SOAL 9 Suatu rangkaian penguat satu tingkat dengan bias pembagi tegangan mempunyai data parameter-parameter sebagai berikut : Vcc = 30 volt
R1 = 22 Kohm
R2 = 10 Kohm
RC = 1 ohm
RE = 1 Kohm
RL = 1 Kohm
Vin = 25 mV
Cin = 1 uF
βdc = 100 VBE = 0,7 V RS = 1 Kohm Cout = 1 uF
CE = 1 uF
Tugas : a. Gambar Gambar rangkaia rangkaian n lengkap lengkap dan rangka rangkaian ian analisa analisa DC DC b. Tent entukan ukan IB, IC, IE , αdc, dan VCE c. IC saturasi, VCE cut off dan garis beban DC Penyelesaian : a. Gamb Gambar ar Rang Rangka kaia ian n
Rangkaian Analisa DC
b. Vth
= Vc Vcc*R2/(R1+R2) = 9,375 volt
Rth
= (R1*R2)/(R1+R2) = 6875 Ohm
Ib
= (Vth – VBE)/(R B+(1+ βdc)RE) = (9,375-0,7)/(6875+(1 (9,375-0,7)/(6875+(1+100)1K) +100)1K) = 8,675/107875 = 0,080417 mA
Ic = βdc x IB = 100 x 8,0417 mA = 8,0417 mA
IE = IB+IC = 0,080417 mA +8,0417 mA = 8,122 mA
αdc
= IC / IE = 8,0417 mA /8,122 mA = 0,9901
VCE = VCC – IC (RC +RE / αdc) = 30 – 8,0417mA (1+1K/0,9901) = 21,869 V
IC sat
= VCC/(RC + RE /αdc) = 30 / (1+1K/0,9901) = 29,673 mA
Vcut off = VCC = 30 V
c. Gari Garis s Beb Beban an DC
IB Q point
SOAL 10 Suatu rangkaian penguat satu tingkat dengan umpan balik kolektor mempunyai data parameter-parameter sebagai berikut : Vcc = 25 volt
RB = 240 Kohm
βdc = 120
RC = 2 Kohm
VBE = 0,72 V
RE = 1 Kohm
RL = 1 Kohm
RS = 2 Kohm
CE = 2,5 uF
Vin = 30 mV
Cin = 1,5 uF
Cout = 2,1 uF
Tugas : a. Gambar Gambar rangkaia rangkaian n lengkap lengkap dan rangka rangkaian ian analisa analisa DC DC b. Tent entukan ukan IB, IC, IE, αdc, dan VCE c. IC saturasi, VCE cut off dan garis beban DC d. Ulangi Ulangi pertanyaa pertanyaan n a,b, dan dan c bila RE tidak dipas dipasang ang
Penyelesaian : •
Dengan RE a. Gamb Gambar ar Ran Rangk gkai aian an
Rangkaian Analisa DC
b. Ib
= (Vcc – VBE)/(RB+(1+ βdc)(RC + RE ) = (25-0,72)/(240K+(1+120 (25-0,72)/(240K+(1+120)(2K+1K)) )(2K+1K)) = 24,28/603K = 0,0402 mA
Ic
= βdc x IB = 120 x 0,0402 mA = 4,831 mA
IE
= IB+IC = 0,0402 mA + 4,831 mA = 4,8712 mA
αdc = IC / IE = 4,831 mA /4,8712 mA = 0,9917
VCE = VCC – (IC (RC + RE )/ αdc)
= 25 – (4,831 mA (2K+1K)/ 0,9917) = 25 – 14,614 V = 10,385 V
IC saturasi
= αdc x VCC / (RC + RE ) = 0,9917 x 25 / 3k = 8,264 mA
Vcut off
= VCC = 25 V
c. Gari Garis s Beb Beban an DC
8,2 64
IB Q point
4,8 31
•
Tanpa RE
a. Gamb Gambar ar Ran Rangk gkai aian an
Rangkaian Analisa DC
b. Ib
= (Vcc – VBE)/(RB+(1+ βdc)RC ) = (25-0,72)/(240K+(1+120 (25-0,72)/(240K+(1+120)2K) )2K) = 24,28/482K = 0,05037 mA
Ic
= βdc x IB = 120 x 0, 0,05037 mA = 6,044 mA
IE
= IB+IC = 0,05037 mA + 6,044 mA = 6,0951 mA
αdc = IC / IE = 6,044 mA /6,0951 mA = 0,9916
VCE = VCC – (IC RC/ αdc) = 25 – (6,044 mA * 2K/ 0,9916) = 25 – 12,1903 V = 12,8096 V
IC saturasi
= αdc x VCC / RC = 0,9916 x 25 / 2k = 12,395 mA
Vcut off
= VCC = 25 V
c. Gari Garis s Beb Beban an DC
IB
12,395 Q point
6,044
12,8096
