# Sinusoidal Oscillator/Sinusoidal Oscillator Page 2

Sinusoidal Oscillator Page 2

SAQ-1

1.(a) is the right answer, as positive feedback increases the gain of amplifier to large enough with the correct phase
(b), (c) and (d): If your answer is either of these three, you have got it wrong, as except positive feedback, other feedback does not produce large signal with correct phase

2. False. In oscillator, positive feedback is used to supply its own input signal for its operation.

SAQ-2

1. Barkhausen

2.(d) is the right answer, as to start oscillation, the total phase shift of an oscillator should be either 0 or 360 degree
(a), (b) and (c): If your answer is either of these three, you have got it wrong, as Barkhausen condition does not satisfy.

SAQ-3

1.(b) is the right answer, at room temperature the resistor acts as small ‘ac’ voltage source. When power supply is turned on, for the first time this small ac noise voltage gets amplified and appears at the output terminal.
(a) and (d): If your answer is either of these two, you have got it wrong, as it does not contribute to produce thermal noise
(c) If your answer this, you have got it wrong, as in oscillator external input signal is not present

2. At room temperature free electrons present in resistor move randomly and generate a noise voltage across the resistor due to collisions. The resistor acts as small ‘ac’ voltage source. This voltage is known as thermal noise voltage.
When power supply is turned on, for the first time this small ac noise voltage gets amplified and appears at the output terminal. This amplified output is applied to feedback circuit and output of feedback circuit is fed back to the amplifier as an input. It is again amplified by amplifier and fed back to input through feedback circuit. This process is repeated and at one particular frequency, circuit satisfies the necessary conditions to start oscillation.

SAQ-4

1. I-c), II-d), III-b), IV-a)

2.(c) is the right answer. From positive feedback equation, to get overall infinity gain, ‘Aβ’ is equal to 1, but practically to get sustained oscillations, at the first time when the circuit is turned on, the loop gain must be slightly greater than one. This will ensure that oscillations build up in the circuit.
(b) For ideal situation, it is right answer
(a) and (d): If your answer is either of these two, you have got it wrong, as loop gain condition does not satisfied at the first time when the circuit is turned on.

SAQ-5

Sinusoidal oscillators:
1.RC oscillators: suitable for low (audio range) and moderate frequency applications (5Hz to 1MHz).
2.LC oscillators: suitable for radio frequency(1 to 500 MHz)
3.Crystal oscillator: suitable for radio frequency applications with greater degree of stability and accuracy

SAQ-6

1.This RC combination performs the dual function. It acts as feedback network as well as frequency determining network of the oscillator.

2. (b) is the right answer. In RC phase shift oscillator, the feedback network produces 180 degrees phase shift as additional 180° phase shift is given by an amplifier to get total 360 degrees phase shift
(a), (c) and (d): If your answer is either of these three, you have got it wrong, as ‘total phase shift’ Barkhausen condition does not satisfy.

3. True. To get total phase shift of 360º.

SAQ-7

1.Lag circuit consists of single resistor in series branch and single capacitor in shunt branch as shown figure [Student must draw diagram for lag circuit] and it is often called RC bypass circuit. The name lag circuit is given because the output voltage lags the input voltage at higher frequencies. That means if input voltage has a phase angle of 0°, output voltage shows phase angle of 0°. But if you increase the frequency of input signal, then output voltage has a phase angle variation between 0° and -90°. At very high frequency, the phase angle ‘f’ becomes -90°. [ Student may write formulae but it is optional]
Lead circuit consists of single capacitor in series branch and single resistor in shunt branch as shown figure [Student must draw diagram for lag circuit] and this circuit is also known as coupling circuit. As this circuit responds to high frequency signals, it is also known as high pass filter. Here, if input voltage has a phase angle of 0° and if you increase the frequency of it, then the output voltage has a phase angle between 0°and +90°. At very high frequency the phase angle ‘f’ becomes +90°. °. In short, as output voltage leads to input voltage hence called as lead circuit. [Student may write formulae but it is optional]

At high frequencies, the reactance of capacitor $C_1$ and $C_2$ approaches zero and acts as a short. Hence, the output voltage Vo will be zero since output is taken across $R_2$ and $C_2$ combination. At high frequencies, circuit acts as a 'lag circuit'
At low frequencies, both capacitors act as open because capacitor offers very high reactance. Again output voltage will be zero because the input signal is dropped across the $R_1$ and $C_1$ combination. Here, the circuit acts like a 'lead circuit'
Basically, lead-lag circuit acts like a resonant circuit. But at one particular frequency between the two extremes, the output voltage reaches to the maximum value. At this frequency only, resistance value becomes equal to capacitive reactance and gives maximum output. Hence, this particular frequency is known as resonant frequency or oscillating frequency. It is given by, $f = \frac{1}{2{\Pi}R\,C\,}$

3. (c) is the right answer as if you increase the frequency of input signal, then output voltage has a phase angle variation between 0° and -90°.
(a), (b) and (c): If your answer is either of these three, you have got it wrong, as in lag circuit, at higher frequency output voltage has a phase angle variation between 0° and -90°.

4. (b) is the right answer as a Lead circuit consists of single capacitor in series branch and single resistor in shunt branch and responds to high frequency signals.
(a), (b) and (c): If your answer is either of these three, you have got it wrong, as only Lead circuit acts a coupling circuit.

SAQ-8

1. Working of Wien bridge oscillators with neat block diagram:
Refer circuit diagram

• Wien bridge consists of four arms
• Two arms : Purely resistive
• Two arms : Frequency sensitive (Lead-Lag circuit)
• To start oscillations
• Total phase shift must be 0° or 360°
• Magnitude of loop gain must be greater than one,
• To satisfy phase condition
• Use Noninverting amplifier: Positive Feedback between output and noninverting terminal
• To satisfy mangnitude condition
• Negative Feedback between output and inverting terminal
• For sustained oscillations product of A and β must be one

To satisfy above condition
Amplifier Gain A must be 3, Feedback gain β must be 1/3. This gives,$R_3=2R_4$
The frequency of oscillation is decided by resistor R and capacitor C, given by $f = \frac{1}{2{\Pi}R\,C\,}$

2.Role of feedbacks in Wien Bridge Oscillator:
Positive Feedback to satisfy the essential phase condition to start the oscillations- the total phase shift of the circuit must be 360°. Noninverting amplifier is used, as Wien bridge provides zero phase shift.
Negative Feedback to satisfy the essential magnitude condition to start the oscillations- the magnitude of loop gain must be greater than one. The loop gain must be slightly greater than one when circuit is turned on for the first time. For Wien bridge oscillator, the gain of the amplifier must be greater than three (A>3), which will ensure that sustained oscillations build up in the circuit. The gain is decided by resistor $R_3$ and $R_4$. For sustained oscillations, resistor $R_3$ must be twice of resistor $R_4$.

3.Gain Reduction at higher output voltage

• Use of Tungsten Lamp in place of resistor $R_4$

4. (d) is the right answer as for sustained oscillation, the gain required of a Wien bridge oscillator is 3.
(a), (b) and (c): If your answer is either of these three, you have got it wrong, as for sustained oscillation, the gain required of a Wien bridge oscillator is not satisfy.

SAQ-9

1. Gain components selection

For noninverting amplifier, gain is given by,

$A = 1 + \frac{R_3}{R_4}$

This gives, $R_3 = 2 R_4\,$

Frequency components selection

For maximum output, $R = X_c\,$

This gives, $f = \frac{1}{2{\Pi}\sqrt{R_1\,R_2\,C_1\,C_2}}$

If $R_1 = R_2 = R\,$ and $C_1 = C_2 = C\,$ then

$f = \frac{1}{2{\Pi}R\,C\,}$

2. Frequency components

Using frequency of oscillation formula $f = \frac{1}{2{\Pi}R\,C\,}$

Let $C= 0.05 uF$ (Micro farad), using above formula, R comes out as 3.3KΩ
Gain components

Let $R_4= 12k{\Omega}$, using gain formula

$A = 1 + \frac{R_3}{R_4}$
This gives, $R_3 = 2R_4 = 24k{\Omega}$

3. (b) is the right answer as the frequency expression of a Wien bridge oscillator consists of two resistors and two capacitors.
(a), (c) and (d): If your answer is either of these three, you have got it wrong, as the frequency expression of a Wien bridge oscillator consists of two resistors and two capacitors.

Critical Thinking Question

Give in detail the technique used to reduce the loop gain at higher output voltage. Hints: refer the graph of the resistance versus voltage of the tungsten lamp When power is first switched on, at room temperature, Lamp offers low resistance and provides less negative feedback. As the voltage across the lamp increases, the resistance of tungsten lamp also increases, oscillations build up. As no current flows through the OPAMP, this increased current flows through tungsten lamp. Therefore, the filament of the lamp gets heated and the resistance of the lamp increases. This results in more negative feedback and reduces the gain of the amplifier. At one particular frequency, the loop gain (Aβ) becomes exactly equal to one and it avoids clipping of the output waveform.

• Botkar, K. R. (1994). Integrated Circuits . Khanna Publisher.
• Gayakwad, R. A. (1993). Op-Amps and Linear Integrated Circuits (3rd ed.). New Delh: PHI.
• Malvino. Electronics Principles (6th ed.). Tata McGraw-Hill. Available at http://www.malvino.com

End of Unit

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