Q: Modified series inverter: attainable output frequency In a modified series inverter (using forced or auxiliary commutation extensions), how does the achievable output frequency compare with the natural damped resonant frequency of the L–C commutatin…

Introduction / Context: Series inverters use resonant L–C commutation. Modified series inverters employ auxiliary switches/paths so that commutation is not limited strictly by the natural resonant period, enabling higher fundamental output frequency. Given Data / Assumptions: Presence of a modified (assisted) commutation network. Switches and diodes ideal for conceptual analysis. Load is compatible with resonant current pulses. Concept / Approach: A basic series inverter's period is largely set by L and C, giving a damped oscillation with natural frequency f_d. By adding auxiliary commutation, the devices can be turned off and re-fired earlier than the natural half-cycle, effectively increasing the repetition rate of power pulses and therefore the output frequency. Step-by-Step Solution: Define natural frequency: f_d ≈ 1 / (2π * sqrt(L * C)) (ignoring damping).In a simple series inverter, switching follows this half-sinusoid → output frequency ties to f_d.Modified inverter introduces controlled commutation so that the next cycle can begin before the resonant current completely decays.Thus, achievable f_out can exceed f_d when designed to commutate early. Verification / Alternative check: Waveforms of modified circuits show shortened current pulses and increased repetition based on gating strategy rather than pure LC decay. Why Other Options Are Wrong: Less than / cannot exceed: True only for unassisted basic series inverters. Any of the above: Over-broad and contradicts known capability of modified topologies. Common Pitfalls: Assuming natural resonance always caps frequency; in practice, controlled commutation can reshape intervals within device limits. Final Answer: the output frequency can be even higher than the frequency of damped oscillations

Q: Single-phase cycloconverter with centre-tapped input transformer For a single-phase cycloconverter that uses a centre-tapped input transformer, how many thyristors are required to realize both positive and negative converters?

Introduction / Context: A single-phase cycloconverter synthesizes low-frequency AC directly from a higher-frequency AC source by phase-controlling segments of the input. With a centre-tapped transformer, separate controlled paths are used for the positive and negative groups. Given Data / Assumptions: Centre-tapped secondary providing two half-windings. Need for controlled positive and negative output half-cycles. Use of line-commutated thyristors (SCRs). Concept / Approach: Each converter group (positive and negative) requires two controlled devices when using a centre-tap, mirroring a full-wave controlled rectifier structure. One group conducts to build the positive half-cycle, the other for the negative half-cycle. Step-by-Step Solution: Count devices per group → 2 SCRs (centre-tap equivalent full-wave).Two groups (positive and negative) → total devices = 2 + 2 = 4.Thus, the minimum practical implementation requires four thyristors. Verification / Alternative check: Draw the centre-tapped rectifier pair for each polarity; duplicating for the opposite polarity yields four controlled elements. Why Other Options Are Wrong: 2 thyristors: Insufficient for independent control of both polarities. 8 thyristors: Excessive for single-phase centre-tapped case. Either 2 or 4: Ambiguous; correct is 4 for full functionality. Common Pitfalls: Confusing bridge-type cycloconverters (which use more devices) with centre-tapped arrangements. Final Answer: 4 thyristors

Q: Quadrant claim for Class B chopper Evaluate the statement: “A Class B chopper operates in the third quadrant.” Determine whether it is correct and justify based on voltage and current polarities.

Introduction / Context: Chopper classes are assigned by the voltage and current signs at the DC output. This determines the quadrant of operation and the direction of power flow (to the load or back to the source). Given Data / Assumptions: Class B chopper is a regenerative step-up type. Voltage reference is positive at the load; current positive into the load. Concept / Approach: Quadrants: Q1 (+V, +I), Q2 (+V, −I), Q3 (−V, −I), Q4 (−V, +I). Class B returns energy from the load to the source with a positive load voltage and negative current (current flowing from load to source), which is Quadrant II. Step-by-Step Solution: Identify Class B behavior → regeneration with V > 0 and I < 0.Match to quadrant → Q2, not Q3.Therefore, the statement “operates in third quadrant” is incorrect. Verification / Alternative check: Energy sign: P = V*I is negative for Class B, consistent with power flowing back to the source, and with Q2 coordinates. Why Other Options Are Wrong: Marking it True would imply V < 0, I < 0 (Q3), which is not the case for standard Class B implementations. Common Pitfalls: Mixing up Class B (single-quadrant regeneration in Q2) with multi-quadrant choppers (Class C or D) that can access Q3 in specific configurations. Final Answer: False

Q: Duty cycle of a step-down (buck) chopper Select the correct definition of duty cycle for an ideal step-down chopper switching with period T and on-time Ton. Also infer its relation to the output/input voltage ratio.

Introduction / Context: The duty cycle of a chopper determines the average output voltage. In a buck (step-down) converter, the control variable is the fraction of time the switch stays on within each switching period. Given Data / Assumptions: Ideal components (no drops, no ripple for average relation). Switching period T with on-time Ton and off-time Toff. Concept / Approach: By volt-second balance on the inductor in steady state, the average output voltage equals the input voltage multiplied by the duty cycle. Hence, duty cycle is defined as Ton/T and directly scales the output. Step-by-Step Solution: Define D = Ton / T.Apply inductor volt-second balance: average inductor voltage over one period is zero.During on: V_L ≈ Vin − Vout; during off: V_L ≈ −Vout.Set D*(Vin − Vout) + (1 − D)*(−Vout) = 0 → Vout = D * Vin. Verification / Alternative check: Simulation or measured average output of a PWM buck confirms proportionality to D for continuous conduction. Why Other Options Are Wrong: Toff/T: That is 1 − D, not D. Vin/Vout or inverse powers: Dimensional/physical mismatch. f_s/Vout: Nonsense dimensionally (Hz/Volt). Common Pitfalls: Confusing duty cycle with conduction ratio in discontinuous mode (where Vout relation deviates), or ignoring non-ideal drops that slightly reduce Vout. Final Answer: D = Ton / T (and ideally Vout/Vin = D)

Q: Junction biasing in a reverse-biased SCR When a standard PNPN thyristor (SCR) is reverse biased (anode negative with respect to cathode), which junctions are reverse/forward biased?

Introduction / Context: An SCR consists of a PNPN structure with three junctions (J1, J2, J3). Understanding which junctions are forward or reverse biased in forward-blocking and reverse-blocking states explains why the device blocks voltage until turned on. Given Data / Assumptions: Conventional SCR with anode at outer P layer and cathode at outer N layer. Reverse bias means anode potential is lower than cathode. Concept / Approach: In forward blocking (anode positive), J1 and J3 are forward biased, J2 is reverse biased. In reverse blocking (anode negative), the biasing swaps such that J1 and J3 become reverse biased and J2 becomes forward biased. The two reversed junctions enforce blocking. Step-by-Step Solution: Identify layers: P1–N1–P2–N2 with junctions J1 (P1–N1), J2 (N1–P2), J3 (P2–N2).Apply reverse bias (anode negative): junctions at outer terminals oppose current flow.Outcome: J1 and J3 reverse biased; J2 forward biased → overall reverse blocking. Verification / Alternative check: Two-transistor model (PNP–NPN) under reverse bias also indicates both outer junctions block, preventing conduction. Why Other Options Are Wrong: All three reverse/forward: Not possible with PNPN under simple reverse polarity. One reverse and two forward (option c): Describes forward-blocking state, not reverse-blocking. Common Pitfalls: Forgetting the swap between forward and reverse blocking arrangements or mislabeling junction order. Final Answer: two junctions are reverse biased and one junction is forward biased.

Question 1

DIAC device leads How many external terminals (leads) does a DIAC have?

Accepted Answer

Introduction / Context: A DIAC (Diode for Alternating Current) is a bidirectional trigger device widely used to initiate conduction in TRIAC-based phase control circuits (e.g., lamp dimmers and speed controllers). Knowing its terminal count is essential for correct wiring and understanding its symmetric behavior. Given Data / Assumptions: DIAC behaves like a bilateral trigger diode. Breakover voltage is similar in both polarities. No gate terminal exists (unlike TRIAC or SCR). Concept / Approach: Because the DIAC must operate symmetrically for positive and negative half-cycles, it is constructed to have two main terminals (commonly marked MT1 and MT2 or simply A1 and A2). There is no control terminal; conduction begins when the applied voltage reaches the breakover level in either polarity. Step-by-Step Solution: Identify DIAC function: bilateral triggering element.Being a trigger diode, it naturally has two terminals.Therefore the correct count of leads is two. Verification / Alternative check: Datasheets from major vendors depict a two-lead, bidirectional symbol and specify equal breakover voltages for positive and negative directions. Why Other Options Are Wrong: Three leads would imply a gate/control terminal; four or six leads correspond to more complex devices (e.g., opto-triacs, multi-pole components). Common Pitfalls: Confusing DIAC (two-lead trigger) with TRIAC (three-lead controllable bidirectional device) or with SCR (three-lead unidirectional device). Final Answer: 2

Question 2

Ideal rectifier performance criteria Which set of characteristics best describes an ideal rectifier and its output quality?

Accepted Answer

Introduction / Context: An ideal rectifier converts AC to DC with perfect effectiveness and produces a pure DC output with no AC components (harmonics). Understanding these ideal metrics helps benchmark practical rectifiers and assess filters and power-factor correction requirements. Given Data / Assumptions: Idealized conversion (no losses, no device drops). Pure DC output (no ripple, no AC harmonics). Unity displacement factor (current in-phase with voltage). Concept / Approach: Efficiency is output DC power divided by input power; in the ideal case, it is 100%. Ripple factor measures the AC component relative to DC; for an ideal DC output, it is zero. Harmonic factor quantifies distortion components; with a pure DC output, it is zero. Displacement factor is the cosine of the phase angle between fundamental current and voltage; with perfect in-phase operation, it is unity. Step-by-Step Solution: Define efficiency: η = P_out / P_in → 1.0 ideally.Define ripple factor: rf = I_ac,rms / I_dc → 0 for pure DC.Harmonic factor: based on harmonic RMS relative to fundamental → 0 if no harmonics.Displacement factor: cos φ → 1 when φ = 0°. Verification / Alternative check: Any nonzero ripple or harmonic immediately contradicts the ideal definition; any phase displacement lowers the displacement factor below unity. Why Other Options Are Wrong: Unity ripple factor means significant AC content; unity harmonic factor is non-ideal; zero displacement factor implies 90° phase shift, not acceptable for ideal conversion. Common Pitfalls: Confusing ripple factor with power factor; overlooking that practical rectifiers always have some ripple and distortion unless perfectly filtered. Final Answer: 100% efficiency, zero ripple factor, zero harmonic factor, unity displacement factor

Question 3

Turning off an SCR (thyristor) To ensure an SCR turns off and regains blocking capability, its current must be reduced below which characteristic level?

Accepted Answer

Introduction / Context: An SCR is a latching device: once triggered into conduction, it remains on as long as the anode current exceeds a certain threshold. Understanding the correct current level needed for turn-off is essential for commutation design and fault clearing in power converters and motor drives. Given Data / Assumptions: Latching behavior applies after the device has been successfully turned on. Device parameters include latching current (for initial turn-on) and holding current (minimum to stay on). Commutation or current zero techniques are applicable. Concept / Approach: After conduction is established, the SCR remains on while I_A ≥ I_H (holding current). To turn it off, the anode current must be forced below I_H for a sufficient time so that internal carrier recombination occurs and the device recovers its blocking capability. Trigger current relates to the gate during turn-on, and breakover current relates to exceeding the device’s blocking capability—neither defines the needed off condition. Step-by-Step Solution: During normal operation: ensure I_A falls below I_H (by natural current zero or forced commutation).Maintain reverse or zero bias for the device’s specified turn-off time (t_q).After t_q, the device is ready to block again. Verification / Alternative check: Datasheets specify I_H and t_q. Circuits like class C/D commutation explicitly reduce current below I_H to achieve turn-off. Why Other Options Are Wrong: Trigger current is a gate requirement for turn-on; breakover current is a breakdown phenomenon, not a turn-off condition; “none” ignores the standard definition. Common Pitfalls: Confusing latching current with holding current; assuming removal of gate drive alone will turn off an SCR—this is false once conduction is established. Final Answer: Holding current

Question 4

Half-wave controlled rectifier with continuous firing in positive half-cycle Assertion (A): In the shown circuit, if the SCR is continuously gated during the positive half-cycle, it will conduct for the full 180° of that half-cycle. Reason (R): A th…

Accepted Answer

Introduction / Context: Assertion–Reason questions test both conceptual correctness and causal linkage. In a half-wave controlled rectifier, continuous gating during the positive half-cycle can make the SCR conduct from near the start of that half-cycle until the natural current zero. The reason statement addresses a necessary condition for conduction (forward bias) but does not fully explain the sufficiency for full 180° conduction. Given Data / Assumptions: Half-wave AC source, resistive or RL load with current naturally reaching zero at ωt = π. SCR is forward biased only during the positive half-cycle. Gate drive is applied continuously in that interval. Concept / Approach: With continuous gate drive, the SCR triggers as soon as forward bias appears (near ωt ≈ 0+), then remains on until the current naturally falls to zero at the end of the positive half-cycle. Thus, conduction spans approximately 180°. The reason statement is true—forward bias is required—but forward bias alone does not guarantee conduction without gate triggering (unless breakover occurs). Hence, R does not wholly explain why conduction lasts the entire half-cycle under continuous gating. Step-by-Step Solution: At positive zero crossing: device becomes forward biased.Gate present: SCR turns on immediately and carries load current.At ωt = π: current crosses zero → SCR turns off → conduction interval ≈ 180°. Verification / Alternative check: Classic half-wave control waveforms show full half-cycle conduction when firing angle α → 0°, achievable with continuous gating. Why Other Options Are Wrong: Claiming R explains A overlooks the role of gate triggering; stating A is wrong contradicts standard operation; saying R is wrong denies the forward-bias prerequisite. Common Pitfalls: Assuming forward bias alone turns on an SCR; forgetting that without a gate pulse, the device may not conduct unless the voltage rises to breakover. Final Answer: Both A and R are correct but R is not correct explanation of A

Question 5

Effect of source inductance (overlap) in a three-phase fully controlled bridge converter What is the principal effect on the output when finite source inductance causes commutation overlap?

Accepted Answer

Introduction / Context: In controlled rectifiers, source (or supply) inductance prevents instantaneous current transfer between device pairs, creating an overlap interval during which both outgoing and incoming thyristors conduct. This overlap reduces the effective output voltage because a portion of the line-to-line voltage is “lost” in the commutation process. Given Data / Assumptions: Three-phase fully controlled bridge (six-pulse) under continuous current. Finite source inductance causes a commutation angle μ > 0. Device drops neglected for first-order understanding. Concept / Approach: During overlap, two line phases simultaneously feed the DC link, effectively shorting part of the line-to-line voltage across source inductances. The DC output sees a lower instantaneous voltage than the ideal case. Averaging across the cycle yields a DC output smaller than the ideal value (at the same firing angle). Current generally is determined by the load (e.g., E–L–R), not directly increased by overlap. Step-by-Step Solution: Ideal average voltage (no overlap): V_d0 = (3√2/π) * V_phase_peak * cos α.With overlap μ: V_d = V_d0 − ΔV(μ), where ΔV is the commutation-induced reduction.Hence the principal, observable effect is a decreased average output voltage. Verification / Alternative check: Measured waveforms show flattened notches during commutation and reduced DC average proportional to μ; analytical models compute ΔV ≈ (ωL_s) * I_d * (constant), confirming the drop. Why Other Options Are Wrong: Claimed increases in output voltage contradict physics of overlap; statements about increased current are not generally true without specifying load dynamics. Common Pitfalls: Assuming overlap only adds ripple; ignoring that it also lowers mean output voltage and can impact control range near inversion. Final Answer: Decrease in the average output voltage

Question 6

DC–DC chopper quadrants (Class A vs. Class B) In power electronics, identify the operating quadrants of Class A and Class B choppers on the voltage–current (V–I) plane. Clarify which quadrant each class occupies and how the load power flow direction…

Accepted Answer

Introduction / Context: Choppers (DC–DC converters) are categorized by the quadrant of operation on the voltage–current (V–I) plane. This question tests recognition of the quadrants for Class A (step-down) and Class B (regenerative) choppers and the associated power flow direction. Given Data / Assumptions: Idealized converters and passive loads. Voltage polarity defined positive from source to load; current positive into load. Steady switching within continuous conduction where applicable. Concept / Approach: Quadrants are defined by the signs of voltage and current: Q1 (+V, +I), Q2 (+V, −I), Q3 (−V, −I), Q4 (−V, +I). Class A delivers power from source to load (V > 0, I > 0). Class B performs regenerative braking/charging (V > 0, I < 0), returning energy from the load back to the source. Step-by-Step Solution: Identify Class A topology → conventional step-down with freewheel path, average V_out positive, load current positive.Map to quadrant → (+V, +I) → first quadrant.Identify Class B operation → source sees positive terminal while load forces current back to source during switching intervals.Map to quadrant → (+V, −I) → second quadrant. Verification / Alternative check: Energy flow: Class A has P = VI > 0 (to load). Class B has P = VI < 0 (from load to source). Signs match Q1 and Q2 respectively. Why Other Options Are Wrong: Operate in first quadrant / second quadrant (alone): Each describes only one class, not both. Either quadrant: Too vague; classes have specific quadrants. Class A in Q2 and Class B in Q1: Reverses the correct mapping. Common Pitfalls: Confusing Class B (regenerative) with Class C or D multi-quadrant arrangements, and mixing sign conventions of current direction. Final Answer: class A chopper operates in first quadrant while class B operates in second quadrant

Question 7

Modified series inverter: attainable output frequency In a modified series inverter (using forced or auxiliary commutation extensions), how does the achievable output frequency compare with the natural damped resonant frequency of the L–C commutatin…

Accepted Answer

Introduction / Context: Series inverters use resonant L–C commutation. Modified series inverters employ auxiliary switches/paths so that commutation is not limited strictly by the natural resonant period, enabling higher fundamental output frequency. Given Data / Assumptions: Presence of a modified (assisted) commutation network. Switches and diodes ideal for conceptual analysis. Load is compatible with resonant current pulses. Concept / Approach: A basic series inverter's period is largely set by L and C, giving a damped oscillation with natural frequency f_d. By adding auxiliary commutation, the devices can be turned off and re-fired earlier than the natural half-cycle, effectively increasing the repetition rate of power pulses and therefore the output frequency. Step-by-Step Solution: Define natural frequency: f_d ≈ 1 / (2π * sqrt(L * C)) (ignoring damping).In a simple series inverter, switching follows this half-sinusoid → output frequency ties to f_d.Modified inverter introduces controlled commutation so that the next cycle can begin before the resonant current completely decays.Thus, achievable f_out can exceed f_d when designed to commutate early. Verification / Alternative check: Waveforms of modified circuits show shortened current pulses and increased repetition based on gating strategy rather than pure LC decay. Why Other Options Are Wrong: Less than / cannot exceed: True only for unassisted basic series inverters. Any of the above: Over-broad and contradicts known capability of modified topologies. Common Pitfalls: Assuming natural resonance always caps frequency; in practice, controlled commutation can reshape intervals within device limits. Final Answer: the output frequency can be even higher than the frequency of damped oscillations

Question 8

Single-phase cycloconverter with centre-tapped input transformer For a single-phase cycloconverter that uses a centre-tapped input transformer, how many thyristors are required to realize both positive and negative converters?

Accepted Answer

Introduction / Context: A single-phase cycloconverter synthesizes low-frequency AC directly from a higher-frequency AC source by phase-controlling segments of the input. With a centre-tapped transformer, separate controlled paths are used for the positive and negative groups. Given Data / Assumptions: Centre-tapped secondary providing two half-windings. Need for controlled positive and negative output half-cycles. Use of line-commutated thyristors (SCRs). Concept / Approach: Each converter group (positive and negative) requires two controlled devices when using a centre-tap, mirroring a full-wave controlled rectifier structure. One group conducts to build the positive half-cycle, the other for the negative half-cycle. Step-by-Step Solution: Count devices per group → 2 SCRs (centre-tap equivalent full-wave).Two groups (positive and negative) → total devices = 2 + 2 = 4.Thus, the minimum practical implementation requires four thyristors. Verification / Alternative check: Draw the centre-tapped rectifier pair for each polarity; duplicating for the opposite polarity yields four controlled elements. Why Other Options Are Wrong: 2 thyristors: Insufficient for independent control of both polarities. 8 thyristors: Excessive for single-phase centre-tapped case. Either 2 or 4: Ambiguous; correct is 4 for full functionality. Common Pitfalls: Confusing bridge-type cycloconverters (which use more devices) with centre-tapped arrangements. Final Answer: 4 thyristors

Question 9

Quadrant claim for Class B chopper Evaluate the statement: “A Class B chopper operates in the third quadrant.” Determine whether it is correct and justify based on voltage and current polarities.

Accepted Answer

Introduction / Context: Chopper classes are assigned by the voltage and current signs at the DC output. This determines the quadrant of operation and the direction of power flow (to the load or back to the source). Given Data / Assumptions: Class B chopper is a regenerative step-up type. Voltage reference is positive at the load; current positive into the load. Concept / Approach: Quadrants: Q1 (+V, +I), Q2 (+V, −I), Q3 (−V, −I), Q4 (−V, +I). Class B returns energy from the load to the source with a positive load voltage and negative current (current flowing from load to source), which is Quadrant II. Step-by-Step Solution: Identify Class B behavior → regeneration with V > 0 and I < 0.Match to quadrant → Q2, not Q3.Therefore, the statement “operates in third quadrant” is incorrect. Verification / Alternative check: Energy sign: P = V*I is negative for Class B, consistent with power flowing back to the source, and with Q2 coordinates. Why Other Options Are Wrong: Marking it True would imply V < 0, I < 0 (Q3), which is not the case for standard Class B implementations. Common Pitfalls: Mixing up Class B (single-quadrant regeneration in Q2) with multi-quadrant choppers (Class C or D) that can access Q3 in specific configurations. Final Answer: False

Question 10

Duty cycle of a step-down (buck) chopper Select the correct definition of duty cycle for an ideal step-down chopper switching with period T and on-time Ton. Also infer its relation to the output/input voltage ratio.

Accepted Answer

Introduction / Context: The duty cycle of a chopper determines the average output voltage. In a buck (step-down) converter, the control variable is the fraction of time the switch stays on within each switching period. Given Data / Assumptions: Ideal components (no drops, no ripple for average relation). Switching period T with on-time Ton and off-time Toff. Concept / Approach: By volt-second balance on the inductor in steady state, the average output voltage equals the input voltage multiplied by the duty cycle. Hence, duty cycle is defined as Ton/T and directly scales the output. Step-by-Step Solution: Define D = Ton / T.Apply inductor volt-second balance: average inductor voltage over one period is zero.During on: V_L ≈ Vin − Vout; during off: V_L ≈ −Vout.Set D*(Vin − Vout) + (1 − D)*(−Vout) = 0 → Vout = D * Vin. Verification / Alternative check: Simulation or measured average output of a PWM buck confirms proportionality to D for continuous conduction. Why Other Options Are Wrong: Toff/T: That is 1 − D, not D. Vin/Vout or inverse powers: Dimensional/physical mismatch. f_s/Vout: Nonsense dimensionally (Hz/Volt). Common Pitfalls: Confusing duty cycle with conduction ratio in discontinuous mode (where Vout relation deviates), or ignoring non-ideal drops that slightly reduce Vout. Final Answer: D = Ton / T (and ideally Vout/Vin = D)

Question 11

Junction biasing in a reverse-biased SCR When a standard PNPN thyristor (SCR) is reverse biased (anode negative with respect to cathode), which junctions are reverse/forward biased?

Accepted Answer

Introduction / Context: An SCR consists of a PNPN structure with three junctions (J1, J2, J3). Understanding which junctions are forward or reverse biased in forward-blocking and reverse-blocking states explains why the device blocks voltage until turned on. Given Data / Assumptions: Conventional SCR with anode at outer P layer and cathode at outer N layer. Reverse bias means anode potential is lower than cathode. Concept / Approach: In forward blocking (anode positive), J1 and J3 are forward biased, J2 is reverse biased. In reverse blocking (anode negative), the biasing swaps such that J1 and J3 become reverse biased and J2 becomes forward biased. The two reversed junctions enforce blocking. Step-by-Step Solution: Identify layers: P1–N1–P2–N2 with junctions J1 (P1–N1), J2 (N1–P2), J3 (P2–N2).Apply reverse bias (anode negative): junctions at outer terminals oppose current flow.Outcome: J1 and J3 reverse biased; J2 forward biased → overall reverse blocking. Verification / Alternative check: Two-transistor model (PNP–NPN) under reverse bias also indicates both outer junctions block, preventing conduction. Why Other Options Are Wrong: All three reverse/forward: Not possible with PNPN under simple reverse polarity. One reverse and two forward (option c): Describes forward-blocking state, not reverse-blocking. Common Pitfalls: Forgetting the swap between forward and reverse blocking arrangements or mislabeling junction order. Final Answer: two junctions are reverse biased and one junction is forward biased.

Question 12

Definition reference for thyristor turn-off time (tq) From which instant is the turn-off time tq of a thyristor measured in natural or forced commutation analysis?

Accepted Answer

Introduction / Context: The turn-off time tq is a key dynamic parameter of SCRs. It is the time required after current extinction before the device can again block a forward voltage without retriggering. Given Data / Assumptions: Standard SCR definition used in datasheets. Natural or forced commutation brings anode current to zero. Concept / Approach: Internally, charge stored in the device must recombine or be swept out after current goes to zero. tq is measured from the instant the anode current falls to zero (not the voltage zero) until the SCR regains forward blocking capability. Step-by-Step Solution: Identify current zero crossing due to commutation.Start timing tq at i_A = 0.End timing when device can withstand specified forward voltage without turning on. Verification / Alternative check: Manufacturer switching waveforms annotate tq beginning at the current zero and ending at the recovery of blocking state. Why Other Options Are Wrong: Anode voltage zero: Not the reference used; voltage may not be zero at current zero in forced commutation. Both zero: Unnecessary and often not simultaneous. Gate pulse removal: SCR is a latching device; gate removal does not define tq. Common Pitfalls: Confusing reverse recovery of diodes with tq of SCRs; the definitions are analogous but not identical. Final Answer: when anode current becomes zero

Question 13

Device count in a single-phase fully controlled bridge rectifier How many thyristors are used in a single-phase fully controlled bridge (full converter) topology?

Accepted Answer

Introduction / Context: A single-phase fully controlled bridge rectifier replaces all four diodes of a standard bridge with controllable devices to regulate the average DC output via firing angle control. Given Data / Assumptions: Single-phase AC source. Full converter (not half-controlled). Concept / Approach: The full bridge consists of four arms. In a fully controlled design, each arm contains one SCR. Two SCRs conduct at a time (one from the top leg and one from the opposite bottom leg), commutating every half cycle. Step-by-Step Solution: Count arms in the bridge → 4.Replace each diode with an SCR → 4 SCRs total.Operation uses pairs conducting for 180° intervals with 90° phase shift in gating. Verification / Alternative check: Standard full converter schematics confirm four controllable devices. Why Other Options Are Wrong: 2: Would be half-controlled (M-2) at best. 8 or 16: Excess for single-phase full bridge; such counts occur in multipulse or parallel arrangements. Common Pitfalls: Confusing single-phase with three-phase bridges (which use six devices). Final Answer: 4

Question 14

Number of conducting devices in a 3-phase bridge inverter (120° conduction mode) At any instant during 120° conduction mode operation of a three-phase bridge inverter, how many thyristors (main devices) conduct simultaneously?

Accepted Answer

Introduction / Context: Three-phase inverters can operate in 180° or 120° conduction modes. The instantaneous number of conducting devices affects line-to-line voltage waveforms and harmonic content. Given Data / Assumptions: Voltage-source three-phase bridge with ideal switches and antiparallel diodes. Operation in 120° conduction mode. Concept / Approach: In 120° mode, each device conducts for 120° of the electrical cycle, and conduction intervals are arranged so that only one device from the upper group and one from the lower group conduct at any time, producing a quasi-square line-to-line voltage. Step-by-Step Solution: Assign sequential 120° conduction blocks to the six devices.Ensure that devices from the same leg do not conduct simultaneously.At any instant: exactly two devices ON (one on the positive bus, one on the negative bus), delivering current to a pair of phases. Verification / Alternative check: State tables and space-vector diagrams for 120° mode confirm two active devices at a time, yielding six non-zero voltage vectors per cycle. Why Other Options Are Wrong: 1: Cannot form a closed current path for three-phase load. 3: Applies to 180° conduction mode. 4 or 6: Overconstrained, not typical for basic 120° mode. Common Pitfalls: Confusing device count in conduction with the number of active phases carrying current (which is also two in 120° mode). Final Answer: 2

Question 15

Purpose and composition of an SCR snubber network Identify the standard passive snubber used with an SCR to limit dv/dt and share turn-off stress. What elements and connection are used?

Accepted Answer

Introduction / Context: SCRs are sensitive to high dv/dt which can cause false triggering. A passive snubber network shapes transient voltages and currents during switching, improving reliability and device sharing in series strings. Given Data / Assumptions: Line-commutated or forced-commutated SCR applications. Objective: limit dv/dt and dissipate snubber energy safely. Concept / Approach: The classic snubber is an RC series network placed across the device. The capacitor limits dv/dt by momentarily diverting current, and the series resistor damps oscillations and dissipates stored energy, preventing excessive current spikes. Step-by-Step Solution: Connect C in series with R across anode–cathode.During a sudden voltage step, current flows into C, reducing dv/dt across SCR.R discharges C between events and damps ringing with circuit stray L. Verification / Alternative check: Waveforms with RC snubbers show reduced voltage slope and controlled recovery transients compared with unsnubbed operation. Why Other Options Are Wrong: C alone in parallel: Can cause high peak currents and ringing without damping. C in series with device: Disrupts conduction and is not a snubber. Inductor alone: Limits di/dt, not dv/dt, and is used in series for current control. Common Pitfalls: Choosing R and C without considering line impedance and device ratings; excessive C increases losses, too little R fails to damp oscillations. Final Answer: a resistor and capacitor in series connected across thyristor

Question 16

Dual converter (circulating-current mode) In a dual converter operated in circulating-current mode, when does the internal circulating current flow?

Accepted Answer

Introduction / Context: A dual converter consists of two fully controlled bridges connected in antiparallel on the DC side. In the circulating-current mode, both bridges are kept active so that instantaneous reversal of DC voltage polarity is possible without discontinuity. Understanding when and why circulating current flows is essential for correct reactor sizing and protection. Given Data / Assumptions: Two line-commutated bridges share the same DC terminals. Bridges are set to provide slightly different average DC voltages to establish bias. A series reactor is present to limit circulating current. Concept / Approach: In circulating-current mode, one bridge operates as a rectifier while the other is biased as an inverter. Their instantaneous EMFs are not exactly equal at all times. Any small difference drives current through the interposed reactor, creating a closed loop between the two bridges even if the external load current is momentarily zero. Step-by-Step Solution: Identify that both bridges are gated continuously to enable four-quadrant capability.Recognize that the bridges’ instantaneous DC voltages are out of phase, producing a voltage difference.This voltage difference drives a loop current through the reactor and bridges, independent of external load.Therefore, circulating current exists even at zero load current. Verification / Alternative check: Waveform sketches show finite reactor current during no-load intervals; turning off one bridge (non-circulating mode) removes this internal current, but sacrifices instantaneous reversal. Why Other Options Are Wrong: “when load current is zero” is incomplete; the key is that it flows even with zero load current. “when load current is non-zero” is not the defining condition. “only during load transients” ignores steady-state internal loop current. “none of the above” is incorrect since the correct condition is listed. Common Pitfalls: Assuming circulating current equals load current; it is an internal loop current limited by the reactor. Under-sizing the reactor, leading to excessive losses and device stress. Final Answer: even when load current is zero

Question 17

Assertion–Reason (HVDC thyristor triggering) Assertion (A): Light triggering is very suitable for HVDC transmission valves. Reason (R): Light triggering provides complete electrical isolation of the gate drive from high-voltage stacks.

Accepted Answer

Introduction / Context: Large HVDC converters use series strings of high-power thyristors. Triggering each device safely and reliably at very high potentials demands exceptional isolation and noise immunity. Light-triggered thyristors (LTTs) solve this by using fiber-optic light pulses instead of electrical gate leads. Given Data / Assumptions: HVDC valve stacks operate at tens to hundreds of kilovolts. Fiber optics provide galvanic isolation and immunity to electromagnetic interference. Light-triggered device includes an integrated photosensitive gate structure. Concept / Approach: Electrical gate-drive distribution up a high-voltage stack is complex, requires robust insulation, and can pick up noise. Fiber-optic links transmit light pulses to each device, eliminating conductive paths and thereby achieving complete electrical isolation between control and power circuits. Step-by-Step Solution: Recognize HVDC valves require precise, isolated triggering.Fiber optics deliver light pulses directly to the thyristor gate region.Isolation is intrinsic; there is no galvanic connection to control ground.Thus, light triggering is well suited; the reason correctly explains the assertion. Verification / Alternative check: Modern HVDC projects commonly employ fiber-optic gating and LTTs to reduce complexity, improve reliability, and enhance immunity to dv/dt and EMI. Why Other Options Are Wrong: (b) denies the causal link—incorrect. (c) or (d) contradict established HVDC practice. (e) both statements are well-supported by industry practice. Common Pitfalls: Confusing light triggering with opto-coupled electrical gates; LTTs integrate optical sensitivity. Ignoring need for per-device timing alignment in long strings. Final Answer: Both A and R are correct and R is correct explanation of A

Question 18

Jones chopper commutation network In a Jones chopper, which elements form the commutating circuit that provides current reversal for turn-off?

Accepted Answer

Introduction / Context: Jones chopper is a class of resonant-commutated DC choppers. It uses a resonant network to apply a reverse current or reverse voltage across the main thyristor, thereby turning it off without relying on line commutation. Given Data / Assumptions: DC supply with a thyristor main switch. Auxiliary switch and a resonant commutation branch. Commutation requires a controlled reverse current or reverse voltage pulse. Concept / Approach: Resonant commutation employs an LC network. When the auxiliary path is fired, the pre-charged capacitor and inductor exchange energy sinusoidally. This creates a transient that opposes the main thyristor current, forcing it to zero and providing reverse bias to complete turn-off. Step-by-Step Solution: Charge the commutation capacitor to a defined polarity during conduction.Trigger the auxiliary device to insert the LC across the main thyristor.LC resonance produces a reverse current pulse.Main device current goes to zero and receives reverse bias → successful turn-off. Verification / Alternative check: Waveforms show an oscillatory current in the LC branch superimposed on the main current, with the net current through the main thyristor driven to zero. Why Other Options Are Wrong: A capacitor alone cannot produce the required resonant current; an inductor is essential. An autotransformer is not a standard element of Jones commutation. “None of the above” is incorrect since LC is correct. Common Pitfalls: Mis-timing the auxiliary firing, causing commutation failure. Undersized LC leading to insufficient reverse current. Final Answer: a capacitor and an inductor

Question 19

Lead count comparison What are the total numbers of external leads for SCR, DIAC, and TRIAC respectively?

Accepted Answer

Introduction / Context: Proper device identification and pin-out knowledge are foundational in power electronics. SCRs, DIACs, and TRIACs are common devices used for switching and control in AC and DC circuits. Given Data / Assumptions: SCR: Silicon Controlled Rectifier (unidirectional). DIAC: Bidirectional trigger device with no gate terminal. TRIAC: Bidirectional thyristor used in AC control. Concept / Approach: SCRs have three terminals: anode (A), cathode (K), and gate (G). DIACs are two-terminal devices that switch at breakover in either polarity. TRIACs have three terminals: main terminal 1 (T1), main terminal 2 (T2), and gate (G). Step-by-Step Solution: Count terminals for SCR: A, K, G → 3.Count terminals for DIAC: two main terminals only → 2.Count terminals for TRIAC: T1, T2, G → 3.Hence, the sequence is 3, 2, 3. Verification / Alternative check: Datasheets and standard symbols confirm the terminal counts. Why Other Options Are Wrong: Any option giving DIAC or TRIAC four terminals is wrong; DIAC has two, TRIAC has three. Swapping counts does not match device definitions. Common Pitfalls: Confusing DIAC with TRIAC; DIAC is a trigger device, not a power switch. Final Answer: 3, 2, 3

Question 20

Forward dv/dt rating of an SCR Which definition correctly describes the forward dv/dt rating (forward rate of rise of anode voltage) of an SCR?

Accepted Answer

Introduction / Context: An SCR can turn on spuriously when subjected to a high rate of rise of forward anode-cathode voltage, due to displacement current through the device capacitances. The forward dv/dt rating defines the immunity to such unintended turn-on in the absence of a gate drive. Given Data / Assumptions: Gate is un-driven during dv/dt test condition. Device junction capacitance and internal structure determine dv/dt susceptibility. Snubber networks are commonly used to limit dv/dt. Concept / Approach: Displacement current i = Cj * dv/dt may flow into the gate-cathode equivalent and trigger the device. The forward dv/dt rating is therefore stated as the maximum dv/dt that will not cause turn-on when the gate is open (no intentional gate current). Step-by-Step Solution: Recognize dv/dt-induced triggering risk.Apply definition: maximum dv/dt tolerated without unintended conduction in the absence of gate drive.Hence, option describing “no gate signal applied” is correct. Verification / Alternative check: Manufacturer datasheets list dv/dt ratings under specified gate conditions (usually gate open). Why Other Options Are Wrong: (a) contradicts the intent; rating is for immunity, not for intentional triggering. (b) is irrelevant; with a gate applied, device may conduct by design. (d) cannot be correct because (b) is incorrect. (e) confuses dv/dt rating with commutation capability. Common Pitfalls: Ignoring snubber needs in inductive applications. Assuming dv/dt rating ensures safe turn-off; it does not. Final Answer: the maximum dv/dt that will not trigger the SCR if no gate signal is applied

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