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• LuisCabreraOrosco

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TABLA 2

Para P1=0Ω Por dato ya que el transistor TR85 está hecho de GERMANIO, entonces su VBE= 0.2v y β=90. Sabemos: Rb= Rb=

R1×R2 R1+R2 56K×22K (56+22)K

Rb=15.794kΩ V= V=

R2×Vcc R1+R2 22k×(−12) (56+22)𝑘

V= -3.384v

Hallando Ib

Ib= Ib=

V−VBE Rb+(β+1)Re −3.384−0.2 15.794×103 +(90+1)330

Ib= -78.212µA

Se sabe que

Ic=Ib×β Ic=-78.212µ×90 Ic= -7.0391mA

VE = Re×Ic

VE = 330(-7.0391× 10−3) VE =-2.323v Vcc= Ic×Rc + VCE + Ic×Re VCE=Vcc – Ic (Rc+Re) VCE=12 – (-7.0391×10−3 )(1000+330) VCE =21.362v

Hallando Icmax(VCE=0v) Icmax=-12/(1000+330) Icmax= -9.023mA

TABLA 3 Cambiamos el valor de R1 = 68kΩ.

Sabemos: Rb= Rb=

R1×R2 R1+R2 68K×22K (68+22)K

Rb=16.622kΩ V= V=

R2×Vcc R1+R2 22k×(−12) (68+22)𝑘

V= -2.933v Hallando Ib

Ib= Ib=

V−VBE Rb+(β+1)Re −2.933−0.2 1.622×103 +(90+1)330

Ib=-98.983µA

Se sabe que

Ic=Ib×β Ic=-98.983×10−6×90 Ic=-8.9085mA

VE = Re×Ic VE = 330×(-8.9085) ×10−3 VE = -2.939v

Vcc= Ic×Rc + VCE + Ic×Re VCE=Vcc – Ic (Rc+Re) VCE=-12 – (-8.9085×10−3 )(1000+330) VCE =23.848v

Icmax=12/(1000+330) Icmax= 9.0226mA

TABLA 5 Para P1=100kΩ. Sabemos: Rb= Rb=

(R1+P1)×R2 R1+P1+R2 (68+100)K×22K (68+100+22)K

Rb=19.4526kΩ V= V=

R2×Vcc R1+P1+R2 22k×(−12) (68+100+22)𝑘

V= -1.3894v

Hallando Ib

Ib= Ib=

V−VBE Rb+(β+1)Re −1.3894−0.2 19.4526×103 +(90+1)330

Ib=-32.1023µA

Se sabe que

Ic=Ib×β Ic=-32.1023×10−6×90 Ic=-2.891mA

Vcc= Ic×Rc + VCE + Ic×Re

VCE=Vcc – Ic(Rc+Re) VCE=-12 – (-2.891×10−3)(1000+330) VCE =-8.155v

Para P1=250kΩ Sabemos: Rb= Rb=

(R1+P1)×R2 R1+P1+R2 (68+250)K×22K (68+250+22)K

Rb=20.576kΩ

V= V=

R2×Vcc R1+P1+R2 22k×(−12) (68+250+22)𝑘

V= -0.776v

Hallando Ib

Ib= Ib=

V−VBE Rb+(β+1)Re −0.776−0.2 20.576×103 +(90+1)330

Ib=-19.286µA

Se sabe que

Ic=Ib×β Ic=-19.286×10−6×90 Ic=-1.736mA

Vcc= Ic×Rc + VCE + Ic×Re VCE=Vcc – Ic(Rc+Re) VCE=-12 – (-1.736×10−3 )(1000+330) VCE =-9.691v

Para P1=500K Sabemos: Rb= Rb=

(R1+P1)×R2 R1+P1+R2 (68+500)K×22K (68+500+22)K

Rb=21.179kΩ V= V=

R2×Vcc R1+P1+R2 22k×(−12) (68+500+22)𝑘

V=-0.447v Hallando Ib

Ib= Ib=

V−VBE Rb+(β+1)Re −0.447−0.2 21.179×103 +(90+1)330

Ib=-12.634µA

Se sabe que

Ic=Ib×β Ic=-12634×10−6×90 Ic=-1.137mA

Vcc= Ic×Rc + VCE + Ic×Re VCE=Vcc – Ic(Rc+Re) VCE=-12 – (-1.137×10−3(1000+330) VCE =-10.488v

Para P1=1M Sabemos: Rb= Rb=

(R1+P1)×R2 R1+P1+R2 (68+1000)K×22K (68+1000+22)K

Rb=21.555kΩ V= V=

R2×Vcc R1+P1+R2 22k×12 (68+1000+22)𝑘

V=-0.2422v

Hallando Ib

Ib= Ib=

V−VBE Rb+(β+1)Re −0.2422−0.2 21.555×103 +(90+1)330

Ib=-8.572µA

Se sabe que

Ic=Ib×β

Ic=-8.572×10−6 ×90 Ic=-0.772mA

Vcc= Ic×Rc + VCE + Ic×Re VCE=Vcc – Ic(Rc+Re) VCE=-12 – (-0.772×10−3 )(1000+330) VCE =-10.973v