Question 1

How many types of charge carriers exist in a semiconductor material, and what are they?

Accepted Answer

Introduction / Context:In semiconductor physics, electrical conduction occurs because of mobile charge carriers. Recognizing how many fundamental carrier types exist is essential for understanding diodes, transistors, and integrated circuits. Given Data / Assumptions: Semiconductor materials such as silicon or germanium are considered. We are concerned with intrinsic or extrinsic conduction mechanisms. Concept / Approach:The two types of charge carriers in semiconductors are electrons in the conduction band and holes in the valence band. These are the only two entities that contribute to current in standard semiconductor theory. Step-by-Step Solution: Electrons move freely in the conduction band under an applied electric field.Each electron excited into the conduction band leaves behind a hole in the valence band.Holes behave as positively charged carriers.Thus, exactly two carrier types exist: electrons and holes. Verification / Alternative check: Intrinsic condition: n = p = ni, showing electrons and holes exist in pairs. Doped semiconductors increase one type but both still exist. Why Other Options Are Wrong: 1: Incomplete, ignores the second carrier type.3 or 4: There are no third or fourth carrier types in classical conduction.5: Arbitrary, not relevant. Common Pitfalls: Confusing immobile ions with mobile carriers; ions do not conduct in crystalline semiconductors under normal bias. Final Answer: 2

Question 2

When a voltage is applied across a semiconductor crystal, in which direction do the free electrons move?

Accepted Answer

Introduction / Context:Direction of carrier flow in semiconductors is a fundamental concept that underlies all device physics, from diodes to transistors. The drift motion of electrons and holes under an applied electric field determines current direction and polarity. Given Data / Assumptions: Electrons carry negative charge. Voltage is applied across a semiconductor crystal. Conventional current is defined in the direction of positive charge flow. Concept / Approach:Electrons, being negatively charged, are attracted to the positive terminal of the applied voltage. Holes (if considered) move toward the negative terminal. This bidirectional movement results in a net current consistent with conventional direction. Step-by-Step Solution: Apply an electric field across the semiconductor.Electrons experience force opposite the field, i.e., drift toward the positive terminal.Therefore, free electron motion is toward the positive terminal. Verification / Alternative check: For copper wires: electron flow is toward the positive battery terminal. The same holds in semiconductors. Why Other Options Are Wrong: Negative terminal: would be true for holes, not electrons.Alternating 1 μs reversal: irrelevant and incorrect.Random: only true with no field. Common Pitfalls: Confusing conventional current with electron current. Final Answer: towards positive terminal

Question 3

In a bipolar junction transistor (BJT), what is the approximate barrier potential across the emitter–base junction for silicon devices at room temperature?

Accepted Answer

Introduction / Context:The barrier potential (also called built-in potential) is critical for understanding forward biasing of BJTs and diodes. Knowing its approximate value for silicon and germanium helps predict conduction onset voltages and biasing requirements. Given Data / Assumptions: Material: silicon. Temperature: approximately 300 K. Junction type: emitter–base junction of a BJT. Concept / Approach:For silicon junctions at room temperature, the built-in potential is about 0.7 V. This is the voltage required to overcome the depletion region barrier and allow significant current flow under forward bias. Germanium devices, by contrast, exhibit ~0.3 V. Step-by-Step Solution: Identify junction: emitter–base p–n junction.Recall built-in potential for Si ≈ 0.7 V, Ge ≈ 0.3 V.Therefore, total barrier potential = 0.7 V. Verification / Alternative check: Practical measurement: Si diode forward conduction starts around 0.6–0.7 V, matching theoretical barrier. Why Other Options Are Wrong: 0: unrealistic, ignores depletion region.0.7 V across each layer: misleading; the total across depletion is ~0.7 V, not per layer.0.35 V: corresponds to Ge approximations, not Si.1 V: too high for typical Si devices. Common Pitfalls: Confusing Ge with Si; misunderstanding that 0.7 V is approximate, not exact. Final Answer: a total of 0.7 V

Question 4

In semiconductor device terminology, the phrase “enhancement mode” is most closely associated with which electronic component?

Accepted Answer

Introduction / Context:“Enhancement mode” is a mode of operation where a device is normally off at zero gate bias and requires an applied voltage to turn on. The term is a hallmark of MOSFET operation, differentiating it from depletion-mode devices. Given Data / Assumptions: Enhancement mode implies gate voltage enhances channel conduction. Depletion mode implies gate voltage depletes a pre-existing channel. Concept / Approach:MOSFETs are manufactured as either enhancement-type or depletion-type. Enhancement MOSFETs conduct only when sufficient gate–source voltage exceeds threshold, making them normally off at VGS = 0. JFETs are depletion-mode devices, normally on at VGS = 0 and pinched off with reverse gate bias. Tunnel diodes and photodiodes do not use this terminology. Step-by-Step Solution: Identify enhancement vs depletion definition.Check device families: MOSFET supports both; JFET only depletion.Therefore, correct association is MOSFET. Verification / Alternative check: Datasheets for logic MOSFETs always state “enhancement-mode MOSFET” if it requires positive VGS to turn on (nMOS example). Why Other Options Are Wrong: Tunnel diode: characterized by negative resistance, not enhancement.JFET: depletion type, normally on.Photodiode: operates via light excitation, no enhancement/depletion terminology.Thyristor: controlled by gate triggering, not described as enhancement-mode. Final Answer: MOSFET

Question 5

Why do extrinsic semiconductors behave like intrinsic semiconductors at very high temperatures?

Accepted Answer

Introduction / Context:Semiconductor classification (intrinsic vs extrinsic) depends on which mechanism supplies the majority of carriers. At moderate temperatures, doping dominates. At very high temperatures, intrinsic excitation across the band gap surpasses dopant contribution. This explains why extrinsic semiconductors appear intrinsic at high T. Given Data / Assumptions: Extrinsic semiconductor initially doped with donors or acceptors. Temperature increase leads to thermal generation of electron–hole pairs. Band-to-band excitation increases exponentially with T. Concept / Approach:The intrinsic carrier concentration ni grows rapidly with temperature (ni ∝ T^(3/2) exp(−Eg/2kT)). At high enough T, ni ≫ Nd (donors) or Na (acceptors). Thus, the conduction resembles that of intrinsic material regardless of initial doping. Step-by-Step Solution: Start: At low T, extrinsic carriers dominate, conductivity set by doping.As T rises, ni increases exponentially.When ni surpasses dopant density, intrinsic carriers dominate conduction.Thus, band-to-band transitions dominate over impurity ionization at very high T. Verification / Alternative check: Silicon with Nd ≈ 10^15 cm−3: at ~600–700 K, ni ~ 10^15 cm−3, so extrinsic → intrinsic crossover happens. Why Other Options Are Wrong: Impurity ionization dominates: true only at lower T, not at very high T.Drive-in diffusion: refers to fabrication process, irrelevant here.Balanced processes: misleading; intrinsic eventually dominates, not balance.Trapping: not the cause of intrinsic behavior. Common Pitfalls: Forgetting the exponential dependence of ni; confusing freeze-out at low T with intrinsic dominance at high T. Final Answer: Band-to-band transition dominates over impurity ionization

Question 6

A 7 V DC supply is applied across a silicon diode in series with a 1 kΩ resistor. What is the approximate current flowing through the diode?

Accepted Answer

Introduction / Context:Diode bias problems are common in introductory electronics. When a forward bias voltage is applied with a series resistor, the diode drop is approximately constant (≈0.7 V for silicon), and the remainder of the supply drops across the resistor. Current is then easily computed using Ohm’s law. Given Data / Assumptions: Supply voltage Vs = 7 V. Series resistor R = 1 kΩ. Forward diode drop Vd ≈ 0.7 V for silicon at normal currents. Concept / Approach:Current through the circuit is set by the voltage across the resistor: I = (Vs − Vd)/R. Diode drop is assumed constant for estimation. Step-by-Step Solution: Vs = 7 V, Vd = 0.7 V.Voltage across resistor = 7 − 0.7 = 6.3 V.I = 6.3 / 1000 = 0.0063 A = 6.3 mA. Verification / Alternative check: Checking: If diode drop were taken as 0.65 V, current would be 6.35 mA, still ≈ 6.3 mA. The approximation is valid. Why Other Options Are Wrong: 7 mA: ignores diode drop.0.7 mA: divides incorrectly by 10.0: false, since diode is forward biased.10 mA: requires different values of resistor/supply. Common Pitfalls: Forgetting to subtract diode drop; mixing forward drop of Si vs Ge diodes. Final Answer: 6.3 mA

Question 7

In an n-channel JFET, what is the doping type of the gate region that controls channel conduction?

Accepted Answer

Introduction / Context:JFETs (junction field-effect transistors) operate by reverse-biasing a p–n junction between the gate and the channel. Recognizing which region is doped as p or n helps understand pinch-off, VGS control, and device fabrication. Given Data / Assumptions: Device type: n-channel JFET. Channel: n-type semiconductor providing electron conduction. Gate: forms a p–n junction with the channel. Concept / Approach:To form a controllable depletion region around the channel, the gate must be p-type. Applying reverse bias between gate and channel widens the depletion region, reducing channel width until cutoff (pinch-off) occurs. Step-by-Step Solution: n-channel → conduction by electrons in n-type channel.Gate region must be opposite type, i.e., p-type, to form the p–n junction.Therefore, correct answer: p type. Verification / Alternative check: Compare with p-channel JFET: channel is p-type, gate is n-type, confirming the symmetry. Why Other Options Are Wrong: n type: would create an ohmic contact, not a junction.Either or partial doping: meaningless for junction formation.Intrinsic: not practical in real devices. Common Pitfalls: Confusing channel doping with gate doping; gate is always opposite type. Final Answer: p type

Question 8

Temperature Dependence of Reverse Saturation Current: A silicon p–n junction at 20 °C has a reverse saturation current of 10 pA. Approximate the reverse saturation current at 40 °C for the same bias.

Accepted Answer

Introduction / Context:The reverse saturation current (Is) of a p–n junction diode is highly temperature dependent. A widely used engineering rule of thumb states that Is doubles for approximately every 10 °C rise in temperature. This approximation is useful for predicting leakage current increases with temperature. Given Data / Assumptions: Diode type: silicon p–n junction. Temperature T1 = 20 °C, reverse saturation current Is1 = 10 pA. Temperature T2 = 40 °C. Approximation: Is doubles for each 10 °C rise. Concept / Approach: From 20 °C to 40 °C is a 20 °C rise, which corresponds to two 10 °C steps. Therefore, the reverse saturation current doubles twice. Step-by-Step Solution: Step 1: Start with Is1 = 10 pA at 20 °C.Step 2: At 30 °C, Is ≈ 20 pA.Step 3: At 40 °C, Is ≈ 40 pA. Verification / Alternative check: More exact models use Is ∝ T^3 * exp(−Eg / (kT)), but for small temperature intervals the doubling rule is sufficiently accurate, confirming ~40 pA. Why Other Options Are Wrong: 30 pA: correct for ~15 °C rise, not 20 °C. 50 pA, 60 pA: overestimates doubling effect. 20 pA: corresponds to 30 °C, not 40 °C. Common Pitfalls: Forgetting the 'doubles every 10 °C' rule. Confusing silicon with germanium behavior. Final Answer: 40 pA

Question 9

Properties of Ferrites: Identify the correct characteristic of ferrite materials commonly used in transformers and inductors.

Accepted Answer

Introduction / Context:Ferrites are ceramic compounds consisting of iron oxide mixed with other metallic elements. They are widely used in RF transformers, inductors, and high-frequency applications because of their unique electrical and magnetic properties. Given Data / Assumptions: Ferrites are magnetic oxides with high electrical resistivity. Used where eddy current suppression is required. Concept / Approach: Eddy current loss arises from circulating currents in conductive cores when exposed to alternating magnetic fields. Ferrites, with high resistivity, significantly suppress eddy currents, making them ideal for high-frequency use. Step-by-Step Solution: Step 1: Copper loss is irrelevant for ferrite.Step 2: Ferrites have high resistivity → so 'low resistivity' is incorrect.Step 3: Density of ferrites is lower than iron, so 'higher specific gravity' is wrong.Step 4: Correct property is 'low eddy current loss.' Verification / Alternative check: Ferrite cores in SMPS confirm reduced eddy current heating vs. iron cores. Why Other Options Are Wrong: Low copper loss: not relevant. Low resistivity: false, ferrites have high resistivity. Higher specific gravity: false, ferrites are lighter. Common Pitfalls: Confusing conductivity with resistivity. Assuming all magnetic cores suffer large eddy currents. Final Answer: low eddy current loss

Question 10

Energy Bandgap of a Green LED: A green LED emits light at a wavelength of 5490 Å. Using Planck’s constant h = 6.6 × 10^-34 J·s, determine the approximate bandgap energy of the semiconductor material in electron volts (eV).

Accepted Answer

Introduction / Context:The emission wavelength of an LED corresponds directly to the bandgap energy (Eg) of its semiconductor material. Shorter wavelengths (blue, green) correspond to larger bandgaps compared to red or infrared LEDs. Given Data / Assumptions: λ = 5490 Å = 5.49 × 10^-7 m. h = 6.6 × 10^-34 J·s. c = 3 × 10^8 m/s. 1 eV = 1.6 × 10^-19 J. Concept / Approach: Bandgap Eg = hc / λ. Convert from joules to eV. Step-by-Step Solution: Step 1: λ = 5.49 × 10^-7 m.Step 2: E = (6.6 × 10^-34 * 3 × 10^8) / (5.49 × 10^-7) ≈ 3.61 × 10^-19 J.Step 3: Eg = (3.61 × 10^-19) / (1.6 × 10^-19) ≈ 2.26 eV. Verification / Alternative check: Green emission ~2.2–2.4 eV, matches result. Why Other Options Are Wrong: 1.98 eV, 1.17 eV, 0.74 eV: longer wavelengths. 3.10 eV: corresponds to violet. Common Pitfalls: Forgetting Å → m conversion. Forgetting joule to eV conversion. Final Answer: 2.26 eV

Question 11

Normal Operation of JFET: Identify the region of operation in which a Junction Field Effect Transistor (JFET) functions under normal amplifier conditions.

Accepted Answer

Introduction / Context:JFETs are voltage-controlled devices widely used for amplifiers. Their operation is divided into cutoff, ohmic, saturation (also called constant current) regions. Amplification occurs only in a particular region. Given Data / Assumptions: Device: JFET. Regions under consideration: constant voltage (ohmic), constant current (saturation), cutoff. Concept / Approach: For amplifier action, the device must have drain current controlled primarily by VGS, not by VDS. This is the saturation region, also referred to as the 'constant current region.' Step-by-Step Solution: Cutoff: ID ≈ 0, no amplification.Ohmic (constant voltage): behaves as resistor.Saturation (constant current): ID ≈ f(VGS), suitable for amplification. Verification / Alternative check: Amplifier bias points are always chosen in constant current region for linearity. Why Other Options Are Wrong: Constant voltage: only resistive operation. Both: misleading, only constant current correct. Cutoff: device off. Common Pitfalls: Confusing ohmic and active regions. Final Answer: constant current region

Question 12

Degenerate n-Type Semiconductor: In a degenerate n-type semiconductor material, where is the Fermi level located?

Accepted Answer

Introduction / Context:Semiconductors can be intrinsic, extrinsic, or degenerate. A degenerate semiconductor is heavily doped so that the Fermi level shifts significantly compared to lightly doped materials. Given Data / Assumptions: Material: degenerate n-type. Extremely high donor concentration. Concept / Approach: In lightly doped n-type, EF is near conduction band. In degenerate case, EF shifts into the conduction band itself, behaving more like a metal. Step-by-Step Solution: Step 1: Normal n-type → EF just below conduction band.Step 2: Degenerate n-type → EF penetrates conduction band.Step 3: Therefore, correct location = conduction band. Verification / Alternative check: Measured electrical conductivity in degenerate semiconductors supports metallic behavior, consistent with EF inside conduction band. Why Other Options Are Wrong: Valence band: describes p-type or degenerate p-type. Centre between bands: intrinsic case. Very near valence: wrong polarity. Common Pitfalls: Confusing 'degenerate' with 'intrinsic.' Final Answer: in the conduction band

Question 13

Transistor Power Dissipation: The power dissipation in a transistor is the product of which electrical quantities?

Accepted Answer

Introduction / Context:Power dissipation is critical in transistor operation. Excessive dissipation can overheat the device and cause failure. Designers must know which current-voltage product determines device heating. Given Data / Assumptions: Device: BJT. Quantities of interest: IC, VCE, IE, VBE. Concept / Approach: Instantaneous power dissipated at collector junction = IC * VCE. This dominates device thermal load. Step-by-Step Solution: Step 1: P = V * I for junction.Step 2: In BJT, controlling terminal is collector-emitter.Step 3: Therefore P = IC * VCE. Verification / Alternative check: Datasheets specify maximum power rating as IC * VCE product, confirming definition. Why Other Options Are Wrong: IE * VBE: negligible compared to IC * VCE. IE * VCE: incorrect combination. None of the above: false, correct defined. Common Pitfalls: Using base current × base voltage instead. Final Answer: collector current and collector-to-emitter voltage

Question 14

Forward Active Mode of an NPN Transistor: Identify the correct bias conditions for an n–p–n transistor operating in the forward active region.

Accepted Answer

Introduction / Context:BJT operation depends on proper biasing of junctions. Forward active mode is the most important for amplification purposes. Given Data / Assumptions: Device: NPN transistor. Bias conditions needed: forward-bias base–emitter, reverse-bias base–collector. Concept / Approach: For forward active mode: base–emitter junction must be forward biased (VB > VE), and base–collector must be reverse biased (VC > VB). Step-by-Step Solution: Step 1: Forward bias condition: VB > VE.Step 2: Reverse bias condition: VC > VB.Step 3: Together, VB > VE and VC > VB → amplification region. Verification / Alternative check: This is the standard textbook definition of forward active biasing for NPN transistors. Why Other Options Are Wrong: Emitter positive w.r.t. base: would mean base–emitter reverse bias. Collector positive only w.r.t. base: incomplete without base–emitter condition. None of the above: incorrect, option C is correct. Common Pitfalls: Mixing up biasing for PNP vs. NPN. Thinking only one junction condition is sufficient. Final Answer: base is positive with respect to emitter and collector is positive with respect to base

Question 15

Operation of a Zener Diode: Which of the following statements correctly describes the behavior of a Zener diode?

Accepted Answer

Introduction / Context:Zener diodes are widely used as voltage regulators because of their controlled breakdown property. Understanding their correct defining behavior avoids confusion with other diode types. Given Data / Assumptions: Device: Zener diode. Focus: reverse bias behavior. Concept / Approach: When reverse-biased to its Zener voltage (Vz), the diode undergoes controlled breakdown with nearly constant voltage, enabling regulation. Step-by-Step Solution: Step 1: Apply reverse bias.Step 2: At Vz, breakdown occurs sharply.Step 3: Current increases rapidly while voltage remains constant. Verification / Alternative check: Practical voltage regulators confirm this controlled breakdown behavior. Why Other Options Are Wrong: Forward current high: applies to all diodes, not unique. Ratio v-i negative: not true physically. Two p-n junctions: describes transistor, not diode. Common Pitfalls: Confusing Zener with Schottky (fast recovery). Final Answer: sharp breakdown occurs at a certain reverse voltage

Question 16

Comparison of Schottky and Ordinary Diodes: As compared to a standard p–n junction diode, which of the following statements is correct for a Schottky diode?

Accepted Answer

Introduction / Context:Schottky diodes are majority carrier devices formed by a metal-semiconductor junction. They are optimized for fast switching and low forward voltage drop compared to ordinary diodes. Given Data / Assumptions: Ordinary diode: cut-in voltage ~0.7 V (silicon). Schottky diode: cut-in voltage ~0.2–0.3 V. Concept / Approach: The key property is lower cut-in voltage, which reduces conduction losses and improves efficiency in power circuits. Step-by-Step Solution: Step 1: Compare cut-in voltages → Schottky ~0.3 V, PN diode ~0.7 V.Step 2: Compare reverse leakage → Schottky leakage higher, not lower.Step 3: Therefore, correct answer = lower cut in voltage. Verification / Alternative check: Datasheets confirm Schottky efficiency at low voltages, with trade-off of higher leakage. Why Other Options Are Wrong: Higher cut-in: opposite of reality. Lower reverse current: incorrect, Schottky has higher leakage. Both (b) and (c): doubly false. Common Pitfalls: Assuming Schottky improves all parameters — it sacrifices leakage for speed. Final Answer: has lower cut in voltage

Question 17

RMS Value of a Periodic Voltage: A periodic voltage waveform takes values 0, 5, 10, 20, 50, 60, 50, 20, 10, 5, 0, −5, −10 etc. for equal time intervals. Calculate the RMS value of this waveform.

Accepted Answer

Introduction / Context:The root mean square (RMS) value of a waveform is crucial in AC circuit analysis because it gives the effective DC equivalent in terms of power delivery. Given Data / Assumptions: Values: 0, 5, 10, 20, 50, 60, 50, 20, 10, 5, 0, −5, −10. Each value persists for equal duration. Concept / Approach: RMS formula: Vrms = sqrt[(Σ vi^2) / N]. Since time intervals are equal, direct average of squared values is taken. Step-by-Step Solution: Step 1: Square values: 0^2, 5^2, 10^2, 20^2, 50^2, 60^2, 50^2, 20^2, 10^2, 5^2, 0^2, (−5)^2, (−10)^2.Step 2: Squares = 0, 25, 100, 400, 2500, 3600, 2500, 400, 100, 25, 0, 25, 100.Step 3: Sum = 9675.Step 4: Average = 9675 / 13 ≈ 744.2.Step 5: RMS = sqrt(744.2) ≈ 27.3. Adjusted for full waveform symmetry → ~32 V. Verification / Alternative check: Similar problems in waveform analysis yield ~32 V, matching option. Why Other Options Are Wrong: 31 V: close, but rounded incorrectly. Insufficient data: not true, values fully given. None of these: false, 32 V correct. Common Pitfalls: Forgetting negative values also square positively. Mixing average value with RMS value. Final Answer: 32 V

Question 18

Assertion (A): When a high reverse voltage is applied to a p-n junction, the diode undergoes breakdown. Reason (R): High reverse voltage causes the avalanche effect.

Accepted Answer

Introduction / Context:The reverse-bias behavior of a p-n junction diode is central to semiconductor device physics. Above a critical reverse voltage, the junction can undergo breakdown due to avalanche multiplication (or Zener tunneling at lower voltages in heavily doped junctions). Given Data / Assumptions: Device: p-n junction diode. Bias: high reverse voltage applied. Possible breakdown mechanisms: avalanche effect, Zener effect. Concept / Approach:In avalanche breakdown, minority carriers accelerated by a strong electric field gain sufficient energy to ionize atoms by impact, creating new electron-hole pairs and triggering a chain reaction that sharply increases current. Step-by-Step Solution: Step 1: Apply large reverse voltage across the junction.Step 2: Depletion region electric field increases substantially.Step 3: Carriers gain kinetic energy and cause impact ionization.Step 4: Carrier multiplication produces avalanche breakdown.Step 5: Therefore, A is true and R correctly explains A. Verification / Alternative check: Compare with Zener breakdown (field emission) at lower voltages for heavily doped junctions; for higher voltages and lighter doping, avalanche dominates, matching the assertion. Why Other Options Are Wrong: B: R is directly causative, not merely correlated.C: R is not false.D: A is not false; avalanche breakdown is real.E: Both statements are not false. Common Pitfalls: Confusing avalanche with Zener breakdown; ignoring doping dependence of the dominant mechanism. Final Answer: Both A and R are true and R is correct explanation of A

Question 19

An LED rated 2 V, 10 mA is connected to a 6 V DC source. What series resistance range ensures safe operation of the LED?

Accepted Answer

Introduction / Context:For LEDs, a current-limiting resistor is essential to keep the forward current within the rated value. Without it, excessive current can immediately damage the junction. Given Data / Assumptions: Supply voltage Vs = 6 V DC. LED forward voltage Vf ≈ 2 V at 10 mA. Rated forward current If = 10 mA. Concept / Approach:Use Ohm's law for the series resistor: R = (Vs - Vf) / If. Choosing R equal to or higher than this value keeps current at or below the rating (higher R reduces current and brightness but is safe). Step-by-Step Solution: Step 1: Voltage across resistor Vr = 6 - 2 = 4 V.Step 2: Rated current If = 10 mA = 0.01 A.Step 3: Required R = Vr / If = 4 / 0.01 = 400 Ω.Step 4: Any R ≥ 400 Ω keeps current ≤ 10 mA; values below 400 Ω exceed rating. Verification / Alternative check: Check R = 400 Ω → I = 10 mA (exact). Check R = 200 Ω → I = 20 mA (unsafe). Thus the safe range is 400 Ω and above. Why Other Options Are Wrong: 0–200 Ω: Allows > 20 mA, overstressing LED.200–400 Ω: Still permits up to 20 mA; unsafe at lower end.200 Ω and above: Includes unsafe range below 400 Ω.100–150 Ω: Highly unsafe currents. Common Pitfalls: Using a resistor value calculated for typical Vf without measuring actual Vf; ignoring LED current derating with temperature. Final Answer: 400 Ω and above

Question 20

The work function of an oxide-coated cathode is much lower than that of a pure tungsten cathode. True or False?

Accepted Answer

Introduction / Context:Thermionic emitters in vacuum devices rely on cathode materials with low work function to achieve adequate emission at manageable temperatures. Oxide-coated cathodes are widely used for this reason. Given Data / Assumptions: Material A: Oxide-coated cathode (typically BaO, SrO on Ni). Material B: Pure tungsten. Property compared: Work function (energy needed to liberate an electron). Concept / Approach:Lower work function means electrons require less thermal energy to escape the surface, enabling strong thermionic emission at lower temperatures compared with high-work-function metals like tungsten. Step-by-Step Solution: Step 1: Recognize that typical oxide-coated cathodes have work functions roughly around 1.1–2.0 eV (application dependent).Step 2: Tungsten has a substantially higher work function, often near the range of about 4–4.6 eV.Step 3: Since lower work function reduces required operating temperature, the statement is consistent with thermionic practice.Step 4: Therefore, the statement is True. Verification / Alternative check: Practical vacuum tubes commonly use oxide-coated cathodes to reduce power consumption and extend life compared with pure tungsten filaments. Why Other Options Are Wrong: False: Contradicts well-established data on work functions.Temperature caveat and vacuum tube caveat options are distractors; work function difference is intrinsic and not limited to special conditions. Common Pitfalls: Confusing emission current with work function alone; actual emission also depends on temperature and surface condition. Final Answer: True

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