Oscillators Questions and Answers

Q: Assertion (A): Assertion: A Clapp oscillator is more stable than a Colpitts. Reason (R): Reason: Its extra series capacitor dominates the frequency and swamps the device capacitances. Select the correct answer from the options below.

Both hold and the second is the mechanism: the transistor's junction capacitances drift with temperature and bias, but a small series capacitor with a good coefficient decides the frequency and their contribution is negligible beside it. Making the frequency depend on the most stable available component, rather than on a sum including unstable ones, is a general technique and this is its clearest application in oscillator design.

Q: Assertion (A): Assertion: An astable multivibrator's frequency depends strongly on supply voltage. Reason (R): Reason: Its capacitors charge towards the supply rail. Select the correct answer from the options below.

The reason is true but the assertion is false: a higher supply charges the capacitor faster and also raises the voltage it must reach, so the two effects largely cancel and the period is nearly independent of supply. That cancellation is why the circuit works acceptably without a regulated supply. A mechanism that appears twice with opposite signs is worth tracing before concluding that a quantity matters.

Q: Consider the following statements about the Barkhausen criterion. 1. The loop gain must be unity in magnitude 2. The total phase shift around the loop must be zero or a multiple of a full turn 3. Both conditions must be satisfied at the same frequency Which of the statements given above is or are correct?

All three are correct, and the third is the one most often left unstated even though it does the real work: it is the coincidence of the two conditions at one frequency that picks the oscillation out of the noise. A circuit meeting the gain condition at one frequency and the phase condition at another will not oscillate at either. Stating a criterion completely, including the conditions under which its parts must hold, is what makes it usable rather than merely quotable.

Q: Statement I: A Wien bridge oscillator uses a non-inverting amplifier. Statement II: A phase shift oscillator uses a non-inverting amplifier. In the light of the above statements, choose the correct answer.

The first is true and the second is false: the Wien network gives zero phase shift so a non-inverting stage completes the loop, while the phase shift network gives a hundred and eighty degrees and needs an inverting stage to supply the rest. Whether a network shifts by zero or by a half turn decides which kind of amplifier it must be paired with, and that pairing is the first thing to establish about any oscillator topology.

Q: Match each oscillator with its feedback element. List I A. Hartley B. Colpitts C. Armstrong List II P. Tapped inductor Q. Capacitive divider R. Coupled winding Select the correct match.

The pairing is A-P, B-Q, C-R. All three are the same amplifier and tank with three ways of extracting a fraction of the tank's signal, which is why they share a frequency formula and differ only in how the fraction is set. Recognising a family as one circuit with one element varied is far more economical than learning three circuits.

Q: Match each oscillator with the gain its amplifier must supply. List I A. Wien bridge B. Three-section phase shift C. Emitter follower buffer List II P. Three Q. Twenty-nine R. None; it takes no voltage gain Select the correct match.

The pairing is A-P, B-Q, C-R. Each gain requirement is the reciprocal of its network's transmission at the oscillation frequency, and the buffer is included to make the point that a stage in an oscillator need not contribute gain at all. Computing a requirement from the network's loss is more reliable than recalling it, and it works for a topology never seen before.

Q: A Colpitts uses \( C_{1} = 100 \ \mathrm{pF} \) and \( C_{2} = 1000 \ \mathrm{pF} \). Its feedback fraction is about

The fraction returned is set by the capacitor ratio:\[ \beta = \frac{C_{1}}{C_{2}} = \frac{100}{1000} = 0.1 \]giving 0.1, so the amplifier must supply a gain of at least ten. Inverting the ratio gives ten, which would be a fraction greater than one and is impossible for a passive divider. Checking that a computed fraction lies below one is a quick test that catches the inversion immediately.

101.

Assertion (A): Assertion: An oscillator's loop gain must be exactly one for it to start.

Reason (R): Reason: The Barkhausen criterion requires a loop gain of unity.

Select the correct answer from the options below.

A.A is true but R is false
B.A is false but R is true
C.Both A and R are true, and R is the correct explanation of A
D.Both A and R are true, but R is not the correct explanation of A

Answer: B

The reason states the criterion correctly, but the assertion misapplies it: a loop gain of exactly one would sustain an existing amplitude and could never build one from noise, so a practical oscillator starts with a loop gain above one and is pulled back to unity by a non-linearity. The criterion describes the steady state rather than the starting condition. Distinguishing the condition for sustaining from the condition for growing is the central practical point in oscillator design.

Discuss in Forum

102.

Assertion (A): Assertion: A Clapp oscillator is more stable than a Colpitts.

Reason (R): Reason: Its extra series capacitor dominates the frequency and swamps the device capacitances.

Select the correct answer from the options below.

A.A is true but R is false
B.A is false but R is true
C.Both A and R are true, and R is the correct explanation of A
D.Both A and R are true, but R is not the correct explanation of A

Answer: C

Both hold and the second is the mechanism: the transistor's junction capacitances drift with temperature and bias, but a small series capacitor with a good coefficient decides the frequency and their contribution is negligible beside it. Making the frequency depend on the most stable available component, rather than on a sum including unstable ones, is a general technique and this is its clearest application in oscillator design.

Discuss in Forum

103.

Assertion (A): Assertion: An astable multivibrator's frequency depends strongly on supply voltage.

Reason (R): Reason: Its capacitors charge towards the supply rail.

Select the correct answer from the options below.

A.Both A and R are true, and R is the correct explanation of A
B.Both A and R are true, but R is not the correct explanation of A
C.A is true but R is false
D.A is false but R is true

Answer: D

The reason is true but the assertion is false: a higher supply charges the capacitor faster and also raises the voltage it must reach, so the two effects largely cancel and the period is nearly independent of supply. That cancellation is why the circuit works acceptably without a regulated supply. A mechanism that appears twice with opposite signs is worth tracing before concluding that a quantity matters.

Discuss in Forum

104.

Consider the following statements about the Barkhausen criterion.

1. The loop gain must be unity in magnitude

2. The total phase shift around the loop must be zero or a multiple of a full turn

3. Both conditions must be satisfied at the same frequency

Which of the statements given above is or are correct?

A.1, 2 and 3
B.1 and 2 only
C.2 and 3 only
D.1 and 3 only

Answer: A

All three are correct, and the third is the one most often left unstated even though it does the real work: it is the coincidence of the two conditions at one frequency that picks the oscillation out of the noise. A circuit meeting the gain condition at one frequency and the phase condition at another will not oscillate at either. Stating a criterion completely, including the conditions under which its parts must hold, is what makes it usable rather than merely quotable.

Discuss in Forum

105.

Consider the following statements about crystals.

1. The parallel resonance lies slightly above the series resonance

2. The crystal is inductive between the two resonances

3. A crystal's quality factor is typically a few hundred

Which of the statements given above is or are correct?

A.1, 2 and 3
B.1 and 2 only
C.1 and 3 only
D.2 and 3 only

Answer: B

The first two are correct and the third understates by orders of magnitude: a crystal's quality factor runs from tens of thousands to over a million, while a few hundred is what a good LC tank achieves. That gap is the entire reason crystals are used, so getting the magnitude right matters more than the exact figure. A statement that is qualitatively right and quantitatively wrong by four orders of magnitude is still wrong in the way that counts.

Discuss in Forum

106.

Statement I: A Wien bridge oscillator uses a non-inverting amplifier.

Statement II: A phase shift oscillator uses a non-inverting amplifier.

In the light of the above statements, choose the correct answer.

A.Both Statement I and Statement II are incorrect
B.Statement I is incorrect but Statement II is correct
C.Statement I is correct but Statement II is incorrect
D.Both Statement I and Statement II are correct

Answer: C

The first is true and the second is false: the Wien network gives zero phase shift so a non-inverting stage completes the loop, while the phase shift network gives a hundred and eighty degrees and needs an inverting stage to supply the rest. Whether a network shifts by zero or by a half turn decides which kind of amplifier it must be paired with, and that pairing is the first thing to establish about any oscillator topology.

Discuss in Forum

107.

Match each oscillator with its feedback element.

List I

A. Hartley

B. Colpitts

C. Armstrong

List II

P. Tapped inductor

Q. Capacitive divider

R. Coupled winding

Select the correct match.

A.A-Q, B-P, C-R
B.A-R, B-Q, C-P
C.A-P, B-R, C-Q
D.A-P, B-Q, C-R

Answer: D

The pairing is A-P, B-Q, C-R. All three are the same amplifier and tank with three ways of extracting a fraction of the tank's signal, which is why they share a frequency formula and differ only in how the fraction is set. Recognising a family as one circuit with one element varied is far more economical than learning three circuits.

Discuss in Forum

108.

Match each oscillator with the gain its amplifier must supply.

List I

A. Wien bridge

B. Three-section phase shift

C. Emitter follower buffer

List II

P. Three

Q. Twenty-nine

R. None; it takes no voltage gain

Select the correct match.

A.A-P, B-Q, C-R
B.A-Q, B-P, C-R
C.A-R, B-Q, C-P
D.A-P, B-R, C-Q

Answer: A

The pairing is A-P, B-Q, C-R. Each gain requirement is the reciprocal of its network's transmission at the oscillation frequency, and the buffer is included to make the point that a stage in an oscillator need not contribute gain at all. Computing a requirement from the network's loss is more reliable than recalling it, and it works for a topology never seen before.

Discuss in Forum

109.

Consider the following pairs of oscillator and its typical application.

Wien bridge - Audio signal generator

Pierce - Microcontroller clock

555 astable - Low-cost timing and flashing

Colpitts - Precision frequency standard

How many of the pairs given above are correctly matched?

A.All four
B.Only three
C.Only one
D.Only two

Answer: B

The first three are correctly paired and the fourth is not, so three match. A precision standard needs a crystal or better, since a Colpitts with an LC tank drifts by parts per thousand rather than parts per million. Matching a topology to an application means asking what stability the application demands, and an LC oscillator is two or three orders of magnitude short of a standard.

Discuss in Forum

110.

A Colpitts uses \( C_{1} = 100 \ \mathrm{pF} \) and \( C_{2} = 1000 \ \mathrm{pF} \). Its feedback fraction is about

A.10, being the ratio of the larger to the smaller
B.0.5
C.0.1
D.0.9, being the ratio of the larger to the sum

Answer: C

The fraction returned is set by the capacitor ratio:

\[ \beta = \frac{C_{1}}{C_{2}} = \frac{100}{1000} = 0.1 \]

giving 0.1, so the amplifier must supply a gain of at least ten. Inverting the ratio gives ten, which would be a fraction greater than one and is impossible for a passive divider. Checking that a computed fraction lies below one is a quick test that catches the inversion immediately.

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Oscillators Questions and Answers || Oscillators MCQ Questions and Answers || Oscillators GK Questions and Answers || Oscillators GK MCQ Questions || Oscillators Multiple Choice Questions and Answers || Oscillators GK || GK on Oscillators || Analog Electronics Questions and Answers || Analog Electronics MCQ Questions and Answers || Analog Electronics GK Questions and Answers || GK on Analog Electronics

Oscillators Questions and Answers

Topics

Other Subjects

About Us

Stay informed about all important Dates & Notifications