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Penguat CC (Common Collector)
Disebut juga :
• Penguat Kolektor Bersama
• Pengikut Emiter (Emitter Follower)
VCC = IE.RE + VCE
IC ≈ IE = (VCC – VCE) / RE
vout = ie.RE
vin = ie (RE + re’)
A = vout / vin = RE / (RE + re’)
Jika RE >> re’ maka
A≈1
Contoh 8-1
VTH = 10V.(10K / 20K)
= 5V
VE = 5V – 0,7V
= 4,3V
VC = 10V
IE = V E / RE
= 4,3V / 4,3KΩ
= 1 mA
re’ = 25mV/1mA
= 25Ω
A = 4300 / (4300 + 25)
= 0,994
vout = 0,994 x 1V
= 0,994V
Model AC dari penguat CC
zin(basis) = β(RE + re’)
≈ β.RE
zin = R1 // R2 // zin(basis)
≈ R1 // R2
zout(emiter) = r e’ + (RS // R1 // R2)/β
zout = zout(emiter) // RE
Jika RE sangat besar maka :
zout ≈ zout(emiter)
A = RE / {RE + re’ + (RS // R1 //R2)/β}
Jika β.zin(base) sangat besar maka
zin ≈ R1//R2
Jika RE>>zout(emitter) maka
zout ≈ zout(emitter)
Contoh 8-3
zin = R1//R2
= 5kΩ
re’ = 25Ω (dari soal lalu)
RS//R1//R2 = 2,09kΩ
β = 100
zout(emitter) = 25Ω + 2,09kΩ/100
= 45,9Ω
zout ≈ 45,9Ω
vin = 100mV.(5k/8k6)
= 58,1mV
Jika A = 1 maka
vout = 58,1mV
Penguat Darlington
IE2 = (VTH – 2.VBE)/RE
IE1 = IB2 ≈ IE2/βDC

zin2 = β2.RE
zin1 = β1.β2.RE
Karena ini sangat besar maka :
zin ≈ R1//R2
rth = RS//R1//R2
zout1= re1’ + rth/β1
zout2= re2’ + zout1/β2
= re2’ + (re1’ + rth/β1)/β2
Regulator Seri
VOUT = VZ – VBE
IB = IL/βDC
Zout = re’ + RZ/β
Disipasi Daya :
PD = VCE.IC
VCE = VIN – VOUT
Jika β sangat besar maka
I C ≈ IE
Contoh 8-6
Hitung IZ jika β = 80
IS = (20V – 10V) / 680Ω
= 14,7mA
VOUT = VB – VBE
=10V – 0,7V
= 9,3V
IE = VOUT / RL
= 0,62A
IB = IE/β
= 0,62A / 80
= 7,75 mA
IZ = I S – I B
= 14,7mA – 7,75mA
= 6,95mA

PD = (20V – 9,3V).0,62A
= 6,63W
re’ = 25mV/0,62A
= 0,04Ω
zout = 0,04Ω + 7/100
= 0,11Ω

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10. Penguat Common Colector

  • 1. Penguat CC (Common Collector) Disebut juga : • Penguat Kolektor Bersama • Pengikut Emiter (Emitter Follower)
  • 2.
  • 3. VCC = IE.RE + VCE IC ≈ IE = (VCC – VCE) / RE vout = ie.RE vin = ie (RE + re’) A = vout / vin = RE / (RE + re’) Jika RE >> re’ maka A≈1
  • 4. Contoh 8-1 VTH = 10V.(10K / 20K) = 5V VE = 5V – 0,7V = 4,3V VC = 10V IE = V E / RE = 4,3V / 4,3KΩ = 1 mA re’ = 25mV/1mA = 25Ω A = 4300 / (4300 + 25) = 0,994 vout = 0,994 x 1V = 0,994V
  • 5. Model AC dari penguat CC
  • 6. zin(basis) = β(RE + re’) ≈ β.RE zin = R1 // R2 // zin(basis) ≈ R1 // R2 zout(emiter) = r e’ + (RS // R1 // R2)/β zout = zout(emiter) // RE Jika RE sangat besar maka : zout ≈ zout(emiter) A = RE / {RE + re’ + (RS // R1 //R2)/β}
  • 7.
  • 8. Jika β.zin(base) sangat besar maka zin ≈ R1//R2 Jika RE>>zout(emitter) maka zout ≈ zout(emitter)
  • 9. Contoh 8-3 zin = R1//R2 = 5kΩ re’ = 25Ω (dari soal lalu) RS//R1//R2 = 2,09kΩ β = 100 zout(emitter) = 25Ω + 2,09kΩ/100 = 45,9Ω zout ≈ 45,9Ω vin = 100mV.(5k/8k6) = 58,1mV Jika A = 1 maka vout = 58,1mV
  • 10. Penguat Darlington IE2 = (VTH – 2.VBE)/RE IE1 = IB2 ≈ IE2/βDC zin2 = β2.RE zin1 = β1.β2.RE Karena ini sangat besar maka : zin ≈ R1//R2
  • 11. rth = RS//R1//R2 zout1= re1’ + rth/β1 zout2= re2’ + zout1/β2 = re2’ + (re1’ + rth/β1)/β2
  • 12. Regulator Seri VOUT = VZ – VBE IB = IL/βDC Zout = re’ + RZ/β Disipasi Daya : PD = VCE.IC VCE = VIN – VOUT Jika β sangat besar maka I C ≈ IE
  • 13. Contoh 8-6 Hitung IZ jika β = 80 IS = (20V – 10V) / 680Ω = 14,7mA VOUT = VB – VBE =10V – 0,7V = 9,3V IE = VOUT / RL = 0,62A IB = IE/β = 0,62A / 80 = 7,75 mA IZ = I S – I B = 14,7mA – 7,75mA = 6,95mA PD = (20V – 9,3V).0,62A = 6,63W re’ = 25mV/0,62A = 0,04Ω zout = 0,04Ω + 7/100 = 0,11Ω