Hey friends, welcome to the Kohiki ALL ABOUT ELECTRONICS. In the previous article on the oscillator, we understood the working principle of the oscillator. And based on the type of feedback, we had seen the different types of oscillators. Now, one such oscillator is the RC phase shift oscillator.

So, in this oscillator, the RC circuit is used in the feedback path. Now, this RC phase shift oscillator generates stable sine waves. And usually, they are used in the low-frequency generations. Typically in the range of audio frequencies.

- How does RC phase shift oscillator work?
- How Simple RC Circuit (Phase Lead Circuit)
- Design of RC Shift Oscillator using Op-amp
- What is the frequency of RC phase shift oscillator?
- RC phase shift oscillator derivation
- What are the applications of RC phase shift oscillator?
- What are the advantages of RC phase shift oscillator?
- FAQ’s
- YouTube Video

## How does RC phase shift oscillator work?

So, in this RC phase shift oscillator, the amplifier provides a 180 degree of phase shift. And usually, either the transistor or the inverting op-amp is used for the amplification.

Now, in the previous article, we had seen that to get the sustained oscillations loop gain of the oscillator should be equal to 1. And the phase shift that is introduced by the amplifier and the feedback circuit should be equal to Zero or 360 degrees.

### 1

So, to get the sustained oscillations, in this RC phase shift oscillator, this feedback path should also provide the 180 degrees of phase shift.

### 2

So that the overall phase shift that is introduced by this amplifier and the feedback circuit is equal to zero. And by tuning the gain of this amplifier and the feedback circuit it is possible to achieve the loop gain that is equal to 1.

### 3

So, now let’s understand in this RC phase shift oscillator how this feedback circuit provides the 180 degrees of phase shift. And before we see that, first of all, let’s quickly see the simple RC circuit.

## How Simple RC Circuit (Phase Lead Circuit)

RC Phase Shift Oscillator So, this circuit over here is nothing but the high pass filter. And it is also known as the phase lead circuit. Because in this circuit the output voltage used to lead the input voltage.

So, first of all, let’s find out the transfer function of this RC circuit. So, here this output by input can be given as R divided by R -X Where Xc is nothing but 1/wC.

So, RC Phase Shift Oscillator we can write this expression as R/(R – j/wC))And by dividing the numerator and the denominator by the factor of R, we can write this expression 1/(1- j/wCR)) So, if we find the phase-shift, then the phaseshift phi, can be given as (0 – tan-1(-1/wCR)) That is equal to tan-1( 1/wCR). And as we know, 1/wC is nothing but Xc.

So, we can say that the phase shift phi is equal to tan-1(Xc/R). So, as you can see over here, in this circuit the phase shift depends on the value of R and the value of capacitance C.

So, for a very small value of Xc, the phase shift will be equal to zero. And whenever R is equal to zero, at that time this term Xc/R will become infinite.

So, at that time the phase shift that is provided by the circuit will be equal to 90 degrees. So, in this way, this RC circuit provides the phase shift between zero to 90 degrees. And theoretically, to get 180 degrees of phase shift, we need to cascade the two such circuits.

So, each circuit theoretically should provide a 90 degree of phase shift. But like I said, to get a 90 degree of phase shift, RC Phase Shift Oscillator the value of R used to be very small or theoretically it should be zero. But whenever R is zero, at that time the gain that is provided by the circuit will be equal to zero.

So, practically it is not possible to achieve a 180 degree of phase shift just by using the two RC stages. So, in the practical case, instead of two RC stages, we require more stages. So, using the three stages, it is possible to achieve the 180 degrees of phase shift.

So, here each stage will provide a 60 degree of phase shift. And overall phase shift that is introduced by three stages will be equal to 180 degrees. RC Phase Shift Oscillator But because of the loading effect of the other stages, the actual phase shift that is provided by each stage will be slightly different. But if you see the overall phase shift, then the overall phase shift will be equal to180 degrees.

So, in this way, by using the multiple stages it is possible to achieve the 180 degrees of phase shift. And in practical RC Phase shift oscillator, usually more than 3 stages are used to get a stable phase shift.

For example, if you use 4 stages, then each stage will provide approximately 45 degrees of phase shift. So, combining all the four stages we can get 180 degrees of phase shift.

So, in this RC phase shift oscillator, like I said the amplifier is used in the inverting configuration. So, it provides a 180 degree of phase shift. And through this feedback circuit or through this RC stage remaining 180 degrees of phase shift is introduced.

So, the overall phase shift that is introduced by the amplifier and the feedback circuit will be equal to zero. And by setting the gain of this amplifier and the feedback circuit, it is possible to achieve the loop gain of unity.

## Design of RC Shift Oscillator using Op-amp

So, here is the RC phase shift oscillator circuit which is designed using the op-amp. And the same circuit can be redrawn like this. RC Phase Shift Oscillator So, as you can see over here, this op-amp is configured in the inverting mode. So, it used to provide a 180 degree of phaseshift.

- While in the feedback circuit, these RC stages used to provide a 180 degree of phase shift.
- So, in this way, the overall phase shift that is introduced will be equal to 360 degrees or 0 degrees.
- Now, in this RC phase shift oscillator, the attenuation that is introduced by this feedback stage is equal to 1/29.

- So, to get a unity loop gain, the gain that is provided by this op-amp should be equal to 29.

**What is the frequency of RC phase shift oscillator?**

Now, this circuit will provide unity gain only at one particular frequency. And at only one particular frequency only the overall phase shift of the circuit will be equal to zero. And if three stages are used then this frequency can be given by the expression, 1/2πRC√6.

- So, at this particular frequency only, the loop gain of the circuit will be equal to 1.
- as well as the overall phase shift of the circuit will be equal to zero degrees.
- And if more than three stages are used then in general, for N number of stages this frequency f can be given by the expression, 1/2πRC√(2N)So, if four number RC stages are used, in that case, this expression of ‘f’ will be equal to 1/2πRC√8 And for three stages, it will be equal to1/2πRC√6.
- Now, some of you might be wondering that how we have got this expression of the frequency.
- As well as, why the attenuation of this RC stage is equal to 1/29.
- So, let’s derive these two expressions very quickly.

## RC phase shift oscillator derivation

So, here the same RC stages are drawn separately. Now, here let’s assume that Vin is the input to this RC stage and V out is the output of this RC circuit. Now, in this RC circuit, just by using the mesh analysis method we can find the relationship between the output and input. Or we can also use the nodal analysis method to get the relationship between the output and input.

So, by using any of these two methods we can find the relationship between the output and input. And if you find it, then you will get that this V out/Vin for this RC network will be equal to -1/29. So, let’s quickly derive this relationship between output and input.

So, here let’s assume that the voltage at this node is V1. And the voltage at this node is equal to V2. And I1, I2, and I3 are the currents that are flowing through these three capacitors. So, here what we will do, we will find the value of this V2, V1, I2, and I1. And we will find these four expressions in terms of the output voltage. And using these four expressions, we will find the relationship between the output and input.

So, first of all, let’s concentrate on this node V2. So, at this node V2, we can write this voltageV2 is equal to this voltage Vo plus the drop across the reactance Xc. That is equal to, I3 times (-jXc). Now, if you look over here, this current I3is flowing through this resistor R.

That means this output voltage V out is nothing but I3 times R. Or we can say that this current I3 is equal to Vo/R. And if we put the value of Xc as 1/wC, then we can write this expression of V2 as (Vo/jwCR +Vo)That means V2 is equal to Vo (1 +1/jwCR) So, in this way, we got the expression ofV2 in terms of the V out.

Now, similarly. let’s find out the expression of I2 in terms of Vout. So, if we apply the KCL at this node, then we can write I2 = (V2/R) + I3Now, we already found the value of this V2. And we know that I3 is equal to Vo/R.So, w can write the expression of I2 as (Vo/R)(1+ 1/jwCR) +(Vo/R)And if we simplify it then we can write, Vo/R * ( 2 +1/jwCR)So, in this way, we got the two expressions V2 and I2 in terms of the output voltage. Similarly, at this node V1, let’s find out the expression of V1 and I1 in terms of the output voltage Vout.

So, at this node V1, we can write this voltageV1 as( V2 + I2/jwC) That is the drop across this reactance Xc. Now, we already know the value of this V2and I2. So, if we put the value of this V2 and I2then we can write the expression of V1 as Vo * (1 + 1/jwCR) + (Vo/R)*( 2 + 1/jwCRC)(1/jwC)And if we solve this expression, then the value of this V1 will be equal to Vo * {1 +( 3/jwCR )- (1/w^2*C^2*R^2)}Similarly, by applying the KCL at this node V1, let’s find out the expression of I1 in terms of this output voltage Vout. So, applying KCL at this node we can write, current I1 will be equal to (V1/R)+I2 Now, we already know the value of V1 and I2.

So, let’s put the value of V1 and I2 in this expression. And we can write this current I1 as Vo/R * { 1+(3/jwCR) + (1/w^2*C^2*R^2)} + {Vo/R * (2 + 1/jwCR )}So, if we simplify this expression then the value of I1 will be equal to Vo/R * { 3 +(4/jwCR) – (1/w^2*C^2*R^2) } So, now so far we got four expressions ofV2, I2, V1 and I1 in terms of the output voltage. Now, the only expression that remained is the relationship between the input voltage and output voltage.

So, now let’s write down the expression ofVin in terms of this I1 and V1. So, now we can write this input voltage Vinas V1 plus the drop across this capacitance Xc. That is equal to (I1 /j wC). And let’s put the value of V1 and I1 in this expression. So, the input voltage Vin will be equal to Vo * { 1 + (3/jwCR) – (1/w^2*C^2*R^2)} So, here we have just put the value of V1.

Now, let’s put the value of I1 in this expression. And as we know I1 is Vo/R * (1/jwC) * { 3+ (4/jwCR) – (1/w^2*C^2*R^2)} So, if we simplify this expression then we can write this input voltage Vi as Vo * {1 + (6/jwCR) – (5/w^2*C^2*R^2) – 1/(j*W^3*C^3*R^3)}So, this is the relationship between the input and output voltage.

Now, the output voltage will only have a real part. That means the imaginary part of this expression should be equal to zero. So, by equating the imaginary part to the zero, we can write 6/wCR – 1/( w^3*C^3*R^3)= 0Or we can simplify this expression as w^2*C^2*R^2 = 1/6That means w will be equal to 1 /(√6* RC )And if we write this expression in terms of the frequency then we can say that f will be equal to 1/ ( 2*pi* RC* √6)

So, now let’s find out the gain of the circuit at this particular frequency. Now, once the imaginary part is zero, the expression can be written as Vin = Vo * { 1 – 5/(w^2*C^2*R^2)}Now, in this expression. let’s put w= 1 /(√6* RC ). So, if we put the value of this w, then this input voltage Vin will be equal to Vo* (1-5/(R^2*C^2/6*R^2*C^2)) So, if we simplify it then the input voltage Vin will be equal to Vo (1-30) Or we can say that Vo/Vin is equal to -1/29.

So, in this way, we can say that this RC network used to provide an attenuation of 1/29. And here the negative sign indicates that this RC network used to provide 180 degrees of phase shift.

So, in this way, hereby tuning the value of R and C in the feedback network, it is possible to get an oscillation at a particular frequency. Now, generally in these RC phase shift oscillators, the value of R remains fixed and only the value of the capacitor is changed to get the desired frequency of oscillation.

So, here in this RC phase shift oscillator, all the capacitors are ganged together. So, just by changing the one capacitor, itis possible to tune all the capacitors. And in this way, it is possible to tune the circuit at one particular frequency. So, this is all about the RC phase shift oscillator.

## What are the applications of RC phase shift oscillator?

- This
**part**shift**generator****is employed****to come up with**the signals over**an in-depth****vary**of frequency. They**employed in**musical instruments, GPS units, & voice synthesis. - The applications of this
**part**shift**generator****embody**voice synthesis, musical instruments, and GPS units.

## What are the advantages of RC phase shift oscillator?

- It is useful for frequencies in the audio range.
- Simplicity of the circuit.
- They have a wide range of frequency.
- The
**RC phase shift oscillator**gives good Frequency stability. - The output of this circuit is sinusoidal that is quite distortion free..

**FAQ’s**

### How does RC phase shift oscillator work?

**RC phase**–**shift oscillators** use resistor-capacitor (**RC**) network (Figure 1) to provide the **phase**–**shift** required by the feedback signal. They have excellent frequency stability and can yield a pure sine wave for a wide range of loads. … Further, the circuit also shows three **RC** networks employed in the feedback path.

### What is the frequency of RC phase shift oscillator?

The **phase**–**shift oscillator** circuit consists of a single transistor amplifier section and a **RC phase**–**shift** network. The **phase shift** network in this circuit, consists of three **RC** sections. At the resonant **frequency** f_{o}, the **phase shift** in each **RC** section is 60^{o} so that the total **phase shift** produced by **RC** network is 180^{o}.

### What are the applications of RC phase shift oscillator?

This **phase shift oscillator** is used to generate the signals over an extensive range of frequency. They used in musical instruments, GPS units, & voice synthesis. The **applications** of this **phase shift oscillator** include voice synthesis, musical instruments, and GPS units.

### What are the advantages of RC phase shift oscillator?

It is useful for frequencies in the audio range.

Simplicity of the circuit.

They have a wide range of frequency.

The **RC phase shift oscillator** gives good Frequency stability.

The output of this circuit is sinusoidal that is quite distortion free..

### What is the basic principle of oscillator?

There are many types of electronic **oscillators**, but they all operate according to the same **basic principle**: an **oscillator** always employs a sensitive amplifier whose output is fed back to the input in phase. Thus, the signal regenerates and sustains itself. This is known as positive feedback.

## YouTube Video

So here are a youtube video based on RC phase shift oscillator, And The Video Was Uploaded By Ekeeda

So, I hope in this article, you understood how we can design the oscillator using the op-amp. So, if you have any questions or suggestions, do let me know in the comment section below.

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