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INFORME PREVIO N° 6
TABLA 2
Para P1=0ΩPor dato ya que el transistor AC128 está hecho de GERMANIO, entonces su
VBE= 0.2v y β=90
Sabemos:
Rb=R1×R2R1+R2 V=
R2×VccR1+R2
Rb=56K×22K(56+22)K V=
22k×(12)(56+22)k
Rb=15.794kΩ V= 3.384v
Hallando Ib
Ib=V−V BE
Rb+ (β+1 ) ℜ
Ib=3.384−0.2
15.794×103+(90+1 )330
Ib= 69.483µA
Se sabe que
Ic=Ib×β VE = Re×Ic
Ic= 69.438µA ×90 VE = 330×(6.249×10−3)
Ic= 6.249mA Ve = 2.062 v
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Vcc= Ic×Rc + VCE + Ic×Re
VCE=Vcc – Ic (Rc+Re)
VCE= 12 – (6.249 ×10−3 ¿(1000+330)
Vce = 3.689 v
Hallando Icmax(VCE=0v)
Icmax= 12/(1000+330)
Icmax = 9.022mA
Tabla 3
Cambiamos el valor de R1 = 68kΩ.
Sabemos:
Rb=R1×R2R1+R2 V=
R2×VccR1+R2
Rb=68K ×22K(68+22)K V=
22k×(12)(68+22)k
Rb=16.622kΩ V=2.933v
Hallando Ib
Ib=V−V BE
Rb+ (β+1 ) ℜ
Ib=2.933−0.2
16.622×103+ (90+1 )330
Ib=58.582µA
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Se sabe que
Ic=Ib×β VE = Re×Ic
Ic= 58.582µA ×90 VE = 330×(5.272×10−3)
Ic= 5.272 mA Ve = 1.739 v
Vcc= Ic×Rc + VCE + Ic×Re
VCE=Vcc – Ic (Rc+Re)
VCE= 12 – (5.272 ×10−3 ¿(1000+330)
Vce = 4.988 v
Hallando Icmax(VCE=0v)
Icmax= 12/(1000+330)
Icmax = 9.022mA
TABLA 5
Para P1=100kΩ.
Sabemos:
Rb=(R1+P1)×R2R1+P1+R2
V=R2×Vcc
R1+P1+R2
Rb=(68+100)K ×22K(68+100+22)K V=
22k ×12(68+100+22)k
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Rb=19.4526kΩ V=1.3894v
Hallando Ib
Ib=V−V BE
Rb+ (β+1 ) ℜ
Ib=1.3894−0.2
19.4526×103+(90+1 )330
Ib= 23.992µA
Se sabe que
Ic=Ib×β Vcc= Ic×Rc + VCE + Ic×Re
Ic=23.992×10−6×90 VCE=Vcc – Ic(Rc+Re)
Ic= 2.159mA VCE=12 – (2.159×10−3)(1000+330)
Vce = 9.1285 v
Para P1=250kΩ
Sabemos:
Rb=(R1+P1)×R2R1+P1+R2
V=R2×Vcc
R1+P1+R2
Rb=(68+250)K ×22K(68+250+22)K V=
22k ×12(68+250+22)k
Rb=20.576kΩ V=0.7764v
Hallando Ib
Ib=V−V BE
Rb+ (β+1 ) ℜ
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Ib=0.7764−0.2
20.576×103+(90+1 )330
Ib= 11.389µA
Se sabe que
Ic=Ib×β Vcc= Ic×Rc + VCE + Ic×Re
Ic=11.389×10−6×90 VCE=Vcc – Ic(Rc+Re)
Ic= 1.025mA VCE=12 – (1.025×10−3)(1000+330)
Vce = 10.636 v
Para P1=500K
Sabemos:
Rb=(R1+P1)×R2R1+P1+R2
V=R2×Vcc
R1+P1+R2
Rb=(68+500)K ×22K(68+500+22)K V=
22k ×12(68+500+22)k
Rb=21.179kΩ V=0.447v
Hallando Ib
Ib=V−V BE
Rb+ (β+1 ) ℜ
Ib=0.447−0.2
21.179×103+(90+1 )330
Ib= 4.8233µA
Se sabe que Vcc= Ic×Rc + VCE + Ic×Re
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Ic=Ib×β VCE=Vcc – Ic(Rc+Re)
Ic=4.8233×10−6×90 VCE=12 – (0.434×10−3)(1000+330)
Ic= 0.434mA Vce = 11.4227 v
Para P1=1M
Sabemos:
Rb=(R1+P1)×R2R1+P1+R2
V=R2×Vcc
R1+P1+R2
Rb=(68+1000)K ×22K(68+1000+22)K V=
22k ×12(68+1000+22)k
Rb=21.555kΩ V=0.2422v
Hallando Ib
Ib=V−V BE
Rb+ (β+1 ) ℜ
Ib=0.2422−0.2
21.555×103+(90+1 )330
Ib= 0.8180µA
Se sabe que
Ic=Ib×β Vcc= Ic×Rc + VCE + Ic×Re
Ic=0.8180×10−6×90 VCE=Vcc – Ic(Rc+Re)
Ic= 0.0736mA VCE=12 –(0.0736×10−3)(1000+330)
Vce = 11.9021 v
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INFORME PREVIO N° 7
TABLA 2
Para P1=0ΩPor dato ya que el transistor PN2222A está hecho de GERMANIO, entonces su
VBE= 0.6v y β=200
Sabemos:
Rb=R1×R2R1+R2 V=
R2×VccR1+R2
Rb=56K×22K(56+22)K V=
22k×(12)(56+22)k
Rb=15.794kΩ V= 3.384v
Hallando Ib
Ib=V−V BE
Rb+ (β+1 ) ℜ
Ib=3.384−0.6
15.794×103+(200+1 )330
Ib= 33.89µA
Se sabe que
Ic=Ib×β VE = Re×Ic
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Ic= 33.89µA ×200 VE = 330×(6.778×10−3)
Ic= 6.778mA Ve =2.2367 v
Vcc= Ic×Rc + VCE + Ic×Re
VCE=Vcc – Ic (Rc+Re)
VCE= 12 – (6.778 ×10−3 ¿(1000+330)
Vce = 2.9852 v
Hallando Icmax(VCE=0v)
Icmax= 12/(1000+330)
Icmax = 9.022mA
Tabla 3
Cambiamos el valor de R1 = 68kΩ.
Sabemos:
Rb=R1×R2R1+R2 V=
R2×VccR1+R2
Rb=68K ×22K(68+22)K V=
22k×(12)(68+22)k
Rb=16.622kΩ V=2.933v
Hallando Ib
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Ib=V−V BE
Rb+ (β+1 ) ℜ
Ib=2.933−0.6
16.622×103+ (200+1 )330
Ib=28.124µA
Se sabe que
Ic=Ib×β VE = Re×Ic
Ic= 28.124µA ×200 VE = 330×(5.624×10−3)
Ic= 5.624 mA Ve = 1.8559 v
Vcc= Ic×Rc + VCE + Ic×Re
VCE=Vcc – Ic (Rc+Re)
VCE= 12 – (5.624 ×10−3 ¿(1000+330)
Vce = 4.520 v
Hallando Icmax(VCE=0v)
Icmax= 12/(1000+330)
Icmax = 9.022mA
TABLA 5
Para P1=100kΩ.
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Sabemos:
Rb=(R1+P1)×R2R1+P1+R2
V=R2×Vcc
R1+P1+R2
Rb=(68+100)K ×22K(68+100+22)K V=
22k ×12(68+100+22)k
Rb=19.4526kΩ V=1.3894v
Hallando Ib
Ib=V−V BE
Rb+ (β+1 ) ℜ
Ib=1.3894−0.6
19.4526×103+(200+1 )330
Ib= 9.202µA
Se sabe que
Ic=Ib×β Vcc= Ic×Rc + VCE + Ic×Re
Ic= 9.202×10−6×200 VCE=Vcc – Ic(Rc+Re)
Ic= 1.840mA VCE=12 – (1.840×10−3)(1000+330)
Vce = 9.5528 v
Para P1=250kΩ
Sabemos:
Rb=(R1+P1)×R2R1+P1+R2
V=R2×Vcc
R1+P1+R2
Rb=(68+250)K ×22K(68+250+22)K V=
22k ×12(68+250+22)k
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Rb=20.576kΩ V=0.7764v
Hallando Ib
Ib=V−V BE
Rb+ (β+1 ) ℜ
Ib=0.7764−0.6
20.576×103+(200+1 )330
Ib= 2.0297µA
Se sabe que
Ic=Ib×β Vcc= Ic×Rc + VCE + Ic×Re
Ic=2.0297×10−6×200 VCE=Vcc – Ic(Rc+Re)
Ic= 0.4059mA VCE=12 – (0.4059×10−3)(1000+330)
Vce = 11.460 v
Para P1=500K
Sabemos:
Rb=(R1+P1)×R2R1+P1+R2
V=R2×Vcc
R1+P1+R2
Rb=(68+500)K ×22K(68+500+22)K V=
22k ×12(68+500+22)k
Rb=21.179kΩ V=0.447v
Hallando Ib
Ib=V−V BE
Rb+ (β+1 ) ℜ
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Ib=0.447−0 .6
21.179×103+(200+1 )330
Ib= -1.7483µA
Se sabe que Vcc= Ic×Rc + VCE + Ic×Re
Ic=Ib×β VCE=Vcc – Ic(Rc+Re)
Ic= - 1.7483×10−6×200 VCE=12 – (-0.3496×10−3)(1000+330)
Ic= -0.3496mA Vce = 12.464 v
Para P1=1M
Sabemos:
Rb=(R1+P1)×R2R1+P1+R2
V=R2×Vcc
R1+P1+R2
Rb=(68+1000)K ×22K(68+1000+22)K V=
22k ×12(68+1000+22)k
Rb=21.555kΩ V=0.2422v
Hallando Ib
Ib=V−V BE
Rb+ (β+1 ) ℜ
Ib=0.2422−0.6
21.555×103+(200+1 )330
Ib= - 4.071µA
Se sabe que
Ic=Ib×β Vcc= Ic×Rc + VCE + Ic×Re
Ic= -4.071×10−6×200 VCE=Vcc – Ic(Rc+Re)
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Ic= - 0.8142mA VCE=12 –(-0.8142×10−3)(1000+330)
Vce = - 13.082 v
UNMSM
FACULTAD DE INGENIERIA ELECTRICA , ELECTRONICA Y TELECOMUNICACIONES
APELLIDOS Y NOMBRES CODIGO DE MATRICULA
Castro Bañares Jhon Leonidas 14190120CURSO TEMA
Dispositivos electrónicos Transistor bipolarINFORME FECHA NOTA
previo REALIZADO ENTREGA
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NUMERO 13/02/15 23/02/15
6 y 7
GRUPO /HORA PROFESOR
Grupo 3/viernes de 11am-2pm
Ing. Luis Paretto Quispe