Oscillators Questions and Answers

Q: For a given unijunction transistor (UJT) circuit peak voltage (Vp) is ..........Diode Voltage VD= 0.6VIntrinsic stand of ratio (η) = 0.6Symbol carrier usual meaning.

The peak voltage is 2.4 V.The UJT fires when the emitter voltage exceeds the voltage at the internal junction of its two base resistances by one diode drop. That junction sits at the fraction \(\eta\) (the intrinsic standoff ratio) of \(V_{BB}\):\[ \begin{aligned} V_{p} &= \eta V_{BB} + V_{D} \\ &= 0.6 \times 3 + 0.6 = 2.4\ \text{V} \end{aligned} \]Leaving out the diode drop would give 1.8 V, which is not offered. The other values do not fit \(\eta V_{BB} + V_{D}\) with \(\eta = 0.6\): 3.4 V would need \(\eta \approx 0.93\), while 1.4 V and 1.0 V would need \(\eta\) of only about 0.27 and 0.13.

Q: Which of the following oscillator make use of both positive and negative feedbacks ?

The Wien bridge oscillator uses both positive and negative feedback.Its RC series-parallel network gives positive feedback at the frequency of oscillation, and a resistive divider across the amplifier gives negative feedback, which fixes the gain and stabilises the amplitude and lowers distortion.The Hartley and Colpitts oscillators use only positive feedback through an LC tank, and the phase shift oscillator uses only positive feedback through an RC ladder that supplies a shift of 180 degrees.

Q: What two conditions must a loop satisfy to oscillate?

The Barkhausen criterion demands a loop gain of one in magnitude and a total phase shift around the loop of zero, or equivalently a whole number of full turns. Both must hold at the same frequency, and it is that coincidence which picks out the frequency of oscillation. Positive feedback and a resonant element are how the conditions are usually met rather than the conditions themselves, and bandwidth is a property of the amplifier. Two conditions at one frequency is the whole of oscillator theory, and every topology in this subject is an arrangement for satisfying them.

21.

For a given unijunction transistor (UJT) circuit peak voltage (V_p) is ..........

Diode Voltage V_D= 0.6V
Intrinsic stand of ratio (η) = 0.6

UJT circuit: VBB = 3 V between base B2 and base B1 (B1 grounded) and a supply VE between the emitter E and ground

Symbol carrier usual meaning.

A.3.4V
B.1.4V
C.1.0V
D.2.4V

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The peak voltage is 2.4 V.

The UJT fires when the emitter voltage exceeds the voltage at the internal junction of its two base resistances by one diode drop. That junction sits at the fraction \(\eta\) (the intrinsic standoff ratio) of \(V_{BB}\):

\[ \begin{aligned} V_{p} &= \eta V_{BB} + V_{D} \\ &= 0.6 \times 3 + 0.6 = 2.4\ \text{V} \end{aligned} \]

Leaving out the diode drop would give 1.8 V, which is not offered. The other values do not fit \(\eta V_{BB} + V_{D}\) with \(\eta = 0.6\): 3.4 V would need \(\eta \approx 0.93\), while 1.4 V and 1.0 V would need \(\eta\) of only about 0.27 and 0.13.

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22.

The following are the oscillators :

1. Crystal
2. Weinbridge
3. Colpitt's
4. Tunnel diode oscillator
The correct sequence of decreasing order of the frequency of oscillation will be :

A.4, 1, 3, 2
B.4, 1, 2, 3
C.4, 3, 2, 1
D.4, 2, 3, 1

Exam Year(s)

UGC NET - 2012,

Answer: A

The tunnel diode oscillator works in the gigahertz range, a crystal oscillator reaches tens of megahertz, an LC (Colpitts) oscillator works up to a hundred megahertz or so, and a Wien bridge RC oscillator is used at audio and low radio frequencies, so the order of decreasing frequency is 4, 1, 3, 2 (option A).

Options B and C put the Wien bridge or the Colpitts above the crystal, and D has the Colpitts below the Wien bridge, which is the wrong way round.

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23.

The phase shift introduced by feed network or amplifier in Wein bridge oscillator is

A.0°
B.90°
C.180°
D.270°

Exam Year(s)

UGC NET - 2012,

Answer: A

The phase shift introduced by the feedback network of a Wien bridge oscillator is zero degrees, option (A). At the frequency of oscillation, \(f = 1/(2\pi RC)\), the series-parallel RC network gives a phase shift of 0 degrees and an attenuation of one third. The non-inverting amplifier that goes with it also gives 0 degrees, so the total phase around the loop is 0 and the Barkhausen condition is met when the amplifier gain is 3. A shift of 180 degrees, option (C), is typical of the RC phase-shift oscillator, and 90 and 270 degrees, options (B) and (D), are not needed in the Wien bridge.

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24.

In a Wein bridge oscillator circuit, the value of frequency can be calculated by the following expression :

A.\(f_o = \dfrac{1}{2\pi RC}\)
B.\(f_o = \dfrac{1}{2\pi \sqrt{LC}}\)
C.\(f_o = \dfrac{1}{2\pi \sqrt{L_1 (C_1 + C_2)}}\)
D.\(f_o = \dfrac{1}{2\pi \sqrt{(R_1^{2} + R_2^{2})(C_1^{2} + C_2^{2})}}\)

Exam Year(s)

UGC NET - 2013,

Answer: A

The frequency of a Wein bridge oscillator is \(f_o = \dfrac{1}{2\pi RC}\), the first option, where the two arms of the frequency-selective network use equal resistors R and equal capacitors C. At this frequency the network gives zero phase shift and an attenuation of one third, so an amplifier gain of three sustains oscillation.

The expression with \(\sqrt{LC}\) is for an LC tank in a Colpitts or Hartley type oscillator, and the options with \(L_1\) and with sums of squares of the R and C values do not describe the Wein bridge, which has no inductor.

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25.

Which of the following oscillator make use of both positive and negative feedbacks ?

A.Hartley oscillator
B.Colpit's oscillator
C.Phase shift oscillator
D.Wein-Bridge oscillator

Exam Year(s)

UGC NET - 2013,

Answer: D

The Wien bridge oscillator uses both positive and negative feedback.

Its RC series-parallel network gives positive feedback at the frequency of oscillation, and a resistive divider across the amplifier gives negative feedback, which fixes the gain and stabilises the amplitude and lowers distortion.

The Hartley and Colpitts oscillators use only positive feedback through an LC tank, and the phase shift oscillator uses only positive feedback through an RC ladder that supplies a shift of 180 degrees.

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26.

Hartley oscillator is commonly used in

A.Radio receivers
B.Radio transmitters
C.TV receivers
D.Microwave transmitters

Exam Year(s)

UGC NET - 2013,

Answer: A

The Hartley oscillator uses a tapped inductor, or two inductors, with one capacitor in its tank circuit, and it is easily tuned over a wide range of radio frequencies by a variable capacitor. For this reason it is commonly used as the local oscillator in radio receivers. Transmitters need a stable carrier, for which crystal oscillators are normally used, television receivers use other fixed-frequency oscillators, and microwave transmitters use devices such as klystrons and Gunn diodes, because LC oscillators do not work at those frequencies. The answer is radio receivers.

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27.

In an RC phase shift oscillator the frequency of oscillation is given by

A.\(\frac{0.65}{RC}\)
B.\(\frac{0.065}{RC}\)
C.\(\frac{6.5}{RC}\)
D.\(\frac{1}{2\pi RC}\)

Exam Year(s)

UGC NET - 2014,

Answer: B

The frequency of oscillation is \(\frac{0.065}{RC}\). In an RC phase shift oscillator the amplifier gives a 180 degree shift and a three-section RC ladder gives the other 180 degrees. That happens at \[ f = \frac{1}{2\pi\sqrt{6}\,RC} \] and since \(2\pi\sqrt{6} \approx 15.4\), \[ f \approx \frac{0.065}{RC} \]

The form \(\frac{1}{2\pi RC}\) is the corner frequency of a single RC section and the cut-off of a Wien bridge oscillator, so it does not apply to the three-section phase shift network. The values 0.65 and 6.5 are off by factors of 10 and 100 from the correct constant.

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28.

What two conditions must a loop satisfy to oscillate?

A.Unity loop gain and zero net phase shift
B.Loop gain above unity and ninety degrees of phase
C.Positive feedback and a resonant element somewhere in the loop
D.A gain above one and a bandwidth wider than the frequency wanted

Answer: A

The Barkhausen criterion demands a loop gain of one in magnitude and a total phase shift around the loop of zero, or equivalently a whole number of full turns. Both must hold at the same frequency, and it is that coincidence which picks out the frequency of oscillation. Positive feedback and a resonant element are how the conditions are usually met rather than the conditions themselves, and bandwidth is a property of the amplifier. Two conditions at one frequency is the whole of oscillator theory, and every topology in this subject is an arrangement for satisfying them.

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29.

Why must the loop gain exceed unity at switch-on?

A.The supply takes time to reach its nominal voltage
B.Oscillation must grow from noise to a usable amplitude
C.The amplifier's gain falls once the output is established
D.The phase condition cannot be met at low amplitudes

Answer: B

A loop gain of exactly one would sustain whatever amplitude already existed but could never build one, so a practical oscillator is designed with a loop gain somewhat above one and grows from circuit noise until something limits it. The amplifier's gain is constant until the amplitude limiting acts, the phase condition is amplitude-independent, and the supply's rise is incidental. Designing a loop to be marginally unstable, and then arranging what stops it growing, is the practical reading of the criterion.

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30.

What starts an oscillator when the supply is applied?

A.The resonant circuit's stored energy from the previous run
B.The amplifier's own offset voltage at its input
C.Noise and the switching transient
D.A deliberate pulse injected by the bias network

Answer: C

Every circuit contains thermal and shot noise across a wide band, and the supply's own step excites the resonant circuit as well - the loop selects whatever component lies at its frequency and amplifies it repeatedly until limiting sets in. No injected pulse is needed, nothing is stored between runs, and offset is a direct quantity rather than a signal. That an oscillator needs no starting signal, only the absence of anything preventing growth, is what makes an unwanted oscillation so hard to suppress once a loop meets the conditions.

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