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.

121.

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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122.

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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123.

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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124.

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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125.

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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126.

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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127.

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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