Experiencia 9

Prévia do material em texto

Experiência 9
Amplificador Bootstrapping
Grupo: Jéssica Rangel, Rachel Reuters , Vinicio Mendes
O amplificador bootstrapping funciona semelhante ao coletor comum, com ganho de tensão menor que 1, no entanto, a sua impedância de entrada é maior.
Quando se aumenta a impedância de entrada há o aumento do ganho de corrente e de potência.
>>Transistor BC 547B e f = 400 Hz
RL = 0,11 kΩ e excursão de sinal de saída de 6 Vpp:
Vce = 3 V + Vcesat, Vce = 3 V + 1 V, Vce = 4 V
Análise AC:
Vce = R Icq ,R = (R1//R2) // RE // RL
Supor RE = 0,11 kΩ
	
Análise DC:
0,7 V
R3
Vce
Saída:
Vcc = Vce + RE. Ieq
Ieq = (15 – 4) / 0,11 k, Ieq = 100 mA, Icq = Ieq = 100 mA,
 Icq = 100 mA, hfe = 100
hie = (0,05. hfe) / Icq, hie = 0,05 kΩ
VE = RE Ieq, VE = 11 V
V1 = VE + 0,7 V1 = 11,7 V
V2 = Vcc – V1, V2 = 15 – 11,7 V2 = 3,3 V
I1? I2? R3?
I1 ~ I2 = (Icq / 10) I1 = I2 = 10 mA
R1 = V1 / I1, R1 = 11,7 / 10 R1 = 1,17 kΩ
R2 = V2 / I2, R2 = 3,3 / 10 R2 = 0,33 kΩ
(hie / R3) < 1, R3 > 0,05 kΩ = 50 Ω R3 = 50Ω
Valores Comerciais dos Resistores:
R1 = 1,2 kΩ, R2 = 0,33 kΩ, RE = 0,12 kΩ, R3 = 0,050 kΩ, RL = 0,12 kΩ
hfe = β = 100, hie = 0,05 kΩ, Rg = 0,05 kΩ 
Análise DC:
0,7 V
RE
Vceq
R3
Vcc
Rb
Vbb
Polarização:
Rb = (R1.R2) / (R1 + R2) Rb = 0,26 kΩ
Vbb = (R1.Vcc) / (R1 + R2) Vbb = 11,76 V
Entrada:
Vbb = Rb.Ibq + R3.Ibq + 0,7 + RE.Ieq,
Vbb = Rb.[Ieq / (β +1)] + R3.[Ieq / (β +1)] + RE.Ieq
Ieq = (Vbb – 0,7) / [RE + (Rb + R3) / (β + 1)] 
Ieq = 90 mA Icq = Ieq = 90 mA
Ibq = Ieq / (β + 1) Ibq = 89,1 mA
	
Modelo h – híbrido (Simplificado):
 Ie
(Ii – Ib) →
VL
Ii → B
+
Vi
_
R3
Sendo R = (Rb // RE // RL) R = 50 Ω
R3.(Ii – Ib) = hie.Ib
R3.Ii – R3.Ib = hie.Ib Ib = [1 / (1 + (hie / R3))]. Ii
Vi = (R3 // hie).Ii + R. Ie Ie?
Ie = Ii + Ic = Ii + hfe.Ib Ie = Ii + hfe.Ii.[1 / (1 + (hie / R3))]
Vi = (R3 // hie).Ii + R.Ii + R.hfe.Ii.[1 / (1 + (hie / R3))]
Zi = Vi / Ii, portanto:
Zi = (R3 // hie) + R + R.hfe.[1 / (1 + (hie / R3))]
Zi = (68 Ω // 50 Ω) + 50 Ω + 50.100.[1 / (1 + (50 / 68))]
Zi = 2,96 kΩ
	
VL = R.(Ii + a.hfe), sendo a = [1 / (1 + (hie / R3))], a = 0,576 
Vg = [Rg + (R3 // hie)].Ii + R.(Ii + a.hfe)
Avs = VL / Vg, 
Avs = {1 / [1 + ((Rg + [(R3 //hie) / (R.(1 + a.hfe))]}
Avs = {1 / [1 + ((50 + [(68 // 50) / (50.(1 + 0,576.100))]}
Avs = 0,97
Avi = VL / Vi
Avi = {1 / [1 + [(R3 //hie) / (R.(1 + a.hfe))]}
Avi = {1 / [1 + [(68 //50) / (50.(1 + 0,567.100))]}
Avi = 0,99
Ai = (Zi.Avi) / R
Ai = (2,96k.0,99) / 0,05k Ai = 58,6
Zo = R // Zo`
Zo` = VL`/ Io`
VL`= - Ii.[Rg + (R3 // hie)] Io`= - Ii.[a.(1 + hfe)]
Zo`= [Rg + (R3 // hie)] / [a.(1 + hfe)]
Zo`= [50 + (68 // 50)] / [0,567.(1 + 100)] Zo`= 1,37 Ω
Logo: Zo = 50 Ω // 1,37 Ω Zo = 1,36 Ω
Modelo h – híbrido (Completo):
 Ii → B
hie
(Ii – Ib) →
+
Vi
_
Icq = 9 mA → hre = 1,2.10-4 → hoe = 50 μA/V (datasheet)
Portanto: Icq = 90 mA → hre = 12.10-4 → hoe = 500 μA/V
Vi = (R3 // hie).Ii + R.(Ii + Ie) + hre.Vce Ie?
Ie = Ii + Ic = Ii + (1 + hoe).Ib Ie = Ii + (1 + hoe).Ii.[1 / (1 + (hie / R3))]
Vce = hoe.hfe.Ib Vce = hoe.hfe. Ii.[1 / (1 + (hie / R3))]
Vi = (R3 // hie).Ii + R.Ii + R.(1 + hoe ).Ii.[1 / (1 + (hie / R3))] + hoe.hfe. Ii.[1 / (1 + (hie / R3))]
Zi = Vi / Ii, portanto:
Zi = (R3 // hie) + R + R(1 + hoe ).[1 / (1 + (hie / R3))] + hoe.hfe. .[1 / (1 + (hie / R3))]
Zi = 2,93 kΩ
Avi = VL / Vi 
VL = R.Ii.[a + a.hfe.(1 + hoe)] sendo a = [1 / (1 + (hie / R3))]
Avi = {R.[a + a.hfe.(1 + hoe)]} / {[(R3 // hie) + hre.hoe.hfe.a + R.a.[1 + (1 + hoe).hfe]]}
Avi = 0,99
Ai = (Zi /R). Avi Ai = (2930 Ω / 50 Ω). 0,99 Ai = 58
Avs = VL /Vg
Avs = {R.Ii.[a + a.hfe.(1 + hoe)]} / {Rg +(R3 // hie).Ii + R.Ii + R.(1 + hoe ).Ii.[1 / (1 + (hie / R3))] + hoe.hfe. Ii.[1 / (1 + (hie / R3))]}
Avs = 0,97
Zo = R // Zo`
Zo` = VL`/ Io`
VL`= - Ii.[Rg + (R3 // hie) + hre.hoe.hfe.a] Io`= - Ii.[a + a.hfe(1 + hoe)]
Zo`= [Rg + (R3 // hie) + hre.hoe.hfe.a ] / [a + a.hfe(1 + hfe)]
Zo`= 1,3 Ω
Logo: Zo = 50 Ω // 1,3 Ω Zo = 1,26 Ω
Modelo π – híbrido (Completo):
	
Ie 
(Ii – Ib) 
+
Vi
_
Ii B
Ro
gm.Vπ
Rπ
Rπ = Vt / Ibq = 25 mV / 89,1 mA Rπ = 0,28 Ω
Ro = Vce / Icq = 4,2 V / 90 mA Ro = 46,7 Ω
gm = Icq / Vt = 90 mA / 25 mV gm = 3,6 A/V
Ib = [1 / (1 + (Rπ / R3))]. Ii
Ic = Vπ.(gm + Vπ / Ro), sendo Vπ = Rπ.Ib
Ic = Ib. Rπ.(gm + 1/Ro) 
Vi = (R3 // Rπ ).Ii + R.(Ii + Ic)
Vi = (R3 // Rπ).Ii + R.Ii + R. Rπ.Ii.[1 / (1 + (Rπ / R3).(gm + 1/Ro))]
Zi = Vi / Ii
Zi = (R3 // Rπ) + R + R. Rπ.[1 / (1 + (Rπ / R3).(gm + 1/Ro))]
Zi = 50,27 + 50,7 Zi = 100,1 Ω
Avi = VL / Vi
VL = R.(Ii + b.Rπ), sendo b = [1 / (1 + (Rπ / R3))], b = 1
Vg = [Rg + (R3 // Rπ)].Ii + R.(Ii + b.Rπ)
Avs = VL / Vg, 
Avs = {1 / [1 + ((Rg + [(R3 //Rπ) / (R.(1 + b.Rπ))]}
Avs = 0,66
Avi = VL / Vi
Avi = {1 / [1 + [(R3 //Rπ) / (R.(1 + b. Rπ))]}
Avi = 0,99
Ai = (Zi.Avi) / R
Ai = (100,1.0,99) / 50 Ai = 2
Zo = R // Zo`
Zo` = VL`/ Io`
VL`= - Ii.[Rg + (R3 // Rπ)] Io`= - Ii.[b.(1 + Rπ.(gm + 1/Ro))]
Zo`= [Rg + (R3 // hie)] / [b.(1 + Rπ.(gm + 1/Ro))]
Zo`= 25 Ω
Logo: Zo = 50 Ω // 25 Ω Zo = 17 Ω
Modelo π – híbrido (Simplificado):
Ie 
+
Vi
_
(Ii – Ib) 
Ii B 
gm.Vπ
Rπ
Rπ = 0,28 Ω gm = 3,6 A/V
Ib = [1 / (1 + (Rπ / R3))]. Ii
Ic = gm .Vπ, sendo Vπ = Rπ.Ib Ic = gm. Rπ.Ib
Ic = gm. Rπ. [1 / (1 + (Rπ / R3))]. Ii
Vi = (R3 // Rπ ).Ii + R.(Ii + Ic)
Vi = (R3 // Rπ).Ii + R.Ii + R. Rπ.gm.Ii.[1 / (1 + (Rπ / R3))]
Zi = Vi / Ii
Zi = (R3 // Rπ) + R + R. Rπ.gm.[1 / (1 + (Rπ / R3))]
Zi = 99,3 Ω
Avi = VL / Vi
VL = R.(Ii + b.Rπ.gm), sendo b = [1 / (1 + (Rπ / R3))], b = 1
Vg = [Rg + (R3 // Rπ)].Ii + R.(Ii + b.Rπ.gm)
Avs = VL / Vg, 
Avs = {1 / [1 + ((Rg + [(R3 //Rπ) / (R.(1 + b.Rπ.gm))]}
Avs = 0,68
Avi = VL / Vi
Avi = {1 / [1 + [(R3 //Rπ) / (R.(1 + b. Rπ.gm))]}
Avi = 0,98
Ai = (Zi.Avi) / R
Ai = (99,3.0,98) / 50 Ai = 1,94
Zo = R // Zo`
Zo` = VL`/ Io`
VL`= - Ii.[Rg + (R3 // Rπ)] Io`= - Ii.[b.(1 + Rπ.gm)]
Zo`= [Rg + (R3 // Rπ)] / [b.(1 + Rπ.gm)]
Zo`= 25,1 Ω
Logo: Zo = 50 Ω // 25,1 Ω Zo = 16,7 
Modelo re ou T (Completo):
Ie 
(Ii – Ib) 
Ii B
Ro
gm.Vbe
+
Vi
_
re
re = Vt / Ieq = 25 mV / 90 mA re = 0,27 Ω Ro = 46,7 Ω gm = 3,6 A/V
Vbe = Vt = 25 mA R = (RE // RL // Rb) R = 50 Ω
Ib = [1 / (1 + (re / R3))]. Ii
Ic = gm.Vbe + Vbe / Ro Vbe = re.Ib
Ic = re.Ib.[gm + (1 / Ro)]
Vi = (R3 // re ).Ii + R.(Ii + Ic)
Vi = (R3 // re).Ii + R.Ii + R. re.Ii.[1 / (1 + (re / R3).(gm + 1/Ro))]
Zi = Vi / Ii
Zi = (R3 // re) + R + R. re.[1 / (1 + (re / R3).(gm + 1/Ro))]
Zi = 50,27 + 48,87 Zi = 99,1 Ω
Avi = VL / Vi
VL = R.(Ii + c.re), sendo c = [1 / (1 + (re / R3))], c = 1
Vg = [Rg + (R3 // re)].Ii + R.(Ii + c.re)
Avs = VL / Vg, 
Avs = {1 / [1 + ((Rg + [(R3 // re) / (R.(1 + c.re))]}
Avs = 0,66
Avi = VL / Vi
Avi = {1 / [1 + [(R3 // re) / (R.(1 + c.re))]}
Avi = 0,99
Ai = (Zi.Avi) / R
Ai = (99,1.0,99) / 50 Ai = 1,96
Zo = R // Zo`
Zo` = VL`/ Io`
VL`= - Ii.[Rg + (R3 // re)] Io`= - Ii.[b.(1 + re.(gm + 1/Ro))]
Zo`= [Rg + (R3 // re)] / [b.(1 + re.(gm + 1/Ro))]
Zo`= 25 Ω
Logo: Zo = 50 Ω // 25 Ω Zo = 17 Ω
Modelo re ou T (Simplificado):
Ie 
+
Vi
_
Ii B
(Ii – Ib) 
gm.Vbe
re
Ib = [1 / (1 + (re / R3))]. Ii
Ic = gm .Vbe, sendo Vbe = re.Ib Ic = gm. re.Ib
Ic = gm. re. [1 / (1 + (re / R3))]. Ii
Vi = (R3 // Rπ ).Ii + R.(Ii + Ic)
Vi = (R3 // re).Ii + R.Ii + R. re.gm.Ii.[1 / (1 + (re / R3))]
Zi = Vi / Ii
Zi = (R3 // re) + R + R. re.gm.[1 / (1 + (re / R3))]
Zi = 98,7 Ω
Avi = VL / Vi
VL = R.(Ii + c.re.gm), sendo c = [1 / (1 + (re / R3))], c = 1
Vg = [Rg + (R3 // re)].Ii + R.(Ii + c.re.gm)
Avs = VL / Vg, 
Avs = {1 / [1 + ((Rg + [(R3 //re) / (R.(1 + c.re.gm))]}
Avs = 0,67
Avi = VL / Vi
Avi = {1 / [1 + [(R3//re) / (R.(1 + c.re.gm))]}
Avi = 1
Ai = (Zi.Avi) / R
Ai = (98,7.1) / 50 Ai = 1,97
Zo = R // Zo`
Zo` = VL`/ Io`
VL`= - Ii.[Rg + (R3 // re)] Io`= - Ii.[c.(1 + re.gm)]
Zo`= [Rg + (R3 // re)] / [c.(1 + re.gm)]
Zo`= 25,5 Ω
Logo: Zo = 50 Ω // 25,5 Ω Zo = 16,8 Ω
Retas de Carga para valores comerciais dos resistores:
 Ieq=0:5:180
 Vceq=15-0.12.*Ieq
 plot (Ieq,Vceq,'k')
 hold on
 title('DC x AC') 
 xlabel('Ieq (mA)') 
 ylabel('Vceq(V)') 
 Vce=4.39-0.025.*Ieq
 plot (Ieq,Vce,'r')
 text(80,6, 'DC')
 text(5,5, 'AC')