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Kerala Plus Two Electronics (DHSE) practice questions

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Question 1Basic Electronics and Semiconductor Devices

In a pure silicon crystal at room temperature, the number of free electrons is equal to the number of holes. This equality arises because

  1. the crystal lattice repels holes towards the surface
  2. donor impurities supply matching numbers of electrons and holes
  3. each broken covalent bond creates one electron and one hole Correct answer.
  4. thermal energy injects electrons from the metal contacts

Explanation

In an intrinsic semiconductor, thermal energy breaks covalent bonds and each broken bond releases one electron plus one vacancy (hole), so the two carrier concentrations stay equal. This is the defining property of intrinsic conduction, and it explains why ni equals pi.

Question 2Basic Electronics and Semiconductor Devices

The forbidden energy gap of silicon at 300 K is approximately

  1. 1.1 eV Correct answer.
  2. 1.7 eV
  3. 5.5 eV
  4. 0.3 eV

Explanation

Silicon has a band gap of about 1.1 eV at room temperature, which is why it conducts only weakly when pure. Germanium is near 0.7 eV and insulators such as diamond exceed 5 eV.

Question 3Basic Electronics and Semiconductor Devices

A pentavalent impurity such as phosphorus is added to silicon. The resulting material is

  1. p-type, because holes become the majority carriers
  2. n-type, because each impurity atom donates a free electron Correct answer.
  3. intrinsic, because the impurity is neutral
  4. insulating, because the impurity disrupts the lattice

Explanation

A pentavalent atom forms four covalent bonds and its fifth valence electron is loosely bound, so it is easily freed and becomes a conduction electron. Such donors make electrons the majority carriers, giving n-type material.

Question 4Basic Electronics and Semiconductor Devices

In a p-n junction diode, the depletion region contains

  1. a high density of majority carriers on both sides
  2. equal numbers of free electrons and holes
  3. only free electrons supplied by the p side
  4. immobile ions with no free carriers Correct answer.

Explanation

Diffusion of carriers across the junction leaves behind fixed ionised donor and acceptor atoms, and this space-charge region is swept free of mobile carriers. The resulting ions create the built-in potential barrier.

Question 5Basic Electronics and Semiconductor Devices

For a silicon diode, the forward voltage at which appreciable current begins to flow is about

  1. 0.7 V Correct answer.
  2. 0.1 V
  3. 0.3 V
  4. 1.5 V

Explanation

A silicon diode needs roughly 0.7 V forward bias to overcome the barrier potential and conduct appreciably. Germanium diodes conduct from about 0.3 V instead.

Question 6Basic Electronics and Semiconductor Devices

A diode is connected in series with a resistor and a battery in reverse bias. The current in the circuit is

  1. equal to the forward current for the same voltage
  2. nearly zero, being only the small reverse saturation current Correct answer.
  3. oscillating because the diode switches on and off
  4. large, limited only by the resistor

Explanation

In reverse bias the depletion region widens and the barrier increases, so only a tiny minority-carrier leakage current flows. This reverse saturation current is almost independent of the applied voltage until breakdown.

Question 7Basic Electronics and Semiconductor Devices

The output of a half-wave rectifier using a single diode fed from an AC supply is

  1. a full-wave DC with two pulses per cycle
  2. a steady DC with no ripple
  3. an AC waveform of double the input frequency
  4. a pulsating DC with one pulse per AC cycle Correct answer.

Explanation

A half-wave rectifier conducts only during the half cycle that forward-biases the diode, so the load sees one pulse per input cycle. The result is pulsating DC whose ripple frequency equals the supply frequency.

Question 8Basic Electronics and Semiconductor Devices

In a common-emitter transistor amplifier, the phase difference between the input signal at the base and the output signal at the collector is

  1. 0 degrees
  2. 90 degrees
  3. 180 degrees Correct answer.
  4. 360 degrees

Explanation

An increase in base current raises collector current and so increases the drop across the collector resistor, lowering the collector voltage. The output therefore moves opposite to the input, giving a 180-degree phase inversion.

Question 9Basic Electronics and Semiconductor Devices

A zener diode is to be used as a voltage regulator. It must be connected

  1. in reverse bias, in series with the load
  2. in forward bias, in parallel with the load
  3. in reverse bias, in parallel with the load Correct answer.
  4. in forward bias, in series with the load

Explanation

Regulation uses the zener breakdown region, which is reached under reverse bias, and the diode is placed across the load so it clamps the voltage. The series resistor drops the excess supply voltage.

Question 10Basic Electronics and Semiconductor Devices

A transistor has base current 40 microampere and collector current 4 milliampere. Its current gain beta is

  1. 0.01
  2. 10
  3. 1000
  4. 100 Correct answer.

Explanation

Beta equals collector current divided by base current, so 4 mA divided by 40 microampere gives 100. This is the common-emitter DC current gain, a dimensionless ratio.

Question 11Basic Electronics and Semiconductor Devices

In a full-wave bridge rectifier, the number of diodes conducting during each half cycle is

  1. three
  2. two Correct answer.
  3. four
  4. one

Explanation

In a bridge rectifier, the AC source drives current through one diagonal pair of diodes in each half cycle. The other two are reverse biased, so exactly two conduct at any instant.

Question 12Basic Electronics and Semiconductor Devices

The main purpose of a capacitor filter connected across the load of a rectifier is to

  1. convert AC into DC by blocking one half cycle
  2. smooth the pulsating output by charging and discharging Correct answer.
  3. step up the AC input voltage
  4. increase the ripple voltage

Explanation

The capacitor charges near the peak of each pulse and discharges slowly through the load between pulses, filling in the troughs. This reduces ripple and raises the average output voltage.

Question 13Basic Electronics and Semiconductor Devices

The current gain of a transistor in common-base configuration is 0.98. The corresponding common-emitter current gain beta is approximately

  1. 2
  2. 0.98
  3. 98
  4. 49 Correct answer.

Explanation

Beta equals alpha divided by (1 minus alpha), so 0.98 divided by 0.02 gives 49. This shows how a common-base gain close to unity corresponds to a much larger common-emitter gain.

Question 14Basic Electronics and Semiconductor Devices

A transistor is biased so that its quiescent collector current is zero. The transistor is operating in the

  1. active region
  2. cut-off region Correct answer.
  3. saturation region
  4. breakdown region

Explanation

Cut-off occurs when the base-emitter junction is not forward biased enough to turn the transistor on, so collector current falls to leakage level. In this state the transistor behaves like an open switch.

Question 15Basic Electronics and Semiconductor Devices

The input resistance of a common-collector (emitter follower) amplifier is high and its voltage gain is approximately unity. Its main use is therefore as

  1. a rectifier for AC supplies
  2. a buffer to match a high-impedance source to a low-impedance load Correct answer.
  3. an oscillator to generate carrier waves
  4. a voltage amplifier with large gain

Explanation

Because the emitter follower has high input impedance, near-unity voltage gain and low output impedance, it isolates a weak source from a heavy load. It acts as a buffer stage rather than a voltage amplifier.

Question 16Basic Electronics and Semiconductor Devices

A light-emitting diode emits light when

  1. the diode is heated above its rated temperature
  2. the junction is illuminated by an external source
  3. electrons and holes recombine across the junction in forward bias Correct answer.
  4. the diode is reverse biased beyond breakdown

Explanation

In forward bias, electrons injected into the p region recombine with holes and release energy as photons. The photon energy is set by the band gap, which determines the colour.

Question 17Basic Electronics and Semiconductor Devices

A photodiode is normally operated in reverse bias because

  1. forward bias makes it emit light
  2. the reverse leakage current varies strongly with illumination Correct answer.
  3. reverse bias reduces the depletion capacitance to zero
  4. forward current is too small to measure

Explanation

In reverse bias, thermally generated minority carriers form a small leakage current that increases almost in proportion to the light falling on the junction. This makes the device a light-controlled current source.

Question 18Basic Electronics and Semiconductor Devices

A silicon diode with a 0.7 V drop carries 20 mA. The power dissipated in the diode is

  1. 7 mW
  2. 14 mW Correct answer.
  3. 0.35 mW
  4. 28 mW

Explanation

Diode power dissipation equals forward voltage multiplied by forward current, so 0.7 V times 20 mA gives 14 mW. This heating must be kept within the device rating.

Question 19Basic Electronics and Semiconductor Devices

In a transistor, the base region is made very thin and lightly doped mainly to

  1. allow most injected carriers to reach the collector Correct answer.
  2. raise the collector breakdown voltage
  3. increase the base current so gain falls
  4. make the emitter junction reverse biased

Explanation

A thin, lightly doped base means few injected carriers recombine there, so nearly all reach the collector. This keeps alpha close to unity and gives a large common-emitter current gain.

Question 20Basic Electronics and Semiconductor Devices

A half-wave rectifier is fed from a 50 Hz supply. The ripple frequency in its output is

  1. 50 Hz Correct answer.
  2. 25 Hz
  3. 100 Hz
  4. 150 Hz

Explanation

A half-wave rectifier passes one pulse per input cycle, so the output ripple repeats at the supply frequency of 50 Hz. A full-wave circuit would double this to 100 Hz.

Question 21Basic Electronics and Semiconductor Devices

A bridge rectifier delivers a peak output of 12 V to a load drawing 60 mA. Each conducting diode drops 0.7 V. The peak voltage available from the secondary is

  1. 12.7 V
  2. 13.4 V Correct answer.
  3. 10.6 V
  4. 12.0 V

Explanation

In a bridge rectifier two diodes conduct in series with the load during each half cycle, so the secondary peak must supply the load peak plus twice the diode drop. Adding 1.4 V to 12 V gives 13.4 V.

Question 22Basic Electronics and Semiconductor Devices

A transistor with beta of 100 is used in a circuit whose base current is 50 microampere. If the collector-emitter voltage is 8 V, the power dissipated in the transistor is approximately

  1. 0.4 mW
  2. 4 mW
  3. 40 mW Correct answer.
  4. 400 mW

Explanation

Collector current equals beta times base current, giving 5 mA, and power is collector-emitter voltage times collector current. Multiplying 8 V by 5 mA yields 40 mW.

Question 23Basic Electronics and Semiconductor Devices

A zener regulator uses a 10 V zener with a 220 ohm series resistor fed from a 15 V supply, and the load draws no current. The current through the zener is approximately

  1. 68 mA
  2. 22.7 mA Correct answer.
  3. 10 mA
  4. 45.5 mA

Explanation

The series resistor carries the full current when the load is open, and the voltage across it is 15 V minus 10 V, or 5 V. Dividing 5 V by 220 ohm gives about 22.7 mA, which flows through the zener.

Question 24Basic Electronics and Semiconductor Devices

Under reverse bias, a diode shows a sudden large increase in current at a fixed voltage below the normal breakdown rating of comparable diodes. This behaviour indicates

  1. the zener breakdown mechanism, where a strong field frees carriers directly Correct answer.
  2. excessive heating of the junction by forward current
  3. a manufacturing fault that prevents any current flow
  4. the normal forward conduction knee

Explanation

Zener breakdown occurs in heavily doped junctions where the intense electric field across a narrow depletion layer pulls electrons out of covalent bonds. The result is a sharp, reversible rise in reverse current at a well-defined low voltage.

Question 25Basic Electronics and Semiconductor Devices

A common-emitter amplifier has a voltage gain of 50 and feeds a 1 kilohm load. If the input signal is 20 mV peak, the output power delivered to the load is approximately

  1. 100 mW
  2. 50 mW Correct answer.
  3. 25 mW
  4. 500 mW

Explanation

The output peak voltage is 50 times 20 mV, or 1 V, and power for a sine wave is peak voltage squared divided by twice the load resistance. This gives 1 divided by 2000, or 50 mW.

Question 26Basic Electronics and Semiconductor Devices

A p-type semiconductor sample is formed by adding acceptor impurities to pure silicon. The majority charge carriers in this sample are

  1. positive acceptor ions
  2. negative donor ions
  3. holes Correct answer.
  4. free electrons

Explanation

Trivalent acceptor atoms create holes in the silicon crystal lattice, so holes become the majority carriers in a p-type material. The fixed acceptor ions are bound in the lattice and cannot carry current, so they are not charge carriers.

Question 27Basic Electronics and Semiconductor Devices

A silicon diode is forward biased and the current through it is doubled from 10 mA to 20 mA. The forward voltage drop across the diode

  1. increases only slightly, by a few tens of millivolts Correct answer.
  2. remains fixed at exactly 0.7 V
  3. decreases to half its earlier value
  4. doubles from 0.7 V to 1.4 V

Explanation

Beyond the knee, the forward characteristic of a silicon diode is very steep, so the voltage rises only a little for a large increase in current. The 0.7 V figure is a convenient approximation, not a constant value, and it does not scale in proportion to current.

Question 28Basic Electronics and Semiconductor Devices

A transistor is used as a switch to drive a relay. For the transistor to behave as a closed switch with the smallest possible voltage across it, it should be driven into

  1. saturation Correct answer.
  2. cut-off
  3. reverse breakdown of the base-emitter junction
  4. active region

Explanation

In saturation both junctions are forward biased and the collector-emitter voltage falls to a small value, so the transistor acts almost like a closed switch. Cut-off gives an open switch, while the active region is used for amplification, not switching.

Question 29Basic Electronics and Semiconductor Devices

The output of a full-wave rectifier is passed through a capacitor-input filter. If the load resistance is made larger while the capacitance stays the same, the ripple in the output voltage

  1. remains unchanged, since capacitance is unaltered
  2. decreases, because the capacitor discharges more slowly Correct answer.
  3. increases, because the load draws more current
  4. becomes zero, because the capacitor never discharges

Explanation

The ripple depends on the time constant formed by the capacitor and the load resistance. A larger load resistance lengthens the discharge time constant, so the capacitor voltage sags less between successive charging peaks and the ripple amplitude falls.

Question 30Basic Electronics and Semiconductor Devices

A common-emitter amplifier has a voltage gain of 40 with no load. When a load resistance equal to the amplifier's own output resistance is connected, the overall voltage gain becomes approximately

  1. 80
  2. 40
  3. 0
  4. 20 Correct answer.

Explanation

The output resistance and the external load form a voltage divider at the output. When the load equals the output resistance, half of the internally generated signal voltage is lost across the output resistance, so the gain halves to about 20.

Question 31Power Supplies and Rectifiers

In a half-wave rectifier using a single silicon diode, the input is a 50 Hz sinusoidal supply. What is the ripple frequency of the output waveform?

  1. 150 Hz
  2. 50 Hz Correct answer.
  3. 25 Hz
  4. 100 Hz

Explanation

A half-wave rectifier conducts only on one half of every input cycle, so exactly one output pulse appears per input cycle. The ripple frequency therefore equals the supply frequency, 50 Hz. A full-wave rectifier would double this value.

Question 32Power Supplies and Rectifiers

The output of a full-wave bridge rectifier is fed directly to a capacitor-input filter. Which quantity determines the amount by which the output voltage falls between successive charging peaks?

  1. The peak inverse voltage rating of the diodes
  2. The transformer turns ratio alone
  3. The load current and the filter capacitance Correct answer.
  4. The forward resistance of the diodes

Explanation

Between peaks the capacitor discharges through the load, and the droop equals I·t/C, where I is load current, t the time between peaks and C the capacitance. Larger capacitance or smaller load current therefore gives a smaller ripple. Diode ratings and turns ratio set the peak level, not the droop.

Question 33Power Supplies and Rectifiers

A transformer secondary delivers 12 V rms to a full-wave rectifier. Ignoring diode drops, what is the approximate peak value of the unfiltered output voltage?

  1. 24 V
  2. 17 V Correct answer.
  3. 12 V
  4. 8.5 V

Explanation

The peak of a sine wave is √2 times the rms value, so 12 × 1.414 ≈ 17 V. The rectified output reaches this peak during each conduction interval. The average value would be lower, about 0.637 of the peak.

Question 34Power Supplies and Rectifiers

Why is a bleeder resistor sometimes connected across the output capacitor of a high-voltage power supply?

  1. To increase the ripple frequency
  2. To raise the peak inverse voltage of the rectifier
  3. To convert the supply into a full-wave rectifier
  4. To discharge the capacitor after the supply is switched off Correct answer.

Explanation

A charged filter capacitor can retain a dangerous voltage long after mains power is removed. A bleeder resistor provides a controlled discharge path so the terminal voltage falls to a safe level. It draws a small standing current, which is its main penalty.

Question 35Power Supplies and Rectifiers

In a centre-tap full-wave rectifier, each diode must withstand a peak inverse voltage. What is its magnitude in terms of the peak secondary voltage Vm measured from one end of the secondary to the centre tap?

  1. 2Vm Correct answer.
  2. Vm
  3. Vm divided by √2
  4. 0.5Vm

Explanation

When one diode conducts, the other is reverse biased by the full voltage across the two outer ends of the secondary, which is twice the end-to-centre value. Hence the non-conducting diode sees 2Vm. This is why the bridge circuit, needing only Vm per diode, is often preferred.

Question 36Power Supplies and Rectifiers

Which component is normally placed immediately after the filter capacitor in a regulated DC power supply to hold the output constant as load current varies?

  1. A fuse in the primary winding
  2. A step-up transformer
  3. A voltage regulator IC such as the 7805 Correct answer.
  4. A second rectifier diode

Explanation

A three-terminal regulator such as the 7805 holds its output near 5 V despite changes in input and load, within its dropout limit. The transformer, rectifier and filter only produce a rough unregulated DC level. Regulation is the function that follows filtering.

Question 37Power Supplies and Rectifiers

A zener diode of nominal voltage 6.2 V is used as a shunt regulator with a series resistor from an unregulated supply. The load is disconnected while the supply remains on. What happens to the zener current?

  1. It falls to zero because no load current flows
  2. It remains equal to the previous load current
  3. It reverses direction through the zener
  4. It rises to the full current permitted by the series resistor Correct answer.

Explanation

With the load removed, all the current set by the series resistor must pass through the zener. That current must stay below the zener's maximum rating or the device will overheat. The series resistor is chosen with this worst case in mind.

Question 38Power Supplies and Rectifiers

What is the main reason for using a bridge rectifier instead of a centre-tap rectifier in a low-voltage supply?

  1. It removes the need for any filter capacitor
  2. It needs a centre-tapped transformer and gives lower ripple
  3. It uses the whole secondary winding and needs only half the peak inverse voltage per diode Correct answer.
  4. It produces a higher ripple frequency than the centre-tap circuit

Explanation

The bridge uses the entire secondary in both half cycles, so for the same output it needs a smaller, cheaper transformer without a centre tap. Each diode also sees only Vm of reverse voltage rather than 2Vm. Both circuits give the same ripple frequency.

Question 39Power Supplies and Rectifiers

The ripple factor of a rectifier output is defined as which ratio?

  1. Rms ripple voltage to average DC voltage Correct answer.
  2. Peak voltage to rms voltage
  3. Average DC voltage to rms ripple voltage
  4. Rms ripple current to load resistance

Explanation

Ripple factor compares the unwanted AC content with the useful DC content, so it is the rms value of the ripple divided by the average DC value. A smaller figure means a smoother output. An ideal rectifier would have a ripple factor of zero.

Question 40Power Supplies and Rectifiers

In a half-wave rectifier, the diode is reverse biased for most of the negative half cycle. During that interval, what is the approximate voltage across the diode if the peak secondary voltage is 20 V?

  1. Zero
  2. About 0.7 V
  3. Up to about 20 V Correct answer.
  4. About 10 V

Explanation

While the diode is off, the full secondary voltage appears across it because the load carries no current and drops no voltage. The reverse voltage therefore rises to the peak secondary value, about 20 V. This peak reverse voltage is the rating the diode must survive.

Question 41Power Supplies and Rectifiers

A capacitor-input filter is added to a full-wave rectifier. How does increasing the capacitance affect the output?

  1. Ripple frequency doubles
  2. Average DC voltage falls and ripple rises
  3. Average DC voltage rises and ripple falls Correct answer.
  4. Both average voltage and ripple rise

Explanation

A larger capacitor discharges more slowly between peaks, so the output stays nearer the peak value and the average DC rises. The smaller voltage droop also means less ripple. Ripple frequency, however, is unchanged.

Question 42Power Supplies and Rectifiers

A 7805 regulator is supplied from a filtered input of 7.5 V. Under full load the input falls to 6.5 V. What problem arises?

  1. The regulator may lose regulation because the dropout voltage is exceeded Correct answer.
  2. The output rises above 5 V
  3. The output ripple frequency doubles
  4. The regulator delivers double the rated current

Explanation

A 7805 needs roughly 2 V between input and output to regulate, so an input of 6.5 V leaves only 1.5 V of headroom. Below the dropout limit the output sags with the input and ripple appears at the output. The filter or transformer must be sized to keep the minimum input above about 7 V.

Question 43Power Supplies and Rectifiers

Which statement correctly describes the current in the secondary winding of a centre-tap full-wave rectifier?

  1. Each half of the winding carries current in alternate half cycles Correct answer.
  2. The secondary carries only DC
  3. The whole winding carries current in both half cycles
  4. No current flows in the secondary during conduction

Explanation

In a centre-tap circuit the two diodes conduct on alternate half cycles, each using one half of the secondary. Each half winding therefore carries unidirectional pulses in turn. The magnetic effects in the two halves partly cancel in the core.

Question 44Power Supplies and Rectifiers

A rectifier output shows a ripple voltage of 2 V peak-to-peak on an average DC level of 20 V. What is the approximate rms ripple voltage if the ripple is treated as a triangular wave?

  1. 2 V
  2. 1.15 V
  3. 0.58 V Correct answer.
  4. 0.71 V

Explanation

For a triangular waveform the rms value is the peak amplitude divided by √3. The peak amplitude here is half the peak-to-peak value, 1 V, so the rms ripple is about 0.58 V. Treating the ripple as a sine wave would give a different, larger figure.

Question 45Power Supplies and Rectifiers

What is the purpose of the transformer in a conventional linear DC power supply?

  1. To step the mains voltage down and provide isolation from the mains Correct answer.
  2. To regulate the output against load changes
  3. To rectify the alternating input
  4. To smooth the pulsating DC

Explanation

The transformer converts the mains AC to a lower AC level suited to the rectifier and, crucially, isolates the secondary from the mains. Rectification, smoothing and regulation are done by later stages. Isolation is a key safety function.

Question 46Power Supplies and Rectifiers

In a full-wave bridge rectifier, how many diodes conduct during each half cycle of the input?

  1. Three
  2. Two Correct answer.
  3. Four
  4. One

Explanation

The bridge is arranged so that diagonally opposite diodes conduct together on each half cycle, giving two conducting diodes at a time. The other two are reverse biased. This means the output is always two diode drops below the secondary peak.

Question 47Power Supplies and Rectifiers

A student observes that the output of a full-wave rectifier has a ripple frequency of 100 Hz while the mains is 50 Hz. What happens to the ripple frequency if the mains frequency is raised to 60 Hz?

  1. It becomes 30 Hz
  2. It stays at 100 Hz
  3. It becomes 120 Hz Correct answer.
  4. It becomes 60 Hz

Explanation

A full-wave rectifier gives two output pulses per input cycle, so the ripple frequency is twice the supply frequency. At 60 Hz the ripple therefore becomes 120 Hz. The relationship is fixed by the circuit topology.

Question 48Power Supplies and Rectifiers

A series resistor of 220 Ω feeds a 5.1 V zener from a 12 V unregulated source. Ignoring zener current, what is the approximate current through the series resistor?

  1. 12 mA
  2. 23 mA
  3. 55 mA
  4. 31 mA Correct answer.

Explanation

The resistor drops the difference between the source and the zener voltage, 12 − 5.1 = 6.9 V. Dividing by 220 Ω gives about 31 mA. This current splits between the zener and the load, so the zener must be rated for the no-load case.

Question 49Power Supplies and Rectifiers

Why does a choke-input filter produce a lower output voltage than a capacitor-input filter for the same secondary voltage?

  1. The choke adds extra diode drops
  2. The choke reduces the ripple frequency
  3. The output approaches the average value of the rectified waveform rather than its peak Correct answer.
  4. The choke reverses the rectifier action

Explanation

A capacitor-input filter charges to near the peak, so the DC output is close to Vm. A choke-input filter smooths the current and the output settles near the average of the rectified waveform, roughly 0.637 Vm for full wave. Hence the lower DC level but better regulation.

Question 50Power Supplies and Rectifiers

A fuse is placed in the primary of the mains transformer of a DC power supply. What is its main protective function?

  1. To reduce the ripple in the output
  2. To increase the current rating of the secondary
  3. To limit the DC output voltage
  4. To interrupt the mains supply if the transformer or rectifier develops a fault Correct answer.

Explanation

A primary fuse protects against excessive mains current caused by shorted diodes, a failed capacitor or a faulty transformer. It isolates the equipment from the mains before overheating or fire occurs. It does not regulate or smooth the output.