In the previous RC Oscillator tutorial we saw that a number of resistors and capacitors can be connected together with an inverting amplifier to produce an oscillating circuit. One of the simplest sine wave oscillators which uses a RC network in place of the conventional LC tuned tank circuit to produce a sinusoidal output waveform, is called a **Wien Bridge Oscillator**.

The **Wien Bridge Oscillator** is so called because the circuit is based on a frequency-selective form of the Wheatstone bridge circuit. The Wien Bridge oscillator is a two-stage RC coupled amplifier circuit that has good stability at its resonant frequency, low distortion and is very easy to tune making it a popular circuit as an audio frequency oscillator but the phase shift of the output signal is considerably different from the previous phase shift **RC Oscillator**.

The **Wien Bridge Oscillator** uses a feedback circuit consisting of a series RC circuit connected with a parallel RC of the same component values producing a phase delay or phase advance circuit depending upon the frequency. At the resonant frequency ƒr the phase shift is 0^{o}. Consider the circuit below.

The above RC network consists of a series RC circuit connected to a parallel RC forming basically a High Pass Filter connected to a Low Pass Filter producing a very selective second-order frequency dependant Band Pass Filter with a high Q factor at the selected frequency, ƒr.

At low frequencies the reactance of the series capacitor (C1) is very high so acts like an open circuit and blocks any input signal at Vin. Therefore there is no output signal, Vout. At high frequencies, the reactance of the parallel capacitor, (C2) is very low so this parallel connected capacitor acts like a short circuit on the output so again there is no output signal. However, between these two extremes the output voltage reaches a maximum value with the frequency at which this happens being called the *Resonant Frequency*, (ƒr).

At this resonant frequency, the circuits reactance equals its resistance as Xc = R so the phase shift between the input and output equals zero degrees. The magnitude of the output voltage is therefore at its maximum and is equal to one third (1/3) of the input voltage as shown.

It can be seen that at very low frequencies the phase angle between the input and output signals is “Positive” (Phase Advanced), while at very high frequencies the phase angle becomes “Negative” (Phase Delay). In the middle of these two points the circuit is at its resonant frequency, (ƒr) with the two signals being “in-phase” or 0^{o}. We can therefore define this resonant frequency point with the following expression.

- Where:
- ƒr is the Resonant Frequency in Hertz
- R is the Resistance in Ohms
- C is the Capacitance in Farads

We said previously that the magnitude of the output voltage, Vout from the RC network is at its maximum value and equal to one third (1/3) of the input voltage, Vin to allow for oscillations to occur. But why one third and not some other value. In order to understand why the output from the RC circuit above needs to be one-third, that is 0.333xVin, we have to consider the complex impedance (Z = R ± jX) of the two connected RC circuits.

We know from our AC Theory tutorials that the real part of the complex impedance is the resistance, R while the imaginary part is the reactance, X. As we are dealing with capacitors here, the reactance part will be capacitive reactance, Xc.

If we redraw the above RC network as shown, we can clearly see that it consists of two RC circuits connected together with the output taken from their junction. Resistor R_{1} and capacitor C_{1} form the top series network, while resistor R_{2} and capacitor C_{2} form the bottom parallel network.

Therefore the total impedance of the series combination (R_{1}C_{1}) we can call, Z_{S} and the total impedance of the parallel combination (R_{2}C_{2}) we can call, Z_{P}. As Z_{S} and Z_{P} are effectively connected together in series across the input, V_{IN}, they form a voltage divider network with the output taken from across Z_{P} as shown.

Lets assume then that the component values of R_{1} and R_{2} are the same at: 12kΩ, capacitors C_{1} and C_{2} are the same at: 3.9nF and the supply frequency, ƒ is 3.4kHz.

The total impedance of the series combination with resistor, R_{1} and capacitor, C_{1} is simply:

We now know that with a supply frequency of 3.4kHz, the reactance of the capacitor is the same as the resistance of the resistor at 12kΩ. This then gives us an upper series impedance Z_{S} of 17kΩ.

For the lower parallel impedance Z_{P}, as the two components are in parallel, we have to treat this differently because the impedance of the parallel circuit is influenced by this parallel combination.

The total impedance of the lower parallel combination with resistor, R_{2} and capacitor, C_{2} is given as:

At the supply frequency of 3400Hz, or 3.4kHz, the combined resistance and reactance of the RC parallel circuit becomes 6kΩ (R||Xc) and their parallel impedance is therefore calculated as:

So we now have the value for the series impedance of: 17kΩ’s, ( Z_{S} = 17kΩ ) and for the parallel impedance of: 8.5kΩ’s, ( Z_{S} = 8.5kΩ ). Therefore the output impedance, Zout of the voltage divider network at the given frequency is:

Then at the oscillation frequency, the magnitude of the output voltage, Vout will be equal to Zout x Vin which as shown is equal to one third (1/3) of the input voltage, Vin and it is this frequency selective RC network which forms the basis of the **Wien Bridge Oscillator** circuit.

If we now place this RC network across a non-inverting amplifier which has a gain of 1+R1/R2 the following basic wien bridge oscillator circuit is produced.

The output of the operational amplifier is fed back to both the inputs of the amplifier. One part of the feedback signal is connected to the inverting input terminal (negative feedback) via the resistor divider network of R1 and R2 which allows the amplifiers voltage gain to be adjusted within narrow limits. The other part is fed back to the non-inverting input terminal (positive feedback) via the RC Wien Bridge network.

The RC network is connected in the positive feedback path of the amplifier and has zero phase shift a just one frequency. Then at the selected resonant frequency, ( ƒr ) the voltages applied to the inverting and non-inverting inputs will be equal and “in-phase” so the positive feedback will cancel out the negative feedback signal causing the circuit to oscillate.

The voltage gain of the amplifier circuit MUST be equal too or greater than three “Gain = 3” for oscillations to start because as we have seen above, the input is 1/3 of the output. This value, ( Av ≥ 3 ) is set by the feedback resistor network, R1 and R2 and for a non-inverting amplifier this is given as the ratio 1+(R1/R2).

Also, due to the open-loop gain limitations of operational amplifiers, frequencies above 1MHz are unachievable without the use of special high frequency op-amps.

Determine the maximum and minimum frequency of oscillations of a **Wien Bridge Oscillator** circuit having a resistor of 10kΩ and a variable capacitor of 1nF to 1000nF.

The frequency of oscillations for a Wien Bridge Oscillator is given as:

A *Wien Bridge Oscillator* circuit is required to generate a sinusoidal waveform of 5,200 Hertz (5.2kHz). Calculate the values of the frequency determining resistors R_{1} and R_{2} and the two capacitors C_{1} and C_{2} to produce the required frequency.

Also, if the oscillator circuit is based around a non-inverting operational amplifier configuration, determine the minimum values for the gain resistors to produce the required oscillations. Finally draw the resulting oscillator circuit.

The frequency of oscillations for the Wien Bridge Oscillator was given as 5200 Hertz. If resistors R_{1} = R_{2} and capacitors C_{1} = C_{2} and we assume a value for the feedback capacitors of 3.0nF, then the corresponding value of the feedback resistors is calculated as:

For sinusoidal oscillations to begin, the voltage gain of the Wien Bridge circuit must be equal too or greater than **3**, ( Av ≥ 3 ). For a non-inverting op-amp configuration, this value is set by the feedback resistor network of R3 and R4 and is given as:

If we choose a value for resistor R3 of say, 100kΩ’s, then the value of resistor R4 is calculated as:

While a gain of 3 is the minimum value required to ensure oscillations, in reality a value a little higher than that is generally required. If we assume a gain value of **3.1** then resistor R4 is recalculated to give a value of 47kΩ’s. This gives the final Wien Bridge Oscillator circuit as:

Then for oscillations to occur in a **Wien Bridge Oscillator** circuit the following conditions must apply.

- With no input signal a Wien Bridge Oscillator produces continuous output oscillations.
- The Wien Bridge Oscillator can produce a large range of frequencies.
- The Voltage gain of the amplifier must be greater than 3.
- The RC network can be used with a non-inverting amplifier.
- The input resistance of the amplifier must be high compared to R so that the RC network is not overloaded and alter the required conditions.
- The output resistance of the amplifier must be low so that the effect of external loading is minimised.
- Some method of stabilizing the amplitude of the oscillations must be provided. If the voltage gain of the amplifier is too small the desired oscillation will decay and stop. If it is too large the output will saturate to the value of the supply rails and distort.
- With amplitude stabilisation in the form of feedback diodes, oscillations from the Wien Bridge oscillator can continue indefinitely.

In our final look at Oscillators, we will examine the Crystal Oscillator which uses a quartz crystal as its tank circuit to produce a high frequency and very stable sinusoidal waveform.

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The calculation for parallel RC circuit needed to be checked…

Z (for parallel RC circuit)= 1/Y=1/(sq. root(G*G+B*B))…

Otherwise, good explanation about Wein Bridge Oscillator…

When you calculate series impedance – you use root of sum of squares – it’s right. But when you calculte admitance of the parallel circuit (Y) you sum two elements. This is incorrect, I think. This is AC circuit, and Y = square (G*G+B*B) , where G = 1/R, and B = 2*pi*freq*C (without complex to simplify calculations).

Please verify your solution. Thanks.

In a parallel circuit, admittance is the inverse, or reciprocal, of impedance. Admittance Y = 1/Z. Conductance (G) is the reciprocal of the resistance, 1/R and Susceptance (B) is the reciprocal of reactance, 1/X. Susceptance can be inductive, BL = -1/XL or capacitive BC = XC, and all are measured in Siemens (S).

The relationship between the admittance, the conductance, and the susceptance is the same as the relationship between the impedance, the resistance, and the reactance, giving Y = 1/Z = G + jB = 1/(R + jX). Susceptances (B) in parallel and conductances (G) in parallel are added.

Good job

Very useful. Really appreciate your work

superbbb

This is very useful, helped me a lot in a part of my report. Really appreciate your work *thumbs up*

this was awesome. keep up

I want Viva questions

please can post a working digram of this when bridge oscillator using a know values of components and with Transistor?

good and very simple discussion, appreciate your way.