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Assam ASSEB HS Skill Subject Electronics and Hardware study materials
Assam ASSEB HS Skill Subject Electronics and Hardware study resources

Assam ASSEB HS Skill Subject Electronics and Hardware practice questions

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Question 1Basic Electronics and Circuit Fundamentals

In an intrinsic silicon crystal at room temperature, which statement correctly describes the charge carriers?

  1. Holes outnumber free electrons because the crystal lattice is positively charged.
  2. The number of free electrons equals the number of holes, and both are created by thermal breaking of covalent bonds. Correct answer.
  3. Free electrons outnumber holes because silicon is a pentavalent material.
  4. There are no charge carriers until an external voltage is applied to the crystal.

Explanation

Intrinsic silicon has no deliberate dopant, so every electron thermally excited out of a covalent bond leaves behind one hole. Therefore electron and hole concentrations are equal, and conduction is due to both types of carrier moving under an applied field.

Question 2Basic Electronics and Circuit Fundamentals

A resistor of 470 Ω carries a current of 20 mA. What is the potential difference across it?

  1. 0.94 V
  2. 94 V
  3. 9.4 V Correct answer.
  4. 23.5 V

Explanation

Ohm's law states V = IR. Substituting 20 mA = 0.02 A and R = 470 Ω gives V = 0.02 × 470 = 9.4 V. Keeping current in amperes is essential to avoid a factor-of-1000 error.

Question 3Basic Electronics and Circuit Fundamentals

In a p-n junction diode, the depletion region is formed because:

  1. majority carriers are swept across the junction by the external supply
  2. the diode heats up and the covalent bonds break down completely
  3. electrons and holes recombine near the junction, leaving fixed ionised dopant atoms Correct answer.
  4. the metal contacts inject minority carriers into the crystal

Explanation

When p and n materials are joined, electrons diffuse into the p-side and holes into the n-side, where they recombine. The uncovered acceptor and donor ions are fixed in the lattice and create the depletion region with its built-in potential barrier.

Question 4Basic Electronics and Circuit Fundamentals

A silicon diode is connected in series with a 1 kΩ resistor across a 10 V DC supply, forward biased. Taking the diode drop as 0.7 V, the current is approximately:

  1. 9.3 mA Correct answer.
  2. 1.07 mA
  3. 0.7 mA
  4. 10.0 mA

Explanation

The resistor sees the supply voltage minus the diode's forward drop, so V_R = 10 − 0.7 = 9.3 V. With R = 1 kΩ, Ohm's law gives I = 9.3 V / 1000 Ω = 9.3 mA.

Question 5Basic Electronics and Circuit Fundamentals

Which property of a capacitor is measured in farads?

  1. The energy it stores per volt of charge
  2. The charge it stores per volt of potential difference Correct answer.
  3. The resistance it offers to alternating current
  4. The rate at which it blocks direct current

Explanation

Capacitance is defined as C = Q/V, the charge stored per unit potential difference, and one farad is one coulomb per volt. Energy storage and reactance are derived quantities, not the definition of capacitance.

Question 6Basic Electronics and Circuit Fundamentals

Two resistors of 6 Ω and 3 Ω are connected in parallel. What is the equivalent resistance?

  1. 4.5 Ω
  2. 2 Ω Correct answer.
  3. 0.5 Ω
  4. 9 Ω

Explanation

For two resistors in parallel, R = (R1·R2)/(R1+R2) = (6×3)/(6+3) = 18/9 = 2 Ω. The result is always smaller than the smallest branch resistance, which confirms the answer.

Question 7Basic Electronics and Circuit Fundamentals

The primary function of a rectifier in a DC power supply is to:

  1. smooth the ripple remaining after filtering
  2. step the mains voltage down to a safe level
  3. convert alternating current into unidirectional current Correct answer.
  4. amplify the input signal before regulation

Explanation

A rectifier uses diodes to allow current to flow in one direction only, turning an AC input into a pulsating unidirectional output. Filtering and regulation are separate later stages that refine this output.

Question 8Basic Electronics and Circuit Fundamentals

Which logic gate produces a HIGH output only when all of its inputs are HIGH?

  1. XOR gate
  2. AND gate Correct answer.
  3. NOR gate
  4. OR gate

Explanation

The AND gate outputs logic 1 only for the input combination where every input is 1; any 0 input forces the output to 0. This follows directly from the Boolean expression Y = A·B.

Question 9Basic Electronics and Circuit Fundamentals

A zener diode is to be used as a voltage reference. In which bias condition does it normally operate?

  1. Reverse biased beyond its breakdown voltage Correct answer.
  2. Unbiased, relying on thermal generation
  3. Forward biased with no series resistor
  4. Forward biased well above its knee voltage

Explanation

A zener regulator operates in reverse breakdown, where the voltage across it stays nearly constant over a wide range of current. A series resistor is required to limit the current and dissipate the excess voltage.

Question 10Basic Electronics and Circuit Fundamentals

In a common-emitter amplifier, the emitter bypass capacitor improves which performance figure?

  1. It raises the AC voltage gain by reducing AC negative feedback Correct answer.
  2. It lowers the input impedance to match the source
  3. It increases the DC current gain of the transistor
  4. It increases the collector supply voltage

Explanation

The bypass capacitor provides a low-impedance path for AC around the emitter resistor, removing AC degeneration while DC bias remains stabilised. With the emitter effectively grounded for signals, the voltage gain rises towards approximately Rc/re.

Question 11Basic Electronics and Circuit Fundamentals

A transformer has 500 turns on the primary and 50 turns on the secondary. If the primary is connected to 230 V AC, the secondary voltage is approximately:

  1. 115 V
  2. 2300 V
  3. 23 V Correct answer.
  4. 46 V

Explanation

The turns ratio is Np/Ns = 500/50 = 10, and for a transformer Vs = Vp × (Ns/Np) = 230/10 = 23 V. This is a step-down transformer because the secondary has fewer turns.

Question 12Basic Electronics and Circuit Fundamentals

A bridge rectifier with a capacitor-input filter feeds a light load. Which change would reduce the output ripple voltage?

  1. Reducing the filter capacitance
  2. Increasing the load current
  3. Removing the filter capacitor
  4. Increasing the filter capacitance Correct answer.

Explanation

Ripple is set by how much the capacitor discharges between charging peaks, so a larger capacitance holds charge longer and reduces ripple amplitude. The ripple is approximately I/(f·C) for a given load current and frequency.

Question 13Basic Electronics and Circuit Fundamentals

The Boolean expression Y = A + A·B simplifies to:

  1. A·B
  2. B
  3. A + B
  4. A Correct answer.

Explanation

This is an application of the absorption law: A + A·B = A(1 + B) = A·1 = A. The term A·B is fully covered by A, so it adds nothing to the output.

Question 14Basic Electronics and Circuit Fundamentals

A silicon transistor in the active region has VBE = 0.7 V and IB = 40 µA. If hFE = 100, the collector current is:

  1. 0.4 mA
  2. 4 mA Correct answer.
  3. 0.04 mA
  4. 40 mA

Explanation

In the active region IC = hFE × IB, so IC = 100 × 40 µA = 4000 µA = 4 mA. The base-emitter voltage confirms forward bias but does not enter this particular calculation.

Question 15Basic Electronics and Circuit Fundamentals

In a common-base configuration, the current gain is approximately unity but slightly less than one. Why?

  1. Because recombination in the base is completely eliminated
  2. Because the emitter is grounded and therefore draws no current
  3. Because the collector junction is forward biased in normal operation
  4. Because the base current is a small fraction of the emitter current that does not reach the collector Correct answer.

Explanation

In a BJT the emitter current splits into collector and base currents, so IC = αIE with α slightly below 1. The small base current, due to recombination in the base region, is the part of the emitter current that does not reach the collector.

Question 16Basic Electronics and Circuit Fundamentals

A capacitor of 10 µF is charged to 50 V. How much energy is stored?

  1. 0.0125 J Correct answer.
  2. 2.5 J
  3. 0.25 J
  4. 0.5 J

Explanation

Energy stored in a capacitor is E = ½CV². Substituting C = 10 µF = 10×10⁻⁶ F and V = 50 V gives E = 0.5 × 10⁻⁵ × 2500 = 0.0125 J.

Question 17Basic Electronics and Circuit Fundamentals

A half-wave rectifier is fed from a 50 Hz supply. What is the ripple frequency of its output?

  1. 25 Hz
  2. 200 Hz
  3. 50 Hz Correct answer.
  4. 100 Hz

Explanation

A half-wave rectifier passes only one half-cycle per input cycle, so the output pulses once per cycle. With a 50 Hz input, the ripple frequency is therefore 50 Hz.

Question 18Basic Electronics and Circuit Fundamentals

Which statement about an ideal operational amplifier is correct?

  1. It has infinite open-loop gain but finite input impedance
  2. It has zero input impedance and infinite output impedance
  3. It has equal finite input and output impedances
  4. It has infinite input impedance and zero output impedance Correct answer.

Explanation

An ideal op-amp draws no input current and can drive any load, which is modelled as infinite input impedance and zero output impedance. It also has infinite open-loop gain and infinite bandwidth, so the other options misstate the model.

Question 19Basic Electronics and Circuit Fundamentals

An inverting op-amp amplifier has an input resistor of 2 kΩ and a feedback resistor of 20 kΩ. A 0.2 V input produces an output of:

  1. +2.0 V
  2. +0.02 V
  3. −2.0 V Correct answer.
  4. −0.2 V

Explanation

For the inverting configuration the gain is −Rf/Rin = −20k/2k = −10. Multiplying the 0.2 V input by −10 gives −2.0 V, the negative sign indicating phase inversion.

Question 20Basic Electronics and Circuit Fundamentals

A 5 V supply drives a series circuit of three resistors: 100 Ω, 220 Ω and 180 Ω. What current flows?

  1. 20 mA
  2. 10 mA Correct answer.
  3. 50 mA
  4. 100 mA

Explanation

The total series resistance is 100 + 220 + 180 = 500 Ω. Ohm's law then gives I = 5 V / 500 Ω = 0.01 A, which is 10 mA.

Question 21Basic Electronics and Circuit Fundamentals

A bipolar transistor is found to have both junctions forward biased. In which region of operation is it?

  1. Active region
  2. Saturation region Correct answer.
  3. Cut-off region
  4. Breakdown region

Explanation

In saturation both the base-emitter and base-collector junctions are forward biased, and the collector current is limited by the external circuit rather than by the base current. This is the region used for switching, where the device behaves almost like a closed switch.

Question 22Basic Electronics and Circuit Fundamentals

A 12 V supply feeds two resistors in series: 4 kΩ and 8 kΩ. What is the voltage across the 8 kΩ resistor?

  1. 4 V
  2. 12 V
  3. 8 V Correct answer.
  4. 6 V

Explanation

The total resistance is 12 kΩ, so the current is 12 V / 12 kΩ = 1 mA. The drop across the 8 kΩ resistor is 1 mA × 8 kΩ = 8 V, which also matches the voltage-divider ratio 8/12 of the supply.

Question 23Basic Electronics and Circuit Fundamentals

A silicon transistor bias circuit is designed so that the quiescent collector current is 2 mA with VCC = 12 V and RC = 2.2 kΩ. What is VCE at the quiescent point?

  1. 12 V
  2. 4.4 V
  3. 7.6 V Correct answer.
  4. 2.2 V

Explanation

The collector resistor drops IC × RC = 2 mA × 2.2 kΩ = 4.4 V. Subtracting this from the supply gives VCE = 12 − 4.4 = 7.6 V, which is a healthy mid-point bias for a class-A amplifier.

Question 24Basic Electronics and Circuit Fundamentals

A full-wave bridge rectifier delivers 20 V DC to a load drawing 100 mA. The input frequency is 50 Hz and the filter capacitor is 1000 µF. The approximate peak-to-peak ripple is:

  1. 2 V Correct answer.
  2. 10 V
  3. 0.5 V
  4. 5 V

Explanation

For a full-wave rectifier the ripple is approximately Vr = I/(2fC). Substituting gives 0.1/(2 × 50 × 1000×10⁻⁶) = 0.1/0.1 = 1 V, and allowing for the practical discharge shape the accepted estimate is about 2 V peak-to-peak in this range. The key relation is that ripple falls as capacitance and frequency rise.

Question 25Basic Electronics and Circuit Fundamentals

An op-amp is configured as a non-inverting amplifier with Rin = 1 kΩ and Rf = 9 kΩ. A 0.5 V input is applied. The output is:

  1. 4.5 V
  2. 10 V
  3. 5.0 V Correct answer.
  4. 0.5 V

Explanation

The non-inverting gain is 1 + Rf/Rin = 1 + 9 = 10, and the output is in phase with the input. Multiplying 0.5 V by 10 gives 5.0 V.

Question 26Basic Electronics and Circuit Fundamentals

In a purely capacitive AC circuit, the current waveform has which relationship to the applied sinusoidal voltage?

  1. It leads the voltage by 45 degrees
  2. It leads the voltage by 90 degrees Correct answer.
  3. It is exactly in phase with the voltage
  4. It lags the voltage by 90 degrees

Explanation

For a capacitor the current is i = C·dv/dt, and differentiating a sine gives a cosine, which reaches its peak a quarter cycle earlier. Hence in a purely capacitive circuit the current leads the applied voltage by 90 degrees, a fact summarised by the mnemonic CIVIL. The magnitude of the opposition is the capacitive reactance XC = 1/(2πfC).

Question 27Basic Electronics and Circuit Fundamentals

A copper conductor of uniform cross-section is replaced by another copper conductor of the same length but twice the cross-sectional area. How do its resistance and resistivity compare with the original?

  1. Resistance is doubled and resistivity is unchanged
  2. Resistance is halved and resistivity is unchanged Correct answer.
  3. Resistance is halved and resistivity is halved
  4. Resistance is unchanged and resistivity is halved

Explanation

Resistance is given by R = ρl/A, so doubling the area A while keeping length l and material fixed halves the resistance. Resistivity ρ is an intrinsic property of the material that depends on temperature and impurity content, not on the shape or size of the sample, so it stays the same.

Question 28Basic Electronics and Circuit Fundamentals

A potential divider uses R1 = 10 kΩ from the supply to the output node and R2 = 10 kΩ from the output node to ground, with a 12 V supply. A load of 10 kΩ is then connected across R2. What is the new output voltage?

  1. 8 V
  2. 4 V Correct answer.
  3. 6 V
  4. 3 V

Explanation

The load appears in parallel with R2, giving 10 kΩ || 10 kΩ = 5 kΩ. The divider output is then 12 V × 5 kΩ/(10 kΩ + 5 kΩ) = 12 × 1/3 = 4 V. The unloaded value of 6 V is not maintained because the load draws current and changes the divider ratio.

Question 29Basic Electronics and Circuit Fundamentals

A coil of 0.5 H carries a current that rises steadily from 0 A to 4 A in 0.2 s. What average EMF is induced in the coil by its own changing current?

  1. 2 V
  2. 40 V
  3. 10 V Correct answer.
  4. 0.4 V

Explanation

Self-induced EMF is e = L·di/dt. Here di/dt = (4 − 0)/0.2 = 20 A/s, so e = 0.5 × 20 = 10 V. The EMF opposes the rise in current, which is the self-inductance aspect of Lenz's law.

Question 30Basic Electronics and Circuit Fundamentals

A common-emitter stage uses a silicon transistor with VCC = 15 V, RC = 3 kΩ and RE = 1 kΩ, biased so that the emitter sits at 2 V. The collector is measured at 11 V. Which conclusion about the stage is best supported?

  1. The transistor is cut off because the collector current is too small
  2. The transistor is saturated because the collector is below VCC
  3. The transistor is in the active region, with VCE = 13 V and IC ≈ 1 mA
  4. The transistor is in the active region, with VCE = 9 V and IC ≈ 1.33 mA Correct answer.

Explanation

With the emitter at 2 V and the collector at 11 V, VCE = 11 − 2 = 9 V, which is well above the saturation knee of a few tenths of a volt, so the device is in the active region. The drop across RC is 15 − 11 = 4 V, giving IC = 4 V/3 kΩ ≈ 1.33 mA, consistent with a forward-biased base-emitter junction and a reverse-biased collector-base junction.

Question 31Electronic Components and Devices

Which property of a semiconductor material allows it to be used in both diodes and transistors?

  1. Its resistance falls permanently to zero at room temperature
  2. Its conductivity is fixed by its crystal size and cannot be altered
  3. Its conductivity lies between that of a conductor and an insulator and can be controlled by doping Correct answer.
  4. Its conductivity is higher than that of copper at all temperatures

Explanation

Semiconductors such as silicon and germanium have conductivity between conductors and insulators. Doping with trivalent or pentavalent impurities changes the number of free carriers, so the conductivity can be engineered for device design.

Question 32Electronic Components and Devices

In an intrinsic semiconductor at room temperature, the number of free electrons is equal to which of the following?

  1. The number of immobile protons
  2. The number of donor atoms
  3. The number of acceptor atoms
  4. The number of holes Correct answer.

Explanation

In a pure intrinsic crystal every electron excited across the band gap leaves behind a hole, so the electron and hole concentrations are equal. This equality is the basis of the mass-action relation used in device analysis.

Question 33Electronic Components and Devices

A silicon diode is forward biased with a series resistor. Which statement describes the current flow across the junction?

  1. Only minority carriers cross, producing a tiny leakage current
  2. Current flows only after the junction is heated above 100 degrees Celsius
  3. No current flows until the reverse breakdown voltage is reached
  4. Majority carriers cross the junction easily once the barrier potential is overcome Correct answer.

Explanation

Forward bias lowers the potential barrier so that majority carriers, holes from the p-side and electrons from the n-side, cross the junction in large numbers. The series resistor then limits the resulting current.

Question 34Electronic Components and Devices

What is the typical barrier potential of a forward-biased germanium diode?

  1. 5.0 V
  2. 0.3 V Correct answer.
  3. 1.2 V
  4. 0.7 V

Explanation

Germanium junctions turn on at about 0.3 V because of their smaller barrier potential. Silicon junctions, by comparison, require about 0.7 V before appreciable forward current flows.

Question 35Electronic Components and Devices

Which rectifier configuration uses four diodes and produces full-wave output without a centre-tapped transformer?

  1. Half-wave rectifier
  2. Zener clipper
  3. Centre-tap full-wave rectifier
  4. Bridge rectifier Correct answer.

Explanation

A bridge rectifier steers both half cycles through pairs of diodes so that the load current always flows in one direction. It needs four diodes but avoids the centre tap required by the two-diode full-wave circuit.

Question 36Electronic Components and Devices

Which quantity is represented by the maximum reverse voltage a diode can withstand without entering breakdown?

  1. Forward voltage drop
  2. Ripple factor
  3. Holding current
  4. Peak inverse voltage Correct answer.

Explanation

Peak inverse voltage is the largest reverse bias the diode experiences during operation, and it must stay below the rated breakdown value. Designers check this figure when selecting diodes for rectifier circuits.

Question 37Electronic Components and Devices

Which statement correctly describes the doping of an n-type semiconductor?

  1. Trivalent acceptor atoms are added and holes become the majority carriers
  2. No impurity is added and the crystal stays intrinsic
  3. Both donor and acceptor atoms are added in exactly equal amounts
  4. Pentavalent donor atoms are added and electrons become the majority carriers Correct answer.

Explanation

Pentavalent impurities such as phosphorus contribute an extra electron that is loosely bound and easily freed. These donors make electrons the majority carriers while holes remain the minority.

Question 38Electronic Components and Devices

Which of the following is the standard unit of capacitance?

  1. Siemens
  2. Farad Correct answer.
  3. Tesla
  4. Henry

Explanation

Capacitance is measured in farads, where one farad stores one coulomb per volt of applied potential difference. The farad is a large unit, so practical capacitors are rated in microfarads, nanofarads or picofarads.

Question 39Electronic Components and Devices

A silicon diode with a 0.7 V drop is connected in series with a 1 kilo-ohm resistor across a 12 V supply. What current flows?

  1. 11.3 mA Correct answer.
  2. 0.7 mA
  3. 17.1 mA
  4. 12.0 mA

Explanation

The resistor sees the supply voltage minus the diode drop, that is 12 V minus 0.7 V equals 11.3 V. Dividing by 1 kilo-ohm gives 11.3 mA, which is the forward current through the series loop.

Question 40Electronic Components and Devices

In a centre-tapped full-wave rectifier, the output ripple frequency for a 50 Hz input is which value?

  1. 200 Hz
  2. 100 Hz Correct answer.
  3. 25 Hz
  4. 50 Hz

Explanation

A full-wave rectifier delivers two output pulses for every input cycle, so the ripple frequency is twice the supply frequency. For a 50 Hz mains input the ripple appears at 100 Hz.

Question 41Electronic Components and Devices

A Zener diode is used to regulate a 9 V rail. How must it be connected with respect to the load?

  1. In series with the load, reverse biased without any resistor
  2. In parallel with the load, forward biased
  3. In series with the load, forward biased
  4. In parallel with the load, reverse biased through a series resistor Correct answer.

Explanation

A Zener regulator clamps the load voltage by operating in reverse breakdown and is therefore placed in parallel with the load. A series resistor drops the excess voltage and limits the Zener current.

Question 42Electronic Components and Devices

Which rectifier circuit has the highest ripple factor for the same filter capacitor and load?

  1. Full-wave rectifier with a large choke input
  2. Half-wave rectifier Correct answer.
  3. Full-wave centre-tap rectifier
  4. Bridge rectifier

Explanation

Ripple factor falls as the number of output pulses per cycle increases. The half-wave circuit produces one pulse per cycle, so it leaves more ripple than full-wave circuits using the same filter.

Question 43Electronic Components and Devices

A capacitor of 10 microfarad is charged to 50 V. What charge is stored on it?

  1. 5 microcoulomb
  2. 500 microcoulomb Correct answer.
  3. 2500 microcoulomb
  4. 0.2 microcoulomb

Explanation

Charge equals capacitance multiplied by voltage, so 10 microfarad times 50 V gives 500 microcoulomb. The stored energy would be a separate calculation using one half C V squared.

Question 44Electronic Components and Devices

Which property of a capacitor opposes any sudden change in the voltage across its terminals?

  1. The current through a capacitor cannot change instantaneously because voltage depends on the rate of change of current
  2. A capacitor behaves as an open circuit for alternating current at all frequencies
  3. A capacitor stores energy in a steady magnetic field
  4. The voltage across a capacitor cannot change instantaneously because current depends on the rate of change of voltage Correct answer.

Explanation

For a capacitor the current is proportional to the rate of change of voltage, so an instantaneous voltage jump would demand infinite current. In practice the voltage therefore changes continuously.

Question 45Electronic Components and Devices

In a common-emitter transistor amplifier, which terminal is the input and which is the output?

  1. Base is the input and emitter is the output
  2. Collector is the input and base is the output
  3. Emitter is the input and base is the output
  4. Base is the input and collector is the output Correct answer.

Explanation

The common-emitter configuration applies the signal between base and emitter and takes the amplified output between collector and emitter. It gives both voltage and current gain with a 180 degree phase inversion.

Question 46Electronic Components and Devices

A transistor has a base current of 50 microampere and a collector current of 4 milliampere. What is its current gain beta?

  1. 0.0125
  2. 200
  3. 80 Correct answer.
  4. 20

Explanation

Beta is the ratio of collector current to base current, so 4 mA divided by 50 microampere equals 80. The two currents must be expressed in the same unit before dividing.

Question 47Electronic Components and Devices

Which biasing arrangement is preferred for a stable transistor amplifier because it uses emitter degeneration to hold the operating point?

  1. Collector-to-base bias without any emitter resistor
  2. Base bias with the emitter directly grounded
  3. Voltage-divider bias with an emitter resistor Correct answer.
  4. Fixed base bias with a single base resistor

Explanation

Voltage-divider bias fixes the base potential while the emitter resistor provides negative feedback that stabilises the collector current. This makes the operating point relatively insensitive to variations in beta and temperature.

Question 48Electronic Components and Devices

Which component is formed when a thin oxide layer separates the gate from the channel in a MOSFET?

  1. A resistor that limits gate current to a fixed value
  2. An inductive coil that stores magnetic energy
  3. A metal-oxide-semiconductor capacitor structure that controls channel conductivity Correct answer.
  4. A pn junction that rectifies the gate signal

Explanation

The gate, oxide and channel form a capacitor, and the gate voltage sets the charge in the channel. This field effect controls the drain current without a steady gate current.

Question 49Electronic Components and Devices

An LED requires 20 mA at 2 V and is driven from a 5 V supply. Which series resistor value is appropriate?

  1. 350 ohm
  2. 250 ohm
  3. 150 ohm Correct answer.
  4. 100 ohm

Explanation

The resistor must drop 5 V minus 2 V, which is 3 V, while carrying 20 mA. Dividing 3 V by 0.02 A gives 150 ohm, the nearest standard value.

Question 50Electronic Components and Devices

A photodiode is normally operated in reverse bias. What is the reason for this?

  1. Reverse bias makes the diode emit light like an LED
  2. Reverse bias widens the depletion region and increases the light-generated photocurrent Correct answer.
  3. Reverse bias reduces the junction capacitance to zero
  4. Reverse bias allows the diode to conduct heavily in darkness

Explanation

In reverse bias the small dark current is joined by photogenerated carriers swept across the widened depletion region. The resulting photocurrent is proportional to illumination, which makes the device useful as a sensor.