10. penguat common colector
TRANSCRIPT
Penguat CC (Common Collector)
+VCC
VOUT
VIN VCE
+10V
+1,3V
+2V
1KRE
+10V
+2,3V
+3V
1K
Disebut juga :• Penguat Kolektor Bersama • Pengikut Emiter (Emitter Follower)
IC
VCE
IC
VCE
10mA
10V
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
10K
4K310K
+10V
+1V
-1V
0
+6V
+4V
+5V
+5,29V
+3,31V
+4,3V
VTH = 10V.(10K / 20K) = 5VVE = 5V – 0,7V = 4,3VVC = 10VIE = VE / RE
= 4,3V / 4,3K = 1 mAre’ = 25mV/1mA = 25A = 4300 / (4300 + 25) = 0,994vout = 0,994 x 1V = 0,994V
Model AC dari penguat CC
R1
RER2
+VCC
vout
vs
RS
R1 // R2 vin
RS
vs
RS // R1 // R2
vs
re’
RE
ic
vout
R1 // R2 vout
vin
re’ + (RS//R1//R2)/b
zin(basis) = b(RE + re’) ≈ b.RE
zin = R1 // R2 // zin(basis)
≈ R1 // R2
zout(emiter) = r e’ + (RS // R1 // R2)/b
zout = zout(emiter) // RE
Jika RE sangat besar maka :
zout ≈ zout(emiter)
A = RE / {RE + re’ + (RS // R1 //R2)/b}
R1
RER2
+VCC
vout
vs
RS
zin(base)zin zout(emiter) zout
zout
zin
vin vout
A.vin
zout(emitter)
R1//R2
vin vout
vin
Model AC Model AC yang disederhanakan
Jika b.zin(base) sangat besar maka zin ≈ R1//R2
Jika RE>>zout(emitter) maka zout ≈ zout(emitter)
Contoh 8-3
10K
4K310K
+10V
vout
100mV
3K6
zin = R1//R2
= 5kre’ = 25 (dari soal lalu)RS//R1//R2 = 2,09kb = 100zout(emitter) = 25 + 2,09k/100 = 45,9zout ≈ 45,9
vin = 100mV.(5k/8k6) = 58,1mVJika A = 1 makavout = 58,1mV
Penguat Darlington
Q1R1
RER2
+VCC
voutvs
RSQ2
IE2
IE1
IE2 = (VTH – 2.VBE)/RE
IE1 = IB2 ≈ IE2/bDC
zin2 = b2.RE
zin1 = b1.b2.RE
Karena ini sangat besar maka :zin ≈ R1//R2
rth = RS//R1//R2
zout1= re1’ + rth/b1
zout2= re2’ + zout1/b2
= re2’ + (re1’ + rth/b1)/b2
Regulator Seri
VOUT = VZ – VBE
IB = IL/bDC
Zout = re’ + RZ/b
Disipasi Daya :PD = VCE.IC
VCE = VIN – VOUT
Jika b sangat besar makaIC ≈ IE
VZ
RS
RL
VIN VOUT
IB
IZ
IS
IL
Contoh 8-6
10V
680
15
20V VOUT
IB
IZ
IS
IL
Hitung IZ jika b = 80IS = (20V – 10V) / 680 = 14,7mAVOUT = VB – VBE =10V – 0,7V = 9,3VIE = VOUT / RL
= 0,62AIB = IE/b = 0,62A / 80 = 7,75 mAIZ = IS – IB = 14,7mA – 7,75mA = 6,95mA
PD = (20V – 9,3V).0,62A = 6,63Wre’ = 25mV/0,62A = 0,04zout = 0,04 + 7/100 = 0,11