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INFORME-PREVIO-7
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INFORME-PREVIO-7
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James Lizonde Peredo
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Facultad de Ingeniería Electrónica, Eléctrica y Telecomunicaciones T elecomunicaciones
Lizonde Peredo, James
15190167
DISPOSITIVOS ELECTRONICOS
PREVIO 07
02
06 DE JUNIO DEL 2016
LUNES 10am – 12pm 12pm
13 DE JUNIO DEL 2016
Determinar el punto de operación del circuito del experimento. (Valores teóricos Tablas 2, 3 y 5)
POLARIDAD: NPN MATERIAL: GERMANIO (Si) GANANCIA DE CORRIENTE (β) = 130
= 220Ω = 1kΩ = 56K Ω (Tabla 2.) = 68K Ω (Tabla 1.) = 22K Ω. = 12 V
*Por ser de Silicio:
= 0,67
= 130
Hallando el punto Q:
Ic (mA)
Ib (μA)
Β
Vce (v.)
Vbe (v.)
Ve (v.)
7.911
60.853µ
130
2.348
0,67 v
1.753
Valores
(R1=56KΩ) Teóricos
=
× +
=
=
× +
12 × 22 56 + 22
=
56 × 22
= 15.795Ω 56 + 22
−
=
= 3.385
+ ( + 1)
=
3.385 − 0.67
= 60.853µ 15.795 + (130 + 1)220
= × = 60.853µ × 130 = 7.911 = ( + ) = (60.853µ + 7.911)220Ω = 1.753 = − ( + ) = 12 − {7.911(1Ω + 220Ω)} = 2.348
Ic (mA)
Ib (μA)
β
Vce (v.)
Vbe (v.)
Ve (v.)
6.473
49.799µ
130
4.102
0.67
1.435
Valores
(R1=68KΩ) Teóricos
=
× +
=
=
=
× +
12 × 22 = 2.933 68 + 22
=
68 × 22
= 16.622Ω 68 + 22
− + ( + 1)
=
2.933 − 0.67
= 49.799µ 16,622 + (130 + 1)220
= × = 49.799µ × 130 = 6.473 = ( + ) = (49.799µ + 6.473)220Ω = 1.435
= − ( + ) = 12 − {6.473(1Ω + 220Ω)} = 4.102
Para P=100 k Ω:
Al estar unidas en serie las resistencias R1 y P1, hallaremos su resistencia equivalente:
′ = + → ′ = 56 + 100 → ′ = Hallando los siguientes valores:
×
=
′
=
+
=
× +
12 × 22 156 + 22
=
156 × 22 156 + 22
−
=
= 1.483
+ ( + 1)
=
= 19.281Ω
1.483 − 0.67 = 16.901µ 19.281 + {(130 + 1)220}
= × = 16.901µ × 130 = 2.197 = − ( + ) = 12 − 2.197(1Ω + 220Ω) = 9.319
Para P=250 kΩ:
Al estar unidas en serie las resistencias R1 y P1, hallaremos su resistencia equivalente:
′ = + → ′ = 56 + 250 → ′ = Hallando los siguientes valores:
=
× ′ +
=
=
=
× +
12 × 22 306 + 22
=
= 0.805
306 × 22 306 + 22
− + ( + 1)
=
= 20.524Ω
0.805 − 0.67 = 2.735µ 20.524 + {(130 + 1)220}
= × = 2.735µ × 130 = 0.355 = − ( + ) = 12 − {0.355(1Ω + 220Ω)} = 11.566
Para P=500 kΩ:
Al estar unidas en serie las resistencias R1 y P1, hallaremos su resistencia equivalente:
′ = + → ′ = 56 + 500 → ′ = Hallando los siguientes valores:
×
=
′ +
=
=
× +
12 × 22 556 + 22
=
556 × 22 556 + 22
−
=
= 0.457
+ ( + 1)
=
= 21.163Ω 0.457 − 0
= 9.143µ 21.163 + (130 + 1)220
= × = 9.143µ × 130 = 1.188 = − ( + ) = 12 − {1.188(1Ω + 220Ω)} = 10.550
Para P= 1MΩ:
Al estar unidas en serie las resistencias R1 y P1, hallaremos su resistencia equivalente:
′ = + → ′ = 56 + 1000 → ′ = Hallando los siguientes valores:
=
× ′ +
=
=
=
× +
12 × 22
= 0.245 1056 + 22
=
1056 × 22 = 21.551Ω 1056 + 22
− + ( + 1)
=
0.245 − 0
= 4.863µ 21.551 + (130 + 1)220
= × = 4.836µ × 130 = 0.632 = − ( + ) = 12 − {0.632(1Ω + 220Ω) = 11.228
Ω
Ω
Ω
Ω
Ic(mA)
2.197
0.355
1.188
0.632
Ib(µA)
16.901µ
2.735µ
9.143µ
4.863µ
Vce
9.319
11.566
10.550
11.228
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