Given, VCE = 5 V and IC = 2 mA VCC = 2VCE = 2 X 5 = 10 V Let VRE = 10% VCC = 1 V RE = VRE / ( IC + IB ) IB = IC / β = 2 mA / 100 = 20 µA RE = 1 / ( 2 m = 20 µ ) = 495 Ω Choose RE = 470 Ω Apply KVL to collector loop VCC – IC RC – VCE - VE = 0
RC = ( VCC – VCE – VE ) / IC = ( 10 – 5 – 1 ) / 2 m RC = 2 kΩ
Choose Rc = 2.2 kΩ
Let IR1 = 10 IB = 10 X 20 µA VR2 = VBE + VE = 0.6 + 1 ( Since transitor is silicon make VBE = 0.6 ) R2 = 8.8 kΩ choose R2 = 10 kΩ R1 = ( VCC – VR2 ) / IR1 = ( 10 – 1.6 ) / 200 µA R1 = 42 kΩ
choose R1 = 47kΩ
XCE < < RE XCE = RE / 10 1 / (2πfCE ) = 470 / 10 CE = 33 µF
let f = 100 Hz
choose CE = 47 µF
Procedure: 1. Check the components / Equipments for their working condition. 2. Connections are made as shown in the circuit diagram. 3. Apply the input signals as mentioned in the circuit diagram. 4. Observe the output waveform. 5. Measure the output frequency, it has to be equal to fO = f1 – f2 Hz
Result: f1 = _________Hz f2 = _________Hz fo = _________Hz
