Question: draw and explain the working of transisterized wein bridge oscillator

Subject: Electronic Devices and Circuits II

Topic: Oscillators

Difficulty: MEdium

edc2(34) • 372 views
modified 11 months ago  • written 14 months ago by gravatar for Amruta Padhye Amruta Padhye20

Win bridge oscillator is RC phase shift oscillator

The following figure shows the wein bridge oscillator

enter image description here


It is essentially a two-stage amplifier with an R-C bridge circuit. R-C bridge circuit (Wien bridge) is a lead-lag network.

Wien bridge oscillator consists of two stage amplifier with common emitter voltage divider configuration.

In bridge circuit R1 in series with C1, R3, R4 and R2 in parallel with C2 forms the four arms. The phase shift across this bridge lags with increase in frequency and leads with decrease in frequency In between, there is a resonant frequency fr at which the phase angle equals 0°

this bridge circuit can be used as feedback circuit for oscillator provided that the phase shift through the amplifier is zero. This condition is achieved by employing two transistor each producing phase of 180º thus total phase shift 360º or 0º

The circuit uses positive and negative feedbacks. The positive feedback is through R1 C1 R2, C2 to tran­sistor Q1 and negative feedback is through the voltage divider to the input of transistor Q1. Resistors R3 and R4 are used to stabilize the amplitude of the output.

Circuit has two transistors Q1 and Q2. Transistor Q1 acts as oscillator circuit and Q2 acts as inverter providing additional 180º phase shift. The output of second transistor is supplied to the R-C bridge circuit and the voltage across C2R2 parallel combination is given to transistor Q1.

The amplifier voltage gain, A = R3 + R4 / R4 = R3 / R4 + 1 = 3

Since R3 = 2 R4

The above corresponds with the feedback network attenuation of 1/3. Thus, in this case, voltage gain A, must be equal to or greater than 3, to sustain oscillations.

To have a voltage gain of 3 is not difficult. On the other hand, to have a gain as low as 3 may be difficult. For this reason also negative feedback is essential.


The circuit is set in oscillation by random change in base current of Q1 transistor that may be due to change in dc power supply or due to noise.

The transistor Q1 amplifies this signal with 180º phase shift this signal is input to second transistor Q2 through capacitor C4 which further amplifies the signal with 180º phase shift since the signal has two phase-shift of 180º the o/p signal at the second transistor is in- phase with the input signal.

A part of output signal of second transistor is supplied to bridge circuit at input point (point A-C).

A part of this feedback signal is applied to emitter resistor R4 where it produces negative feedback. Similarly, a part of the feedback signal is applied across the base resistor R2 where it produces positive feedback.

At frequency of oscillation effect of regeneration is made slightly more than that of degeneration so as to obtain sustained oscillations.

The continuous frequency variation in this oscillator can be had by varying the two capacitors C1 and C2 simultaneously. These capacitors are variable air-gang capacitors. We can change the frequency range of the oscillator by switching into the circuit different values of resistors R1 and R2.


It gives low distortion for wide range of frequencies.

Frequency range can be selected by using different resistances

The frequency of oscillation can be change by varying capacitance C1 and C2 simultaneously. The overall gain is high because of two transistors.


The circuit needs two transistors and a large number of other com­ponent

written 11 months ago by gravatar for Amruta Padhye Amruta Padhye20
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