 1.8𝐾
π‘‘π‘ž ′ = 2(232,8𝐾) + 300(80𝐾//(1.8𝐾
+ 300(0.5𝐾))
′
π‘‘π‘ž = 16.4 𝑀𝛺
π‘’π‘œ
𝐴𝑣 = −6 = ′
π‘’π‘œ
𝑍′π‘’π‘ž = 16.4 𝑀𝛺

𝑓𝐻 = 2πœ‹ 𝑍 ′
1
π‘’π‘ž (210 𝑝𝐹)
= 46 𝐻𝑧

𝑖 = 2(10 + β„Žπ‘–π‘’1 )𝑖𝑏 + 3(150)π‘₯3
𝑉𝑖
= 𝑍𝑛 = 472 𝐾𝛺
𝑖𝑏

𝑒′0 = (2𝑖π‘₯) 𝑍 ′ π‘’π‘ž
𝑒′0 = (2𝑖π‘₯)16.4 𝑀𝛺 … . (𝛼)
𝑉𝑖
= 472π‘˜ ; 𝑖π‘₯ = 150𝑖𝑏
𝑖𝑏
𝑉𝑖
𝑉𝑖
= 472π‘˜ ; 𝑖π‘₯ =
𝑖π‘₯
3.14

𝐸𝑛 (∝):
2𝑉𝑖
𝑒′0 = (
)(16.4 𝑀𝛺)
3.14
𝑒0
2
𝑒0
=(
) (16.4 𝑀𝛺)𝑉𝑖 → = πŸ”πŸπŸ“πŸŽπŸŽ = 𝑨𝒗
6
3.14
𝑉𝑖
PROBLEMA 3.
𝑓𝑙 =?,
𝑓𝐻 =?,
𝑓𝑑 =?
𝑄 => πœŒπ‘š = 1.2 π‘š
=> πΆπœ‹ = 0.5 𝑝𝐹
=> πΆπœ‡ = 0.5 𝑝𝐹
Solución:
A) FREQ. BAJA:
𝐴
𝑉
𝝎𝟏 :
𝜏1 = (2πœ‡πΉ)(0,5𝑀) = 1 𝑠
1
π‘Ÿπ‘Žπ‘‘
πœ”1 =
=1
(2πœ‡πΉ)(0,5𝑀)
𝑠
𝑓1 = 0,159 𝐻𝑧
𝝎𝟐 :
π‘Ÿπ‘Žπ‘‘
𝑠
𝑓2 = 3,8 𝐻𝑧
πœ”2 = 23,9
πŽπŸ‘ :
π‘Ÿπ‘Žπ‘‘
𝑠
𝑓3 = 3,18 𝐻𝑧
πœ”πΏ
𝑓𝐿 =
2πœ‹
π‘Ÿπ‘Žπ‘‘
πœ”πΏ = 39
𝑠
∴ 𝑓𝐿 = 6,2 𝐻𝑧
πœ”3 = 20
B) FREQ. MEDIA:
𝐴𝑉 =
π‘”π‘š 𝑅1
1 + π‘”π‘š 𝑅2
∴ 𝐴𝑉 = 0,52
C) FREQ. ALTA:
πΆπœ‹ = βˆ„
πΆπœ‡ =
π‘…π‘žπœ‡∗ = 0,45π‘˜
1
(2𝑝𝐹)(0,45)
π‘€π‘Ÿπ‘Žπ‘‘
πœ”πœ‡ = 1111
<> 177 𝑀𝐻𝑧
𝑠
πœ”πœ‡ =
𝐢𝐿 :
π‘…π‘ž∗ = 0,43π‘˜
π‘€π‘Ÿπ‘Žπ‘‘
πœ”πΏ = 233
<> 37 𝑀𝐻𝑧
𝑠
1
1,15
= ∑πœ”
πœ”π»
𝑋
1,15
1
1
=
+
πœ”π»
1111 233
π‘Ÿπ‘Žπ‘‘
πœ”π» = 221M
𝑠
∴ 𝑓𝐻 = 35.17 𝑀𝐻𝑧
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