Your Answer : (Not Answered)

Correct Answer : B

The frequency is nearly 5 kHz.

For the three-section RC phase shift oscillator, using the standard formula:

\[ f = \frac{1}{2\pi RC\sqrt{6}} \]

\[ \begin{aligned} 2\pi RC\sqrt{6} &= 2\pi(10^{3})(13 \times 10^{-9})(2.449) \\ &\approx 2.0 \times 10^{-4}\ \text{s} \end{aligned} \]

so \(f \approx 5000\) Hz. The \(\sqrt{6}\) arises because the three RC sections must together give 180° of phase shift. Leaving it out gives about 12 kHz, and none of the other options follows from these values.

Your Answer : (Not Answered)

Correct Answer : C

In increasing order of gain the sequence is (iv), (i), (ii), (iii).

Negative feedback reduces gain to \(A/(1 + A\beta)\), so the amplifier with negative feedback has the least gain. The amplifier without feedback has its open-loop gain A. Positive feedback increases gain to \(A/(1 - A\beta)\), larger than A while Aβ is less than 1. An oscillator is the limit where Aβ = 1, so the gain becomes infinite and the circuit produces an output with no input.

The other codes misplace at least one of these, for example by putting the oscillator before the amplifiers or starting from the amplifier without feedback.

Your Answer : (Not Answered)

Correct Answer : B

The average DC voltage is nearly 270 Volts.

With a 300-0-300 V centre-tapped transformer, each diode is fed by 300 V rms, whose peak is \(V_{m} = 300\sqrt{2} = 424\) V. A full-wave rectifier's average output is \(2V_{m}/\pi\):

\[ \begin{aligned} V_{dc} &= \frac{2V_{m}}{\pi} = \frac{2 \times 424.3}{3.1416} \\ &\approx 270\ \text{V} \end{aligned} \]

That is 0.9 times the rms value. 300 V confuses rms with average, 150 V is half the rms value (the half-wave average would be \(V_{m}/\pi = 135\) V), and the load resistance does not affect the average voltage when diode drops are neglected.

Your Answer : (Not Answered)

Correct Answer : A

The mutual conductance is about 0.23 mho.

For a bipolar transistor the small-signal transconductance depends only on the collector current and the thermal voltage:

\[ \begin{aligned} V_{T} &= \frac{kT}{q} \approx 25.9\ \text{mV at } 300\ \text{K} \\ g_{m} &= \frac{I_{C}}{V_{T}} = \frac{6 \times 10^{-3}}{25.9 \times 10^{-3}} \\ &\approx 0.23\ \text{S (mho)} \end{aligned} \]

The current gain h_FE = 120 is not needed for g_m; it would be used to find r_π = β/g_m, about 520 Ω. 0.02 mho is ten times too small, and 2.3 or 4.33 mho would need currents of tens of milliamperes.

Your Answer : (Not Answered)

Correct Answer : C

The Zener current is about 19.6 mA.

Without the Zener the load would see \(18 \times \frac{1000}{1270} \approx 14.2\) V, more than 10 V, so the Zener breaks down and holds the load at 10 V. Then:

\[ \begin{aligned} I_{S} &= \frac{18 - 10}{270} \approx 29.6\ \text{mA} \\ I_{L} &= \frac{10}{1000} = 10\ \text{mA} \\ I_{Z} &= 29.6 - 10 = 19.6\ \text{mA} \end{aligned} \]

29.6 mA is the current in the series resistor, which forgets that 10 mA of it goes to the load. 14.8 mA and 39.6 mA do not follow from the circuit.

Your Answer : (Not Answered)

Correct Answer : D

A relaxation oscillator makes a non sinusoidal wave (a sawtooth or pulse) by charging a capacitor through a resistor until a switching device fires and discharges it. The unijunction transistor (UJT) is the classic device for this: when the capacitor voltage reaches the peak point of the emitter characteristic the UJT turns on and quickly discharges the capacitor, and the cycle repeats.

An SCR is a latching switch used for power control, and a PN junction diode and a photodiode have no negative resistance region to make the circuit switch back and forth. The answer is UJT, option D.

Your Answer : (Not Answered)

Correct Answer : A

This is crossover distortion, worst at small signals where the step is a large fraction of the output. Harmonic and intermodulation distortion are measures of distortion generally rather than mechanisms, and diagonal clipping is a detector defect in an AM receiver. That this distortion grows relatively worse as the signal falls is the reverse of most mechanisms, and it is why class B sounds poor at low volume until the bias is added. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : C

A single pole falls at six decibels per octave, equivalently twenty per decade. Three decibels is the drop at the break itself rather than a rate, and twelve per octave describes two poles. Twenty is right only when the unit is a decade rather than an octave. Two ways of stating one slope, plus one number that is a point rather than a slope, is the confusion this topic turns on most often.

Your Answer : (Not Answered)

Correct Answer : C

The diode's capacitance sits at the summing node, so it works against the feedback resistance to set the dominant pole - which is why receivers use small-area diodes and reverse bias them hard to reduce it. It does not reduce the photocurrent, is at the input rather than the output, and affects noise through a separate mechanism. That the detector rather than the amplifier often sets the speed is worth knowing before a faster amplifier is specified.

Your Answer : (Not Answered)

Correct Answer : A

The reason is true but the assertion is false: a coupling capacitor is in series with the signal, so its size affects only the low-frequency end and a larger one extends the response downward. The high end is set by shunt capacitances, which a bigger coupling capacitor does nothing about. Whether a capacitance is in series or in shunt decides which end of the response it governs, and applying a low-end remedy to a high-end problem is the mistake this item is built on.

Your Answer : (Not Answered)

Correct Answer : A

A follower's output impedance is inductive over part of its range because the reflected source impedance rises with frequency, and that inductance with a capacitive load forms a resonant circuit that can ring or oscillate. The gain does not exceed unity, and the output impedance rises rather than falling. A small series resistor at the output damps it, which is why one is so often found there in practice even though the ideal analysis says it is unnecessary.

Your Answer : (Not Answered)

Correct Answer : D

Channel-length modulation is the depletion region at the drain end growing with drain voltage, shortening the effective channel and raising the current slightly - which gives the output characteristics their finite slope, exactly as the Early effect does in a bipolar device. The other options describe the ohmic region, the body effect and gate capacitance. Recognising an analogous mechanism across the two families saves learning each family's output resistance from scratch.

Your Answer : (Not Answered)

Correct Answer : A

Both are true and the second is the reason: with no gate current, the input impedance is set by whatever bias resistor the designer fits rather than by the device. That is why a field effect input stage can present megohms while a bipolar one presents kilohms. A specification the designer chooses directly is more valuable than one inherited from the device, and this is the clearest example in the subject. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Your Answer : (Not Answered)

Correct Answer : C

The emitter voltage divided by the emitter resistor gives the current at once, and the current is what the design actually specifies. Meter loading is a genuine but secondary consideration, stiffness needs two measurements to assess, and beta needs the base current as well. Measuring at the point where a single reading converts straight into the quantity of interest is worth a moment's thought before probing anything. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Your Answer : (Not Answered)

Correct Answer : B

Beta rises and the base-emitter drop falls, and both changes increase the collector current at a fixed bias - which, since more current means more heat, closes a positive feedback loop. Leakage rising is a third contributor in the same direction. That all the significant temperature effects push the same way is what makes bias stabilisation the central concern of the next topic rather than an afterthought.

Your Answer : (Not Answered)

Correct Answer : A

The capacitor passes a burst of current at each edge, charging and then removing base charge quickly, while the resistor sets the steady drive. It does not reduce the steady current, offers no reverse protection, and cannot raise the available collector current. Giving a circuit a different response to transitions than to steady conditions is a recurring trick, and a capacitor across a resistor is its simplest form.

Your Answer : (Not Answered)

Correct Answer : D

Dynamic resistance in breakdown is a Zener specification, since an ordinary rectifier is never operated there and would be destroyed if it were. The other three are standard rectifier ratings covering reverse blocking, brief forward overload and thermal limits. A parameter that only makes sense in a region the device is not designed for is a reliable sign that it belongs to a different device.

Your Answer : (Not Answered)

Correct Answer : C

Current flows in the secondary only on one half cycle, so a net direct component magnetises the core and pushes it towards saturation, wasting capability and increasing loss. The primary's ripple, the available voltage and the ripple frequency are all separate matters. Any winding carrying unidirectional current biases its core, which is why full-wave and bridge arrangements are preferred at anything above the smallest powers. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Your Answer : (Not Answered)

Correct Answer : C

Dividing the power rating by the operating voltage:

\[ I = \frac{0.5}{10} = 50 \ \mathrm{mA} \]

so 50 mA, and a prudent design would run well below it since the rating assumes a stated ambient temperature. The other options misplace a decade. A power rating converted into a current limit at the actual operating voltage is the form the designer needs, and doing that conversion first avoids a great deal of confusion later.