Q: Assertion–Reason on SCR turn-on behavior Assertion (A): As an SCR turns on and anode current rises, the anode–cathode voltage drop decreases. Reason (R): During turn-on, the reverse-biased junction inside the SCR breaks down in avalanche.

Introduction / Context: Understanding the internal physics of SCR turn-on helps explain observed waveforms: a rapid drop in anode–cathode voltage accompanies a rise in current after triggering. It is important to distinguish regenerative carrier injection from avalanche breakdown. Given Data / Assumptions: Silicon-controlled rectifier (four-layer p-n-p-n structure). Gate-triggered turn-on under normal operating conditions. Concept / Approach: When properly triggered, the SCR turns on via regenerative action between its two internal transistor equivalents. Carrier injection forward-biases the central junction; the device transitions from blocking to conducting with a steep decrease in VAK. This is not due to destructive avalanche breakdown under normal operation. Step-by-Step Solution: Apply gate current, initiating conduction in the p-n junction near the gate.Mutual transistor action causes rapid increase in carriers; the middle junction switches from reverse to effectively forward-biased by injection.The device enters the low on-state resistance region; VAK collapses while IA rises.Therefore A is true (VAK falls as IA rises), but R is false because normal turn-on is not avalanche breakdown. Verification / Alternative check: Typical SCR V–I characteristics and turn-on waveforms show a transition to a low on-state voltage with moderate slope resistance; avalanche is associated with overvoltage, not standard gate turn-on. Why Other Options Are Wrong: (a), (b) would require R to be a correct explanation, which it is not. (d) A is well-known to be true. (e) Both being wrong contradicts basic SCR behavior. Common Pitfalls: Confusing “breakover” with “avalanche”; controlled turn-on uses gate injection below breakover voltage. Final Answer: A is correct but R is wrong

Question 1

Pulse number of a three-phase fully controlled bridge converter A three-phase fully controlled rectifier (bridge) is classified as a __-pulse converter. Select the correct pulse number.

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Introduction / Context: Converters are often identified by their pulse number, which equals the number of DC output ripple pulses per input cycle. This metric affects harmonic content and filtering requirements. Given Data / Assumptions: Three-phase full bridge using six controlled devices (e.g., SCRs). Standard three-phase line voltages with 120° displacement. Concept / Approach: In a three-phase full bridge, each device conducts for 120°, and commutation occurs every 60°. Hence, six pulses of DC ripple are produced each line cycle, giving a 6-pulse converter. Step-by-Step Solution: Identify commutation every 60° → six commutations per cycle.Each commutation produces a distinct ripple pulse on the DC side.Therefore, “6-pulse” classification applies. Verification / Alternative check: Harmonic spectra of 6-pulse converters display characteristic orders at 6k ± 1 on the AC side. Why Other Options Are Wrong: 3-pulse/2-pulse: Associated with single-phase or half-wave arrangements. 12-pulse: Requires two 6-pulse bridges with phase-shifting transformer. Common Pitfalls: Confusing the number of switches with the pulse number; pulse number is tied to commutation per cycle. Final Answer: 6-pulse converter

Question 2

Typical on-state voltage drop of a conducting thyristor When an SCR is in the on state and conducting rated current, the approximate voltage drop across it is:

Accepted Answer

Introduction / Context: The on-state drop of power devices directly influences conduction losses and heat sink sizing. For SCRs, the drop is relatively low compared with transistors at similar currents. Given Data / Assumptions: Thyristor in full conduction (latch-on), steady current. Typical industrial devices at nominal current and temperature. Concept / Approach: SCRs exhibit a quasi-constant on-state drop, often around 1 to 2 V depending on current and temperature, due to their PNPN conduction characteristics. Step-by-Step Solution: Select the closest nominal value used for loss estimates → ~1 V (sometimes 1–2 V).Rule out obviously excessive values like 10 V or more at rated current. Verification / Alternative check: Datasheets list V_T(on) typically in the 1–2 V range for rated currents. Why Other Options Are Wrong: 10–100 V: Would cause huge losses; not realistic for a conducting SCR. 0.1 V: Too small for SCR technology; closer to low-Rds(on) MOSFETs at modest current. Common Pitfalls: Using a single fixed value; real devices vary with current and junction temperature. Final Answer: about 1 V

Question 3

Effect of a freewheeling diode in a controlled rectifier (inductive load) Adding a freewheeling diode across the load in a controlled rectifier primarily improves the shape of which waveform?

Accepted Answer

Introduction / Context: In controlled rectifiers feeding inductive loads, current continuity is crucial. A freewheeling diode provides an alternate path when the supply voltage reverses or the controlled device is turned off. Given Data / Assumptions: Inductive load where current cannot change instantaneously. Freewheeling diode connected across the load (or DC terminals). Concept / Approach: When the source voltage becomes unfavorable, the inductor forces current to continue; the freewheeling diode conducts that current, preventing abrupt current decay and reducing ripple. Step-by-Step Solution: Without diode: current falls sharply at each commutation → high ripple.With diode: current circulates through the diode → smoother, more DC-like load current.Thus, the waveshape of load current is improved (less ripple, more continuity). Verification / Alternative check: Waveform sketches show extended current flow during the freewheel interval leading to reduced torque ripple in motor loads. Why Other Options Are Wrong: Marking this statement as False would contradict standard rectifier practice. Common Pitfalls: Assuming the diode is unnecessary for resistive loads; its benefit is primarily with inductive loads. Final Answer: True

Question 4

Freewheeling diodes in half-bridge inverters In a single-phase half-bridge (voltage source) inverter, freewheeling diodes are required under which load conditions?

Accepted Answer

Introduction / Context: In voltage-source inverters, antiparallel (freewheeling) diodes are used to provide a path for reactive load current when the main switches are off or reverse-biased. Given Data / Assumptions: Half-bridge topology with two main switches and corresponding diodes. We consider which load types necessitate these diodes for safe operation. Concept / Approach: Inductive loads force current continuity. When the applied voltage reverses, the inductor drives current through the antiparallel diode of the opposite switch, preventing overvoltage and current interruption. Step-by-Step Solution: Resistive load → current in phase with voltage; diodes are not strictly required for current continuity.Inductive load → lagging current needs a reverse path when switches commutate → diodes required.Capacitive load leads current; in practice, diode conduction is less critical for current continuity, but device antiparallel diodes are commonly present in VSI modules. Verification / Alternative check: Standard half-bridge power modules incorporate antiparallel diodes specifically for reactive load handling, predominantly for inductive loads. Why Other Options Are Wrong: Always required: Overstated for purely resistive loads. Inductive or capacitive only: The key need is inductive current continuity. Common Pitfalls: Assuming diodes are unnecessary because the load is “mostly resistive”; even small inductance can demand a freewheel path. Final Answer: only when the load is inductive

Question 5

Applications of inverters Which of the following are common real-world applications of inverters?

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Introduction / Context: Inverters convert DC to AC with controlled magnitude and frequency. This function is central to modern power conversion systems across utility, industrial, and consumer sectors. Given Data / Assumptions: Consider standard voltage-source or current-source inverter implementations. Look at typical system-level applications. Concept / Approach: HVDC terminals use line-commutated or voltage-source converters that operate as inverters at one end; UPS produce AC from batteries; variable-speed drives use inverters to synthesize adjustable-frequency outputs for induction or PMSM motors. Step-by-Step Solution: HVDC: DC bulk transmission with AC-grid interfacing requires inverter operation at the receiving end.UPS: Battery DC is inverted to AC for critical loads during outages.Drives: Inverters supply variable frequency/voltage to control motor speed and torque. Verification / Alternative check: Industrial catalogs list inverter-based VFDs, UPS, and HVDC converters as standard products. Why Other Options Are Wrong: Each of A, B, and C is correct individually; hence the combined answer is “all of the above”. Common Pitfalls: Confusing rectifiers (AC to DC) with inverters (DC to AC); many systems contain both stages. Final Answer: all of the above

Question 6

Use of a freewheeling diode in single-phase controlled rectifiers In which of the following single-phase controlled rectifier circuits is it possible and useful to include a freewheeling diode across the DC load?

Accepted Answer

Introduction / Context: A freewheeling diode provides an alternate path for inductive load current when the source voltage reverses or the controlled devices are not conducting. Its inclusion depends on the converter topology. Given Data / Assumptions: Single-phase half-wave controlled rectifier (one controlled device). Single-phase M-2 (half-controlled full-wave) rectifier with two SCRs and two diodes. Inductive load where current continuity is desired. Concept / Approach: In the half-wave circuit, during the negative half-cycle the source cannot support load current; a freewheel diode across the DC load allows current to circulate, reducing ripple. In the M-2 topology, although some freewheeling paths exist internally, adding a dedicated freewheeling diode enhances current continuity and improves power factor. Step-by-Step Solution: Identify intervals when supply is reverse or devices are off → inductor forces current to continue.Provide a diode across the DC output → current circulates locally, sustaining load current and reducing voltage spikes.Thus, both half-wave and M-2 circuits can employ a freewheeling diode effectively. Verification / Alternative check: Standard textbooks show freewheeling diode use in single-phase half-wave and half-controlled bridges to improve current waveform. Why Other Options Are Wrong: Only one of the two: Too restrictive. All controlled rectifiers: Overbroad; some topologies have inherent current paths or different needs. Common Pitfalls: Assuming M-2 always eliminates the need; dedicated freewheeling can still improve performance depending on operating conditions. Final Answer: single-phase half-wave controlled rectifier and single-phase full-wave controlled rectifier (M-2 connection)

Question 7

Power electronics component identification In a TRIAC (a bidirectional AC switch used in AC regulators and dimmers), the three external terminals are conventionally denoted as:

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Introduction / Context: A TRIAC is a three-terminal, bidirectional thyristor device widely used in AC power control (for example, lamp dimmers, heater controls, and single-phase motor speed control). Correctly identifying its terminals is essential for proper gate triggering, snubber placement, and reliable PCB layout. Given Data / Assumptions: Device under discussion is a standard TRIAC. Common datasheet naming conventions are used. Terminal names are functional, not package-pin numbers. Concept / Approach: Unlike an SCR, which has an anode (A), cathode (K), and gate (G), a TRIAC conducts in both directions. Therefore, its two main terminals are not labeled anode/cathode. Instead, they are called main terminal 1 (T1) and main terminal 2 (T2). The third terminal is the gate (G), which can trigger conduction in either polarity quadrant of operation. Step-by-Step Solution: Recognize that bi-directional conduction prevents using anode/cathode naming.Industry-standard notation assigns T1 and T2 to the two power-carrying terminals.The control terminal remains the gate (G).Hence, the correct set is T1, T2, gate. Verification / Alternative check: Consulting any TRIAC datasheet (for example, BTA/BTB series) confirms terminal labels as T1, T2, and G, with triggering specified for four quadrants relative to T1/T2 polarities and gate reference. Why Other Options Are Wrong: anode/cathode terminology applies to unidirectional SCRs, not TRIACs. T1, T2, anode includes a non-existent anode for TRIACs. T1, T2, T3 suggests three power terminals, which is incorrect. MT1/MT2 are synonyms for T1/T2, but “source” is not used for TRIACs. Common Pitfalls: Miswiring T1 and T2 can change gate reference conditions and impair triggering. Assuming SCR anode/cathode naming for TRIACs leads to design errors. Final Answer: T1, T2, gate

Question 8

Assertion–Reason on gate turn-off thyristor (GTO) Assertion (A): A GTO can be turned ON and turned OFF by gate pulses. Reason (R): A GTO can be turned OFF by applying a sufficiently strong negative gate pulse.

Accepted Answer

Introduction / Context: Gate Turn-Off Thyristors (GTOs) are controllable power devices that can be both turned on and turned off using gate signals. This distinguishes them from conventional SCRs, which require forced commutation or line commutation for turn-off. Given Data / Assumptions: A: GTO supports gate-controlled turn-on and turn-off. R: Negative gate current actively extracts carriers to commutate the device off. Concept / Approach: In a GTO, the gate structure is designed to provide substantial reverse gate current to remove stored charge from the device’s base regions. This suppresses the regenerative action responsible for conduction, thereby achieving turn-off without external commutation networks. Step-by-Step Solution: Turn-ON: Apply positive gate current to inject carriers and trigger conduction.Turn-OFF: Apply a strong negative gate pulse; this extracts carriers, disabling the regenerative feedback.Therefore, A is true and R correctly explains how turn-off is achieved. Verification / Alternative check: Manufacturer datasheets specify peak negative gate current requirements and di/dt limits for reliable GTO turn-off, confirming gate-controlled turn-off capability. Why Other Options Are Wrong: (b) Incorrect because R directly explains A. (c) and (d) contradict well-known GTO operation. (e) Both statements are known to be true. Common Pitfalls: Using insufficient negative gate current during turn-off leads to failure to commutate and potential device damage. Confusing GTOs with standard SCRs, which cannot be turned off via gate. Final Answer: Both A and R are correct and R is correct explanation of A

Question 9

Phase control in controlled rectifiers As the firing (triggering) angle of a thyristor in a controlled rectifier is delayed further into the cycle, how does the average DC output voltage change?

Accepted Answer

Introduction / Context: Controlled rectifiers (phase-controlled converters) regulate output voltage by delaying the firing angle of thyristors relative to the AC source. Understanding the relationship between firing angle and average output voltage is fundamental in drives and DC power supplies. Given Data / Assumptions: Single-phase or three-phase line-commutated converter. Ideal source and device drops neglected for the basic trend. Firing angle denoted by alpha (α), measured from the natural commutation point. Concept / Approach: In a controlled rectifier, the average DC output voltage is proportional to the cosine of the firing angle. For example, for a single-phase full-wave bridge, Vdc = (2*Vm/π) * cos α (ideal). As α increases (firing delayed), cos α decreases; beyond 90°, Vdc becomes negative (inversion). Step-by-Step Solution: At α = 0°, conduction begins immediately at each half-cycle’s voltage peak alignment, producing maximum positive Vdc.As α increases, conduction starts later, reducing the area under the rectified waveform and lowering average Vdc.Therefore, delaying firing angle decreases the average output voltage; at α > 90°, the converter can enter inversion with Vdc < 0. Verification / Alternative check: Plot Vdc against α using Vdc = K * cos α; the monotonic decrease with α is evident from 0° to 180°. Why Other Options Are Wrong: “False” would imply Vdc rises with α, contradicting the cosine law of phase control. Common Pitfalls: Confusing instantaneous output with average output; instantaneous may be high at peaks, but average falls with greater α. Ignoring device drops—these shift values slightly but not the trend. Final Answer: True

Question 10

Assertion–Reason on SCR turn-on behavior Assertion (A): As an SCR turns on and anode current rises, the anode–cathode voltage drop decreases. Reason (R): During turn-on, the reverse-biased junction inside the SCR breaks down in avalanche.

Accepted Answer

Introduction / Context: Understanding the internal physics of SCR turn-on helps explain observed waveforms: a rapid drop in anode–cathode voltage accompanies a rise in current after triggering. It is important to distinguish regenerative carrier injection from avalanche breakdown. Given Data / Assumptions: Silicon-controlled rectifier (four-layer p-n-p-n structure). Gate-triggered turn-on under normal operating conditions. Concept / Approach: When properly triggered, the SCR turns on via regenerative action between its two internal transistor equivalents. Carrier injection forward-biases the central junction; the device transitions from blocking to conducting with a steep decrease in VAK. This is not due to destructive avalanche breakdown under normal operation. Step-by-Step Solution: Apply gate current, initiating conduction in the p-n junction near the gate.Mutual transistor action causes rapid increase in carriers; the middle junction switches from reverse to effectively forward-biased by injection.The device enters the low on-state resistance region; VAK collapses while IA rises.Therefore A is true (VAK falls as IA rises), but R is false because normal turn-on is not avalanche breakdown. Verification / Alternative check: Typical SCR V–I characteristics and turn-on waveforms show a transition to a low on-state voltage with moderate slope resistance; avalanche is associated with overvoltage, not standard gate turn-on. Why Other Options Are Wrong: (a), (b) would require R to be a correct explanation, which it is not. (d) A is well-known to be true. (e) Both being wrong contradicts basic SCR behavior. Common Pitfalls: Confusing “breakover” with “avalanche”; controlled turn-on uses gate injection below breakover voltage. Final Answer: A is correct but R is wrong

Question 11

On-state voltage behavior of a conducting thyristor When a thyristor is conducting (on-state), how does the anode–cathode voltage drop vary with increasing load current?

Accepted Answer

Introduction / Context: Designers often approximate a conducting thyristor as a small fixed drop (like a diode). In reality, the on-state voltage has a fixed component plus a component proportional to current, described by the device’s on-state dynamic resistance. Given Data / Assumptions: Thyristor is fully turned on and remains in conduction region. Temperature and device type affect exact values but not the trend. Concept / Approach: The on-state characteristic of an SCR can be modeled as V_on ≈ V_T0 + r_on * I, where V_T0 is the threshold-like component and r_on is the slope (dynamic) resistance. As current rises, the r_on * I term increases the total drop slightly. Step-by-Step Solution: Assume V_on = V_T0 + r_on * I.For higher I, the r_on * I term grows.Therefore, V_on shows a slight upward trend with current. Verification / Alternative check: Manufacturer V–I curves confirm a non-zero slope in the conduction region; tabulated on-state voltage is typically specified at a rated current (for example, 1.6 V at I_T). Why Other Options Are Wrong: Absolutely constant ignores dynamic resistance. Decreasing voltage with more current contradicts observed characteristics. “None of the above” is invalid because (c) is standard behavior. Common Pitfalls: Using a pure constant-drop model in loss calculations underestimates conduction losses at high current. Final Answer: It increases slightly with increasing current (due to on-state slope resistance)

Question 12

Equalizing networks for series-connected thyristors For static (steady-state) voltage sharing across a string of series-connected thyristors, which equalization network is appropriate?

Accepted Answer

Introduction / Context: When connecting thyristors in series to withstand higher DC voltages, device parameter tolerances cause unequal voltage sharing. Proper equalizing networks are required to prevent over-stressing an individual device. Given Data / Assumptions: Concern is steady-state (static) off-state voltage distribution. Dynamic equalization during transients is considered separately. Concept / Approach: Static inequality arises from variations in leakage currents. Placing a relatively large-value resistor in parallel with each device provides a defined sharing current that dominates leakage variations, equalizing the DC voltage division. Step-by-Step Solution: For static sharing: use parallel resistors (R) across each device.Choose R so that balancing current >> worst-case leakage difference.For dynamic sharing (not asked), add capacitors (RC) to address dv/dt and transient splits. Verification / Alternative check: Application notes show “R-only” networks for static sharing and “R//C” networks for both static and dynamic sharing in high dv/dt environments. Why Other Options Are Wrong: Series resistors do not correct off-state voltage sharing. RC across each device is a dynamic (and static) solution but is more than needed if only static sharing is asked. A single shunt across the whole string does not enforce per-device sharing. Common Pitfalls: Selecting R too large, giving negligible balancing current and poor sharing. Omitting RC when high dv/dt transients are present. Final Answer: One resistor in parallel with each thyristor

Question 13

Modified series inverter with two inductors and two capacitors Which is the primary advantage of the modified series inverter topology that employs two inductors and two capacitors?

Accepted Answer

Introduction / Context: Series inverters rely on resonant commutation using L–C components. The modified series inverter introduces an additional inductor and capacitor to improve current continuity from the DC source and to mitigate some drawbacks of the basic configuration. Given Data / Assumptions: Topology: modified series inverter with two inductors and two capacitors. Goal: understand its practical advantage compared to the basic series inverter. Concept / Approach: In the classical series inverter, the DC source may see intermittent or oscillatory current because the resonant loop periodically isolates the source. The modified arrangement shapes the commutation so that the source current is more continuous or at least less intermittently interrupted, reducing source stress and improving usable operating range. Step-by-Step Solution: Identify limitation: in basic series inverter, the source current can become discontinuous due to the resonant path taking over during commutation.Modified topology inserts an extra L–C path, providing alternate current routes that keep the source from repeatedly opening the current path.Result: the DC source avoids intermittent operation, easing filtering and improving behavior under varying loads. Verification / Alternative check: Typical waveforms for the modified series inverter show reduced ripple at the source terminals and less severe current zeros compared to the basic form. Why Other Options Are Wrong: (b) Efficiency can improve in some regimes but is not universally “more” at all operating points. (c) Output frequency is set by control and resonant choices; it is not inherently “low”. (d) Fails because (c) is not generally true. (e) Commutation still relies on resonance; networks are required. Common Pitfalls: Assuming the modified topology guarantees higher efficiency under any load. Equating modified series inverter with current-source inverter—these are distinct. Final Answer: The intermittent operation of the battery (DC source) is avoided

Question 14

Operating quadrants of a fully controlled bridge converter A line-commutated, fully controlled bridge feeding an inductive DC load (current unidirectional) can operate in which voltage–current quadrants?

Accepted Answer

Introduction / Context: Quadrant capability describes the signs of voltage and current at the DC output. A fully controlled bridge (all thyristors) can both rectify and, when firing angle exceeds 90°, invert power back to AC, provided the DC current remains in the same direction due to inductance. Given Data / Assumptions: Unidirectional DC current (due to series inductance and device orientation). Firing angle α variable from 0° to > 90°. Line-commutated operation (no forced commutation). Concept / Approach: With α < 90°, the average DC voltage Vdc is positive, current is positive → Quadrant I (V > 0, I > 0). With α > 90°, the average DC voltage becomes negative while current remains positive (still flowing into the load due to continuity) → power flows from DC side back to AC (inversion), which is Quadrant II (V < 0, I > 0). Step-by-Step Solution: Use Vdc ∝ cos α for a full bridge.For 0° ≤ α < 90°, cos α > 0 ⇒ Vdc > 0, I > 0 → Quadrant I.For 90° < α < 180°, cos α < 0 ⇒ Vdc < 0, but current remains positive → Quadrant II. Verification / Alternative check: Drive texts on DC motor control show two-quadrant operation (I and II) with a single fully controlled bridge and continuous current. Why Other Options Are Wrong: (c) Quadrant III requires negative current, which this topology does not provide. (d) Quadrant IV also needs negative current; again not available with line-commutated bridge and passive series L. (a) Understates the inversion capability. Common Pitfalls: Confusing motor torque reversal (which may need current reversal) with electrical inversion (voltage reversal at same current direction). Final Answer: 1 and 2

Question 15

TRIAC internal equivalence A TRIAC can be modeled (for understanding conduction paths) as the equivalent of:

Accepted Answer

Introduction / Context: Functionally, a TRIAC conducts current in both directions when appropriately triggered. A helpful mental model is to imagine two SCRs connected back-to-back so that each conducts during opposite half-cycles. Given Data / Assumptions: Equivalence is conceptual, not a literal internal schematic. Focus is on conduction directions and triggering quadrants. Concept / Approach: Two conventional SCRs connected in inverse-parallel allow current flow for both polarities of applied voltage, each SCR handling one polarity. That is the same external behavior that a single TRIAC provides, with the added feature of four-quadrant gate triggering referenced typically to T1. Step-by-Step Solution: Consider positive half-cycle: the “forward” SCR conducts when gated.Consider negative half-cycle: the other SCR conducts when gated.Thus, the pair emulates a bidirectional controlled switch—the TRIAC. Verification / Alternative check: Datasheet quadrant triggering diagrams mirror those of two anti-parallel SCRs with shared gate control strategies. Why Other Options Are Wrong: Series connection would not give bidirectional conduction with control. Thyristor+diode or thyristor+transistor do not reproduce the same four-quadrant characteristics. Common Pitfalls: Assuming a TRIAC is physically two discrete SCR chips; internal structures vary, though the behavior is equivalent. Final Answer: Two thyristors in parallel (inverse-parallel connection)

Question 16

Assertion–Reason on inverter output control by PWM Assertion (A): The output voltage of an inverter can be effectively controlled using pulse-width modulation (PWM). Reason (R): In a current-source inverter (CSI), the input current is constant.

Accepted Answer

Introduction / Context: PWM is the predominant technique for controlling inverter output magnitude and waveform quality, particularly in voltage-source inverters (VSIs). Separately, current-source inverters aim to keep the DC input current nearly constant using a large series inductor. Given Data / Assumptions: PWM varies pulse widths or switching patterns to adjust the fundamental output and reduce harmonics. CSI employs a stiff current source at its input. Concept / Approach: Statement A: True. By modulating pulse widths (sinusoidal PWM, space-vector PWM), an inverter’s fundamental RMS output can be linearly controlled while managing harmonic content. Statement R: True in principle for CSIs; a large DC-side inductance imposes quasi-constant input current. However, that constancy does not explain why PWM controls an inverter’s output—the explanation pertains to switching pattern control rather than current-source characteristics. Step-by-Step Solution: Accept A as true: PWM directly sets effective output voltage waveform.Accept R as true: CSI design targets nearly constant input current.Causality check: R does not logically explain A, so choose “both true, but R not the explanation”. Verification / Alternative check: Control textbooks treat PWM as an inverter modulation method independent of whether the DC link is voltage- or current-fed; CSI constant current is a separate design feature. Why Other Options Are Wrong: (a) would require R to explain the PWM principle, which it does not. (c), (d), (e) contradict established inverter theory. Common Pitfalls: Conflating VSI and CSI control methods; PWM control of voltage does not depend on a CSI’s constant-current input. Final Answer: Both A and R correct but R is not correct explanation of A

Question 17

In power electronics, what is an SUS (Silicon Unilateral Switch)? Choose the description that correctly identifies its internal construction and how it achieves unilateral triggering, while keeping in mind how it differs from an SCR, UJT, and PUT.

Accepted Answer

Introduction / Context: In power electronics and triggering circuits, the Silicon Unilateral Switch (SUS) is a small-signal thyristor-like device used to generate a precise pulse when its breakover condition is met. Understanding its internal construction helps distinguish it from SCRs, UJTs, and PUTs. Given Data / Assumptions: SUS provides unilateral triggering at a preset breakover voltage. Comparison is with SCR, UJT, and PUT. Focus is on internal structure and functional equivalence. Concept / Approach: The SUS can be modeled functionally as a PUT combined with a built-in avalanche reference path. This arrangement establishes a repeatable trigger point without external zeners or elaborate biasing. Step-by-Step Solution: 1) SUS goal: produce a sharp trigger when V reaches a designed breakover.2) PUT alone needs external biasing; adding an avalanche diode between gate–cathode sets a reference.3) When anode–cathode voltage and gate conditions satisfy breakover, the device snaps on unilaterally.4) Hence, a practical equivalence is “PUT + avalanche diode (anti-parallel across gate–cathode)”. Verification / Alternative check: Datasheets often state SUS as a PUT derivative with an integral avalanche junction to establish V_BO precisely. That differentiates it from an SCR (power switch) or UJT (two-layer device with emitter injection behavior). Why Other Options Are Wrong: Exactly similar to an SCR: incorrect; SCR is a power device with different structure and bilateral gate physics. Similar to a UJT: incorrect; SUS behaves closer to PUT than UJT. SCR + avalanche diode in parallel: not the way SUS is realized; parallel across anode–cathode is not the SUS reference path. None of the above: incorrect because the correct construction is known. Common Pitfalls: Confusing SCR family devices; SUS is low-power triggering, not a high-power controller. Thinking the avalanche diode is across anode–cathode instead of gate–cathode. Final Answer: A PUT with an avalanche diode in anti-parallel between gate–cathode (Option C).

Question 18

In a single-phase half-bridge inverter, two main thyristors (or switches) operate with complementary gating. What is the phase displacement between their gate pulses for ideal 50% duty operation?

Accepted Answer

Introduction / Context: Half-bridge inverters are foundational DC–AC conversion circuits. They produce a two-level output by alternately turning on two complementary switches, creating a square wave across the load. The required phase displacement between gate signals ensures non-overlap conduction and correct waveform generation. Given Data / Assumptions: Single-phase half-bridge inverter with two complementary devices. Ideal devices and dead-time negligible for the concept. Target: 50% duty each to synthesize a square wave. Concept / Approach: For a half-bridge, when the top device is ON, the output is approximately +V/2 (with a split DC bus). When the bottom device is ON, the output is approximately -V/2. To achieve this alternating pattern, the gate pulses must be 180° apart in phase so that one is ON while the other is OFF. Step-by-Step Solution: 1) A full electrical cycle is 360°.2) Each switch should conduct for approximately 180° to achieve a 50% duty square wave.3) Therefore, the turn-on instants of the two switches differ by 180°. Verification / Alternative check: Observe the output waveform polarity: positive half-cycle when the top device conducts, negative half-cycle when the bottom device conducts. These halves must be sequential and not overlapping, hence 180° separation. Why Other Options Are Wrong: 60°, 90°, 120°: These would distort the duty ratio and either force overlap or gaps, not delivering the intended two-level square wave. None of the above: incorrect because 180° is the standard for complementary operation. Common Pitfalls: Confusing half-bridge with full-bridge timing where legs are also 180° apart but applied to two legs. Ignoring dead-time; while needed in practice, it does not change the nominal 180° separation. Final Answer: 180° (Option D).

Question 19

In semiconductor diodes, the typical cut-in (threshold) voltages are different for germanium and silicon. Identify the correct pair for germanium and silicon, respectively, keeping common textbook values in mind for forward conduction onset.

Accepted Answer

Introduction / Context: Cut-in (or knee/threshold) voltage is the approximate forward voltage where a diode begins conducting appreciable current. Germanium and silicon have different bandgaps, so their forward characteristics differ noticeably in low-current circuits and rectifiers. Given Data / Assumptions: Room temperature operation. Common small-signal/rectifier diode behavior, not Schottky or special-purpose diodes. We are after typical nominal values. Concept / Approach: Due to a smaller bandgap, germanium diodes typically conduct at a lower forward voltage than silicon. Textbook nominal values widely used are about 0.3 V for Ge and about 0.7 V for Si under modest current levels. Step-by-Step Solution: 1) Identify material: Ge vs. Si.2) Recall typical thresholds: Ge ≈ 0.3 V, Si ≈ 0.7 V.3) Match with the option set. Verification / Alternative check: Datasheets and lab experiments consistently show lower forward drops for Ge. Silicon’s higher drop is standard in rectifier and logic diode tutorials. Why Other Options Are Wrong: 0.7 V and 0.3 V: reversed order. 0.6 V and 0.9 V: too high for Si and unusually high for Ge. 0.5 V and 0.7 V: Ge value is too high. 0.2 V and 0.4 V: unrealistically low for typical room-temperature conduction points. Common Pitfalls: Confusing Schottky diodes (lower drops) with Ge/Si PN diodes. Assuming constant drop; actual forward voltage depends on current and temperature. Final Answer: 0.3 V (Ge) and 0.7 V (Si) (Option A).

Question 20

A cycloconverter is often described as a frequency changer built from controlled rectifier sections. Decide whether the statement “A cycloconverter is a group of controlled rectifiers” is correct, and briefly reason about its line-commutated operati…

Accepted Answer

Introduction / Context: Cycloconverters directly convert AC of one frequency to a lower frequency AC without an intermediate DC link. They are widely used in large drives where low-frequency, high-torque operation is needed, such as rolling mills or kilns. Given Data / Assumptions: Traditional thyristor-based cycloconverters (line-commutated). Output frequency typically a fraction of input frequency. Device topology relies on phase-controlled rectifier bridges. Concept / Approach: A cycloconverter is realized by combining multiple controlled rectifier bridges (thyristor bridges). By modulating firing angles, the circuit pieces together positive and negative segments to synthesize a low-frequency output waveform directly from the mains. Step-by-Step Solution: 1) Recognize that each “arm” is a controlled rectifier capable of shaping the segment polarity and magnitude.2) By alternately gating the bridges, the output polarity swaps, creating AC at a lower frequency.3) Hence, structurally and functionally, a cycloconverter is indeed a group of controlled rectifiers coordinated by a controller. Verification / Alternative check: Standard power electronics texts show single-phase and three-phase cycloconverter schematics consisting of anti-parallel controlled rectifier sections feeding a load. Why Other Options Are Wrong: False: contradicted by the well-known thyristor-bridge construction. True but only for DC outputs: incorrect; cycloconverters produce AC outputs. Transistor-only: not historically correct for classic industrial cycloconverters. Cannot be determined: it can; the construction is standard knowledge. Common Pitfalls: Confusing cycloconverters with DC link inverters (AC–DC–AC). Assuming PWM transistor inverters; those are different modern topologies. Final Answer: True (Option A).

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