How do you design an RC phase shift oscillator using transistors?

Answer 1

TRANSISTOR PHASE SHIFT OSCILLATOR

CIRCUIT DIAGRAM:

THEORY:

The Transistor Phase Shift Oscillator produces a sine wave of desired designed frequency. The RC combination will give a 60° phase shift totally three combination will give a 180° phase shift. . The BC107 is in the common emitter configuration. Therefore that will give a 180° phase shift totally a 360° phase shift output is produced. The capacitor value is designed in order to get the desired output frequency. Initially the C and R are connected as a feedback with respect to input and output and this will maintain constant sine wave output. CRO is connected at the output.

DESIGN:

Given : Vcc = 12V , fo = 1 KHz,C = 0.01µF; IE = 5mA.; Stability factor = 10

f = 1/ 2πRC Find R

R1 = (Ri - R)

R >> Rc

Βeta = -1 / 29

Amplifier Design :

Gain formula is given by

Av = -hfe RLeff / hie ( Av = 29, design given )

Assume, VCE = Vcc / 2

RLeff = R c | | RL

re = 26mV / IE

hie = β re where re is internal resistance of the transistor.

hie = hfe re

VE = Vcc / 10

On applying KVL to output loop,

Vcc = IcRc + VCE + IERE

VE = IERE

Rc = ?

Since IB is very small when compared with Ic

Ic approximately equal to IE

RE = VE / IE = ?

VB = VBE + VE

VB = VCC . RB2 / RB1 + RB2

S = 1+ RB / RE

RB =?

RB = RB1|| RB2

Find RB1 & RB2

Input Impedance, Zi = (RB || hie )

Coupling and bypass capacitors can be thus found out.

Input coupling capacitor is given by , Xci = Z i / 10

Xci = 1/ 2pf Ci

Ci = ?

output coupling capacitor is given by ,

Xc0 = 1/ 2pf Co

Co =?

By-pass capacitor is given by, XCE = 1/ 2pf CE

CE =?