Amplificador emisor comun
Páginas: 3 (505 palabras)
Publicado: 29 de marzo de 2012
comun Diseño de Amplificador de potencia clase A |
Vin P=100m [V]VOP=5 [V]RL=4.7 [KΩ] | |
VCE MAX=40 [V]IC MAX=200 [mA]PD MAX=625 [mW] | |
Resistencia del colector:
VOP+≤IC*RC∥RLIC≥VOP+RC∥RL
VRCRC≥VOP+RC∥RL
VRC≥VOP+*RCRC∥RL
Tengo 3 opciones
RC≪RLRC=470 [Ω] | RC=RLRC=4700 [Ω] | RC≫RLRC=47000 [Ω] |
VRC≥5*470470∥4700VRC≥5.5 [V]VRC=6.6 [V] | VRC≥VOP+*RCRC∥RLVRC≥10 [V]VRC=12 [V] |VRC≥VOP+*RCRC∥RLVRC≥55 [V]VRC=66 [V] |
IC=14.04 [mA] | IC=2.55 [mA] | IC=1.4 [mA] |
Elijo la opción 2
Resistencias del emisor:
Vin P-≤IE*RE1IE≥Vin P-RE1VERE≥Vin P-RE1VE≥RERE1Vin P-VE≥2[V]RE=VEIE=2[V]2.55 [mA]=784.31 [Ω] | AV=-RC*RLRC+RLre+RE1=5RE1=15RC*RLRC+RL-reRE1=152350-26 [mV]2.55 [mA]=467.96 [Ω]≈470 [Ω]RE2=RE-RE1=784.31-470=314.31 [Ω]≈330 [Ω] |
Compruebo que con estos valores de resistenciasse cumpla la condición:VE≥2VVE≥RERE1Vin P-2≥1.7 Si cumple |
Resistencias de la base:
I2≫IB MAXIB MAX=ICβMIN=2.55 [mA]100=0.0255 [mA]I2=0.255 [mA]RB2=VE-VBEI2=3.2 [V]0.255 [mA]=12549.02 [Ω]≈10 [KΩ]| VCE≤Vin P++VSAT+VOP-VCE≤1+2+5VCE≤8 [V]VCE=9.6 [V]VCC=VRC+VCE+VE=12+9.6+2=23.6 V≈24[V]RB2=VCC-VBI2+IB=24-2.70.255+0.0255=75.94 [KΩ]≈68 [KΩ] |
capacitores:RTH=RB1*RB2RB1+RB2=68000*1000068000+10000=8717.95 [Ω]
rE=26 [mV]IE=26 [mV]2.55 mA=10.2 [Ω]
Zin=RTH*hfemin+1rE+RE1RTH+hfemin+1rE+RE1=8717.95*10110.2+4708717.95+10110.2+470=7389.65 [Ω]
XCB MAX≪Zin
12πfMINCB≪Zin
CB≫12πfMINZinCB≫12π*1000*7389.65
CB=0.22 [uF]
CB≫12π*100*10^6*7389.65
CB=0.22 [pF]
XCC MAX≪RL
CC≫12πfMINRL
CC≫12π*1000*4700
CC≫33.86 [nF]
CC=0.3 [uF]
AV=RC∥RLrE+RE1+RE2∥XCESe tiene 2 opciones: |XCE≪RE2XCE≪330 | XCE≪rE+RE1XCE≪480.2 |
Se escoge la primera opciónCE≫12πfMINXCECE≫12π*1000*330CE≫0.48 [uF] CE=10 [uF | CE≫12πfMINXCECE≫12π*100x10^6*330CE≫0.48 [uF] CE=10 [pF] |
Calculando elequivalente Thévenin:
VTH=RB1RB1+RB2*VCC=1000068000+10000*24=3.08 [V]
RTH=RB1*RB2RB1+RB2=68000*1000068000+10000=8717.95 [Ω]
VE=IE*RE=2.79 mA*800Ω=2.23 [V] | VB=VBE+VE=0.7+2.23=2.93 [V]...
Vin P=100m [V]VOP=5 [V]RL=4.7 [KΩ] | |
VCE MAX=40 [V]IC MAX=200 [mA]PD MAX=625 [mW] | |
Resistencia del colector:
VOP+≤IC*RC∥RLIC≥VOP+RC∥RL
VRCRC≥VOP+RC∥RL
VRC≥VOP+*RCRC∥RL
Tengo 3 opciones
RC≪RLRC=470 [Ω] | RC=RLRC=4700 [Ω] | RC≫RLRC=47000 [Ω] |
VRC≥5*470470∥4700VRC≥5.5 [V]VRC=6.6 [V] | VRC≥VOP+*RCRC∥RLVRC≥10 [V]VRC=12 [V] |VRC≥VOP+*RCRC∥RLVRC≥55 [V]VRC=66 [V] |
IC=14.04 [mA] | IC=2.55 [mA] | IC=1.4 [mA] |
Elijo la opción 2
Resistencias del emisor:
Vin P-≤IE*RE1IE≥Vin P-RE1VERE≥Vin P-RE1VE≥RERE1Vin P-VE≥2[V]RE=VEIE=2[V]2.55 [mA]=784.31 [Ω] | AV=-RC*RLRC+RLre+RE1=5RE1=15RC*RLRC+RL-reRE1=152350-26 [mV]2.55 [mA]=467.96 [Ω]≈470 [Ω]RE2=RE-RE1=784.31-470=314.31 [Ω]≈330 [Ω] |
Compruebo que con estos valores de resistenciasse cumpla la condición:VE≥2VVE≥RERE1Vin P-2≥1.7 Si cumple |
Resistencias de la base:
I2≫IB MAXIB MAX=ICβMIN=2.55 [mA]100=0.0255 [mA]I2=0.255 [mA]RB2=VE-VBEI2=3.2 [V]0.255 [mA]=12549.02 [Ω]≈10 [KΩ]| VCE≤Vin P++VSAT+VOP-VCE≤1+2+5VCE≤8 [V]VCE=9.6 [V]VCC=VRC+VCE+VE=12+9.6+2=23.6 V≈24[V]RB2=VCC-VBI2+IB=24-2.70.255+0.0255=75.94 [KΩ]≈68 [KΩ] |
capacitores:RTH=RB1*RB2RB1+RB2=68000*1000068000+10000=8717.95 [Ω]
rE=26 [mV]IE=26 [mV]2.55 mA=10.2 [Ω]
Zin=RTH*hfemin+1rE+RE1RTH+hfemin+1rE+RE1=8717.95*10110.2+4708717.95+10110.2+470=7389.65 [Ω]
XCB MAX≪Zin
12πfMINCB≪Zin
CB≫12πfMINZinCB≫12π*1000*7389.65
CB=0.22 [uF]
CB≫12π*100*10^6*7389.65
CB=0.22 [pF]
XCC MAX≪RL
CC≫12πfMINRL
CC≫12π*1000*4700
CC≫33.86 [nF]
CC=0.3 [uF]
AV=RC∥RLrE+RE1+RE2∥XCESe tiene 2 opciones: |XCE≪RE2XCE≪330 | XCE≪rE+RE1XCE≪480.2 |
Se escoge la primera opciónCE≫12πfMINXCECE≫12π*1000*330CE≫0.48 [uF] CE=10 [uF | CE≫12πfMINXCECE≫12π*100x10^6*330CE≫0.48 [uF] CE=10 [pF] |
Calculando elequivalente Thévenin:
VTH=RB1RB1+RB2*VCC=1000068000+10000*24=3.08 [V]
RTH=RB1*RB2RB1+RB2=68000*1000068000+10000=8717.95 [Ω]
VE=IE*RE=2.79 mA*800Ω=2.23 [V] | VB=VBE+VE=0.7+2.23=2.93 [V]...
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