Dse Práctica De Aula (2009/03/02) Amp. Monoetapa

Diseinu eta Simulazio Elektronikoa

PRÁCTICA DE AULA 2009-03-02 Amp. Monoetapa

ANÁLISIS EN CONTINUA (calculamos el punto Q)

Q es un BJT de tipo NPN I E = I B + IC Suponemos Q en ACTIVA VBE = VBE ,ON = 0.7v I C = β ⋅ I B ⇒ I E = I B + β ⋅ I B = (1 + β ) ⋅ I B

I C ? ⇒ I B ? ⇒ Malla de entrada − 9.1I B − VBE ,ON − 9.2 I E − (− 10 ) = 0 − 9.1I B − VBE ,ON − 9.2 ⋅ (1 + β ) ⋅ I B + 10 = 0 − 9.1I B − 0.7 − 9.2 ⋅ (1 + 100 ) ⋅ I B + 10 = 0 I B = 10 µA ⇒ I C = β ⋅ I B = 100 ⋅ 0.01 = 1mA VCE ? ⇒ Malla de salida 10 − 0.68 I C − VCE − 9.2 I E − (− 10 ) = 0 10 − 0.68 I C − VCE − 9.2 ⋅ (I B + I C ) + 10 = 0 10 − 0.68 ⋅1 − VCE − 9.2 ⋅ (0.01 + 1) + 10 = 0 VCE = 10v > 0.2v = VCE , SAT ⇒ Suposición correcta

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Diseinu eta Simulazio Elektronikoa

ANÁLISIS EN ALTERNA (calculamos ∆v, Rent y Rsal ) RI

RO

Rent

Rsal

V1

V2

RL’

RS’

∆v EC = −

h fe ⋅ RL' hie

=−

RL' = RC RL =

hie =

∆v =

Vo

100 ⋅ 0.63 = −24.23 2 .6

0.68 ⋅10 = 0.63K 0.68 + 10

VT 0.026 = = 2 .6 K I BQ 0.01

VL V2 V1 V = ⋅ = ∆v EC ⋅ 1 = (− 24.23) ⋅ 0.95 = −23 VS V1 VS VS

Rs 100Ω

Vs

Rent Vo

VS ⋅ Rent V1 RS + Rent Rent 2 = = = = 0.95 VS VS RS + Rent 0.1 + 2

Rent = RB RI EC =

9 .1 ⋅ 2 .6 = 2K 9 .1 + 2 .6

RI EC = hie = 2.6 K Rsal = RC ROEC = RC = 0.68 K ROEC = ∞

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