Contoh Soal RANGKAIAN ELEKTRONIKA-penguat Satu Tingkat [PDF]

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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 rangkaian lengkap dan rangkaian analisa DC b. Tentukan IB, IC, IE , αdc, dan VCE c. IC saturasi, VCE cut off dan garis beban DC Penyelesaian : a. Gambar Rangkaian



Rangkaian Analisa DC



b. Vth



= Vcc*R2/(R1+R2) = 9,375 volt



Rth



= (R1*R2)/(R1+R2) = 6875 Ohm



Ib



= (Vth – VBE)/(RB+(1+ βdc)RE) = (9,375-0,7)/(6875+(1+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. Garis Beban 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 rangkaian lengkap dan rangkaian analisa DC b. Tentukan IB, IC, IE, αdc, dan VCE c. IC saturasi, VCE cut off dan garis beban DC d. Ulangi pertanyaan a,b, dan c bila RE tidak dipasang



Penyelesaian : •



Dengan RE a. Gambar Rangkaian



Rangkaian Analisa DC



b. Ib



= (Vcc – VBE)/(RB+(1+ βdc)(RC + RE ) = (25-0,72)/(240K+(1+120)(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. Garis Beban DC



8,2 64



IB Q point



4,8 31



•



Tanpa RE



a. Gambar Rangkaian



Rangkaian Analisa DC



b. Ib



= (Vcc – VBE)/(RB+(1+ βdc)RC ) = (25-0,72)/(240K+(1+120)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. Garis Beban DC



IB



12,395 Q point 6,044



12,8096