Q: Synchronous generators – brushless excitation concept In the brushless excitation system used on modern alternators, what is the type of the main exciter that feeds the rotating rectifier assembly to supply direct current to the rotor field?

Introduction / Context: Brushless excitation eliminates slip rings and carbon brushes by generating field current entirely on the rotor side. Understanding which machine serves as the main exciter is fundamental to alternator excitation systems used in power plants and large gensets. Given Data / Assumptions: Main exciter mechanically coupled to the alternator shaft. Output must be rectified on the rotor by a rotating diode wheel to feed the field winding with direct current. No brushes or slip rings are employed in brushless systems. Concept / Approach: The main exciter is typically a rotating armature alternator (AC generator). Its three-phase AC output is immediately rectified by the rotating rectifier (diode bridge) mounted on the same shaft, producing DC for the main machine field. A static or pilot PMG (permanent magnet generator) and AVR regulate exciter field to control alternator voltage. Step-by-Step Solution: Identify requirement: rotor-side DC without brushes → need on-rotor rectification.Rectifiers convert AC to DC → the exciter should produce AC.Therefore, the main exciter is an AC generator mounted on the rotor shaft. Verification / Alternative check: Standard OEM brushless diagrams show PMG → AVR → exciter field → exciter armature (AC) → rotating diodes → main field (DC). This chain confirms AC main exciter. Why Other Options Are Wrong: dc generator: would not need a rotating rectifier; contradicts typical brushless practice. either dc or ac generator: brushless scheme specifically uses AC plus rotating diodes. induction generator: requires external reactive support and is not used as the main exciter here. Common Pitfalls: Confusing the pilot PMG (a small alternator) with the main exciter; both are AC machines but serve different roles. Final Answer: ac generator

Q: DC chopper (step-down, time-ratio control) – average and RMS load current A step-down chopper feeds a purely resistive load R from input voltage V using duty cycle α (0 < α < 1). Under ideal switching, determine the average load current and the RMS …

Introduction / Context: Choppers control the average power delivered to loads by switching the supply rapidly. For a step-down (buck) chopper with a resistive load, recognizing the on/off voltage pattern allows direct computation of average and RMS current values, crucial for device and thermal design. Given Data / Assumptions: Input voltage V (constant, ideal source). Duty cycle α = T_on / T, with period T. Ideal switch, no ripple smoothing element (pure R load). Load voltage is V during T_on, and 0 during T_off. Concept / Approach: For a resistive load, i(t) = v(t)/R instantaneously. Average values depend on time-averaging of square wave segments, while RMS relies on the square root of the time-averaged square of the instantaneous values. Step-by-Step Solution: During T_on: i = V/R; during T_off: i = 0.Average current: I_avg = (1/T) * [(V/R)T_on + 0T_off] = (V/R)(T_on/T) = (αV)/R.RMS current: I_rms = √{ (1/T)[(V/R)^2T_on + 0T_off] } = (V/R)√(T_on/T) = (V/R)√α. Verification / Alternative check: Dimension check: both expressions have amperes. Limiting cases: α → 1 gives I_avg = I_rms = V/R (DC), α → 0 gives both → 0, as expected. Why Other Options Are Wrong: Swapping average and RMS leads to incorrect scaling with √α. Statements with both equal to V/R ignore duty cycle. Both equal to αV/R incorrectly treat RMS as an average. Common Pitfalls: Confusing average with RMS; forgetting that RMS of a chopped square wave scales with √α, not α. Final Answer: I_avg = (αV)/R, I_rms = (V/R)*√α

Q: Four-quadrant DC drive requirement For a DC motor drive that must operate in all four quadrants (motoring and braking in both directions), which converter arrangement is typically required?

Introduction / Context: Four-quadrant operation implies the drive can deliver positive/negative torque and handle positive/negative speed, enabling forward/reverse motoring and regenerative braking. Converter topology directly determines achievable quadrants. Given Data / Assumptions: Armature voltage reversal and/or current reversal are needed. Regeneration capability desired for braking in both directions. Standard line-commutated DC converters considered. Concept / Approach: A dual converter places two full converters back-to-back (anti-parallel) across the DC motor armature, enabling reversal of DC voltage polarity with proper current control and circulating-current management, thus providing all four quadrants. Step-by-Step Solution: Identify required functionality: both voltage and current can change sign.Single full converter provides two quadrants (motoring and regen in one direction); semi-converter is only one quadrant.Dual converter (two full converters) → four-quadrant control. Verification / Alternative check: Control charts of dual converters show Vdc sweeping positive and negative with controllable current, covering all quadrants. Why Other Options Are Wrong: Full converter alone: usually two quadrants. Semi-converter: typically one quadrant only. None of the above: incorrect—dual converter is standard. Common Pitfalls: Assuming mechanical reversal alone suffices; electrical four-quadrant control still needs dual converter or H-bridge chopper. Final Answer: dual converter

Q: DC chopper control of a separately excited DC motor A separately excited DC motor is fed from a DC source through a chopper. At 600 rpm and rated torque, the chopper duty cycle is 0.8. At the same 600 rpm but at half the rated torque, what is the mo…

Introduction / Context: For a DC motor under chopper control, average armature voltage V_avg ≈ E + I_aR_a at a given speed. At constant speed, back EMF E is approximately constant. Reducing torque reduces armature current I_a, which reduces the resistive drop I_aR_a, hence the required V_avg and duty cycle decrease moderately. Given Data / Assumptions: Speed held at 600 rpm in both cases → back EMF E nearly unchanged. Rated torque → I_a,r; half torque → I_a ≈ 0.5I_a,r. V_avg = αV_s = E + I_aR_a. R_a drop is smaller than E but not negligible. Concept / Approach: Duty cycle scales with required V_avg. If current halves, the I_aR_a term halves while E is constant; thus α reduces slightly from 0.8, not drastically. A modest reduction to about 0.75 aligns with typical ratios when E dominates the sum. Step-by-Step Solution: Case 1: α₁V_s = E + I_rR_a (α₁ = 0.8).Case 2: α₂V_s = E + (0.5 I_r)R_a.Subtract equations to eliminate E: (α₁ − α₂)V_s = 0.5 I_rR_a → α₂ = α₁ − (0.5 I_rR_a)/V_s.Since (I_rR_a)/V_s is modest, α₂ is slightly less than 0.8, around 0.75. Verification / Alternative check: In the limit R_a → 0, α would be unchanged at 0.8 (since E dominates). For large R_a, α would drop more than a little. Hence 0.75 is a reasonable mid-case. Why Other Options Are Wrong: 0.9: would increase voltage despite reduced torque—illogical. 0.8: exact equality would imply zero R_a drop, which is unlikely per the prompt. 0.6: indicates excessive drop; inconsistent with “R_a drop smaller than E”. Common Pitfalls: Assuming duty cycle scales directly with torque; it actually depends on both E and I_aR_a. Final Answer: 0.75

Q: Three-phase full converter with firing angle α < 60° – instantaneous DC output polarity In a 3-phase full converter (line-commutated) supplying a DC load, if the firing angle is less than 60°, how does the instantaneous DC output voltage behave?

Introduction / Context: The instantaneous output of a 3-phase full converter is formed from segments of line-to-line voltages selected by controlled thyristors. The firing angle α sets when each device conducts and thus determines polarity and average value of the output voltage. Given Data / Assumptions: Line-commutated 3-phase full converter. Firing angle α 90°). Step-by-Step Solution: At small α, device triggering occurs near natural commutation points, capturing positive line-to-line segments.With continuous current, commutation overlap does not reverse the output polarity for α < 60°.Hence the instantaneous output remains positive throughout. Verification / Alternative check: Waveforms from standard texts show unipolar positive DC for α below about 60°–90°, with average Vdc proportional to cos α and still positive for α 90°) with instantaneous polarity at small firing angles. Final Answer: will have positive part only

Question 1

Load-commutated chopper: commutating elements In a load-commutated (auxiliary commutated) thyristor chopper, which set of components typically forms the commutation network to force turn-off?

Accepted Answer

Introduction / Context: Thyristors cannot turn off by gate command; current must be forced to zero for a finite time. In DC–DC choppers, where line commutation is unavailable, forced (auxiliary) commutation networks are employed. Understanding the typical commutation components is central to chopper design and classification. Given Data / Assumptions: Thyristor-based chopper (not self-commutated IGBT/MOSFET). Load-commutated topology using an auxiliary path. Intent is to create a reverse or zero current through the main SCR for its turn-off time. Concept / Approach: Classic forced commutation employs an LC network pre-charged via switching. At the desired instant, an auxiliary SCR connects the charged capacitor so that reverse current flows through the main SCR, reducing its current to zero and applying reverse bias during the turn-off interval. Step-by-Step Solution: Charge a commutating capacitor to a known voltage.Fire an auxiliary thyristor to connect the capacitor in the proper polarity.The LC path drives reverse current through the main SCR → forced turn-off. Verification / Alternative check: Standard commutation classes (A–E for choppers) show auxiliary SCR plus LC for many load-commutated schemes. Why Other Options Are Wrong: A capacitor alone cannot provide controlled resonant current; an LC without the auxiliary device lacks the switching action; transformer-only is not standard forced-commutation here. Common Pitfalls: Assuming gate control is enough for SCR turn-off; ignoring the need for reverse bias time (t_q). Final Answer: A capacitor, an inductor, and an auxiliary thyristor

Question 2

Transistor vs thyristor conduction drop: assertion–reason Assertion (A): During conduction, a power transistor typically has a lower voltage drop than a thyristor (SCR). Reason (R): Transistors are manufactured with very high voltage and current rat…

Accepted Answer

Introduction / Context: Conduction loss is a major efficiency factor. Typical on-state drops: SCRs exhibit a relatively fixed on-state drop (approximately 1–2 V for many devices), whereas power transistors (BJT/IGBT/MOSFET) show conduction characteristics that can yield lower drop at comparable currents (e.g., V_CE(sat) for BJTs, IR_DS(on) for MOSFETs). The reason behind lower drop is device physics, not simply the available ratings. Given Data / Assumptions: Comparable current levels and well-sized devices. Thyristor in full conduction (on-state). Transistor operated within safe area and proper drive. Concept / Approach: Transistors can exhibit lower instantaneous conduction voltage due to either low saturation voltage (BJTs) or low on-resistance (MOSFETs). Thyristors, while efficient at very high power, have a relatively fixed on-state drop that may be higher for the same current. The reason in the prompt, however, speaks to manufacturing ratings, which do not directly explain the lower conduction drop. Step-by-Step Solution: Assess A: Lower conduction drop for transistors → generally true at many operating points.Assess R: “High ratings” are not the cause of lower drop → incorrect explanation.Therefore: A true, R false. Verification / Alternative check: Datasheets show typical SCR on-state drops around 1–2 V; MOSFET drops are IR_DS(on) (often much less for moderate currents); BJTs show V_CE(sat) < 1 V in many cases at rated currents. Why Other Options Are Wrong: (a) Links lower drop to ratings incorrectly; (b) says R true, but it does not explain A; (d) A is not wrong; (e) both wrong is inconsistent with data. Common Pitfalls: Assuming device ratings dictate conduction loss; overlooking the strong dependence on current density and temperature. Final Answer: A is correct but R is wrong

Question 3

Effect of source inductance (overlap) in an n-pulse rectifier An n-pulse line-commutated rectifier is fed from a source with inductance L. If the average load current is I0, what is the average output voltage drop attributable to commutation overlap?

Accepted Answer

Introduction / Context: In line-commutated converters, finite source inductance causes a commutation overlap angle during which two valves conduct simultaneously. This overlap reduces the average DC output voltage versus the ideal value. Quantifying this drop is essential for accurate performance prediction of rectifiers, drives, and HVDC terminals. Given Data / Assumptions: n-pulse rectifier (n = 6 for three-phase bridge, n = 12 for series/parallel 6-pulse pairs, etc.). Source inductance L represents commutating inductance per phase path. Average DC load current is I0 and assumed reasonably ripple-free. Steady operating point; small-signal effects and resistance neglected. Concept / Approach: During overlap, part of the line-to-line voltage is dropped across the source inductances rather than appearing at the DC output. Classical analysis integrates the line voltage lost while current transfers from the outgoing to the incoming valve. The result is a proportional reduction in average DC voltage that scales with n, ω, L, and I0. Step-by-Step Solution: Ideal average output: Vd,ideal = Vd0 * cos(α) (for controlled case).With overlap μ, the practical average is Vd = Vd0 * cos(α) − ΔVoverlap.Classical result: ΔVoverlap = (n * ω * L * I0) / π. Verification / Alternative check: For a 6-pulse bridge, set n = 6 → ΔV = 6 ω L I0 / π, which matches standard textbook expressions. Increasing L or I0 increases the drop; higher pulse number increases commutations per cycle, thus larger total loss per electrical period. Why Other Options Are Wrong: (a) and (b) omit the dependence on pulse number n; (c) contains an extra 2 in the denominator leading to a lower, incorrect value; (e) has inverse dependence on n, which contradicts the physics (more commutations → more loss). Common Pitfalls: Confusing overlap-caused drop with resistive drop; ignoring that L here is the effective commutating inductance per path; forgetting the linear dependence on n. Final Answer: ΔV = (n ω L I0) / π

Question 4

Synchronous generators – brushless excitation concept In the brushless excitation system used on modern alternators, what is the type of the main exciter that feeds the rotating rectifier assembly to supply direct current to the rotor field?

Accepted Answer

Introduction / Context: Brushless excitation eliminates slip rings and carbon brushes by generating field current entirely on the rotor side. Understanding which machine serves as the main exciter is fundamental to alternator excitation systems used in power plants and large gensets. Given Data / Assumptions: Main exciter mechanically coupled to the alternator shaft. Output must be rectified on the rotor by a rotating diode wheel to feed the field winding with direct current. No brushes or slip rings are employed in brushless systems. Concept / Approach: The main exciter is typically a rotating armature alternator (AC generator). Its three-phase AC output is immediately rectified by the rotating rectifier (diode bridge) mounted on the same shaft, producing DC for the main machine field. A static or pilot PMG (permanent magnet generator) and AVR regulate exciter field to control alternator voltage. Step-by-Step Solution: Identify requirement: rotor-side DC without brushes → need on-rotor rectification.Rectifiers convert AC to DC → the exciter should produce AC.Therefore, the main exciter is an AC generator mounted on the rotor shaft. Verification / Alternative check: Standard OEM brushless diagrams show PMG → AVR → exciter field → exciter armature (AC) → rotating diodes → main field (DC). This chain confirms AC main exciter. Why Other Options Are Wrong: dc generator: would not need a rotating rectifier; contradicts typical brushless practice. either dc or ac generator: brushless scheme specifically uses AC plus rotating diodes. induction generator: requires external reactive support and is not used as the main exciter here. Common Pitfalls: Confusing the pilot PMG (a small alternator) with the main exciter; both are AC machines but serve different roles. Final Answer: ac generator

Question 5

DC chopper (step-down, time-ratio control) – average and RMS load current A step-down chopper feeds a purely resistive load R from input voltage V using duty cycle α (0 < α < 1). Under ideal switching, determine the average load current and the RMS …

Accepted Answer

Introduction / Context: Choppers control the average power delivered to loads by switching the supply rapidly. For a step-down (buck) chopper with a resistive load, recognizing the on/off voltage pattern allows direct computation of average and RMS current values, crucial for device and thermal design. Given Data / Assumptions: Input voltage V (constant, ideal source). Duty cycle α = T_on / T, with period T. Ideal switch, no ripple smoothing element (pure R load). Load voltage is V during T_on, and 0 during T_off. Concept / Approach: For a resistive load, i(t) = v(t)/R instantaneously. Average values depend on time-averaging of square wave segments, while RMS relies on the square root of the time-averaged square of the instantaneous values. Step-by-Step Solution: During T_on: i = V/R; during T_off: i = 0.Average current: I_avg = (1/T) * [(V/R)T_on + 0T_off] = (V/R)(T_on/T) = (αV)/R.RMS current: I_rms = √{ (1/T)[(V/R)^2T_on + 0T_off] } = (V/R)√(T_on/T) = (V/R)√α. Verification / Alternative check: Dimension check: both expressions have amperes. Limiting cases: α → 1 gives I_avg = I_rms = V/R (DC), α → 0 gives both → 0, as expected. Why Other Options Are Wrong: Swapping average and RMS leads to incorrect scaling with √α. Statements with both equal to V/R ignore duty cycle. Both equal to αV/R incorrectly treat RMS as an average. Common Pitfalls: Confusing average with RMS; forgetting that RMS of a chopped square wave scales with √α, not α. Final Answer: I_avg = (αV)/R, I_rms = (V/R)*√α

Question 6

Four-quadrant DC drive requirement For a DC motor drive that must operate in all four quadrants (motoring and braking in both directions), which converter arrangement is typically required?

Accepted Answer

Introduction / Context: Four-quadrant operation implies the drive can deliver positive/negative torque and handle positive/negative speed, enabling forward/reverse motoring and regenerative braking. Converter topology directly determines achievable quadrants. Given Data / Assumptions: Armature voltage reversal and/or current reversal are needed. Regeneration capability desired for braking in both directions. Standard line-commutated DC converters considered. Concept / Approach: A dual converter places two full converters back-to-back (anti-parallel) across the DC motor armature, enabling reversal of DC voltage polarity with proper current control and circulating-current management, thus providing all four quadrants. Step-by-Step Solution: Identify required functionality: both voltage and current can change sign.Single full converter provides two quadrants (motoring and regen in one direction); semi-converter is only one quadrant.Dual converter (two full converters) → four-quadrant control. Verification / Alternative check: Control charts of dual converters show Vdc sweeping positive and negative with controllable current, covering all quadrants. Why Other Options Are Wrong: Full converter alone: usually two quadrants. Semi-converter: typically one quadrant only. None of the above: incorrect—dual converter is standard. Common Pitfalls: Assuming mechanical reversal alone suffices; electrical four-quadrant control still needs dual converter or H-bridge chopper. Final Answer: dual converter

Question 7

DC chopper control of a separately excited DC motor A separately excited DC motor is fed from a DC source through a chopper. At 600 rpm and rated torque, the chopper duty cycle is 0.8. At the same 600 rpm but at half the rated torque, what is the mo…

Accepted Answer

Introduction / Context: For a DC motor under chopper control, average armature voltage V_avg ≈ E + I_aR_a at a given speed. At constant speed, back EMF E is approximately constant. Reducing torque reduces armature current I_a, which reduces the resistive drop I_aR_a, hence the required V_avg and duty cycle decrease moderately. Given Data / Assumptions: Speed held at 600 rpm in both cases → back EMF E nearly unchanged. Rated torque → I_a,r; half torque → I_a ≈ 0.5I_a,r. V_avg = αV_s = E + I_aR_a. R_a drop is smaller than E but not negligible. Concept / Approach: Duty cycle scales with required V_avg. If current halves, the I_aR_a term halves while E is constant; thus α reduces slightly from 0.8, not drastically. A modest reduction to about 0.75 aligns with typical ratios when E dominates the sum. Step-by-Step Solution: Case 1: α₁V_s = E + I_rR_a (α₁ = 0.8).Case 2: α₂V_s = E + (0.5 I_r)R_a.Subtract equations to eliminate E: (α₁ − α₂)V_s = 0.5 I_rR_a → α₂ = α₁ − (0.5 I_rR_a)/V_s.Since (I_rR_a)/V_s is modest, α₂ is slightly less than 0.8, around 0.75. Verification / Alternative check: In the limit R_a → 0, α would be unchanged at 0.8 (since E dominates). For large R_a, α would drop more than a little. Hence 0.75 is a reasonable mid-case. Why Other Options Are Wrong: 0.9: would increase voltage despite reduced torque—illogical. 0.8: exact equality would imply zero R_a drop, which is unlikely per the prompt. 0.6: indicates excessive drop; inconsistent with “R_a drop smaller than E”. Common Pitfalls: Assuming duty cycle scales directly with torque; it actually depends on both E and I_aR_a. Final Answer: 0.75

Question 8

Three-phase full converter with firing angle α < 60° – instantaneous DC output polarity In a 3-phase full converter (line-commutated) supplying a DC load, if the firing angle is less than 60°, how does the instantaneous DC output voltage behave?

Accepted Answer

Introduction / Context: The instantaneous output of a 3-phase full converter is formed from segments of line-to-line voltages selected by controlled thyristors. The firing angle α sets when each device conducts and thus determines polarity and average value of the output voltage. Given Data / Assumptions: Line-commutated 3-phase full converter. Firing angle α < 60°. Typical RL load with sufficiently continuous current. Concept / Approach: For α in the rectification range (0° to < 90°), the converter selects the most positive line-to-line voltages at appropriate intervals, producing a unidirectional (positive) instantaneous DC voltage. Negative instantaneous segments arise as α increases well beyond 60–90° or under inverter operation (α > 90°). Step-by-Step Solution: At small α, device triggering occurs near natural commutation points, capturing positive line-to-line segments.With continuous current, commutation overlap does not reverse the output polarity for α < 60°.Hence the instantaneous output remains positive throughout. Verification / Alternative check: Waveforms from standard texts show unipolar positive DC for α below about 60°–90°, with average Vdc proportional to cos α and still positive for α < 90°. Why Other Options Are Wrong: “Both positive and negative parts” occurs near inversion (α large) or with discontinuous current. “May have both” is too vague for α < 60° under stated assumptions. “Negative part if load is inductive” is incorrect; inductance sustains current but does not flip instantaneous polarity at small α. Common Pitfalls: Confusing average negative voltage (α > 90°) with instantaneous polarity at small firing angles. Final Answer: will have positive part only

Question 9

Power MOSFET versus BJT – loss characteristics Compared to a bipolar junction transistor (BJT), a power MOSFET generally exhibits which combination of switching and conduction losses (for similar voltage/current ratings)?

Accepted Answer

Introduction / Context: Selecting between MOSFETs and BJTs hinges on their dynamic and static loss mechanisms. BJTs are current-driven devices with charge storage; MOSFETs are voltage-driven with resistive conduction paths characterized by Rds_on. Given Data / Assumptions: Similar voltage and current range. Hard-switched applications. Room-temperature comparisons (Rds_on increases with temperature). Concept / Approach: MOSFETs switch faster with negligible charge storage tail, leading to lower switching losses. However, conduction loss is I^2Rds_on, which can exceed BJT conduction loss (≈ IVce(sat)) in some ranges, especially at higher currents or elevated temperatures. Step-by-Step Solution: Switching: MOSFET → low charge storage → lower switching loss than BJT.Conduction: MOSFET uses resistive channel → higher I^2*R losses than BJT’s quasi-constant Vce(sat) in many cases.Therefore: lower switching but higher conduction losses. Verification / Alternative check: Loss curves in datasheets show MOSFET superiority at high frequency, while BJTs can be advantageous for conduction at certain currents if switching frequency is low. Why Other Options Are Wrong: Higher switching losses for MOSFET are generally untrue. Lower conduction losses for MOSFET are not typical at high current unless Rds_on is very low. Both lower is unrealistic across wide ranges. Common Pitfalls: Ignoring temperature rise on Rds_on; not accounting for gate charge in driver design, which affects efficiency but not conduction loss directly. Final Answer: lower switching losses and higher conduction losses

Question 10

Pulse-width modulation (PWM) strategy in choppers In PWM control of a step-down chopper, how are the switching period T and the on-time T_on typically managed to regulate the output?

Accepted Answer

Introduction / Context: PWM is the most common control method for DC choppers and switch-mode supplies. It modulates the duty ratio to control the average output while operating at a fixed switching frequency to simplify filtering and EMI design. Given Data / Assumptions: Step-down (buck) chopper topology. Objective: regulate average output voltage/current. Practical constraint: fixed switching frequency for magnetics and compliance. Concept / Approach: Holding T constant fixes the switching frequency f_s = 1/T. Varying T_on changes duty cycle α = T_on/T, thereby controlling the average output proportionally (ideally V_out = αV_in for resistive load). Step-by-Step Solution: Set T = constant (fixed f_s).Adjust T_on to achieve desired α (0–1).Output responds linearly (ideal case): V_avg ∝ α. Verification / Alternative check: Many controllers use sawtooth carrier versus error-amplified signal; the comparator creates variable T_on pulses at fixed T. Why Other Options Are Wrong: Varying T complicates magnetics and EMI; not standard in PWM. Varying both is unnecessary and adds complexity. “Either” is too broad; classic PWM uses fixed T, variable T_on. Common Pitfalls: Confusing PWM with frequency modulation (PFM), which varies T instead of duty at fixed T. Final Answer: T is kept constant and T_on is varied

Question 11

Three-phase bridge inverter – device current in 180° vs 120° conduction All other parameters being the same, how does the peak thyristor current in a 3-phase bridge inverter operating in 180° conduction mode compare with that in 120° conduction mode?

Accepted Answer

Introduction / Context: Inverters can be modulated for 120° or 180° conduction per device. The conduction interval and number of devices simultaneously conducting determine how phase and line currents split among devices, influencing peak device current ratings. Given Data / Assumptions: Same DC link voltage and load conditions. Bridge inverter using thyristors (or ideal switches). Comparison strictly between conduction schemes. Concept / Approach: With 180° conduction, two devices in each leg can conduct, and three devices are typically on at any instant, changing the current sharing relative to 120° mode (two devices on at a time). The distribution leads to higher peak current demands on individual devices in 180° mode for the same load current envelope. Step-by-Step Solution: Identify conduction pattern: 120° → each device conducts 120°; 180° → each conducts 180° and overlaps more segments.For the same output, current per device peaks higher in 180° mode due to overlap and sharing patterns.Hence, peak thyristor current is higher in 180° conduction. Verification / Alternative check: Device current waveforms in standard references show increased peak in 180° mode for identical load conditions when compared to 120° mode. Why Other Options Are Wrong: “Same” ignores conduction overlap differences. “Lower” contradicts typical current sharing outcomes. “Any of the above” is non-deterministic and incorrect. Common Pitfalls: Conflating RMS and peak values; depending on load nature, RMS may change differently, but the question asks about peak device current. Final Answer: has higher peak thyristor current than in 120° mode of operation

Question 12

Chopper classes – operating quadrants According to the standard DC chopper classification, a Class D chopper operates in which quadrants of the voltage–current (V–I) plane?

Accepted Answer

Introduction / Context: DC choppers are categorized by which quadrants they can realize for load voltage and current. This determines whether the converter can handle motoring/braking and directionality of current and voltage. Given Data / Assumptions: Standard chopper types A, B, C, D. Ideal switching and appropriate freewheel paths assumed. Concept / Approach: Class A: 1st quadrant only. Class B: 2nd quadrant only (regenerative). Class C: 1st and 2nd quadrants (two-quadrant). Class D: 1st and 4th quadrants (two-quadrant with voltage reversal capability), often implemented as an H-bridge arrangement for bidirectional voltage with unidirectional current each path. Step-by-Step Solution: Identify Class D topology → allows load voltage polarity reversal while controlling current direction paths.Hence, operation occurs in quadrants I and IV. Verification / Alternative check: Typical V–I locus diagrams in textbooks confirm Class D covers I and IV, enabling forward motoring and reverse motoring (depending on implementation). Why Other Options Are Wrong: First only or second only correspond to Class A or B, not D. All quadrants require a four-quadrant topology beyond simple Class D. Common Pitfalls: Mixing up Class C (I and II) with Class D (I and IV). Final Answer: can operate in either first or fourth quadrant

Question 13

Sinusoidal PWM (SPWM) for inverters – pulse widths In sinusoidal PWM control of voltage-source inverters, are the widths of the individual gating pulses within a half cycle all equal?

Accepted Answer

Introduction / Context: Sinusoidal PWM (SPWM) shapes the fundamental output and controls harmonics by comparing a high-frequency triangular carrier with a low-frequency sinusoidal modulating signal. The resulting pulse widths convey the sinusoidal information to the load after filtering. Given Data / Assumptions: Voltage-source inverter using carrier-based SPWM. Constant switching frequency carrier (triangular wave). Sinusoidal reference for desired fundamental. Concept / Approach: Where the sinusoidal reference exceeds the carrier, the switch turns on; where it is lower, it turns off. Because the sinusoid varies over the half-cycle, the intersections occur at non-uniform points, creating pulses whose widths increase near the sine peak and decrease toward the zero crossings. Step-by-Step Solution: Compare sinusoidal modulating wave v_m(t) with triangular carrier v_c(t).When v_m(t) > v_c(t), gate = on; otherwise, gate = off.Since v_m(t) changes continuously, each pulse width differs, following the sine envelope. Verification / Alternative check: SPWM spectra show a controllable fundamental with sidebands around multiples of the carrier; equal pulse width would correspond to square-wave or uniform PWM, not sinusoidal PWM. Why Other Options Are Wrong: “True” would imply uniform pulse width, which is not how SPWM encodes the sinusoid. Common Pitfalls: Confusing SPWM with fixed-width, variable-position schemes, or with selective harmonic elimination where a few notches are placed deliberately. Final Answer: False

Question 14

SCR in forward-blocking (off-state) When a silicon-controlled rectifier (SCR) is forward-biased but still blocking (no gate trigger applied), how is the externally applied forward voltage distributed across its internal PN junctions?

Accepted Answer

Introduction / Context: An SCR (PNPN device) in forward-blocking condition has its anode positive with respect to cathode, but no gate current is applied. Understanding which junction blocks the forward voltage is foundational for device physics and protection design. Given Data / Assumptions: Device: SCR with junctions J1 (P–N), J2 (N–P), J3 (P–N) from anode to cathode. Forward-blocking: anode positive, gate open (no trigger), device non-conducting. Quasi-static conditions, neglecting leakage and transient effects. Concept / Approach: Under forward bias without gate triggering, junctions J1 and J3 are forward biased, whereas the central junction J2 is reverse biased. The reverse-biased junction sustains almost the entire applied forward voltage, keeping the device in the blocking state until breakover or gate triggering occurs. Step-by-Step Solution: Identify bias: anode positive → J1 forward, J2 reverse, J3 forward.Reverse-biased J2 supports the applied forward voltage to prevent conduction.Therefore, the applied forward voltage effectively appears across one junction, namely J2. Verification / Alternative check: Two-transistor model equivalence shows insufficient regeneration without gate injection; thus central junction remains reverse-biased and holds the voltage. Why Other Options Are Wrong: Equal sharing among two or three junctions (options b, c) contradicts the biasing state (only J2 is reverse-biased). Option d is vague and not representative of steady forward-blocking physics. Common Pitfalls: Confusing forward-blocking with reverse-blocking, and assuming uniform voltage sharing irrespective of bias polarity. Final Answer: Appears across only one junction

Question 15

3-phase bridge inverter with star-connected resistive load A three-phase bridge inverter is supplied from a 400 V DC source (battery). The load is star-connected with resistance 10 Ω per phase. What is the peak value of the thyristor current under s…

Accepted Answer

Introduction / Context: Six-step three-phase bridge inverters apply stepped line-to-line voltages to a three-phase load. For a star-connected resistive load, current is instantaneously proportional to phase voltage. Knowing the peak device current is essential for device selection and thermal design. Given Data / Assumptions: DC link Vdc = 400 V. Load: star-connected, purely resistive, R_phase = 10 Ω, no neutral. Ideal switches, six-step (square) operation, 120° conduction. Concept / Approach: In a 3-phase bridge inverter without neutral, the phase voltage waveform takes levels of ±(2/3)Vdc and ±(1/3)Vdc relative to the load neutral, derived from the instantaneous pole potentials. For a resistive load, the phase current follows the phase voltage. The peak phase voltage magnitude is 2Vdc/3, so the peak current is that voltage divided by the phase resistance. Step-by-Step Solution: V_phase,peak = 2/3 * Vdc = 2/3 * 400 = 266.67 V.I_phase,peak = V_phase,peak / R_phase = 266.67 / 10 = 26.667 A.Therefore, the peak thyristor current ≈ 26.7 A. Verification / Alternative check: Device conduction path in each 60° interval includes one upper and one lower switch; each device current equals the phase current during its conduction interval, so peak device current equals peak phase current for a resistive load. Why Other Options Are Wrong: 20 A and 24 A underestimate the current because they assume lower phase voltage levels; 30 A overestimates by assuming Vdc/ R_phase or ignoring the 2/3 factor. Common Pitfalls: Confusing line-to-line with phase voltage, and using Vdc directly instead of 2Vdc/3 for the peak phase level in a six-step inverter without a neutral connection. Final Answer: 26.7 A

Question 16

Maximum junction temperature rating of SCRs What is the typical specified maximum junction temperature (approximate) for power thyristors (SCRs) under data-sheet limits?

Accepted Answer

Introduction / Context: The junction temperature limit of an SCR dictates safe operating area, heat-sink sizing, and protection strategies. Exceeding this temperature risks thermal runaway and device failure. Given Data / Assumptions: General-purpose and fast SCRs used in industrial power converters. Typical commercial and industrial device ratings from standard data sheets. Concept / Approach: Power devices have an absolute maximum junction temperature, Tj(max). Many silicon SCRs specify Tj(max) around 110–125°C (often 125°C nominal), though special devices may allow up to 150°C. A conservative typical value to remember for standard SCRs is about 120°C. Step-by-Step Solution: Review typical data sheets: Tj(max) ≈ 110–125°C (sometimes 150°C).Choose the standard approximate rating among options → 120°C. Verification / Alternative check: Design norms for heat-sinking usually target junction temperatures well below Tj(max), providing margin for transients and ambient variation. Why Other Options Are Wrong: 50°C and 80°C are far too low for silicon power devices; 200°C exceeds common silicon SCR ratings and approaches levels suitable only for specialized technologies. Common Pitfalls: Confusing case temperature with junction temperature; relying on ambient temperature limits rather than the actual junction rating. Final Answer: 120°C

Question 17

3-phase full-wave AC regulator with star-connected resistive load A three-phase full-wave AC voltage regulator feeds a star-connected purely resistive load. If the firing angle is 75°, how many thyristors conduct simultaneously (at a given instant)?

Accepted Answer

Introduction / Context: Three-phase full-wave AC regulators control the RMS line-to-load voltage by phase-angle firing. The instantaneous number of conducting devices affects input current shape, harmonic content, and device current stress. Given Data / Assumptions: 3-phase, star-connected resistive load (no neutral). Full-wave regulator using anti-parallel SCR pairs per phase (six SCRs total). Firing angle α = 75°, steady-state operation. Concept / Approach: For a resistive 3-phase load without neutral, conduction occurs in phases where the fired SCRs connect the appropriate line potentials to maintain KCL at the isolated star point. For small α (typically α ≤ 60°), three devices may conduct. When α exceeds 60°, conduction intervals shrink and the star point potential adjusts such that only two SCRs conduct at a time over most instants. Step-by-Step Solution: Identify regime: α = 75° > 60°.In this regime, instantaneous conduction generally involves two SCRs (one from the positive group and one from the negative group).Therefore, the correct count is two devices conducting simultaneously. Verification / Alternative check: Phase-plane and star-point potential analyses for α > 60° show two-device conduction is dominant for resistive star loads. Why Other Options Are Wrong: 1 device cannot satisfy KCL without neutral; 3 devices are typical only for α ≤ 60°; 4 devices do not occur in this configuration with a star-connected resistive load. Common Pitfalls: Assuming three-device conduction independent of α, or confusing delta vs. star load behavior. Final Answer: 2

Question 18

Cycloconverter frequency control principle In a cycloconverter, can the output frequency be changed by adjusting the firing angle control strategy?

Accepted Answer

Introduction / Context: A cycloconverter synthesizes a low-frequency AC output directly from a higher-frequency AC input by phase-controlled firing of SCRs. Frequency control is central to its operation in large variable-speed drives and low-frequency furnaces. Given Data / Assumptions: Line-commutated cycloconverter using phase-controlled converter groups. Objective: change output frequency relative to the supply frequency. Concept / Approach: By modulating the firing angle sequences of the positive and negative converter groups, the cycloconverter constructs segments of the input waveform to approximate a lower-frequency sinusoid. The timing pattern of gating pulses—effectively the average voltage over controlled windows—dictates the instantaneous output, hence the output frequency. Step-by-Step Solution: Define desired output frequency f_out.Schedule firing pulses so that successive positive/negative segments follow a sinusoidal envelope at f_out.Varying the firing angle trajectory versus time changes both the effective amplitude and the rate of alternation → hence the output frequency. Verification / Alternative check: Practical cycloconverter control schemes (cosine-wave crossing methods) implement time-varying phase angles to generate low-frequency outputs well below the input frequency. Why Other Options Are Wrong: Stating that frequency is fixed ignores the essence of cycloconverter modulation by controlled firing sequences. Common Pitfalls: Confusing amplitude control (RMS) with frequency synthesis; forgetting the constraints of line commutation that limit maximum attainable f_out (typically f_out ≤ about one-third of supply frequency). Final Answer: True

Question 19

SCR triggering terminology The minimum gate current that reliably turns on an SCR under specified conditions is called:

Accepted Answer

Introduction / Context: Accurate terminology for SCR parameters is crucial for gate-drive design and interpreting data sheets. Given Data / Assumptions: Standard SCR parameters: I_GT (gate trigger current), I_H (holding current), V_BO or I_BO (breakover), etc. Turn-on achieved via gate injection at specified junction temperature and anode-cathode voltage. Concept / Approach: Gate trigger current (I_GT) is the minimum gate current that causes the SCR to switch from the forward-blocking state to the on-state under specified conditions. This is distinct from holding current I_H, the minimum anode current required to keep the SCR latched after turn-on. Step-by-Step Solution: Identify the parameter definition from data-sheet conventions.Match the description “minimum gate current to turn on” → I_GT = trigger current. Verification / Alternative check: Manufacturers specify I_GT across quadrants, temperature, and dv/dt; it is the design basis for gate drive amplitude. Why Other Options Are Wrong: Holding current relates to maintaining conduction; breakover current pertains to turning on without gate via exceeding breakover; “junction current” is not a standard parameter name. Common Pitfalls: Confusing trigger current with latching or holding currents; designing a gate drive too close to I_GT without margin. Final Answer: Trigger current

Question 20

Parallel inverter switching frequency per device In a classic parallel inverter (class-B type commutation), how many times is each thyristor turned on during each output cycle?

Accepted Answer

Introduction / Context: A parallel inverter uses commutating components in a parallel resonant path to turn SCRs off and generate an AC output across the load. Understanding device gating counts per cycle is useful for gate-drive timing and loss estimation. Given Data / Assumptions: Idealized parallel inverter producing a quasi-square AC output. Two main SCRs that conduct alternately each half-cycle. Concept / Approach: In each half cycle of the output, one SCR conducts while the commutation network prepares to transfer conduction to the other device. Thus, over one full output cycle (two half cycles), each SCR is gated once and conducts for one half cycle. Step-by-Step Solution: Half-cycle 1 → SCR1 on; half-cycle 2 → SCR2 on.Therefore, per full cycle, SCR1 and SCR2 are each turned on once. Verification / Alternative check: Typical timing diagrams for parallel inverters show one gate pulse per device per cycle, with natural commutation via the resonant components. Why Other Options Are Wrong: Options suggesting more than one turn-on per device per cycle do not reflect the standard two-device, square-wave operation without PWM. Common Pitfalls: Confusing parallel inverters with high-frequency PWM inverters where devices may commutate multiple times per fundamental cycle. Final Answer: Each thyristor is turned on once during each cycle

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