Dse Seminario (2009/03/04) Amp. Monoetapa

Diseinu eta Simulazio Elektronikoa

SEMINARIO 2009-03-04 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 )

∆v =

VL VL VI V = ⋅ = ∆v CC ⋅ I = 0.995 ⋅ 0.989 = 0.984 VS VI VS VS

∆v CC =

(h

fe

+ 1) ⋅ R L'

hie + (h fe + 1) ⋅ R

' L

=

R L' = R E R L =

hie =

(100 + 1) ⋅ 4.8 = 0.995 2.6 + (100 + 1) ⋅ 4.8

9.2 ⋅ 10 = 4 .8 K 9.2 + 10

VT 0.026 = = 2 .6 K I BQ 0.01

Rs 100Ω

Vs

Rent Vo

VS ⋅ Rent RS + Rent Rent VI 8.93 = = = = 0.989 VS VS RS + Rent 0.1 + 8.93

Rent = R B R I CC =

9.1 ⋅ 487.4 = 8.93K 9.1 + 487.4

R I CC = hie + (h fe + 1) ⋅ R L' = 2.6 + (100 + 1) ⋅ 4.8 = 487.4 K

Rsal = R E ROCC =

ROCC =

hie ⋅ RS' 2.6 ⋅ 0.099 = = 25Ω h fe + 1 100 + 1

RS' = RS R B =

2

9.2 ⋅ 0.025 = 25Ω 9.2 + 0.025

0 .1 ⋅ 9 .1 = 99Ω 0 .1 + 9 .1

Diseinu eta Simulazio Elektronikoa 10v

Rsal

1kΩ C3

Rent C1

Q1

|IDS|=10mA |VT|=3v

1kΩ

20kΩ

50kΩ Vi

ANÁLISIS EN CONTINUA (calculamos el punto Q) JFET de canal n

10v

VGS=0
SATURACIÓN

ID

 V I D = I DS ⋅ 1 − GS  VT

1kΩ

2

  

2

0   I D = 10 ⋅ 1 −  = 10mA  −3 Q1

50kΩ

VDS

VGS

VDS ? ⇒ Malla de salida 20 − 1 ⋅10 − VDS = 0 VDS = 10v Comprobamos que el transitor está en saturación VGS + VDS > VT 0 + 10 > 3

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VL

Diseinu eta Simulazio Elektronikoa

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

∆v =

VO VI V ⋅ = (∆vSC ) ⋅ I = (− 6.33) ⋅ 0.98 = −6.2 VS VI VS ∆v SC = − g m ⋅ RL' = −6.66 ⋅ 0.95 = −6.33

gm =

2 VT

I DS ⋅ I DQ =

RL' = 1 20 =

2 10 ⋅10 = 6.66 mA V −3 1 ⋅ 20 = 0.95 K 1 + 20

Rs 100Ω

Vs

Rent Vo

VS ⋅ Rent RS + Rent Rent VI 50 = = = = 0.98 VS VS RS + Rent 1 + 50 Rent = 50 RI SC = 50 K RI SC = ∞ Rsal = 1 ROSC = 1K ROSC = ∞

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