Oscillators Questions and Answers

Q: How is the frequency of oscillation selected out of the noise?

The frequency-determining network gives the right phase shift at exactly one frequency, and the gain condition is arranged to be satisfied there, so only that component of the noise grows. The amplifier's bandwidth is far wider, the transient is broadband, and the non-linearity acts on amplitude rather than selecting frequency. It is the coincidence of the two conditions that does the selecting, which is why the sharpness of the phase characteristic decides how precisely the frequency is defined.

Q: Why does a high quality factor improve an oscillator's frequency stability?

A sharp resonance means a small change in frequency produces a large change in phase, so any phase disturbance elsewhere in the loop is corrected by only a tiny shift in frequency. That is why a crystal, with a quality factor in the tens of thousands, holds a frequency so much better than an LC tank at a few hundred. Stored energy and low losses are what produce the high quality factor rather than the mechanism of stability. Frequency stability is always a question of how steep the phase characteristic is at the operating point.

Q: A Wien bridge uses \( R = 10 \ \mathrm{k\Omega} \) and \( C = 10 \ \mathrm{nF} \). Its frequency is about

The centre frequency is where the two arms' phases cancel:\[ f = \frac{1}{2\pi R C} \approx 1592 \ \mathrm{Hz} \]giving about 1592 Hz. The value 650 Hz would follow from the phase shift oscillator's formula with its root six. That the same components give quite different frequencies in the two topologies is worth noticing, and it is entirely due to the different phase conditions being solved.

11.

What is the effect of relying on clipping to limit amplitude?

A.The amplitude becomes dependent on temperature
B.The output acquires substantial harmonic distortion
C.The frequency of oscillation shifts with supply voltage
D.The oscillation fails to start reliably at switch-on

Answer: B

Clipping holds the amplitude by flattening the waveform's peaks, so the output is no longer a sine and carries strong odd harmonics. It is perfectly acceptable where a squarewave or a clock is wanted and unacceptable in a signal generator. Frequency does shift a little with supply, but that is a separate mechanism. The choice of limiting mechanism is the choice of output purity, and it is decided by what the oscillator is for rather than by any theoretical consideration.

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12.

How is the frequency of oscillation selected out of the noise?

A.The supply's switching transient excites only that frequency
B.The non-linearity suppresses every other frequency present
C.Only at one frequency are both conditions met
D.The amplifier's bandwidth admits only that frequency

Answer: C

The frequency-determining network gives the right phase shift at exactly one frequency, and the gain condition is arranged to be satisfied there, so only that component of the noise grows. The amplifier's bandwidth is far wider, the transient is broadband, and the non-linearity acts on amplitude rather than selecting frequency. It is the coincidence of the two conditions that does the selecting, which is why the sharpness of the phase characteristic decides how precisely the frequency is defined.

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13.

Why does a high quality factor improve an oscillator's frequency stability?

A.The resonant circuit stores more energy between cycles
B.The amplifier needs less gain to sustain oscillation
C.The losses in the tank circuit are reduced
D.Phase changes rapidly with frequency near resonance

Answer: D

A sharp resonance means a small change in frequency produces a large change in phase, so any phase disturbance elsewhere in the loop is corrected by only a tiny shift in frequency. That is why a crystal, with a quality factor in the tens of thousands, holds a frequency so much better than an LC tank at a few hundred. Stored energy and low losses are what produce the high quality factor rather than the mechanism of stability. Frequency stability is always a question of how steep the phase characteristic is at the operating point.

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14.

How much phase shift must the RC network of a phase shift oscillator provide?

A.One hundred and eighty degrees
B.Ninety degrees
C.Three hundred and sixty degrees
D.Forty-five degrees

Answer: A

The common-emitter amplifier already inverts, contributing one hundred and eighty degrees, so the network must supply the other hundred and eighty to bring the total round to zero or a full turn. Ninety degrees would leave the loop short, and three hundred and sixty from the network alone would need the amplifier to be non-inverting. Working out what the amplifier already contributes, and requiring the network to make up the difference, is how every oscillator topology is read.

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15.

Why does a phase shift oscillator need three RC sections rather than two?

A.Three sections reduce the gain the amplifier must supply
B.A single section cannot reach ninety degrees
C.Three sections give a sharper frequency selection
D.Two sections would invert the signal twice over

Answer: B

Each RC section approaches but never reaches ninety degrees, so two can never total the hundred and eighty required and at least three are needed, each contributing sixty at the oscillation frequency. Selectivity improves as a side effect, and the gain requirement actually rises with more sections rather than falling. A limit that is approached asymptotically but never attained is what forces the third section, and recognising that saves attempting the two-section version.

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16.

What minimum gain must a three-section phase shift oscillator's amplifier provide?

A.Ten
B.One hundred
C.Twenty-nine
D.Three

Answer: C

The three RC sections attenuate by a factor of twenty-nine at the frequency where they give a hundred and eighty degrees, so the amplifier must supply at least that much gain for the loop gain to reach unity. Three is the Wien bridge's requirement, where the network attenuates by only three. The attenuation and the gain requirement are two statements of one fact, and computing the network's loss at its phase-shift frequency is what gives the number.

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17.

A phase shift oscillator uses \( R = 10 \ \mathrm{k\Omega} \) and \( C = 10 \ \mathrm{nF} \) in each section. Its frequency is about

A.1592 Hz
B.265 Hz
C.1000 Hz
D.650 Hz

Answer: D

The three-section network oscillates where its phase reaches one hundred and eighty degrees:

\[ f = \frac{1}{2\pi R C \sqrt{6}} \approx 650 \ \mathrm{Hz} \]

giving about 650 Hz. The value 1592 Hz omits the root six and is the Wien bridge's frequency for the same components. The root six is what distinguishes the two topologies numerically, and it comes directly from solving the three-section network's phase condition.

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18.

What phase shift does the Wien bridge network provide at its centre frequency?

A.Zero
B.Ninety degrees
C.One hundred and eighty degrees
D.Forty-five degrees

Answer: A

The series and parallel RC arms cancel each other's phase at one frequency, so the network passes the signal with zero shift and a non-inverting amplifier completes the loop. That is why a Wien bridge uses a non-inverting configuration while a phase shift oscillator uses an inverting one. Whether a network shifts by zero or by a hundred and eighty degrees decides which kind of amplifier it must be paired with, and that pairing is the first thing to establish about any oscillator.

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19.

What gain must a Wien bridge oscillator's amplifier provide?

A.Ten
B.Three
C.Twenty-nine
D.One

Answer: B

The Wien network attenuates by a factor of three at its centre frequency, so the amplifier must supply a gain of exactly three for the loop gain to be one. That is easily arranged with a two-to-one resistor ratio in a non-inverting stage, which is one reason the topology is so popular. Twenty-nine is the phase shift oscillator's requirement. A modest and easily set gain requirement is a practical advantage, since the amplitude control then has only a small range to work over.

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20.

A Wien bridge uses \( R = 10 \ \mathrm{k\Omega} \) and \( C = 10 \ \mathrm{nF} \). Its frequency is about

A.159 Hz
B.3184 Hz
C.1592 Hz
D.650 Hz

Answer: C

The centre frequency is where the two arms' phases cancel:

\[ f = \frac{1}{2\pi R C} \approx 1592 \ \mathrm{Hz} \]

giving about 1592 Hz. The value 650 Hz would follow from the phase shift oscillator's formula with its root six. That the same components give quite different frequencies in the two topologies is worth noticing, and it is entirely due to the different phase conditions being solved.

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