- Aim To Design A Ce Transistor Amplifier Objectives 1 To Calculate Voltage Divider Bias Configuration Components 2 1 (69.04 KiB) Viewed 67 times
- Aim To Design A Ce Transistor Amplifier Objectives 1 To Calculate Voltage Divider Bias Configuration Components 2 2 (45.86 KiB) Viewed 67 times
- Aim To Design A Ce Transistor Amplifier Objectives 1 To Calculate Voltage Divider Bias Configuration Components 2 3 (60.14 KiB) Viewed 67 times
- Aim To Design A Ce Transistor Amplifier Objectives 1 To Calculate Voltage Divider Bias Configuration Components 2 4 (43.02 KiB) Viewed 67 times
Example: Design a CE voltage-divider bias configuration BJT amplifier of gain 100, and lower cut off frequency of 200HZ where Rc = 1 KQ and DC Power Supply voltage is 15V, used the network shown in Figure 2. VCC 915V R1 W SRC 1ΚΩ Сс Je Q1 2N2222A Cb 2. Zo Vi 10m Vpk 1kHz 0° Zi R2 w RE ce w Figure 2. CE BJT voltage divider circuit.
Design Steps: We know that: 1. To have a stable DC bias R2 Rc so we choose Re - 2200 2. A, !GRC 26mu to have voltage gain of 100. lc > 2.6 mA Hence: 3. RTH = Ri||R2 Bmin = 100 BR R+R 5 0.1BRE So, we choose Rru 2.2KO (assume h 4. Vrv = V = VE+IgRg = 1.27V = 5. The fact that RTH Ra Ra = 2.2Kn and TH VecR; = 1.27V. If we solve these Ry+R Ri+R two equation for R2. R22.40, we choose a standard value for R2 from our available resistors: R2 = 2.2k. 6. Since the aim is to calculate values for R1 and R2 to fulfil the DC bias requirement, which is linked to the AC voltage gain. Vw= VROR Equation RR2
MUST be used to calculate a value for R1. R23.78K, we choose the closes available standard value for R1. So, R1 = 22K0. 15x2.2 0.7 VCC 915V TE Z2+2.2 0.220 R1 22ka le =3.02ma RC 180 Сс HG Q1 2N2222A Cb 1 Za VI 10mVpk 1kHz ZI R2 2.2k ce RE 2200