Q: In carrier-based PWM for power converters, if A_r is the peak amplitude of the reference (modulating) signal and A_c is the peak amplitude of the carrier (triangular) signal, the modulation index M is defined as:

Introduction / Context: Carrier-based PWM (pulse-width modulation) compares a low-frequency reference (modulating) signal with a high-frequency triangular carrier to generate switching pulses. The modulation index M quantifies how strongly the reference modulates the carrier, thereby controlling the fundamental output voltage of inverters or converters. Given Data / Assumptions: A_r: peak amplitude of the reference signal. A_c: peak amplitude of the triangular carrier. Linear modulation region is assumed (no overmodulation). Concept / Approach: In standard sinusoidal PWM, the modulation index is defined as the ratio of the reference amplitude to the carrier amplitude (both peaks). This ratio governs the fundamental component of the output voltage for a given DC bus. Step-by-Step Solution: 1) Identify A_r as the modulating signal amplitude.2) Identify A_c as the carrier amplitude.3) Use the definition M = A_r / A_c. Verification / Alternative check: Textbooks and application notes present M = A_r / A_c for sinusoidal PWM; fundamental output is proportional to M in the linear region. Why Other Options Are Wrong: A_c / A_r: inverted ratio. Product or difference/sum forms: not the standard PWM modulation index definition. A_r^2 / A_c: dimensionally inconsistent for the definition. Common Pitfalls: Confusing peak, RMS, or average definitions; M uses peak values. Applying the definition outside linear range where distortion occurs. Final Answer: M = A_r / A_c.

Q: In single-phase DC drives, a 'dual converter' is used for reversible operation. Which connection accurately describes a single-phase dual converter?

Introduction / Context: Dual converters provide four-quadrant DC drive capability by allowing power flow and motor current reversal without mechanical contactors. Understanding their connection is essential for regenerative braking and smooth reversals. Given Data / Assumptions: Single-phase system. Reversible voltage polarity at the DC output is required. Concept / Approach: A single-phase dual converter comprises two fully controlled bridges connected in antiparallel on the DC side. One bridge provides positive polarity DC, the other provides negative polarity DC. Interlocking and circulating/non-circulating modes manage current reversal and harmonics. Step-by-Step Solution: 1) Identify need: reversible DC output polarity.2) Use two controlled bridges; one is assigned positive polarity, the other negative.3) Connect antiparallel so only one bridge conducts at a time (or both in circulating mode with reactors). Verification / Alternative check: Drive manuals depict antiparallel fully controlled bridges; semiconverters are not used for full four-quadrant operation. Why Other Options Are Wrong: Parallel same polarity cannot reverse output. Semiconverters do not provide full four-quadrant functionality. Center-tapped single bridge does not enable polarity reversal without additional hardware. Common Pitfalls: Confusing antiparallel with anti-series; polarity reversal requires antiparallel arrangement. Ignoring circulating current management between bridges. Final Answer: Two fully controlled converters in antiparallel.

Q: In a single-phase half-wave AC voltage regulator supplying a purely resistive load via a series thyristor (phase control on one half-cycle only), which statement best describes the RMS load current sign?

Introduction / Context: Half-wave AC regulators use a single series-controlled device to pass a portion of one half-cycle of the source to a resistive load. The RMS (root mean square) value is a magnitude measure of heating effect and cannot be negative by definition. Given Data / Assumptions: Resistive (R) load. Half-wave phase control (typically on the positive half-cycle only). Standard definition of RMS. Concept / Approach: Instantaneous current follows the applied half-cycle waveform scaled by R. Although instantaneous current is zero during blocked intervals and positive during conduction, the RMS value computed over the full period is a non-negative magnitude reflecting equivalent heating in a resistor. Step-by-Step Solution: 1) Instantaneous current i(t) = v(t)/R when thyristor is ON; i(t) = 0 when OFF.2) RMS is sqrt( (1/T) * ∫ i(t)^2 dt ) over one period T.3) Because i(t)^2 ≥ 0 at all times, the RMS result is non-negative; with any conduction the RMS is strictly positive. Verification / Alternative check: Even if conduction angle is small, the RMS remains a positive magnitude proportional to the square-root of mean power delivered. Why Other Options Are Wrong: Negative RMS has no physical meaning; RMS is a magnitude. Zero RMS would imply no conduction at all (thyristor never fires). “Alternates sign” pertains to average or instantaneous values, not RMS. Common Pitfalls: Mixing up average current (which can be zero for symmetric waveforms) with RMS. Confusing instantaneous polarity with RMS sign. Final Answer: It is always positive.

Question 1

Load characteristics for a single-phase full-bridge inverter What type of load can a single-phase full-bridge inverter supply?

Accepted Answer

Introduction / Context: Voltage-source inverters (VSIs) such as the single-phase full-bridge are widely used to supply a variety of loads. Knowing the permissible load characteristics helps with filter design, protection, and commutation considerations. Given Data / Assumptions: Full-bridge VSI with appropriate device ratings. Switching strategy may be square-wave or PWM. Snubbers or filters may be added as needed. Concept / Approach: A VSI controls the output voltage waveform; the load determines the resulting current. Provided the inverter and protection are properly designed, it can supply R, RL, RC, or RLC loads. The terms “underdamped” and “overdamped” describe transient response of RLC circuits, not restrictions on inverter applicability. Step-by-Step Solution: Recognize VSI output is a voltage source with finite impedance.Any practical load (R, RL, RLC, even nonlinear) may be supplied with due design considerations.Therefore, the correct general statement is that it may be any type of load. Verification / Alternative check: Industrial VSIs run induction motors (RL), heaters (R), and power-factor-corrected loads (RLC) with appropriate control and filtering. Why Other Options Are Wrong: “must be underdamped/overdamped” incorrectly constrains the topology. “only purely resistive” is needlessly restrictive. Common Pitfalls: Ignoring necessary snubbers for high dv/dt on capacitive loads. Underestimating required freewheeling paths for inductive loads. Final Answer: may be any type of load

Question 2

Operation of a single-phase dual converter (circulating-current mode) When a single-phase dual converter operates in circulating-current mode, which statement is correct about the two converters?

Accepted Answer

Introduction / Context: Dual converters provide reversible DC voltage and current. In the circulating-current mode, both bridges are gated, allowing instantaneous reversal without gaps in conduction and improved dynamic response. Given Data / Assumptions: Two fully controlled bridges connected antiparallel on the DC side. A series reactor limits circulating current. Firing angles are set such that one bridge rectifies while the other inverts. Concept / Approach: To create a controllable average DC voltage around zero, one converter is biased to produce positive DC voltage (rectifier) while the other is biased to produce negative DC voltage (inverter). Their slight voltage difference creates circulating current through the reactor while enabling instantaneous polarity transitions at the load. Step-by-Step Solution: Set Converter 1 with α < 90° → positive Vdc (rectifier).Set Converter 2 with α > 90° → negative Vdc (inverter).Operate both simultaneously; the reactor carries the circulating current.Hence, one rectifies while the other inverts. Verification / Alternative check: Phase control equations Vdc ∝ cos α confirm opposite polarities with complementary firing angles. Why Other Options Are Wrong: Both rectifying or both inverting would not provide the necessary polarity balance. “only one converter conducts” describes non-circulating mode, not circulating mode. Common Pitfalls: Insufficient reactor inductance causing excessive ripple and stress. Mis-synchronization leading to large loop currents. Final Answer: one converter works as a rectifier while the other works as an inverter

Question 3

Disc (press-pack) mounting of SCRs In disc mounting of an SCR, how is the device clamped to the heat sinks?

Accepted Answer

Introduction / Context: High-power SCRs often use press-pack (disc) packages, which rely on axial compression to ensure low thermal resistance and reliable electrical contact, especially in series/parallel stacks. Given Data / Assumptions: Disc-type SCR with metallized contact faces. Dedicated clamping hardware is used to apply uniform pressure. Cooling is achieved through heat sinks on both sides. Concept / Approach: Press-pack mounting sandwiches the SCR between two heat sinks. The pressure ensures intimate contact, minimizing contact resistance and thermal impedance. This approach is robust under thermal cycling and is commonly used in traction, HVDC valves, and industrial drives. Step-by-Step Solution: Prepare two flat, clean heat sink faces.Insert the SCR between them, aligning terminals as specified.Apply and calibrate the specified clamping force.Achieve low thermal and electrical resistance paths through both sides. Verification / Alternative check: Manufacturer application notes specify required clamping forces and flatness tolerances for press-pack packages. Why Other Options Are Wrong: Single-sided clamping is inadequate for disc types. “Low rating; no clamping” is false; ratings are typically high and clamping is essential. Adhesive attachment is not used for power SCRs. Common Pitfalls: Incorrect clamping force causing damage or poor conduction. Dirty or warped surfaces raising thermal resistance. Final Answer: SCR is clamped between two disc-shaped heat sinks (press-pack)

Question 4

Schottky diode versus p–n diode Compared to a conventional p–n junction diode, a Schottky diode typically has:

Accepted Answer

Introduction / Context: Schottky diodes form a metal–semiconductor junction rather than a p–n junction. They are widely used in high-speed switching power supplies due to their low forward drop and negligible reverse recovery charge. Given Data / Assumptions: Forward “cut-in” (knee) voltage of Schottky ~ 0.2–0.4 V (Si), lower than p–n ~ 0.6–0.7 V. Reverse leakage of Schottky is typically higher than p–n of the same area and rating. Concept / Approach: The barrier at a metal–semiconductor contact is lower than the built-in potential of a p–n junction, yielding a lower forward drop. However, the same structure provides fewer mechanisms to block reverse current tightly, resulting in higher leakage, especially at elevated temperatures. Step-by-Step Solution: Compare forward conduction: Schottky has lower Vf → lower cut-in voltage.Compare reverse behavior: Schottky exhibits larger reverse saturation/leakage.Therefore, the correct pair is “lower cut-in voltage and higher reverse saturation current”. Verification / Alternative check: Datasheets for power Schottky rectifiers specify low Vf and relatively high IR compared to ultrafast p–n diodes. Why Other Options Are Wrong: (a) claims low leakage, which is not typical. (b) and (d) contradict the hallmark low forward drop. (e) reverse current is nonzero in real devices. Common Pitfalls: Using Schottky diodes at high temperature without accounting for increased leakage. Assuming zero reverse recovery—there is very low charge, but not strictly zero dynamics. Final Answer: lower cut-in voltage and higher reverse saturation current

Question 5

BJT operating conditions (active region) Which statement correctly describes a Bipolar Junction Transistor (BJT) in its forward-active region?

Accepted Answer

Introduction / Context: Understanding BJT biasing is fundamental for analog amplifiers and switching applications. The forward-active region is the usual operating region for linear amplification. Given Data / Assumptions: NPN or PNP BJT used in common configurations (CE, CB, CC). Forward-active region biasing. Neglecting second-order effects like Early effect for the basic statement. Concept / Approach: In forward-active region, the emitter–base junction is forward biased to inject carriers from emitter into base, while the collector–base junction is reverse biased to sweep these carriers into the collector. This yields collector current approximately IC ≈ β * IB (or IC ≈ α * IE) with moderate dependence on VCE (Early effect). Step-by-Step Solution: Set VBE ≈ 0.7 V (Si) to forward bias E–B junction.Maintain C–B junction reverse biased (VCB > 0 for NPN).Carriers injected from emitter traverse thin, lightly doped base and are collected by the reverse field at C–B junction.Hence, option describing “E–B forward, C–B reverse” is correct. Verification / Alternative check: Small-signal hybrid-π models and textbook biasing diagrams reflect this junction polarity in forward-active operation. Why Other Options Are Wrong: (b) ignores the Early effect; dependence is small but not zero. (c) BJTs are bipolar devices; minority carrier storage is central. (d) Emitter is heavily doped (for high injection efficiency); base is lightly doped and very thin. (e) Both junctions forward-biased corresponds to saturation, not active region. Common Pitfalls: Confusing active region with saturation region. Assuming perfect independence of IC from VCE. Final Answer: The emitter–base junction is forward biased, and the collector–base junction is reverse biased

Question 6

In electric drives using power electronics, thyristor-based converters and choppers can be used to control the speed/torque of different direct-current (DC) motors. Identify all motor types for which thyristor control is commonly applied in practice.

Accepted Answer

Introduction / Context: Thyristor-based power converters (controlled rectifiers, choppers) are widely used for DC motor control. They provide adjustable armature voltage and, in some topologies, field control to regulate speed and torque. This question checks recognition of which DC motor types can be managed with thyristor control. Given Data / Assumptions: Motors considered: separately excited, shunt, and series DC motors. Armature voltage and/or field control achievable via thyristor converters. Standard industrial practices and classic drive schemes are implied. Concept / Approach: By phase-controlling AC–DC converters or using DC–DC choppers, the average armature voltage is varied, thus changing speed. With appropriate feedback and protection, all mainstream DC motor types can be controlled. Field weakening can extend speed range for shunt and separately excited machines. Step-by-Step Solution: 1) For separately excited DC motors, armature voltage control via a thyristor rectifier or chopper gives smooth speed control.2) For shunt motors, similar armature control applies; additional field control refines speed regulation.3) For series motors, thyristor control is also practical but requires attention to torque–speed characteristics and current surges.4) Therefore, all listed motor types are controllable with thyristors when using suitable topologies and protections. Verification / Alternative check: Classic DC drive textbooks and application notes show thyristor drives for all three types, with nuances in feedback and protection differing by motor type. Why Other Options Are Wrong: Any single motor option alone omits other valid types. None of the above: contradicted by extensive industrial usage. Common Pitfalls: Assuming series motors cannot be thyristor-controlled; they can, with proper design. Confusing control limitations with impossibility; practical issues do not preclude control. Final Answer: All of the above.

Question 7

Assertion–Reason (DC choppers): A class C chopper (two-quadrant) can operate in the first and second quadrants. Reason: A class B chopper operates only in the second quadrant (regenerative quadrant). Select the correct option.

Accepted Answer

Introduction / Context: DC choppers are classified by the quadrant(s) in which the voltage–current pair (V, I) operates. Understanding quadrant operation clarifies power flow and regenerative capability. This question checks knowledge of Class B and Class C choppers and their quadrant behavior. Given Data / Assumptions: Class B chopper: second-quadrant operation (V > 0, I < 0), energy returns to the source. Class C chopper: combination of Class A and B choppers. Concept / Approach: Class C is effectively a two-quadrant chopper combining Class A (first quadrant: motoring) and Class B (second quadrant: regenerative braking), enabling current reversal or power flow back to source depending on load and control. Step-by-Step Solution: 1) Class B chopper permits energy to flow from load to source with V positive and I negative.2) Class C chopper is a combination of Class A and Class B, hence supports both first and second quadrants.3) Therefore, the reason directly supports the assertion. Verification / Alternative check: Standard chopper classification charts show Class C = A + B (two-quadrant operation). Why Other Options Are Wrong: Options stating R is not the explanation are incorrect because knowledge of Class B behavior explains why Class C covers both quadrants. Options claiming A or R is wrong contradict accepted definitions. Common Pitfalls: Confusing Class C with Class D (third and fourth quadrants). Assuming Class B allows forward current; it does not. Final Answer: Both A and R are correct and R correctly explains A.

Question 8

In carrier-based PWM for power converters, if A_r is the peak amplitude of the reference (modulating) signal and A_c is the peak amplitude of the carrier (triangular) signal, the modulation index M is defined as:

Accepted Answer

Introduction / Context: Carrier-based PWM (pulse-width modulation) compares a low-frequency reference (modulating) signal with a high-frequency triangular carrier to generate switching pulses. The modulation index M quantifies how strongly the reference modulates the carrier, thereby controlling the fundamental output voltage of inverters or converters. Given Data / Assumptions: A_r: peak amplitude of the reference signal. A_c: peak amplitude of the triangular carrier. Linear modulation region is assumed (no overmodulation). Concept / Approach: In standard sinusoidal PWM, the modulation index is defined as the ratio of the reference amplitude to the carrier amplitude (both peaks). This ratio governs the fundamental component of the output voltage for a given DC bus. Step-by-Step Solution: 1) Identify A_r as the modulating signal amplitude.2) Identify A_c as the carrier amplitude.3) Use the definition M = A_r / A_c. Verification / Alternative check: Textbooks and application notes present M = A_r / A_c for sinusoidal PWM; fundamental output is proportional to M in the linear region. Why Other Options Are Wrong: A_c / A_r: inverted ratio. Product or difference/sum forms: not the standard PWM modulation index definition. A_r^2 / A_c: dimensionally inconsistent for the definition. Common Pitfalls: Confusing peak, RMS, or average definitions; M uses peak values. Applying the definition outside linear range where distortion occurs. Final Answer: M = A_r / A_c.

Question 9

In single-phase DC drives, a 'dual converter' is used for reversible operation. Which connection accurately describes a single-phase dual converter?

Accepted Answer

Introduction / Context: Dual converters provide four-quadrant DC drive capability by allowing power flow and motor current reversal without mechanical contactors. Understanding their connection is essential for regenerative braking and smooth reversals. Given Data / Assumptions: Single-phase system. Reversible voltage polarity at the DC output is required. Concept / Approach: A single-phase dual converter comprises two fully controlled bridges connected in antiparallel on the DC side. One bridge provides positive polarity DC, the other provides negative polarity DC. Interlocking and circulating/non-circulating modes manage current reversal and harmonics. Step-by-Step Solution: 1) Identify need: reversible DC output polarity.2) Use two controlled bridges; one is assigned positive polarity, the other negative.3) Connect antiparallel so only one bridge conducts at a time (or both in circulating mode with reactors). Verification / Alternative check: Drive manuals depict antiparallel fully controlled bridges; semiconverters are not used for full four-quadrant operation. Why Other Options Are Wrong: Parallel same polarity cannot reverse output. Semiconverters do not provide full four-quadrant functionality. Center-tapped single bridge does not enable polarity reversal without additional hardware. Common Pitfalls: Confusing antiparallel with anti-series; polarity reversal requires antiparallel arrangement. Ignoring circulating current management between bridges. Final Answer: Two fully controlled converters in antiparallel.

Question 10

In a single-phase half-wave AC voltage regulator supplying a purely resistive load via a series thyristor (phase control on one half-cycle only), which statement best describes the RMS load current sign?

Accepted Answer

Introduction / Context: Half-wave AC regulators use a single series-controlled device to pass a portion of one half-cycle of the source to a resistive load. The RMS (root mean square) value is a magnitude measure of heating effect and cannot be negative by definition. Given Data / Assumptions: Resistive (R) load. Half-wave phase control (typically on the positive half-cycle only). Standard definition of RMS. Concept / Approach: Instantaneous current follows the applied half-cycle waveform scaled by R. Although instantaneous current is zero during blocked intervals and positive during conduction, the RMS value computed over the full period is a non-negative magnitude reflecting equivalent heating in a resistor. Step-by-Step Solution: 1) Instantaneous current i(t) = v(t)/R when thyristor is ON; i(t) = 0 when OFF.2) RMS is sqrt( (1/T) * ∫ i(t)^2 dt ) over one period T.3) Because i(t)^2 ≥ 0 at all times, the RMS result is non-negative; with any conduction the RMS is strictly positive. Verification / Alternative check: Even if conduction angle is small, the RMS remains a positive magnitude proportional to the square-root of mean power delivered. Why Other Options Are Wrong: Negative RMS has no physical meaning; RMS is a magnitude. Zero RMS would imply no conduction at all (thyristor never fires). “Alternates sign” pertains to average or instantaneous values, not RMS. Common Pitfalls: Mixing up average current (which can be zero for symmetric waveforms) with RMS. Confusing instantaneous polarity with RMS sign. Final Answer: It is always positive.

Question 11

In a three-phase bridge inverter, the gating (firing) signals for the three legs are phase-shifted to produce a three-phase output. What is the phase difference between the three sets of gate pulses?

Accepted Answer

Introduction / Context: Three-phase bridge inverters synthesize three sinusoidal (or quasi-sinusoidal) phase voltages that are 120° apart. The gating scheme must reflect this separation to generate balanced three-phase outputs for motors or grids (via filters). Given Data / Assumptions: Standard three-leg inverter topology. Balanced three-phase output desired. Concept / Approach: For a balanced three-phase system, phase voltages (or their fundamental components) must be separated by 120°. Thus, the switching functions (or their reference phases) are typically displaced by 120° to achieve the correct phase relationships. Step-by-Step Solution: 1) Identify three legs: A, B, C.2) Apply modulation with phase shifts of 0°, 120°, and 240° (equivalently −120°).3) The fundamental components then have 120° displacement. Verification / Alternative check: Carrier-based PWM references for three-phase inverters use phase-shifted modulating waves by 120° to obtain a balanced set. Why Other Options Are Wrong: 60°/240° alone do not represent the three mutual separations properly. “Any of the above” is incorrect; 120° is specific to three-phase balance. 90° would create four-phase relationships, not three-phase. Common Pitfalls: Confusing electrical angle between line voltages and switching pattern details. Assuming 60° relates to six-step commutations; the phase relation remains 120°. Final Answer: 120°.

Question 12

In a single-phase cycloconverter built with a center-tapped transformer and controlled rectifiers, do all thyristors necessarily conduct for equal durations over a modulation cycle?

Accepted Answer

Introduction / Context: Cycloconverters synthesize a low-frequency output by selecting segments of the input waveform using controlled rectifiers. In center-tapped configurations, multiple thyristors are used per half-cycle and their conduction windows are determined by the modulation pattern and load current continuity. Given Data / Assumptions: Single-phase cycloconverter with center-tapped transformer. Thyristor conduction depends on firing angles and load current. Concept / Approach: Because the output frequency is a fraction of the input frequency, conduction intervals are apportioned to shape the output. Depending on the required output waveform, load power factor, and current continuity, individual thyristors may conduct for different durations within the same output cycle. Step-by-Step Solution: 1) The controller selects which rectifier group conducts to form positive or negative output segments.2) Firing angles shift dynamically across input cycles to approximate the desired low-frequency output.3) Consequently, each device’s conduction time can vary; equalization is not an inherent requirement. Verification / Alternative check: Waveform plots in cycloconverter texts show unequal conduction periods when constructing non-sinusoidal segments at low output frequencies. Why Other Options Are Wrong: Always equal or must be equal: not generally true; depends on modulation. Cannot be equal: they can coincidentally be equal for special cases. Unity power factor condition: not a guarantee of equality. Common Pitfalls: Assuming symmetry mandates identical conduction intervals. Overlooking dependence on load and target waveform. Final Answer: Durations may or may not be equal.

Question 13

A full-bridge DC–AC inverter supplies an underdamped RLC load. Which statement best describes the typical waveshapes of the inverter output voltage and the load current?

Accepted Answer

Introduction / Context: Voltage-source inverters often generate square or PWM waveforms. When these drive RLC loads, the current response is shaped by the load’s impedance versus frequency, often smoothing toward a sinusoid near resonance or within a narrow band. Given Data / Assumptions: Full-bridge inverter producing a square-wave output voltage. Load: underdamped RLC (dominant reactive behavior). Concept / Approach: The square voltage has many harmonics. The RLC load presents higher impedance to higher-order harmonics while allowing the fundamental component to pass, producing a current that is nearer to sinusoidal than the voltage. This is the basis for using LC filters with inverters. Step-by-Step Solution: 1) Decompose the square wave into Fourier components (fundamental plus odd harmonics).2) The RLC impedance magnitude increases away from resonance, attenuating most harmonics.3) The load current thus largely follows the fundamental, becoming nearly sinusoidal. Verification / Alternative check: Inverter–filter application notes show square-wave voltage and near-sinusoidal current when the load/filter corner frequency is set appropriately. Why Other Options Are Wrong: Both square: ignores filtering effect. Both sinusoidal: contradicts the applied square wave at the inverter terminals. Sinusoidal voltage with square current: not applicable for a square-wave inverter. Common Pitfalls: Assuming the current exactly sinusoidal; it is only nearly sinusoidal. Neglecting phase shift between voltage and current due to reactance. Final Answer: Square voltage, nearly sinusoidal current.

Question 14

Triggering a Silicon Controlled Switch (SCS): which of the following correctly describes how an SCS can be turned ON?

Accepted Answer

Introduction / Context: The Silicon Controlled Switch (SCS) is a thyristor-family device with two gates, enabling greater control flexibility than an SCR. Understanding correct triggering polarities for ON action is a common exam check. Given Data / Assumptions: SCS has an anode gate and a cathode gate. Standard triggering conventions apply. Concept / Approach: Unlike an SCR that is typically triggered by a positive pulse at the gate with respect to cathode, an SCS can be turned ON either by applying a positive pulse at its cathode gate or by applying a negative pulse at its anode gate, depending on device configuration and reference potentials. Step-by-Step Solution: 1) Identify available gates and their reference polarities.2) For ON action: use a positive pulse at cathode gate or negative pulse at anode gate.3) Ensure appropriate gate–cathode/anode biasing and observe device limits. Verification / Alternative check: Manufacturer datasheets and power electronics texts list both methods for SCS triggering. Why Other Options Are Wrong: Anode-only or cathode-only unspecified polarity: incomplete or incorrect. Low constant anode voltage: not a triggering method; it risks device stress. Shorting anode to gate: unsafe and not a defined procedure. Common Pitfalls: Applying SCR gate rules directly to SCS without considering dual gate structure. Ignoring need for proper gate drive source impedance. Final Answer: Positive pulse at cathode gate or negative pulse at anode gate.

Question 15

A fully controlled single-phase bridge converter feeds a continuous-current RLE load (E is back-emf, R is resistance, L large). The source is v(t) = V_m sin(ωt). Derive the firing-angle relation and choose the correct expression for cosα.

Accepted Answer

Introduction / Context: Single-phase fully controlled rectifiers feeding DC motor-type loads (RLE) are classic in drives. For continuous current (large L), the average output voltage relates directly to firing angle α, allowing us to solve for α given load back-emf and drop across R. Given Data / Assumptions: v_s(t) = V_m sin(ωt), single-phase. Fully controlled bridge (four-quadrant gating within rectifier). Continuous current (inductor large), so output ripple is small. Average DC output V_dc = E + I_0 R. Concept / Approach: For a fully controlled bridge with sinusoidal source and continuous current: V_dc = (2 V_m / π) * cosα. This comes from integrating the sinusoid over conduction interval with firing delay α. Then set V_dc equal to the DC load requirement E + I_0 R and isolate cosα. Step-by-Step Solution: 1) Use the known average: V_dc = (2 V_m / π) * cosα.2) KVL on DC side: V_dc = E + I_0 R (inductor drop averages to zero in steady state).3) Equate and solve: (2 V_m / π) * cosα = E + I_0 R.4) Hence cosα = (π / (2 V_m)) * (E + I_0 R). Verification / Alternative check: Dimensional consistency: right-hand side is dimensionless since E and I_0 R are volts and the multiplier is π/(2 V_m) with units 1/V. Why Other Options Are Wrong: π/V_m factor: off by a factor of 2. Replacing + with −: violates KVL unless direction conventions differ (here continuous motoring mode uses +). Division by V_m alone: missing π/2 scaling that arises from averaging. Common Pitfalls: Forgetting that average inductor voltage is zero in steady state. Mixing RMS and peak values; formula requires V_m (peak). Final Answer: cosα = (π / (2 V_m)) * (E + I_0 R).

Question 16

Series RLC circuit – condition for underdamped oscillations For a series RLC circuit driven away from steady-state, under what condition on R, L, and C will the natural response be underdamped (i.e., oscillatory with an exponentially decaying envelo…

Accepted Answer

Introduction / Context: In circuit theory and power electronics, the transient response of a series RLC (resistor–inductor–capacitor) network can be underdamped, critically damped, or overdamped. Recognizing the mathematical condition for each regime is essential for designing commutation networks, filters, snubbers, and resonant converters. Given Data / Assumptions: Series RLC with elements R, L, and C. Natural (homogeneous) response considered. Linear, time-invariant components; no non-ideal losses beyond R. Concept / Approach: The differential equation for a series RLC (current i as the state) leads to the characteristic equation s^2 + (R/L)s + 1/(LC) = 0. The damping depends on the discriminant Δ = (R/L)^2 − 4/(LC). If Δ < 0, complex conjugate roots occur and the response is underdamped (oscillatory). If Δ = 0, it is critically damped; if Δ > 0, it is overdamped. Step-by-Step Solution: Characteristic equation: s^2 + (R/L)s + 1/(LC) = 0Discriminant: Δ = (R/L)^2 − 4/(LC)Underdamped condition: Δ < 0 → (R/L)^2 < 4/(LC)Multiply both sides by L^2C: R^2 C < 4LRearrange: R^2 < 4L/C Verification / Alternative check: Define damping factor α = R/(2L) and undamped natural frequency ω0 = 1/sqrt(LC). Underdamped occurs when α < ω0, i.e., R/(2L) < 1/sqrt(LC) → R^2 < 4L/C, which matches the condition derived above. Why Other Options Are Wrong: R^2 = 4L/C (option b): This is the critically damped boundary, not underdamped. R^2 > 4L/C (option c): Overdamped, producing non-oscillatory exponentials. R = 0 only (option d): R = 0 yields purely sinusoidal (undamped) oscillation, not the general underdamped case with decay. Common Pitfalls: Confusing the undamped (R = 0) case with underdamped; underdamped requires finite R but less than the critical value. Another mistake is mixing series and parallel RLC conditions; their formulas differ. Final Answer: R^2 < 4L/C

Question 17

Parallel thyristors – causes of unequal current sharing When thyristors are connected in parallel to share current, which effects can make the current distribution non-uniform?

Accepted Answer

Introduction / Context: High-current applications often use parallel thyristors to reduce per-device stress. However, unequal current sharing can cause thermal runaway and premature failure. Understanding parasitic effects helps in designing equalizing networks and proper layouts. Given Data / Assumptions: Two or more thyristors in parallel. Interconnects introduce parasitic inductance and capacitance. Devices have tolerances in V–I characteristics and thermal behavior. Concept / Approach: Dynamic sharing during turn-on/turn-off is influenced by stray inductances (affecting di/dt and current division) and capacitances (affecting dv/dt and initial current spikes). Static sharing is influenced by unequal on-state voltage drops and temperature coefficients. Equalizing resistors (static) and saturable/series inductors or RC snubbers (dynamic) mitigate these issues. Step-by-Step Solution: Stray inductance differences → unequal di/dt → one device may conduct more.Stray capacitances and dv/dt → displacement currents that bias devices unevenly.Device tolerances and thermal gradients further worsen sharing without equalization. Verification / Alternative check: Manufacturer application notes recommend both static and dynamic equalizing elements specifically because both inductive and capacitive parasitics disturb current balance. Why Other Options Are Wrong: Inductive only or capacitive only (options a, b) understate the problem; both phenomena are relevant. None of the above (option d) ignores widely documented parasitic effects. Common Pitfalls: Relying solely on matched devices without proper layout or equalization. Long busbars can add significant inductance and imbalance. Final Answer: Both inductive and capacitive effects

Question 18

Single-pulse PWM in a bridge inverter With single-pulse width modulation (one pulse per half-cycle) in a single-phase bridge inverter, the output (fundamental) frequency equals the frequency of which signal?

Accepted Answer

Introduction / Context: Pulse width modulation (PWM) uses a low-frequency reference to shape the output fundamental and a high-frequency carrier to determine pulse placement or width. In single-pulse PWM, exactly one pulse is generated per half-cycle, and the fundamental frequency is set by the reference waveform used for modulation. Given Data / Assumptions: Single-phase bridge inverter. Single-pulse PWM (one pulse per half-cycle). Low-frequency reference compared with high-frequency carrier or timing ramp. Concept / Approach: The output voltage is synthesized by switching the DC link according to comparisons between the reference and a carrier (or by timing logic). The fundamental component of the inverter output tracks the frequency of the reference (modulating) signal; the carrier frequency sets the switching frequency, not the fundamental power frequency delivered to the load. Step-by-Step Solution: Define f_ref = frequency of the sinusoidal (or programmed) modulating reference.Define f_car = frequency of the carrier (triangular/sawtooth) or timing signal.For single-pulse PWM, the output voltage waveform repeats each reference half-cycle → fundamental at f_ref. Verification / Alternative check: Fourier analysis of PWM waveforms shows the fundamental component frequency equals the reference frequency; sidebands and harmonics are clustered around multiples of the carrier and its intermodulations but do not alter the fundamental frequency selection. Why Other Options Are Wrong: Carrier frequency (option b) determines switching ripple, not the delivered fundamental. Options c and d conflict with standard PWM principles. Common Pitfalls: Confusing switching frequency with output fundamental; increasing carrier frequency reduces ripple but does not change f_out. Final Answer: Reference (modulating) signal

Question 19

Shockley diode structure How many semiconductor layers does a Shockley diode (two-terminal thyristor) contain?

Accepted Answer

Introduction / Context: The Shockley diode is the simplest thyristor structure—a two-terminal, four-layer device that exhibits regenerative switching. It is foundational to understanding SCRs (which add a gate terminal for controlled turn-on). Given Data / Assumptions: A Shockley diode is a PNPN device with two terminals (anode and cathode). Exhibits breakover behavior and negative resistance region. Concept / Approach: Its internal structure is equivalent to two interconnected bipolar transistors (pnp and npn). Regeneration occurs when combined current gains approach unity, causing rapid turn-on at the breakover voltage. The four alternating layers (PNPN) form three junctions J1, J2, and J3. Step-by-Step Solution: Identify alternating layering: P–N–P–N → 4 layers.Count junctions: J1, J2, J3 → 3 junctions.Two terminals (no gate) distinguish it from an SCR (which is a gated PNPN). Verification / Alternative check: Device symbols and equivalent transistor models confirm the four-layer architecture with two terminals. Why Other Options Are Wrong: 2 or 3 layers cannot produce the characteristic thyristor breakover behavior. Common Pitfalls: Assuming “diode” implies two layers; the Shockley diode is a historical misnomer for a four-layer switch. Final Answer: 4

Question 20

LASCR internal structure A light-activated SCR (LASCR) contains how many semiconductor layers and junctions?

Accepted Answer

Introduction / Context: The LASCR is an SCR that can be triggered by light impinging on its gate region. Despite the optical triggering method, the internal thyristor structure remains that of a conventional SCR. Given Data / Assumptions: LASCR is a PNPN device (thyristor). Light generates carriers near the gate area to initiate conduction. Concept / Approach: The classic SCR comprises four layers (PNPN) and three junctions (J1, J2, J3). The LASCR leverages the same layered stack; the difference lies only in how turn-on is initiated (photons instead of, or in addition to, electrical gate current). Step-by-Step Solution: SCR layering: P–N–P–N → 4 layers.Junctions: 3 (J1, J2, J3).LASCR: same structure, with an optical window or fiber coupling to gate region. Verification / Alternative check: Datasheets specify LASCR gate sensitivity to light but list the same blocking ratings and junction structure as standard SCRs. Why Other Options Are Wrong: 2- or 3-layer structures cannot provide thyristor-like latch characteristics. “Any of these depending on rating” is incorrect; ratings change geometry, not the fundamental layer count. Common Pitfalls: Confusing photodiodes or phototransistors (different layer counts) with LASCRs. Final Answer: 4 semiconductor layers and 3 junctions

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