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15/01/2015
1
AMPLIFICADORES CON BJT
INTRODUCCIÓN
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CONCEPTOS BASICOS SOBRE AMPLIFICADORES
Red de Dos Puertos
RoZoZTH ==
iVAE VNLTH =
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Circuito Equivalente de la Red de DosPuertos
Input Impedance, Zi: Zi = Ri
Output Impedance, Zo: Zo = Ro
FUENTES DE ALIMENTACION YRENDIMIENTO
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Efectos de la Resistencia de Carga en laGanancia de Voltaje
VNLoL
LA
RR
R
Vi
VoAv
+==
VNLoL
LA
RR
R
Vi
VoAv
+==
Efectos de la Resistencia de Carga en laGanancia de Corriente
Ri
Vi
Zi
ViIi ==
LR
VoIo −=
LR
ZiAvAi −=
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Efectos de la Resistencia de Salida dela Fuente de Señal en la Ganancia de
Voltaje
VNLs ARsRi
Ri
Vs
VoAv
+==
Efectos de la Resistencia de Salida de laFuente de Señal en la Ganancia de
Corriente
RiRs
VsIi
+=
LR
VoIo −=
Ai= -Av((Rs +Ri)/RL)
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Efecto Combinado de Rs y RL en laGanancia de voltaje
VNLoL
L
si
is A
RR
R
RR
R
Vs
VoAv
++==
CIRCUITO EQUIVALENTE EN PEQUEÑA SEÑAL PARA BJT
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PROCEDIMIENTO PARA DETERMINAR EL CIRCUITOEQUIVALENTE EN PEQUEÑA SEÑAL
1. Determinar el punto Q con la polarización encontinua (se abren todos los condensadores).
2. Obtener los parametros en pequeña señal rpi ygm a partir del punto Q.
3. Sustituir el BJT por sus equivalente.4. Hacer cero las fuentes de DC y reemplazar los
capacitores por su modelo en banda media(corto circuito o circuito abierto).
5. Aplicar las leyes de circuitos lineales paradeterminar la variables deseadas.
Amplificador Monoetapa
0
AMPLIFICADOR
Re1k
Vcc=10
R1180k
0
Rc1k
Q2N2222
Ce
10u
R2150k
Cl
10u
VsRL
80
Cs
10u
Rs
1k
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Slide 8
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
CE – Voltage-Divider Bias Configuration
Slide 9
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
re Model
You still need to determine β, re, and ro.
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Slide 10
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Impedance Calculations
Input Impedance:[Formula 8.10][Formula 8.11]
Output Impedance:[Formula 8.12][Formula 8.13]
21
2121
RR
RRR||RR
+==′
er||RZi Β′=
oC r||RZo=
CC 10RroRZo ≥≅
Slide 11
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Gain Calculations
Voltage Gain (Av):[Formula 8.14]
[Formula 8.15]
Current Gain (Ai):[Formula 8.16]
[Formula 8.17]
[Formula 8.18]
Current Gain from Voltage Gain:
[Formula 8.19]
e
oC
r
r||R
Vi
VoAv
−==
Ce
C
10Rror
R
Vi
VoAv ≥−≅=
)rR)(R(r
rR
Ii
IoAi
eCo
o
Β+′+′Β==
Ce 10RrorR
R
Ii
IoAi ≥Β+′
′Β≅=
re10R10RC,roIi
IoAi ≥′≥Β≅=
CR
ZiAvAi −=
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Slide 12
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Phase Relationship
A CE amplifier configuration will always have a phase relationship between input andoutput is 180 degrees. This is independent of the DC bias.
RL100kRe
1k
Rc6.8k
R156k
AMPLIFICADOR
00
0
0
Vcc
C1
10u Vcc12V
C2
10u
Vcc
FIG 1
R28.2k
Q2N2222V1FREQ = 1k
VOFF = 0VAMPL = 100mv
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EJERCICIOS E.C
Slide 19
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Emitter-Follower Configuration
You may recognize this as the Common-Collector configuration. Indeed they are the samecircuit.Note the input is on the base and the output is from the emitter.
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Slide 20
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
re Model
You still need to determine β, re, and ro.
Slide 21
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Impedance Calculations
Input Impedance:[Formula 8.37][Formula 8.38][Formula 8.39][Formula 8.40]
Zb||RZi B=Ee 1)R(rZb +Β+Β=
)R(rZb Ee+Β≅ERZb Β≅
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Slide 22
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Impedance Calculations (cont’d)
Output Impedance:
[Formula 8.42][Formula 8.43]
eE r||RZo =
eEe rRrZo >>≅
Slide 23
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Gain Calculations
Voltage Gain (Av):[Formula 8.44]
[Formula 8.45]
Current Gain (Ai):[Formula 8.46]
Current Gain from Voltage Gain:
[Formula 8.47]
eE
E
rR
R
Vi
VoAv
+==
EeEeE RrR,rR1Vi
VoAv ≅+>>≅=
ZbR
RAi
B
B
+Β−≅
ER
ZiAvAi −=
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Slide 24
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Phase Relationship
A CC amplifier or Emitter Follower configuration has no phase shift between input andoutput.
EJERCICIOSB=225
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EJERCICIOS
B=150
Slide 25
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Common-Base (CB) Configuration
The input (Vi) is applied to the emitter and the output (Vo) is from the collector.
The Common-Base is characterized as having low input impedance and high outputimpedance with a current gain less than 1 and a very high voltage gain.
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Slide 26
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
re Model
You will need to determine α and re.
Slide 27
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Impedance Calculations
Input Impedance:[Formula 8.54]
Output Impedance:[Formula 8.55]
eE r||RZi =
CRZo =
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17
Slide 28
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Gain Calculations
Voltage Gain (Av):
[Formula 8.56]
Current Gain (Ai):
[Formula 8.57]
e
C
e
C
r
R
r
R
Vi
VoAv ≅==
1Ii
IoAi −≅−==
Phase Relationship
A CB amplifier configuration has no phase shift between input and output.
Slide 29
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
EJERCICIOS BASE COMUN
Detemine Av suponiendo que ß=225,Rs=50Ω , RE=,3 k Ω, RC=RL=510 Ω,C1=C2=1uF, VCC=15 V, y .VEE=-15 V.
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Slide 46a
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Summary Table
Slide 46b
Robert BoylestadDigital Electronics
Copyright ©2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
All rights reserved.
Summary Table
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19
Amplificador MultietapaVcc= 15B=100
MULTIETAPA HAMBLEYVCC=15 V.
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Corrimiento de Nivel Dc en Circuitoscon BJTs
R21.5k
R34.7k
R41.4k
Q3
Vout
Q2
0
Q1Vg
R16.2k
VEE=-10v
D1Vz=7.3v
1
2
0 R510k
VCC=10V
Rg
1k
Amplificador en cascada acoplado concapacitores
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Desplazamiento de Nivel de DC conDiodo Zener
Desplazamiento de Nivel de DC conTransistores complementarios
VEE=-10v
D1Vz=8.5V
1
2
VCC=10V
Q1
0
Q4Vout
R16.2k
R34.7k
R41.4k
Rg1k
R21.5k
Q3
Vg
R510k
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Multietapa BiFET
Polarización Multietapa BiFET
BIPOLAR JUNCTION TRANSISTORS
NAME Q_Q1 Q_Q2MODEL Q2N3904 Q2N3904IB 1.09E-05 1.36E-05IC 1.61E-03 2.15E-03VBE 6.77E-01 6.84E-01VBC -4.31E+00 -7.17E+00VCE 4.98E+00 7.86E+00BETADC 1.48E+02 1.58E+02GM 6.10E-02 8.08E-02RPI 2.74E+03 2.19E+03RX 1.00E+01 1.00E+01RO 4.86E+04 3.78E+04CBE 2.49E-11 3.09E-11CBC 2.02E-12 1.76E-12CJS 0.00E+00 0.00E+00BETAAC 1.67E+02 1.77E+02CBX/CBX2 0.00E+00 0.00E+00FT/FT2 3.61E+08 3.94E+08
MOSFETSNAME M_M1MODEL MbreaknID 5.32E-03VGS -1.06E+00VDS 1.06E+01VBS 0.00E+00VTH -2.00E+00VDSAT 9.36E-01Lin0/Sat1 -1.00E+00if -1.00E+00ir -1.00E+00TAU -1.00E+00GM 1.14E-02GDS 8.77E-05GMB 0.00E+00CBD 0.00E+00CBS 0.00E+00CGSOV 0.00E+00CGDOV 0.00E+00CGBOV 0.00E+00CGS 0.00E+00CGD 0.00E+00CGB 0.00E+00
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