Question 1

Single-phase semiconverter with freewheeling diode (purely resistive load) For a single-phase semiconverter (half-controlled bridge) feeding a purely resistive load and equipped with a freewheeling diode, if the firing angle is α, what are the condu…

Accepted Answer

Introduction / Context: Semiconverters are widely used controlled rectifiers. The presence of a freewheeling diode primarily affects inductive loads by providing a path for current continuity when the source reverses. For purely resistive loads, current is zero whenever the instantaneous source voltage is negative or the device is off. Given Data / Assumptions: Single-phase semiconverter, resistive (R) load only. Freewheeling diode is present. Firing angle = α. Concept / Approach: With a purely resistive load, current follows voltage instantly (no energy storage). Therefore, once the input crosses zero or becomes negative, the current drops to zero and a freewheeling path is not required. The thyristor conducts only from the firing instant α to the end of that positive half-cycle at π. Step-by-Step Solution: Conduction in each positive half-cycle: from ωt = α to ωt = π → duration = π − α.Freewheeling interval: none for an R-load → duration = 0. Verification / Alternative check: Waveforms confirm that current becomes zero as soon as the source crosses zero because there is no stored energy to sustain it. Hence the freewheeling diode does not conduct for a purely resistive load. Why Other Options Are Wrong: Options invoking a nonzero freewheeling interval (e.g., π − α and α) apply to RL loads. Extended conduction windows such as π + α are not applicable to an R-load with a semiconverter. Common Pitfalls: Assuming the mere presence of a freewheeling diode implies it must conduct; it only does so if energy storage (L) forces current continuity when the source reverses. Final Answer: π − α and 0

Question 2

Definition check: What type of power-electronic conversion does a cycloconverter perform? Choose the correct classification for a cycloconverter.

Accepted Answer

Introduction / Context: Cycloconverters are important in high-power variable-frequency drives for large motors (e.g., rolling mills). They directly change AC at one frequency to AC at another frequency without an intermediate DC link. Given Data / Assumptions: Line-commutated, phase-controlled devices. Output frequency is typically a submultiple of the input frequency. Concept / Approach: By phase-controlling groups of thyristors, cycloconverters synthesize a low-frequency AC waveform directly from a higher-frequency AC source, hence ac-to-ac conversion. This differentiates them from rectifier-inverter DC-link systems. Step-by-Step Solution: Identify input and output: AC in → AC out (adjusted frequency and voltage).No obligatory DC energy storage between stages. Verification / Alternative check: Standard power-electronics textbooks list cycloconverters under ac-ac converters (direct frequency converters). Why Other Options Are Wrong: ac-dc, dc-ac, dc-dc are different categories. ac-dc-ac with a DC link is not a cycloconverter (that is a converter-inverter system). Common Pitfalls: Confusing cycloconverters with matrix converters (also ac-ac but using bidirectional switches) or with DC-link inverters. Final Answer: ac-ac converter

Question 3

SCR operating states: Forward blocking condition During the forward blocking state of an SCR (thyristor), what are the typical anode-cathode voltage and current levels, respectively?

Accepted Answer

Introduction / Context: An SCR exhibits distinct operating states: forward blocking, forward conduction, and reverse blocking. Recognizing the voltage and current levels in each state is vital for protection and gating strategies. Given Data / Assumptions: Anode positive with respect to cathode (forward bias). No gate trigger applied. Device is idealized to illustrate states (ignoring small leakage). Concept / Approach: In forward blocking, the SCR can withstand a high anode-to-cathode voltage but only a very small leakage current flows because the device has not been triggered into conduction. Step-by-Step Solution: Anode positive, gate off → junctions configured so device is non-conducting.Voltage: high (up to rated forward blocking capability).Current: low (leakage only). Verification / Alternative check: Device datasheets specify forward blocking voltage ratings and leakage currents in this state. Why Other Options Are Wrong: High current would imply turn-on. Low voltage/low current corresponds to an off device with no applied forward voltage, not forward blocking capability. Common Pitfalls: Confusing forward blocking with latching/holding current concepts that apply once turned on. Final Answer: high and low

Question 4

Three-phase bridge inverter with resistive star load A 3-phase bridge inverter is supplied by a 400 V DC battery. The load is star-connected with R = 10 Ω per phase (neutral isolated). Assuming square-wave operation, what is the peak value of phase …

Accepted Answer

Introduction / Context: Inverters convert DC to AC. For a three-phase bridge with a star-connected resistive load, the line-to-line voltage steps between ±Vdc, and each phase sees a three-level line-neutral waveform whose maximum magnitude is tied to the DC link voltage. Estimating the peak current helps with device and protection sizing. Given Data / Assumptions: Vdc = 400 V, square-wave inverter, ideal switches. Star-connected resistive load, Rph = 10 Ω, isolated neutral. Neglect device drops and switching transients for simplicity. Concept / Approach: In a standard three-phase bridge, each leg outputs ±Vdc/2 relative to the DC midpoint. The resulting line-neutral phase voltage is a stepped three-level waveform with peak magnitude approximately Vdc/2. For a pure resistor, instantaneous current follows i = v/R, so the peak current equals Vph,peak / Rph. Step-by-Step Solution: Vph,peak ≈ Vdc / 2 = 400 / 2 = 200 V.Rph = 10 Ω → Ipeak = 200 / 10 = 20 A. Verification / Alternative check: Fourier fundamentals yield an RMS current of Ifund,rms = V1,rms / R, but peak instantaneous current is governed by the waveform peak, which is Vdc/2 for the phase voltage in a six-step inverter. Why Other Options Are Wrong: 40 A assumes Vdc across a phase, which is not correct for a star load with isolated neutral. 10 A and 5 A underestimate the peak given Vdc = 400 V and R = 10 Ω. Common Pitfalls: Confusing line-line and line-neutral voltages; forgetting that maximum phase voltage magnitude is Vdc/2 in the six-step sequence. Final Answer: 20 A

Question 5

Half-wave controlled rectifier with RL load and freewheeling diode A single-phase half-wave controlled rectifier supplies an RL load and includes a freewheeling diode. The input is v(t) = Vm sin(ωt). If I0 is the load current during the freewheeling…

Accepted Answer

Introduction / Context: Freewheeling diodes in controlled rectifiers provide a path for inductive current when the source voltage becomes unfavorable, preventing abrupt current interruption and large overvoltages. Writing the correct KVL during the freewheel interval is fundamental for analyzing current decay and ripple. Given Data / Assumptions: Single-phase half-wave controlled rectifier feeding RL. Freewheeling diode across the load. During freewheeling, source is effectively isolated from the load loop. Concept / Approach: When the freewheeling diode conducts, the loop consists only of R, L, and the diode (assumed ideal with negligible drop). Therefore the source does not appear in the loop equation; current decays according to a homogeneous first-order differential equation. Step-by-Step Solution: Apply KVL around the freewheel loop: vL + vR ≈ 0.Replace vL with L * (di/dt) and vR with R * i.Hence, L * (di/dt) + R * i = 0. Verification / Alternative check: Solution i(t) = I0 * exp(−t/τ) with τ = L/R confirms exponential decay independent of source voltage during freewheel. Why Other Options Are Wrong: Options that include Vm sin(ωt) presume the source is in the loop; not true during freewheeling. Signs in (e) are incorrect for passive sign convention. Common Pitfalls: Forgetting to remove the source from the loop during the freewheel interval; neglecting diode drop is acceptable for first-order analysis. Final Answer: L * (di/dt) + R * i = 0

Question 6

Sizing parallel thyristors with derating A 750 A DC circuit uses identical thyristors each rated at 175 A. With a derating factor of 25% (i.e., each device carries only 75% of its nameplate current), how many thyristors must be placed in parallel?

Accepted Answer

Introduction / Context: When paralleling thyristors, current sharing is imperfect due to tolerances and temperature coefficients. A derating factor is applied so each device is loaded below its nominal rating, ensuring reliability. Calculating the required count is a straightforward but essential design step. Given Data / Assumptions: Total circuit current Itotal = 750 A. Device nameplate Irated = 175 A. Derating 25% → usable per device Iusable = 0.75 * Irated. Concept / Approach: Number of devices in parallel n must satisfy n * Iusable ≥ Itotal. Use ceiling to ensure sufficient devices. Step-by-Step Solution: Iusable = 0.75 * 175 = 131.25 A.n ≥ 750 / 131.25 ≈ 5.714...Minimum integer n = 6. Verification / Alternative check: Check margin: 6 * 131.25 = 787.5 A ≥ 750 A → acceptable. Why Other Options Are Wrong: 3, 4, or 5 do not meet the required current with derating; 10 is over-conservative for the stated parameters. Common Pitfalls: Forgetting to apply derating; using rounding down instead of ceiling. Final Answer: 6

Question 7

Triggering series-connected SCR strings Which triggering method for series-connected SCRs commonly uses LASCRs (light-activated SCRs) or fiber-optic links to ensure simultaneous gating and high isolation?

Accepted Answer

Introduction / Context: Series connection of SCRs is used to withstand higher voltages. Equalizing networks and reliable, simultaneous gate triggering are crucial to prevent over-voltage on individual devices. Optical methods (LASCRs or optically coupled gates) offer strong isolation and synchrony. Given Data / Assumptions: High voltage stacks of SCRs. Need for high common-mode isolation and minimal skew. Concept / Approach: Optical triggering delivers identical light pulses to each device or uses LASCRs, ensuring minimal timing mismatch without the problems of transformer parasitics or ground loops. Step-by-Step Solution: Identify requirement: simultaneous, isolated triggers.Match solution: optical triggering with LASCRs or fiber-optic links. Verification / Alternative check: High-power converters frequently adopt fiber-optic gate drivers for IGBTs/SCRs for exactly these reasons. Why Other Options Are Wrong: Pure resistance or sequential triggering does not guarantee simultaneous turn-on across the stack; transformer coupling can work but is not what the question highlights about LASCR usage. Common Pitfalls: Assuming transformer coupling always suffices at very high voltages where isolation and timing skew are critical. Final Answer: Optical triggering

Question 8

Condition for oscillatory commutation in a series inverter (R load) A series inverter supplies a resistive load R with commutating components L and C in series. For the current to be oscillatory (so that natural current zero occurs for thyristor tur…

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Introduction / Context: In a series inverter, commutation relies on the natural oscillation of the series RLC path to force the current through zero. Ensuring underdamped behavior is essential for achieving these current zero crossings within each cycle. Given Data / Assumptions: Series RLC path with load resistance R. Thyristor needs a natural current zero for turn-off. Concept / Approach: The series RLC response is underdamped if the damping factor α = R/(2L) is less than the undamped natural frequency ω0 = 1/√(LC). Under this condition, the current oscillates and crosses zero, enabling reliable natural commutation. Step-by-Step Solution: Compute α = R/(2L).Compute ω0 = 1/√(LC).Require underdamped response: ω0 > α → 1/√(LC) > R/(2L). Verification / Alternative check: Equivalently, R < 2 * √(L/C). Either form ensures an oscillatory response and a natural current zero for commutation. Why Other Options Are Wrong: Equality implies critical damping; the inequality reversed gives overdamping with no oscillation. The other forms do not consistently capture the underdamped criterion for all L, C, R. Common Pitfalls: Forgetting that it is the current zero crossing (not voltage) that commutates the thyristor; choosing components that yield overdamped response will prevent proper turn-off. Final Answer: 1/√(L*C) > R/(2L)

Question 9

Purpose of an RC snubber across a thyristor In power converters, an RC snubber network connected across a thyristor is primarily used to protect the device against what?

Accepted Answer

Introduction / Context: Thyristors are sensitive to rapid changes in voltage (dv/dt) and overvoltage spikes that occur during switching and line disturbances. Snubber networks shape these transients to keep device stresses within safe limits. Given Data / Assumptions: RC snubber placed across the device (often with a series resistor and capacitor). Goal: limit dv/dt, clamp overshoot, and absorb energy from stray inductances. Concept / Approach: The capacitor initially looks like a short to a sudden voltage step, limiting dv/dt across the thyristor; the resistor damps oscillations and dissipates energy. This combination prevents voltage spikes and steep slopes that could cause stress or inadvertent triggering. Step-by-Step Solution: Identify major threat: dv/dt and overvoltage spikes during turn-off/line transients.RC network provides a controlled ramp (reduces dv/dt) and damping.Result: device protected from switching transients. Verification / Alternative check: Waveform measurements with and without snubbers show reduced overshoot and ringing across the thyristor terminals. Why Other Options Are Wrong: While limiting dv/dt can also reduce the chance of false triggering, the primary stated purpose is protection against switching transients. Failure to turn on or commutate involves gating/commutation design, not the snubber’s main role. Common Pitfalls: Choosing snubber values that create excessive losses or inadequate damping; ignoring layout-induced stray inductance that can compromise snubber effectiveness. Final Answer: switching transients

Question 10

Single-phase full-wave AC voltage regulator (purely resistive load): If the RMS load current is 20 A, compare per-thyristor average current and per-thyristor RMS current with 20 A (consider each SCR conducts for alternate half-cycles).

Accepted Answer

Introduction / Context: AC voltage controllers (AC regulators) use antiparallel thyristors to control the RMS voltage applied to a resistive load. Each thyristor conducts on alternate half-cycles. This question probes understanding of how device-level current (per SCR) compares with the total RMS load current reported at the terminals. Given Data / Assumptions: Single-phase, full-wave AC regulator with two antiparallel SCRs. Load is purely resistive; load RMS current IL,rms = 20 A. Symmetrical firing so each SCR conducts on its half-cycles only. We compare a per-device average over a full mains cycle and a per-device RMS over a full cycle. Concept / Approach: Because each SCR conducts for only half of the time (its own polarity half-cycles), the current through each device is time-shared. For a sinusoidal resistive current, distributing conduction across two devices reduces both the per-device average current (over a full cycle) and the per-device RMS current relative to the total load RMS current. Step-by-Step Solution: Let i(t) be the instantaneous load current. For a resistive load, i(t) follows the applied voltage shape within the conduction interval.Over a full cycle, SCR1 conducts during positive half-cycles only; SCR2 conducts during negative half-cycles only.Per-device average current over one full cycle is the average of its half-cycle conduction and zero during the other half → average per device < average of the rectified load current, hence < 20 A.Per-device RMS over one full cycle is RMS of half-cycle conduction combined with zero in the other half. For symmetric conduction, I_rms,device = IL,rms / √2 for sinusoidal halves, which is less than 20 A. Verification / Alternative check: For an ideal full-conduction case on a resistive load: if IL,rms = 20 A, each SCR carries the same waveform but only half the time. Time-sharing lowers both RMS and average when computed over the full period per device. Why Other Options Are Wrong: Option A: Claims device RMS > 20 A, impossible with half-time conduction. Option C: Both higher than 20 A contradicts power balance and time-sharing. Option D: Assigns fixed numbers without basis; device RMS ≠ 20 A for half-cycle conduction averaged over full period. Common Pitfalls: Confusing per-device RMS (over full cycle with zero current for half the cycle) with the instantaneous half-cycle RMS; forgetting that averaging window matters. Final Answer: both the average and rms thyristor currents will be less than 20 A

Question 11

Three-phase AC voltage regulator: How many thyristors are typically used (assume a standard three-phase, bidirectional control of each phase)?

Accepted Answer

Introduction / Context: In a three-phase AC regulator, each phase must be controlled for both positive and negative half-cycles. The common implementation places antiparallel SCR pairs per phase. Knowing the device count is basic to topology understanding and cost/complexity estimation. Given Data / Assumptions: Three independent phases feeding a load (or three single-phase loads). Bidirectional control per phase requires antiparallel devices. No common return path device sharing assumed. Concept / Approach: Each phase needs two controlled unidirectional switches in antiparallel so current of either polarity can be phase-controlled. For three phases, total SCRs = 3 phases * 2 SCRs/phase = 6 SCRs. Step-by-Step Solution: Per phase: need two devices (positive and negative half-cycles).For three phases: 2 * 3 = 6 SCRs total. Verification / Alternative check: Standard three-phase AC regulator schematics universally show six SCRs (or TRIACs for lower power), confirming the count. Why Other Options Are Wrong: 3 thyristors: insufficient; only one per phase cannot control both polarities. 9 or 12: represent other converter topologies or parallel strings, not the basic regulator. Common Pitfalls: Confusing with 6-pulse or 12-pulse rectifiers; here we are not rectifying but regulating AC per phase. Final Answer: 6 thyristors

Question 12

HVDC transmission practice: Which converter pulse number is generally preferred for line-commutated HVDC links to reduce harmonics and smoothing requirements?

Accepted Answer

Introduction / Context: HVDC schemes typically use line-commutated converter bridges. The number of pulses per cycle directly affects characteristic harmonics, filtering needs, and performance. This question targets the standard industry choice for large HVDC links. Given Data / Assumptions: Line-commutated (thyristor) HVDC converters. Harmonic reduction and smoother DC current are desirable. Transformer phase-shifting allows pulse multiplication. Concept / Approach: Two 6-pulse bridges in series/parallel with a 30° phase shift create a 12-pulse converter. This arrangement cancels lower-order characteristic harmonics (e.g., 5th/7th on AC side), lowering filter size and improving DC quality. Hence, 12-pulse is the standard choice for bulk power HVDC. Step-by-Step Solution: Recognize that 12-pulse is formed by phase-shifted 6-pulse bridges.Note harmonic cancellation and smoother DC outcome.Conclude industry preference: 12-pulse. Verification / Alternative check: Most classic HVDC installations (e.g., LCC-HVDC) specify 12-pulse at each end as baseline. Why Other Options Are Wrong: 6-pulse alone has higher harmonics and larger filters. 3-pulse is obsolete for high-power transmission. “Either 12 or 6” is too loose; “generally use” points to 12-pulse. Common Pitfalls: Confusing modular multilevel VSC practice (different pulse concept) with line-commutated HVDC. Final Answer: 12 pulse converters

Question 13

Separately excited DC motor fed via a single-phase semi-converter (constant field): If speed is 600 rpm at firing angle α = 0°, estimate the speed when α = 90° (ignore drop and assume speed ∝ average armature voltage).

Accepted Answer

Introduction / Context: Phase-controlled rectifiers set the average DC voltage applied to a DC motor armature. With constant field, speed is approximately proportional to average armature voltage. A semi-converter has a specific Vdc–α relationship that lets us estimate speed changes as the firing angle increases. Given Data / Assumptions: Semi-converter (half-controlled bridge) on single-phase AC. Constant field flux; neglect armature reaction and losses. Speed ∝ average armature voltage Vdc. Speed at α = 0° is 600 rpm. Concept / Approach: For a single-phase semi-converter: Vdc = (Vm/π) * (1 + cos α), where Vm is the peak of the AC line-to-neutral voltage presented to the converter. Hence Vdc at α = 90° is half of Vdc at α = 0° because cos 90° = 0. Step-by-Step Solution: At α = 0°: Vdc0 = (Vm/π) * (1 + cos 0°) = (Vm/π) * (1 + 1) = 2Vm/π.At α = 90°: Vdc90 = (Vm/π) * (1 + 0) = Vm/π.Ratio: Vdc90 / Vdc0 = (Vm/π) / (2Vm/π) = 1/2.Thus speed at α = 90° ≈ 600 rpm * 1/2 = 300 rpm. Verification / Alternative check: The linear proportion holds well when armature current ripple and drops are small relative to applied voltage. Practical speeds may differ slightly due to commutation overlap and IR drop, but 300 rpm is the correct first-order estimate. Why Other Options Are Wrong: 150 rpm or 75 rpm imply larger than predicted reduction. Zero speed would require α near 180° and torque loss, not 90°. Common Pitfalls: Using the full-converter formula Vdc = (2Vm/π) * cos α instead of the semi-converter expression; mixing up degrees and radians for cos. Final Answer: about 300 rpm

Question 14

Assertion–Reason (inverters and commutation): Assertion (A) “An inverter does not require forced commutation.” Reason (R) “A series inverter is a forced-commutation inverter.” Determine the correct choice.

Accepted Answer

Introduction / Context: Thyristor inverters require means to turn devices off. Depending on topology, commutation can be natural (line-commutated), forced using auxiliary circuits, or inherent via self-commutated switches (IGBT/MOSFET). This item checks clarity on forced commutation in classical SCR inverters and the nature of series inverters. Given Data / Assumptions: Assertion says “inverter does not require forced commutation” (general claim). Reason states “series inverter is a forced-commutation inverter.” Concept / Approach: Many SCR-based inverters away from a stiff AC grid must force current to zero to turn off the SCRs. Series inverters are prime examples using resonant LC paths for forced commutation. Therefore the universal statement in A is wrong; R is true. Step-by-Step Solution: Evaluate A: incorrect as a general claim; numerous inverters require forced commutation.Evaluate R: correct—series inverters use resonant commutation to force SCR current to zero.Mapping to choices → “A wrong, R correct”. Verification / Alternative check: Textbook classifications: line-commutated vs forced-commutated; series/parallel inverters are under forced commutation families. Why Other Options Are Wrong: Options A/B/C assume Assertion is true; it is not. Common Pitfalls: Generalizing from self-commutated IGBT inverters (gate turn-off) to all inverter types; here “forced commutation” is a specific SCR technique. Final Answer: A is wrong but R is correct

Question 15

Application of a single-phase series converter (power electronics): What is the principal use case?

Accepted Answer

Introduction / Context: In converter engineering, series and parallel configurations scale voltage and current capability respectively. Understanding which configuration serves which goal is essential for topology selection and insulation design. Given Data / Assumptions: “Series converter” indicates series connection of converter units/devices. Series connections raise overall blocking voltage capability. No special current-sharing components are implied in the statement. Concept / Approach: Series connection primarily increases voltage withstand and output voltage capability. Conversely, parallel connections are used to raise current capability. Therefore, the primary application of a single-phase series converter is high-voltage output service. Step-by-Step Solution: Identify series connection → voltage scaling.Relate to insulation and device blocking constraints.Select “high voltage output”. Verification / Alternative check: Design guides commonly employ series stacks for HV DC supplies (e.g., test sets, electrostatic precipitators) and parallel units for high current supplies (e.g., plating, drives). Why Other Options Are Wrong: High current: a parallel converter role. High voltage and high current: would need both series and parallel arrangements with complex sharing. “Improving efficiency” is not the defining characteristic of series connection. Common Pitfalls: Assuming series also increases current—without parallelism and sharing networks it does not. Final Answer: high voltage output

Question 16

Average load current in a phase-controlled rectifier: On which parameter(s) does the average load current delivered by a thyristor predominantly depend (assume mains frequency fixed and gate drive above threshold)?

Accepted Answer

Introduction / Context: In AC phase-controlled rectifiers, the average DC output is regulated by delaying the thyristor firing instant within each half-cycle. Understanding which control variable truly sets the average load current helps avoid common misconceptions about gate drive magnitude and repetition rate. Given Data / Assumptions: Supply frequency is fixed (e.g., 50/60 Hz). Gate current amplitude exceeds the necessary trigger threshold. Load is such that conduction follows the phase control behavior. Concept / Approach: The average load current is determined by the average DC output voltage, which depends on firing angle α (e.g., Vdc ∝ cos α in a full converter, or Vdc ∝ (1 + cos α) in a semi-converter). Firing frequency is tied to the mains and not a control variable. Once the gate pulse exceeds threshold, its magnitude does not set the steady-state average load current; it only guarantees reliable triggering. Step-by-Step Solution: Output average voltage is a function of α.Average current I_avg ≈ Vdc / R (or per equivalent) in steady state.Thus the dominant control parameter is α, not gate current magnitude or “firing frequency”. Verification / Alternative check: Converter equations in textbooks express Vdc as functions of α; gate drive amplitude appears only in trigger reliability, not in Vdc formulas. Why Other Options Are Wrong: Firing frequency: locked to mains; not an independent knob. Gate current magnitude: above threshold it does not change Vdc. “All of the above” is incorrect for the reasons above. Common Pitfalls: Assuming stronger gate pulses “force” more current; the converter is voltage-controlled by phase delay, the load sets the current from that voltage. Final Answer: firing angle

Question 17

Silicon Bilateral Switch (SBS): Evaluate the statement “An SBS can conduct in one direction only.” Is it true or false?

Accepted Answer

Introduction / Context: An SBS (Silicon Bilateral Switch) is a low-power triggering device used in lamp dimmers, motor controls, and as a trigger for TRIACs. Understanding whether it conducts bidirectionally is essential for AC triggering circuits. Given Data / Assumptions: Device name explicitly says “bilateral”. Used with AC where both polarities need triggering capability. Concept / Approach: Bilateral devices conduct in both directions once the breakover voltage is reached in either polarity. SBS and DIAC share this fundamental property, making them attractive as symmetrical triggers for TRIACs in AC circuits. Step-by-Step Solution: Identify SBS as a bidirectional trigger device.If one-direction conduction only, it would defeat symmetric AC triggering.Hence the statement is false. Verification / Alternative check: Device datasheets show symmetrical V-I curves with two breakover points of opposite polarity. Why Other Options Are Wrong: “True” contradicts the bilateral characteristic and practical AC applications. Common Pitfalls: Confusing SBS with SCR (unidirectional) or TRIAC (bidirectional power device). SBS is a trigger element, not a power switch. Final Answer: False

Question 18

High-frequency series inverter architecture: To obtain higher power at a given high frequency, a high-frequency series inverter is typically realized as

Accepted Answer

Introduction / Context: Series inverters use resonant commutation and are well suited to high-frequency operation. Practical power scaling often requires combining multiple inverter modules. The arrangement chosen affects current sharing, voltage stress, and reliability. Given Data / Assumptions: Each module is a series-type inverter (resonant commutation). Objective: increase output power at a fixed high frequency. Reasonable current sharing networks and synchronization are provided. Concept / Approach: Power scaling can be done by paralleling identical inverter modules so each shares a portion of the load current while maintaining the same output voltage. Putting series inverters “in series” raises voltage stress and is atypical for power scaling at high frequency. Therefore, multiple series-inverter modules are paralleled for higher current and power capability. Step-by-Step Solution: Select a base series inverter module.Paralleling N identical modules → same voltage, N× current capacity.Hence, “series inverters in parallel”. Verification / Alternative check: Many RF and induction heating supplies parallel resonant inverter modules to meet higher current demands while keeping device voltage within limits. Why Other Options Are Wrong: Series stacking increases voltage, not necessarily desired at HF. “Parallel inverters in parallel/series” does not reflect the base module being a series inverter; the naming in options is to test attention. Common Pitfalls: Equating “series inverter” with “series connection” of modules; the term describes commutation/LC placement, not how modules must be combined. Final Answer: a number of series inverters in parallel

Question 19

Series/parallel connection of thyristors (SCRs): When devices are connected in series and/or parallel for higher ratings, what derating is generally necessary to ensure safe operation?

Accepted Answer

Introduction / Context: To achieve higher system ratings, SCRs are often connected in series (for voltage) and in parallel (for current). Due to inevitable parameter tolerances and dynamic effects, equal sharing is imperfect, so prudent derating is applied in design. Given Data / Assumptions: Series strings require voltage sharing components (static and dynamic). Parallel groups require current sharing measures and thermal management. Device tolerances and transient behavior cannot be eliminated completely. Concept / Approach: In series connection, one device may see more voltage during transients → voltage derating. In parallel connection, one device may carry more current due to lower dynamic resistance or temperature effects → current derating. Therefore, both current and voltage derating are typically specified when combining devices for higher ratings. Step-by-Step Solution: Identify failure modes: overvoltage on one device; overcurrent on another.Mitigate with RC snubbers, balancing resistors/inductors, and derating margins.Conclude both current and voltage derating are necessary. Verification / Alternative check: Manufacturer application notes prescribe derating factors and balancing networks for both series and parallel configurations. Why Other Options Are Wrong: Only current or only voltage derating overlooks the other critical sharing problem. “None” ignores real-world tolerances and transient imbalances. Common Pitfalls: Assuming perfect sharing from idealized SPICE models; practical builds require margins and balancing networks. Final Answer: both current and voltage derating

Question 20

Highly inductive load with thyristor conduction In the circuit shown (load is highly inductive), the average value of the load current is 15 A and current is essentially continuous. What is the peak (maximum instantaneous) current through a conducti…

Accepted Answer

Introduction / Context: In phase-controlled rectifiers feeding highly inductive loads, the inductor smooths current ripple so strongly that the load (and device) current becomes practically constant over a cycle. Understanding how this affects device current ratings is essential for selecting proper thyristors and estimating losses. Given Data / Assumptions: Average load current = 15 A. Load is 'highly inductive', implying continuous, nearly ripple-free current. Conduction is by pairs of thyristors (or one device in midpoint), each device carrying the instantaneous load current during its conduction interval. Concept / Approach: With a very large inductance, di/dt is small and the current waveform is almost flat (constant). In this limiting case, the instantaneous current through the conducting path is essentially equal to the DC (average) value. Hence the peak current through a conducting thyristor equals the nearly constant load current value. Step-by-Step Solution: Model the load current as constant at I ≈ 15 A.During its conduction interval, a thyristor carries the same current as the load (series path).Therefore, i_T(peak) ≈ I = 15 A. Verification / Alternative check: If small ripple exists, the ripple typically oscillates around 15 A by a few percent; the peak would still be very close to 15 A for a 'highly inductive' description. RMS may differ from average, but peak is governed by ripple which is minimal here. Why Other Options Are Wrong: 15√2 A and 30 A assume sinusoidal or pulsed current peaks, not valid for strongly smoothed current. 22.5 A and 12 A are arbitrary and do not follow from the constant-current assumption. Common Pitfalls: Confusing device current with the instantaneous AC source current or transformer secondary current. The thyristor in series with an RLE load carries the load current during conduction, which is near-constant with a large inductor. Final Answer: 15 A

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