FACULTAD DE INGENIERÍA ELECTRÓNICA Y ELÉCTRICA ESCUELA ACADÉMICO PROFESIONAL DE INGENIERÍA ELECTRICA
CURSO:
Circuitos Circuitos electrónicos electrónicos II
TRABAJO: Informe Final Nº2 TEMA:
Amplicador Diferencial Diferencial
ALUMNO:
VILCHE !E"#"$ C%ris &ic%ael '()('')(
PROFESON:
IN*+ CEL", *E-.NI&,
'( de Fe/rero del 2')0
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AMPLIFICADOR DIFERENCIAL DIFERENCIAL 1.-Objeti!"# E1perimentar las propiedades del amplicador diferencial+
$.- C%e"i!&'(i! )(ei! *+%, *+%, '(' '('t te( e(" "ti ti' '" " (e"' (e"'/t /t'& '&te te" " !0(e !0(ee e e/ ')/ ')/i2 i2' '3! 3!(( 3i0e(e&i'/4
"e puede utiliar este amplicador amplicador como un modo in3ersor in3ersor 4La ' salida esta desfasada )5' con respecto respecto a la entrada6 entrada6$o $o modo no in3ersor 4la salida no tiene una desfase con respecto a la entr entrad ada6 a6$$ o modo modo dife diferrenci encial al cuan cuando do util utili ia a los los dos dos modo modoss anteriores+
Este Este am ampl pli ica cado dorr co cont ntie iene ne dos dos entr entrad adas as 7 dos dos sa sali lida das$ s$ La Lass salidas est8n desfasadas una con respecto a la otra en una 8n9ulo de )5''$ o sea ue una esta en fase con la entrada 7 la otra est8 desfasada )5' '+
#na de las caracter;sticas m8s importantes es la simetr;a ue tiene este circuito con respecto a los dos transistores$ ue %ace
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"e usa este amplicador para amplicar las se=ales en medios ruidosos$ o sea el ruido es atenuado en este amplicador 4&odo com>n$ 9anancia 9anancia de 3olta?e peue=a6 7 la se=al es amplicada 4&odo diferencial$ 9anancia de 3olta?e es alta6
$.-E&!&t('( /!" )%&t!" 3e (e)!"! 3e/ ')/i2'3!( 3i0e(e&i'/ ' E5)e(ie&t'( 627%(' 8.' 9 8.b: ': C!& (e"i"te&i' 627%(' 8.':# la
Hacemos el eui3alente del circuito en corriente continua para 9ura @+a$ donde las capacitancias se %acen circuito a/ierto:
Entonces como el circuito es un circuito ito sim
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I B (1K ) + VBE I C
β
+
I E (0.22 K
(1 K ) + VBE +
IC
β
(β
(4.7 K ) − 12v = + 0.05K ) + 2I E (4
+ 1)(0.22 K + 0.05K ) + 2( β + 1)
0.01K ) + 0 .7 .7 + I C (1.01)(0.27) + (9.494) I C I C ( 0. 0.7 + I C (0.2727) + (9.494) I C I C (0.01K ) + 0.
I C (0.01K
0
I C
β
(4.7 K ) −12v
=
0
= 12
= 12
+ 0.2727 K + 9.194 K ) = 12v − 0.7v
(9.77 767 K ) = 11.3 1.3 I C (9.7 11.3
I C
=
I C
= 1.16mA
9.7767 K
ara esta corriente de reposo el transistor tiene una resistencia din8mica ue es: r e
=
26mV 1.16mA
=
22.41Ω
Bam/i
−
V b1
=
V b1
=
Vb1
= −
β (1.16 1.16m)(1 )(1K )
−
100 0.0116v
Bam/i
−
VC
=
I C (7.5 K )
12v
−
VC
=
(1.16m)(7.5 K )
12v
−
V C
=
8 .7
V C
=
12 8.7
VC
=
3.3v
−
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V E
− ( −12) =
V E
+ 12 =
I E (0.27 K ) + 2I E ( 4.7K )
V E
+ 12 =
I E (0.27 K + 9.4K )
V E
+ 12 =
I E (9.67 K )
V E + 12 =
V E + 12 =
I E (0.22 K
+ 0.05K ) + 2I E ( 4.7K )
( β + 1) I C (9.67 K )
β (101)( 101)(1.16 1.16m)(9.67 )(9.67K ) 100
V E + 12 = 11.2 11.24 4 V E = −0.752
Con estos 3alores de 3olta?e tenemos ue: VCE
= VC − V E
V CE
3.3 − (−0.752 0.752)) = 3.3
V CE
3.3 + 0.75 0.752 2 = 3.3
VCE
=
4.052v
Nota: estos puntos de reposo de este amplicador son i9uales para los dos transistores de/ido a la simetr;a del circuito+
b: C!& %&' 0%e&te 3e !((ie&te 627%(' 8.b:# la
Hacemos el eui3alente del circuito en corriente continua para 9ura @+/$ donde las capacitancias se %acen circuito a/ierto:
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Entonces como el circuito ito es un circuito ito sim
4.7k × ( −12 v ) 10k + 4.7k
= −3.83v
Con este dato %allaremos el 3olta?e el en Emisor 2 4VE26: V E 2
= V b 2 −
0 .7 v
= − 4.5v
Hallamos la corriente IC: V E − (−1 2v)
I CQ 2
=
I CQ 2
=
I
= 1.92mA
3.9k −4.5 + 12
3.9 K
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26mV
r e 2
=
r e 2
=
r e 2
= 13.54Ω
IC
26mV 1.92mA
Bam/i
Ic 2 2
=
0.96mA
ara el transistor ) tenemos los si9uientes puntos de reposo: El 3olta?e en el colector ) es 4Vc)6: 12
−
I C 1
Vc1
= 12v −
VC1
=
=
(7.5k ) I C 1
0.96 mA
0.96mA × 7.5k = 4.8v
ara ara la /ase ) tenemos el si9uiente 3olta?e: V B1
= (− I B )(1K )
V B1
=
V B1
=
− I C 1 (1K )
β −0.96 0.96m(1K )
El 3olta?e en el emisor es:
100
=
9.6mV
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VCE1
=
VC1
V CE 1
=
4.8 ( 0.7 0.709) 09)
V CE 1
=
5.509
−
V E1
−
−
ara ara el 3olta?e colector 2 tenemos: 12 − VC 2
=
0.96(7.5 K + 0.22 K + 0.05 K ) + 5 .509
VC 2
= 12 − 0.96(7.5 K + 0.22 K + 0.05 K ) − 5. 509
VC 2
= −0.9682v
El 3olta?e colector emisor es del transistor 2: para 3arios 3alores del potenciómetro de )' tenemos: Consideramos la resistencia de potenciómetro como una 3aria/le 4Vp6 7 %allamos el 3alor de la corriente en función de esta 3aria/le:
Ic Q 2
=
−
59 .69
−
0.7 v × R p
3.9 k × ( R p
+
4.7 k )
+
3.07 mA
ara - )' tenemos$ Ic 2 )+(2 mA ara - 0 tenemos$ Ic2 )+@( mA ara - ' tenemos$ Ic2 G'+)5 mA
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Vemos ue en modo com>n se utilia a una sola fuente para las dos dos entr entrad adas as ue ue rec eci/ i/en en una una co corr rrie ient nte e /ase /ase de la mism misma a ma9nitud+ Hallamos el 3olta?e de salida del circuito es 4colector 7 tierra6: V 0
= − β × Ib1 ×
7.5k
Bam/i
= Ib1 β ×
r e
+ Ib1 β × 0.27k +
2 Ib1 β × 4.7k
or lo tanto anto %all %allar arem emos os la 9ana 9ananc ncia ia en modo modo co com> m>n n del del amplicador diferencial del e1perimento: AC
=
− Ib1 β × 7.5k
Ib1 β × r e
+ Ib1 β × (0.27 k +
9.4 k )
AC
= −0.773
Bam/i
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circuito$ 7 tam/i
Vemos ue en modo com>n se utilia dos fuentes de corriente desfasadas )5'' pero para el an8lisis usaremos solo una fuente para la /ase ) 7 la /ase se pone a tierra$ por lo tanto la 9anancia a modo diferencial la 9anancia de una fuente se le suma la 9anancia de la otra fuente 4superposición6+ 4superposición6+ Bam/i
= Ib1 β × r e + Ib1 β × 0.27k +
2 Ib1 (4.7k // 0.27k + 0.9(1 − α ) + r e )
Hallamos el 3olta?e de salida del circuito es 4colector 7 tierra6: V 0
= α × Ie 2 × 7.5k
Hallaremos el 3olta?e de salida en función de las corrientes de entrada$ entonces la corriente corriente Ie2 es i9ual a:
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V 0
=
7.02 × β × Ib1
or lo tanto %allaremos la 9anancia en modo diferencial del amplicador diferencial del e1perimento: Ad 1
=
7.02 × β × Ib Ib1β × r e
+ Ib1β × 0.27k +
Ad 1
2 Ib1 ( 4.7k // 0.27k + 0.9(1 − α ) + r e )
= −12.31
La 9anancia total en modo diferencial ser;a: Ad 2Ad) 2J+K2 Bam/i
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*;%, e&t'j'" "e !btie&e '/ %ti/i'( %&' 0%e&te 3e !((ie&te e& /%7'( 3e /' (e"i"te&i' 3e ei"!(4 Las 3enta?as de utiliar una fuente de corriente es relacionado a la 9ran impedancia ue esta tiene %aciendo ue:
En modo com>n reduca la 9anancia de 3olta?e: AC
=
− Ib1 β × 7.5k
Ib1 β × r e
+ Ib1 β × ( 0.27k +
2 r 0 )
De/ido De/ido a ue la resi esiste stenci ncia a eui3a eui3alen lente te r' es 9rande la 9anancia en modo com>n es peue=a+
En modo diferencial aumenta la 9anancia de 3olta?e: Ad 1
=
7.02 × β × Ib
Ib1β × r e
+ Ib1β × 0.27k +
2 Ib1 ( r 0 // 0.27k + 0.9(1 − α ) + r e )
Cuando r' est8 en paralelo con una resistencia peue=a no afectando pero aumentando el 3olta?e de salida+
La relación de rec%ao en modo com>n aumenta por lo 7a mencionado anteriormente+ anteriormente+
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1 Fuente DC 1 Transormador Transormador !on toma !entral " Transistores# "$"""" % Resistores &1'( )*# "&+., -* " &(.+ -* "&1/ -* "&1-* " &""/ o0m* &.% -* " Poten!iómetros# 1//o0m. 1/ "Condensadores# "Condensadores# "" uF &1/ 2*
.-PROCEDIMIENTOS# a. Sin aplicar aplicar señal calibrar P1 de de tal manera de obtener obtener para para ambos ambos transistores los mismos puntos de operación:
b. Aplicar Aplicar señal señal hasta hasta obtene obtenerr la la máxima señal de salida sin distorsión (2KH! " dibu#ar en la $ase correcta las si%uientes $ormas de ondas:
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2b1 3 ,%.,1 m2 rms
2!1 3 1.1 2 rms
2e1 3 ,(.4 m2 rms
2b" 3 4/.,1 m2 rms
&'' '&5-
2e" 3 ,,.1 m2 rms
01 0
32 0
2!" 3 %1, m2 rms
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2b" 3 4".,1 m2 rms
2e" 3 4". m2 rms
2!" 3 (%.4, m2 rms
a. ambie ambie la resis resisten tencia cia ,e de de 6.78 a 128 128 " complet complete e la tabla: tabla: Al medir di!0a di!0a resisten!ia nos nos dio el 5alor de Re 3 11.%,6 11.%,6 7
CASO II
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2b" 3 41.,1 m2 m2 rms
&'' '&5-
01 0
2b1 3 4(.11 m2 rms
2e" 3 (%.% m2 m2 rms
2!" 9 1 2
32 0
2!1 3 1%.44 m2 rms rms
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a. ambie ambie la resisten resistencia cia ,e por la $uente $uente de corrient corriente e ($i%. 6.2! 6.2! " calibre hasta encontrar el mism ismo punto de operac aciión para ara ambos transistores. omente la tabla anterior: A!lara!ión# A!lara!ión# :a resisten!ia de .%6 midió en realidad .86
CASO III
&'' )*+,+-)A/
01 0
32 40
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&'' '&5-
01 0
32 0
2b1 3 4(.1% 4(.1% m2 rms 2!1 3 11" u2 rms
2e1 3 4(.+ m2 rms
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ara el C'"! II con la resistencia -e )2:
GANANCIA DIFERENCIAL
1.3v 659 mv
= 1.97
GANANCIA COMN
19.66mv 659 mv
=
0.029
CMRR
1.97 0.029
≈
67.93
ara el C'"! III con la fuente de corriente:
i
Zi
≈ 10.2k Ω
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Ac
=
V 0 V i
40 mV = 0.045 0 .9V !* Rela!i Rela!ión ón de re!0a< re!0a
=
CMR
=
Ad Ac
CMR
=
0.14 0.045
=
3.11
AS' )): (,+ 12 K! a* Modo Modo Die Diere ren! n!ia iall # Ad =
V 0 V i
115 mV
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Ac
=
Ac
=
V 0 V i 70 mV 0.8V
=
0.0175
!* Rela!i Rela!ión ón de re!0a< re!0a
=
Ad Ac
CMR
=
0.1625V 0.0175V
=
0.13
Comentario:
Se =uede 5er >ue a medida >ue se aumente la R? la Ganan!ia en modo Com;n se redu!e @ =or tanto la CMR aumente @ eso es lo >ue se >uiere =ara nuestro am=lii!ador dieren!ial @ !uando se trabaBo !on una uente !orriente la
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Ampli$icador con