Transistor Phase Shift Oscillator

Transistor Phase Shift Oscillator:

At low frequencies (around 100 kHz or less), resistors and capacitors are usually employed to determine the frequency of oscillation. Fig. 1 shows transistorized phase shift oscillator circuit employing RC network. If the phase shift through the common emitter amplifier is 180°, then the oscillation may occur at the frequency where the RC network produces an additional 180° phase shift.

Since a transistor is used as the active element, the output across R of the feedback network is shunted by the relatively low input resistance of the transistor, because input diode is a forward biased diode

Fig. 1

Hence, instead of employing voltage series feedback, voltage shunt feedback is used for a transistor phase shift oscillator. The load resistance R_L is also connected via coupling capacitor. The equivalent circuit using h-parameter is shown in fig. 2.

Fig. 2

For the circuit, the load resistance R_L may be lumped with R_C and the effective load resistance becomes R'_L(= R_C || R_L). The two h-parameters of the CE transistor amplifier, h_oe and h_re are neglected.

The capacitor C offers some impedance at the frequency of oscillation and, therefore, it is kept as it is, while the coupling capacitor behaves like ac short. The input resistance of the transistor is R_i≈h_ie. Therefore the resistance R₃ is selected such that R=R₃+R_i=R₃+h_ie. This choice makes the three R C selections alike and simplifies the calculation. The effect of biasing resistor R₁ , R₂, & R_Eon the circuit operation is neglected.

Since this is a voltage shunt feedback, therefore instead of finding V_R /V_O, we should find the current gain of the feedback loop.

The simplified equivalent circuit is shown in fig. 3.

Fig. 3

Applying KVL,

Since I₃ and I_b must be in phase to satisfy Barkhausen criterion, therefore

Also initially I₃ > I_b, therefore, for oscillation to start,

Therefore, the two conditions must be satisfied for oscillation to start and sustain.

Exampl - 1

(a). Show that the OPAMP phase shifter shown in fig. 4

.

(b) Cascade two identical phase shifters of the type sown in fig. 4. Complete the loop with an inverting amplifier. Show that the system will  oscillate at the frequency f = 1 / 2πRC provided that the amplifier gain exceeds unity.

(c) Show that the circuit produces two quadrature sinusoids (sine wave differing in phase by 90°).

Fig. 4

Solution:

(a)The voltage the non-inverting terminal input of the OPAM is given by

Since the differential input voltage of OPAMP is negligible small, therefore, the voltage at the inverting terminal is also given by

The input impedance of the OPAMP is very large, therefore,

or          V_i - 2V₂ = -V_O

Substituting v₂ in above equation, we get,

Substituting X_C, we get

Therefore,

and the phase angle between V_O and V_i is given by

The magnitude of V_O / V_i is unity for all frequencies and the phase shift provided by this circuit is 0° for R = 0 and –180° for R −> infinity.

(b). If two such phase shifters are connected in cascade and an inverting amplifier with gain ß is connected in the feedback loop, then the net loop gain becomes

Loop gain = Gain of phase shifter 1 x Gain of phase shifter 2 x β
= 1 x 1x β
= β

Therefore, the oscillation takes place if gain β =1, but it is kept >1 so that the losses taking place in the amplifier can be compensated.

The total phase shift around the loop is given by

total phase shift = -2 tan^-1 (ωRC) - 2 tan^-1(ωRC) + 180°

Further, for oscillation to take place the net phase shift around the loop would be 0°.Therefore,

-2 tan^-1 (ωRC) - 2 tan^-1(ωRC) + 180° =0
or            ωRC = 1
or            f = 1 / 2πR C

(c). The phase shift provided by amplifier in the feedback path is 180°, therefore, the phase shift provided by the phase shifters should also be 180° to have 360° or 0° phase shift. Thus ,the phase shift provided by the individual shifter will be 90° as both are identical. Therefore, the sine wave produces by two phase shifters are 90° apart and the circuit produces two quadrature sinusoids.