Q: Half-wave controlled charger for a 50 V battery A single-phase half-wave converter is used to charge a 50 V battery. The thyristor gate is driven continuously (DC gate drive). The input is v(t) = Vm sin(ωt). Over what interval(s) does the thyristor …

Introduction / Context: Battery charging with a phase-controlled rectifier depends on the source instantaneous voltage exceeding the battery EMF (plus small drops). A continuously applied gate drive merely ensures the SCR will trigger whenever forward anode-to-cathode conditions exist; it does not force conduction when the source cannot forward-bias the SCR against the battery voltage. Given Data / Assumptions: Single SCR half-wave converter. Battery emf ≈ 50 V (assume negligible series resistance for the decision threshold). Continuous DC gate drive available at all times. Input v(t) = Vm sin(ωt) on the same polarity half-cycle as the SCR. Concept / Approach: For current to flow into the battery, the instantaneous supply voltage must exceed the battery voltage. With continuous gating, the SCR will turn on immediately once v(t) exceeds 50 V in the forward half-cycle, and will continue to conduct until v(t) falls back below 50 V or current naturally goes to zero near the end of the half-cycle. Step-by-Step Solution: Condition to conduct: v(t) > Vbattery ≈ 50 V.Turn-on occurs at the instant ωt = arcsin(50 / Vm), provided 50 ≤ Vm.Conduction ceases as v(t) falls below 50 V on the trailing part of the half-cycle. Verification / Alternative check: If Vm < 50 V, no conduction occurs (even with continuous gate), validating that the battery EMF sets the threshold. If Vm ≫ 50 V, conduction occupies a central portion of the positive half-cycle only. Why Other Options Are Wrong: 0–180° suggests full positive half-cycle conduction, which is false unless Vm is infinite and the drop negligible. 0–90° arbitrarily limits conduction; actual turn-off occurs after the peak when v drops below 50 V. “When v 50 V

Q: Step-down chopper (buck chopper) average output voltage A step-down (buck) chopper is fed from a DC source of magnitude V. If the switch on-time and off-time per cycle are Ton and Toff respectively, identify what V * (Ton / (Ton + Toff)) represents.

Introduction / Context: DC–DC step-down (buck) choppers reduce the average output voltage by switching a DC source on and off with duty ratio control. Understanding the relationship between duty ratio and average output is fundamental for power electronics design and control. Given Data / Assumptions: Input DC voltage = V (constant). Switch on-time = Ton; off-time = Toff; switching period T = Ton + Toff. Ideal switch and diode; continuous conduction assumed for the average relation. Concept / Approach: The average of a periodic waveform equals the area under one period divided by the period. The chopper applies V to the load during Ton and 0 V during Toff (ignoring the inductor freewheel path for the idealized average calculation), hence the average equals V multiplied by the duty ratio D = Ton/T. This relation is independent of the specific R, L, or E in the load for the average value, provided steady-state averaging is valid. Step-by-Step Solution: Duty ratio D = Ton / (Ton + Toff).Output voltage during on-state ≈ V; during off-state ≈ 0 (ideal).Average V_out(avg) = V * D = V * (Ton / (Ton + Toff)). Verification / Alternative check: Inductor volt-second balance in continuous conduction also yields V_out = D * V for a buck converter. Why Other Options Are Wrong: (b) RMS requires squaring and is not simply V * D. (c) The expression is the average regardless of load type under standard buck operation. (d) Not applicable since it clearly represents the average. Common Pitfalls: Confusing average with RMS; forgetting that ripple filters do not change the average relation V_out = D * V. Final Answer: Average output voltage

Q: SCR internal connection statement check Consider the construction of a silicon-controlled rectifier (SCR). “In a thyristor the gate lead is connected to the n-layer near the p-layer to which the anode is connected.” Is this statement true or false?

Introduction / Context: Understanding the internal layer structure of an SCR (PNPN device) is essential for appreciating how gating works and why the device exhibits latching behavior. The gate connection location determines how minority carrier injection assists turn-on. Given Data / Assumptions: Standard SCR layer order from anode to cathode: P1–N1–P2–N2. Anode connected to outer P1; cathode connected to outer N2. Gate terminal is taken from the inner P2 region (also called p-base) near the cathode junction. Concept / Approach: The gate must effectively modulate the junction nearest the cathode (J2) to initiate regenerative action in the two-transistor equivalent. Hence, the gate is bonded to the inner p-base (P2) region, not to an n-layer near the anode. This allows a small gate current to inject carriers that forward-bias the internal transistor pair and trigger conduction. Step-by-Step Solution: Identify the 4-layer structure: P1 (anode), N1, P2 (gate-connected), N2 (cathode).Gate metallization contacts P2 close to the cathode junction to be effective.Therefore the statement claiming a gate connection to an n-layer near the anode is incorrect. Verification / Alternative check: The two-transistor model (PNP–NPN) shows that injecting into the P2 base region facilitates rapid turn-on via positive feedback. Why Other Options Are Wrong: Marking the statement as true contradicts SCR construction; there is no standard variant where the gate connects to an n-layer near the anode. Common Pitfalls: Confusing the gate-bond region (P2) with the outer P1 anode region or with N-type layers; misreading cross-sectional diagrams. Final Answer: False

Q: Device exhibiting peak and valley points on its V–I characteristic Which device’s characteristic shows a distinct peak point followed by a valley point (negative-resistance region)?

Introduction / Context: Some two-terminal or three-terminal devices exhibit a negative-resistance region where current rises as voltage drops. This region is characterized by a peak point (onset) and a valley point (near the low-voltage high-current state). Recognizing which devices show such behavior is key in triggering and relaxation-oscillator design. Given Data / Assumptions: Comparing SCR, PUT, SCS, and Shockley diode. Focus on the presence of peak and valley points in the emitter/terminal characteristic. Concept / Approach: The classical device known for a peak and valley in its characteristic is the Unijunction Transistor (UJT). A PUT (Programmable UJT) is a PNPN device configured to emulate UJT-like behavior and also exhibits a programmable peak and valley point. The SCR and SCS have breakover and holding characteristics but are not generally described by peak/valley nomenclature in their common terminal curves. Step-by-Step Solution: Identify device with negative-resistance emitter curve: UJT-like behavior → PUT.PUT’s peak current occurs at the programmed peak voltage; thereafter the characteristic drops to a valley region.Hence, the requested device is PUT. Verification / Alternative check: PUT data sheets and application notes (relaxation oscillators) explicitly specify peak current/voltage and valley current/voltage parameters. Why Other Options Are Wrong: SCR/SCS are four-layer switches with latching and holding currents; Shockley diode shows breakover rather than clearly defined peak and valley points used in UJT/PUT terminology. Common Pitfalls: Equating breakover and holding points of an SCR with the peak/valley description used for UJT/PUT emitter characteristics. Final Answer: PUT (Programmable Unijunction Transistor)

Question 1

Single-phase half-wave AC voltage regulator (controller) Which device combination is typically used to realize a single-phase half-wave regulator for a resistive load?

Accepted Answer

Introduction / Context: AC voltage regulators adjust the RMS voltage supplied to a load by phase-controlling conduction. In the single-phase half-wave variant, only one polarity half-cycle is controlled, while the opposite polarity is either blocked or allowed by a passive device to complete the circuit as designed. Given Data / Assumptions: Single-phase supply and a resistive load. Half-wave regulation implies one controlled half-cycle. Low/medium power implementation. Concept / Approach: A common half-wave regulator uses one thyristor to control conduction during one half-cycle and one diode to provide the complementary path during the other half-cycle (when desired). This achieves half-wave control with a minimal parts count and provides a predictable waveform for the load. Step-by-Step Solution: Choose a thyristor for controlled conduction on, say, the positive half-cycle.Use a diode for the negative half-cycle path or to block it according to the topology.Hence, the typical device pair is 1 diode + 1 thyristor. Verification / Alternative check: Standard educational schematics of half-wave AC regulators show a single SCR in anti-parallel with a diode for basic control. Why Other Options Are Wrong: Two thyristors are used for full-wave control (antiparallel), not half-wave. Two diodes provide rectification, not controlled regulation. “2 diodes and 2 thyristors” corresponds to more complex controllers. Common Pitfalls: Confusing half-wave and full-wave AC regulators; mixing rectifier topologies with regulator topologies. Final Answer: 1 diode and 1 thyristor

Question 2

Junction temperature calculation (SCR) A thyristor dissipates 40 W and operates in 20°C ambient. If the overall junction-to-ambient thermal resistance is 1.6°C/W, estimate the junction temperature (°C).

Accepted Answer

Introduction / Context: Accurate junction temperature estimates are essential to ensure device reliability. The thermal resistance model provides a straightforward way to compute Tj from ambient temperature, power, and Rth,ja. Given Data / Assumptions: Ambient temperature Ta = 20°C. Power dissipation P = 40 W. Overall thermal resistance Rth,ja = 1.6°C/W. Steady-state operation; linear thermal model applies. Concept / Approach: Temperature rise above ambient equals power times thermal resistance. Then junction temperature Tj = Ta + P * Rth,ja. Step-by-Step Solution: ΔT = P * Rth,ja = 40 * 1.6 = 64°CTj = Ta + ΔT = 20 + 64 = 84°C Verification / Alternative check: Confirm that Tj is below device maximum (commonly 110–125°C for many SCRs). At 84°C, the device is operating within typical safe limits. Why Other Options Are Wrong: 94°C or 114°C would imply larger Rth or power than given. 164°C exceeds typical device limits and contradicts the calculation. Common Pitfalls: Omitting interface thermal resistances in practice can raise Tj. Measurement should be cross-checked with thermocouples or IR methods during prototyping. Final Answer: 84° C

Question 3

Three-phase semi-converter (average DC voltage) For a certain firing angle in a 3-phase semi-converter, the average DC output voltage equals 50% of the maximum possible value. What is the firing angle α?

Accepted Answer

Introduction / Context: Average DC output voltage in phase-controlled converters depends on firing angle. The three-phase semi-converter (half-controlled bridge) has a characteristic equation involving (1 + cos α). Knowing this helps set α for a desired fraction of maximum DC output. Given Data / Assumptions: Topology: 3-phase semi-converter. Vdc,max occurs at α = 0°. Required Vdc is 50% of Vdc,max. Concept / Approach: For a 3-phase semi-converter, Vdc = (3Vm / 2π) * (1 + cos α). At α = 0°, Vdc,max = (3Vm / π). Setting Vdc to half of Vdc,max gives a simple trigonometric relationship for α. Step-by-Step Solution: Vdc / Vdc,max = [(3Vm / 2π)(1 + cos α)] / (3Vm / π) = (1 + cos α) / 2Set (1 + cos α) / 2 = 1/2 → 1 + cos α = 1 → cos α = 0α = 90° Verification / Alternative check: At α = 90°, the converter’s DC output is exactly half of its maximum, consistent with the formula and standard curves. Why Other Options Are Wrong: 30° or 60° yield cos α > 0, giving more than 50% of maximum. 120° yields cos α < 0, giving less than 50% of maximum. Common Pitfalls: Using the full-converter equation Vdc ∝ cos α (omitting the 1 + term) leads to incorrect α. Always match formula to topology. Final Answer: 90°

Question 4

Half-wave controlled charger for a 50 V battery A single-phase half-wave converter is used to charge a 50 V battery. The thyristor gate is driven continuously (DC gate drive). The input is v(t) = Vm sin(ωt). Over what interval(s) does the thyristor …

Accepted Answer

Introduction / Context: Battery charging with a phase-controlled rectifier depends on the source instantaneous voltage exceeding the battery EMF (plus small drops). A continuously applied gate drive merely ensures the SCR will trigger whenever forward anode-to-cathode conditions exist; it does not force conduction when the source cannot forward-bias the SCR against the battery voltage. Given Data / Assumptions: Single SCR half-wave converter. Battery emf ≈ 50 V (assume negligible series resistance for the decision threshold). Continuous DC gate drive available at all times. Input v(t) = Vm sin(ωt) on the same polarity half-cycle as the SCR. Concept / Approach: For current to flow into the battery, the instantaneous supply voltage must exceed the battery voltage. With continuous gating, the SCR will turn on immediately once v(t) exceeds 50 V in the forward half-cycle, and will continue to conduct until v(t) falls back below 50 V or current naturally goes to zero near the end of the half-cycle. Step-by-Step Solution: Condition to conduct: v(t) > Vbattery ≈ 50 V.Turn-on occurs at the instant ωt = arcsin(50 / Vm), provided 50 ≤ Vm.Conduction ceases as v(t) falls below 50 V on the trailing part of the half-cycle. Verification / Alternative check: If Vm < 50 V, no conduction occurs (even with continuous gate), validating that the battery EMF sets the threshold. If Vm ≫ 50 V, conduction occupies a central portion of the positive half-cycle only. Why Other Options Are Wrong: 0–180° suggests full positive half-cycle conduction, which is false unless Vm is infinite and the drop negligible. 0–90° arbitrarily limits conduction; actual turn-off occurs after the peak when v drops below 50 V. “When v < 50 V” contradicts the forward-bias requirement. Common Pitfalls: Assuming gate drive alone guarantees conduction regardless of source EMF; forgetting that an SCR needs sufficient forward anode voltage. Final Answer: when v > 50 V

Question 5

Assertion–Reason (GTO vs SCR circuitry) Assertion (A): A GTO-based power converter can be more compact than an SCR-based converter. Reason (R): A Gate-Turn-Off thyristor (GTO) can be turned off by applying an appropriate negative gate current pulse.

Accepted Answer

Introduction / Context: Classic SCR inverters require commutation networks (auxiliary capacitors, inductors, and switches) to force device current to zero for turn-off. GTOs, by contrast, can be turned off through the gate, reducing or eliminating bulky commutation circuitry and thus enabling more compact designs (subject to snubber needs). Given Data / Assumptions: SCRs need forced (or line) commutation unless self-commutated devices are used. GTOs accept negative gate current to extract carriers and turn off. Packaging and magnetics volume are major contributors to size. Concept / Approach: By eliminating large commutation components, GTO converters can achieve a smaller footprint for similar ratings. Although GTOs still need snubbers and robust gate drives, the absence of a full commutation network is the primary reason for compactness. Step-by-Step Solution: Establish A: GTO converters can be more compact than SCR converters → generally true.Establish R: GTOs turn off with negative gate current → true statement.Explain linkage: Eliminating commutation hardware directly reduces size → R explains A. Verification / Alternative check: Topological comparisons in textbooks show series/parallel inverters (SCR) with sizeable commutation components versus GTO/VSC drives with smaller magnetics at similar power levels. Why Other Options Are Wrong: Options B–D fail because either R would not explain A or one of the statements would be false. Common Pitfalls: Assuming GTOs never need snubbers; in practice, snubbering and strong gate drives are still required, but the commutation network bulk is avoided. Final Answer: Both A and R are correct and R is correct explanation of A

Question 6

Stud-mounted SCR mechanical/electrical interface In a stud-mounted SCR package, how is the stud normally connected with respect to the heat sink?

Accepted Answer

Introduction / Context: Stud-mounted SCRs are designed for direct mounting to a heat sink through the threaded stud. The stud is typically tied to one device terminal (commonly the cathode), providing both heat removal and an electrical connection through the mounting hardware. Given Data / Assumptions: Standard stud package per common SCR datasheets. Heat sink is metallic and grounded or appropriately connected. No special insulating bushings are installed (unless explicitly required by design). Concept / Approach: The stud forms part of the thermal path and also serves as an electrical terminal. Therefore, by default the stud is in electrical and thermal contact with the heat sink. If electrical isolation is required, insulating hardware and pads are added to break the conductive path, but that is not the default arrangement. Step-by-Step Solution: Identify the stud as both mechanical and electrical interface.Recognize that direct mounting provides low thermal resistance and an electrical path.Therefore, option “electrical and thermal contact” is correct for the normal case. Verification / Alternative check: Package drawings label the stud terminal (often cathode) and indicate torque specifications for thermal/electrical integrity. Why Other Options Are Wrong: “Thermal only” ignores the metal-to-metal electrical path. “Insulated” is only true if special insulating kits are used. “Either/or” is ambiguous and not representative of the default package use. Common Pitfalls: Mounting without understanding which terminal is the stud can cause unintended grounding or shorts. Always follow datasheet polarity and isolation guidance. Final Answer: in electrical and thermal contact with heat sink

Question 7

McMurray–Bedford full-bridge inverter Identify the commutation method primarily associated with the McMurray–Bedford full-bridge inverter topology.

Accepted Answer

Introduction / Context: Classical thyristor inverters employ forced commutation techniques to turn devices off. The McMurray and McMurray–Bedford families use specific capacitor/inductor arrangements to commutate one device using the complementary device or leg, reducing component stress and improving switching behavior for bridge topologies. Given Data / Assumptions: Full-bridge SCR inverter topology. No external line for natural commutation (i.e., stand-alone inverter). Named McMurray–Bedford method. Concept / Approach: Complementary commutation denotes schemes in which the turning-on of one device (or leg) assists in commutating its complementary device (or the opposite leg) off via resonant transfer using commutation capacitors. This is the hallmark of the McMurray–Bedford approach, distinct from generic “auxiliary commutation” that may use separate auxiliary switches without complementary leg interaction. Step-by-Step Solution: Recognize that line-commutation (“natural”) is unavailable in a self-contained inverter.McMurray–Bedford connects commutation capacitors to leverage complementary device action.Therefore, the appropriate label is “complementary commutation”. Verification / Alternative check: Topology descriptions and timing diagrams show commutation capacitors linked across opposite devices; the conduction of one leg reverses voltage across the other, forcing current to zero. Why Other Options Are Wrong: Natural commutation requires an AC grid; not applicable. “Auxiliary commutation” is too broad and does not capture the complementary leg action specific to McMurray–Bedford. “Any of the above” is incorrect because the topology uses a particular method. Common Pitfalls: Confusing broad categories (auxiliary, resonant) with the specific complementary scheme; overlooking that “McMurray–Bedford” is a well-known named method. Final Answer: complementary commutation

Question 8

Single-phase full-wave AC voltage regulator with R–L load If a single-phase full-wave AC voltage controller (full-wave regulator) feeds an R–L load whose load power-factor angle is θ (radians), what is the permissible control range of the firing ang…

Accepted Answer

Introduction / Context: AC voltage controllers (also called AC regulators) adjust the RMS voltage delivered to a load by delaying the firing of thyristors. With an R–L load, current lags voltage by the load power-factor angle θ, which impacts the allowable firing angle range to ensure natural turn-off and avoid commutation overlap into the next half-cycle. Given Data / Assumptions: Single-phase full-wave regulator (anti-parallel thyristor pair or bridge). Load is R–L with power-factor angle θ. Natural (line) commutation; no forced commutation circuits. Objective: avoid overlap between successive half-cycles and ensure current has decayed to zero before the next gating instant. Concept / Approach: In R–L loads, the current lags the applied voltage and persists after the instantaneous voltage crosses zero. If the firing angle α is too small relative to θ, the outgoing current from one half-cycle may still be flowing when the next device should be triggered, risking commutation failure or uncontrolled overlap. To guarantee natural extinction before the next half-cycle firing, α must not be set earlier than the lag-induced extinction boundary. Step-by-Step Solution: Define θ as the load current lag with respect to the source voltage.Positive-half conduction fired at α continues approximately until α + (π − θ) for typical R–L behavior.To ensure current is zero before the next negative-half firing, choose α such that extinction precedes the next gating instant; this leads to α ≥ θ.Maximum delay in a line-commutated controller is α = π. Verification / Alternative check: Textbook conduction diagrams for full-wave R–L regulators show a practical bound α ∈ [θ, π], ensuring turn-off without auxiliary means. Why Other Options Are Wrong: (a) Allows α < θ, risking overlap. (c) 0 to (π − θ) is for different continuity considerations and does not guarantee extinction. (d) Extending above π is not possible in a single-phase line-commutated regulator. Common Pitfalls: Confusing rectifier constraints with AC regulator constraints; ignoring lagging current tail in inductive loads. Final Answer: θ ≤ α ≤ π

Question 9

Step-down chopper (buck chopper) average output voltage A step-down (buck) chopper is fed from a DC source of magnitude V. If the switch on-time and off-time per cycle are Ton and Toff respectively, identify what V * (Ton / (Ton + Toff)) represents.

Accepted Answer

Introduction / Context: DC–DC step-down (buck) choppers reduce the average output voltage by switching a DC source on and off with duty ratio control. Understanding the relationship between duty ratio and average output is fundamental for power electronics design and control. Given Data / Assumptions: Input DC voltage = V (constant). Switch on-time = Ton; off-time = Toff; switching period T = Ton + Toff. Ideal switch and diode; continuous conduction assumed for the average relation. Concept / Approach: The average of a periodic waveform equals the area under one period divided by the period. The chopper applies V to the load during Ton and 0 V during Toff (ignoring the inductor freewheel path for the idealized average calculation), hence the average equals V multiplied by the duty ratio D = Ton/T. This relation is independent of the specific R, L, or E in the load for the average value, provided steady-state averaging is valid. Step-by-Step Solution: Duty ratio D = Ton / (Ton + Toff).Output voltage during on-state ≈ V; during off-state ≈ 0 (ideal).Average V_out(avg) = V * D = V * (Ton / (Ton + Toff)). Verification / Alternative check: Inductor volt-second balance in continuous conduction also yields V_out = D * V for a buck converter. Why Other Options Are Wrong: (b) RMS requires squaring and is not simply V * D. (c) The expression is the average regardless of load type under standard buck operation. (d) Not applicable since it clearly represents the average. Common Pitfalls: Confusing average with RMS; forgetting that ripple filters do not change the average relation V_out = D * V. Final Answer: Average output voltage

Question 10

SCR internal connection statement check Consider the construction of a silicon-controlled rectifier (SCR). “In a thyristor the gate lead is connected to the n-layer near the p-layer to which the anode is connected.” Is this statement true or false?

Accepted Answer

Introduction / Context: Understanding the internal layer structure of an SCR (PNPN device) is essential for appreciating how gating works and why the device exhibits latching behavior. The gate connection location determines how minority carrier injection assists turn-on. Given Data / Assumptions: Standard SCR layer order from anode to cathode: P1–N1–P2–N2. Anode connected to outer P1; cathode connected to outer N2. Gate terminal is taken from the inner P2 region (also called p-base) near the cathode junction. Concept / Approach: The gate must effectively modulate the junction nearest the cathode (J2) to initiate regenerative action in the two-transistor equivalent. Hence, the gate is bonded to the inner p-base (P2) region, not to an n-layer near the anode. This allows a small gate current to inject carriers that forward-bias the internal transistor pair and trigger conduction. Step-by-Step Solution: Identify the 4-layer structure: P1 (anode), N1, P2 (gate-connected), N2 (cathode).Gate metallization contacts P2 close to the cathode junction to be effective.Therefore the statement claiming a gate connection to an n-layer near the anode is incorrect. Verification / Alternative check: The two-transistor model (PNP–NPN) shows that injecting into the P2 base region facilitates rapid turn-on via positive feedback. Why Other Options Are Wrong: Marking the statement as true contradicts SCR construction; there is no standard variant where the gate connects to an n-layer near the anode. Common Pitfalls: Confusing the gate-bond region (P2) with the outer P1 anode region or with N-type layers; misreading cross-sectional diagrams. Final Answer: False

Question 11

Device exhibiting peak and valley points on its V–I characteristic Which device’s characteristic shows a distinct peak point followed by a valley point (negative-resistance region)?

Accepted Answer

Introduction / Context: Some two-terminal or three-terminal devices exhibit a negative-resistance region where current rises as voltage drops. This region is characterized by a peak point (onset) and a valley point (near the low-voltage high-current state). Recognizing which devices show such behavior is key in triggering and relaxation-oscillator design. Given Data / Assumptions: Comparing SCR, PUT, SCS, and Shockley diode. Focus on the presence of peak and valley points in the emitter/terminal characteristic. Concept / Approach: The classical device known for a peak and valley in its characteristic is the Unijunction Transistor (UJT). A PUT (Programmable UJT) is a PNPN device configured to emulate UJT-like behavior and also exhibits a programmable peak and valley point. The SCR and SCS have breakover and holding characteristics but are not generally described by peak/valley nomenclature in their common terminal curves. Step-by-Step Solution: Identify device with negative-resistance emitter curve: UJT-like behavior → PUT.PUT’s peak current occurs at the programmed peak voltage; thereafter the characteristic drops to a valley region.Hence, the requested device is PUT. Verification / Alternative check: PUT data sheets and application notes (relaxation oscillators) explicitly specify peak current/voltage and valley current/voltage parameters. Why Other Options Are Wrong: SCR/SCS are four-layer switches with latching and holding currents; Shockley diode shows breakover rather than clearly defined peak and valley points used in UJT/PUT terminology. Common Pitfalls: Equating breakover and holding points of an SCR with the peak/valley description used for UJT/PUT emitter characteristics. Final Answer: PUT (Programmable Unijunction Transistor)

Question 12

Two-transistor equivalent model of an SCR The well-known two-transistor model of a thyristor (SCR) consists of which pair of transistors?

Accepted Answer

Introduction / Context: The regenerative action in an SCR can be visualized using an equivalent of two coupled bipolar transistors. This model provides intuition for triggering, latching, and holding behavior, and explains why gate injection can rapidly turn the device on. Given Data / Assumptions: SCR layer sequence: P–N–P–N from anode to cathode. Model uses two interlinked BJTs sharing internal junctions. Concept / Approach: The inner junctions of a PNPN structure can be mapped to a p–n–p transistor and an n–p–n transistor connected in positive feedback. When gate current is applied, it increases the current gain of one transistor, which in turn drives the other, rapidly pushing the sum of current gains toward unity and beyond, leading to latching. Step-by-Step Solution: Map PNPN to PNP and NPN with cross-coupled collector-base connections.Apply small gate current to the base–emitter of the NPN side (equivalently the p-base region).Regenerative feedback causes rapid rise of anode current and latching. Verification / Alternative check: Derivations show I_A = I_CPNP + I_CNPN with α_sum approaching 1; holding and latching currents follow naturally from the model. Why Other Options Are Wrong: Two identical polarities (both PNP or both NPN) cannot represent PNPN feedback correctly. A UJT is a different device and not part of the SCR model. Common Pitfalls: Forgetting that both carrier types contribute in the on-state (conductivity modulation), consistent with the complementary transistor pair. Final Answer: One n–p–n and one p–n–p

Question 13

Thyristor commutation classification Which commutation method explicitly uses an external pulse source to force the current to zero in the thyristor (external pulse commutation)?

Accepted Answer

Introduction / Context: Commutation in thyristor converters refers to turning the device off by forcing its current to zero and reverse-biasing its junctions. Various classes categorize how this is achieved. Recognizing the class that uses an external pulse source is important for high-performance or special-purpose converters. Given Data / Assumptions: Standard commutation classes A through F. Device is a line-commutated or force-commutated SCR. Concept / Approach: Class E commutation employs an external pulse (often via a pulse transformer or an auxiliary supply) to apply a reverse current or reverse voltage across the SCR to turn it off, independent of the main load network resonance or the line voltage zero-crossing. Step-by-Step Solution: Identify commutation need: remove charge carriers and restore blocking.Class E: use external pulse source to inject a reverse current or voltage.Device turns off once current remains below holding current for its recovery time. Verification / Alternative check: Commutation taxonomy: A (load), B (resonant), C (auxiliary with complementary SCR), D (impulse or parallel capacitor), E (external pulse), F (line/natural). Why Other Options Are Wrong: (a) and (b) rely on load/auxiliary resonant networks; (d) is natural at line zero-crossings, not via external pulses. Common Pitfalls: Mixing auxiliary commutation (Class C) with external pulse commutation (Class E); the former uses internal commutating components tied to the main supply. Final Answer: Class E (External pulse commutation)

Question 14

Half-bridge inverter output frequency vs. device gating frequency A single-phase half-bridge inverter gates each thyristor at a frequency f with 180° conduction (no PWM). What is the fundamental output frequency delivered to the load?

Accepted Answer

Introduction / Context: In a half-bridge inverter, two complementary switches (thyristors or transistors) connect the output alternately to +Vdc/2 and −Vdc/2. The relationship between the gating frequency per device and the fundamental output frequency is essential for designing control and filters. Given Data / Assumptions: Each device conducts for half a cycle (180° conduction). No PWM or modulation; simple square-wave operation. Complementary gating; dead-time negligible for this ideal discussion. Concept / Approach: When the top device turns on, the output is +Vdc/2 for half a period; when the bottom device turns on, the output is −Vdc/2 for the next half period. Each device turns on once per output cycle. Therefore, the per-device gating repetition rate equals the output fundamental frequency. Step-by-Step Solution: Define device gating frequency per switch = f.One full output cycle requires one ON of the top and one ON of the bottom device.Thus the output fundamental frequency f_out = f. Verification / Alternative check: Waveform analysis: the square wave has a period equal to the time between consecutive ON instants of the same switch, identical to the gating period. Why Other Options Are Wrong: (a) Implies two device turns per period per device, which is not the 180° scheme. (c) and (d) contradict the basic two-step per cycle operation. Common Pitfalls: Confusing total switching events per cycle (two across both devices) with per-device gating frequency. Final Answer: f

Question 15

Series-connected thyristors equalization requirements When thyristors are connected in series to withstand higher blocking voltage, what equalization is required for reliable operation?

Accepted Answer

Introduction / Context: Series connection of thyristors is used to achieve higher voltage ratings. However, device parameter tolerances cause unequal voltage sharing in both steady-state and transient conditions, which can overstress individual devices. Equalizing networks are therefore essential. Given Data / Assumptions: Multiple SCRs connected in series for higher blocking voltage. Devices have tolerances in leakage current, junction capacitance, and recovery characteristics. Concept / Approach: Static equalization (typically high-value resistors across each device) ensures that in steady-state blocking (DC or low frequency), leakage mismatches do not concentrate voltage on one device. Dynamic equalization (typically capacitors and sometimes RC snubbers) ensures that during voltage transients, turn-off, and recovery, the rapid dv/dt is shared so that no single device momentarily sees an excessive voltage. Step-by-Step Solution: Provide parallel resistors for static sharing to match leakage currents.Add capacitors or RC snubbers for dynamic sharing to limit dv/dt per device.Verify sharing under worst-case tolerance and temperature. Verification / Alternative check: Design practice and manufacturer guidelines specify both networks for series stacks, especially in HV drives and HVDC valves. Why Other Options Are Wrong: (a) Addresses only steady-state; (b) addresses only transients; both are needed in real applications. (d) would risk overvoltage on a single device. Common Pitfalls: Relying on matched devices without equalization; ignoring temperature-dependent leakage causing drift in static sharing. Final Answer: Both static and dynamic equalizing circuits

Question 16

Single-phase bridge-type cycloconverter In power electronics, a single-phase bridge cycloconverter (line-commutated, using two anti-parallel bridges) employs how many thyristors in total?

Accepted Answer

Introduction / Context: A single-phase bridge-type cycloconverter directly converts AC at one frequency to AC at a lower frequency without an intermediate DC link. It accomplishes this with two full thyristor bridges connected in anti-parallel to synthesize positive and negative portions of the low-frequency output from the mains supply. Given Data / Assumptions: Configuration: two full-bridge converters connected anti-parallel (one for positive current, one for negative current). Each full bridge uses 4 thyristors (line-commutated SCRs). Gate control determines which bridge conducts at a given instant to shape the output waveform. Concept / Approach: Counting devices is straightforward: a standard single-phase full bridge requires four SCRs arranged as two legs. Two such bridges are required in anti-parallel so that current can be driven in either direction at the output while remaining line-commutated. Therefore, total SCRs = 4 + 4 = 8. Step-by-Step Solution: Identify one full bridge → 4 SCRs.Duplicate for the opposite polarity bridge → another 4 SCRs.Total thyristors used = 4 + 4 = 8. Verification / Alternative check: Most text-book schematics of single-phase bridge cycloconverters clearly show two bridge groups, each with four controlled devices, confirming the count of eight. Why Other Options Are Wrong: Any claim of fewer than eight devices would imply a half-controlled bridge or non-bridge topology; more than eight would correspond to multi-pulse or multi-module designs, not the standard single-phase bridge cycloconverter. Common Pitfalls: Confusing the cycloconverter with a single-bridge AC regulator (which uses antiparallel SCRs or a TRIAC) or with a DC–AC inverter. Cycloconverters specifically need two full bridges for bidirectional current control. Final Answer: True

Question 17

Operating quadrants of a single-phase fully controlled converter A single-phase full converter supplying a load can operate in how many quadrants of the V–I plane (considering rectification and inversion modes with current direction maintained by th…

Accepted Answer

Introduction / Context: A single-phase fully controlled bridge (using four SCRs) can both rectify power from AC to DC and, with appropriate firing angle and suitable load (e.g., back-EMF), return power to the AC source (inversion). This raises the classic question of how many quadrants of operation are achievable. Given Data / Assumptions: Fully controlled bridge with SCRs and proper gating. Unidirectional current in the DC side (typical with RL or EMF-type loads). Firing angle α can be set from near 0 to beyond 90 degrees. Concept / Approach: With SCRs, current through the bridge is unidirectional, but the average DC voltage can be positive (rectification, α < 90°) or negative (inversion, α > 90°) while current keeps the same direction. That corresponds to two quadrants: voltage sign can change, current sign is fixed. Step-by-Step Solution: For α < 90°, Vdc > 0 and Idc > 0 → Quadrant I.For α > 90°, Vdc < 0 but Idc > 0 → Quadrant II or IV depending on chosen axes conventions; in any case, a second quadrant with opposite voltage sign.Current reversal is not allowed by the SCR bridge without additional circuitry → limited to two quadrants. Verification / Alternative check: Standard converter operation charts show two-quadrant capability for fully controlled single-phase bridges with passive DC loads. Why Other Options Are Wrong: 1 quadrant: Ignores inversion capability. 3 or 4 quadrants: Would require current reversal and/or bidirectional DC-side current, which this bridge alone does not provide. Common Pitfalls: Confusing the fully controlled converter (2-quadrant) with dual converters or four-quadrant chopper-inverter systems that can reverse DC current. Final Answer: 2 quadrants

Question 18

Effect of gate current on thyristor forward breakover voltage In an SCR (thyristor), how does the forward breakover voltage vary with the applied gate current?

Accepted Answer

Introduction / Context: The forward breakover voltage of a thyristor is the anode-to-cathode voltage at which the device turns on without a gate signal. In practical use, a gate current is applied to trigger conduction at a lower voltage. Understanding this relationship is essential for reliable gating circuits. Given Data / Assumptions: Standard SCR behavior: gate current aids device turn-on by injecting carriers. Device is forward-biased (anode positive with respect to cathode). Concept / Approach: Increasing gate current reduces the required anode-cathode voltage to reach conduction by preconditioning the device’s internal junctions (regenerative feedback starts earlier). Therefore, the effective forward breakover voltage is lowered as gate current increases. Step-by-Step Solution: At zero gate current, forward breakover is at the datasheet-specified highest value.Apply a small gate current: conduction begins at a lower anode-cathode voltage.As gate current rises further (within allowable limits), the turn-on voltage continues to decrease. Verification / Alternative check: SCR V–I curves with different Ig values show a leftward shift of the triggering point as Ig increases, confirming the decrease in forward breakover voltage. Why Other Options Are Wrong: Constant or increases: Contradicts the fundamental regenerative turn-on aided by gate drive. “May be constant or may depend”: Too vague; in practice it does depend and decreases with higher gate current. Common Pitfalls: Confusing holding or latching current with breakover voltage; those are separate parameters. Final Answer: decreases as gate current is increased

Question 19

Power in a single-phase half-wave controlled rectifier (R load) A single-phase half-wave converter feeds a purely resistive load. Let Pdc be the DC output power computed as Vdc * Idc, and Pac be the total real power actually delivered to the load (i…

Accepted Answer

Introduction / Context: In rectifiers, learners often equate the product Vdc * Idc with the “power in the load.” However, when the load voltage and current are not pure DC but contain ripple, the real (average) power dissipated in a resistive load includes contributions from both the DC component and the AC ripple components. Given Data / Assumptions: Converter: single-phase half-wave controlled rectifier. Load: purely resistive (R), so instantaneous power is v(t) * i(t) and is always non-negative when current flows. Definitions: Pdc = Vdc * Idc; Pac = average of v(t) * i(t) over one cycle. Concept / Approach: For any periodic waveform, the average power in a resistor equals the sum of the powers due to the DC component and each AC harmonic. The DC product Vdc * Idc accounts only for the DC component. Because there is ripple, additional positive power results from the AC components, thus Pac exceeds Pdc. Step-by-Step Solution: Express v(t) and i(t) as DC + AC ripple.Real power in R: P = average{v(t) * i(t)} = Vdc * Idc + average{vac(t) * iac(t)}.Since i(t) in R is in phase with v(t), the ripple term contributes a positive average.Therefore, Pac = Pdc + positive ripple power ⇒ Pdc < Pac. Verification / Alternative check: Fourier analysis shows that for a resistive load, each harmonic contributes In2 * R / 2 to the average power (when voltage and current contain that harmonic), hence total real power surpasses the DC-only estimate. Why Other Options Are Wrong: Pdc = Pac: Only true for pure DC with zero ripple, which never occurs in half-wave control. Conditional statements with angle: Even at α = 0°, half-wave output still has significant ripple; equality never holds. Common Pitfalls: Ignoring ripple contribution to power; confusing apparent and real power; assuming R load eliminates ripple power—which it does not. Final Answer: Pdc < Pac for all firing angles

Question 20

Relaxation oscillators in electronics Which device is most commonly used to build a simple, stable relaxation oscillator for timing and triggering applications?

Accepted Answer

Introduction / Context: Relaxation oscillators generate repetitive waveforms (typically sawtooth or pulse) using a charge–discharge process and a threshold device. They are widely used for timing, triggering SCRs, and generating clock pulses in low-to-medium frequency ranges. Given Data / Assumptions: Devices considered: UJT, tunnel diode, PIN diode. Goal: a simple, economical, and predictable RC-based oscillator. Concept / Approach: The UJT features a distinct negative-resistance region and a well-defined intrinsic stand-off ratio, making it ideal for RC charge–discharge oscillators. When the capacitor voltage reaches the peak point, the UJT conducts rapidly, discharging the capacitor to the valley point, then turns off, and the cycle repeats—creating a relaxation oscillation. Step-by-Step Solution: Use RC network to charge a capacitor toward a supply.When capacitor voltage hits UJT peak voltage (≈ ηVBB + VD), the UJT turns on.Capacitor discharges through the UJT until the valley point, where the UJT turns off.Capacitor starts charging again—completing the oscillation cycle. Verification / Alternative check: Classic SCR trigger circuits and many textbooks demonstrate UJT-based relaxation oscillators due to their simplicity and reliable frequency control via RC. Why Other Options Are Wrong: Tunnel diode: Can form relaxation oscillators, but is less common in basic power electronics labs and triggers compared to UJT. Common Pitfalls: Assuming any diode with nonlinearity will suffice. UJT’s characteristics and three-terminal control make it the go-to device. Final Answer: UJT (uni-junction transistor)

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