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Power Electronics Practice Hub

Practice questions for Power Electronics covering rectifiers, DC-DC converters, inverters, and thyristor circuits for GATE ECE and EEE.

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BeginnerRipple FactorIntermediateBoost Duty CycleAdvancedBuck-Boost Ripple

Hub intro

This hub covers power diode and thyristor circuits, controlled and uncontrolled rectifiers, DC-DC converter topologies, and PWM inverters. The sets build from diode rectifier analysis through converter duty-cycle calculations to GATE-level switch-mode converter problems.

Difficulty levels

BeginnerIntermediateAdvanced

Practice sets

Beginner: Definitions and Direct Formulas
Beginner
Define ripple factor for a rectifier and state its value for a full-wave rectifier with a capacitor filter.
Answer: Ripple factor r = V_r(rms) / V_dc, where V_r(rms) is the rms value of the AC ripple component and V_dc is the DC (average) output. For a full-wave rectifier with a large capacitor filter, r ≈ 1 / (4√3 f C R_L), which is approximately 0.48 for no filter and approaches zero for large C.
Without a filter, a full-wave rectifier has r = 0.483 (ripple factor = 48.3%). With a capacitor filter, ripple reduces as C or R_L increases. A higher capacitor value stores more charge and maintains the output voltage closer to the peak during the discharge phase. The formula r = 1/(4√3 f C R_L) applies when the ripple is small compared to V_dc.
State the voltage conversion ratio M = V_out / V_in for a buck, boost, and buck-boost DC-DC converter in continuous conduction mode (CCM) in terms of duty cycle D.
Answer: Buck (step-down): M = D. Boost (step-up): M = 1/(1-D). Buck-boost (inverting): M = -D/(1-D). All expressions assume ideal components and CCM.
D is the fraction of the switching period during which the switch is closed (0 < D < 1). For the buck converter, V_out = D × V_in: output is always less than input. For the boost converter, as D approaches 1, V_out approaches infinity theoretically; in practice, non-ideal components limit the maximum voltage. The buck-boost output is inverted (negative) relative to the input.
Define the firing angle α for a thyristor and state the range of α for a half-wave controlled rectifier with a resistive load.
Answer: The firing angle α is the delay, measured in degrees from the natural firing point (zero crossing of the supply voltage for resistive loads), at which the gate trigger pulse is applied to turn the thyristor on. For a half-wave resistive load, 0 ≤ α ≤ 180°.
At α = 0, the thyristor conducts from the positive zero crossing for the full half cycle, giving maximum output. At α = 90°, conduction starts at the peak of the supply and the average output is reduced. At α = 180°, there is no conduction. For an inductive load, the thyristor can conduct beyond 180° into the negative half cycle due to the energy stored in the inductor, and the range extends up to α = 90° for continuous conduction.
Intermediate: Multi-step Problems
Intermediate
A single-phase fully controlled bridge rectifier has V_s = 230 V rms at 50 Hz. The firing angle α = 60°. Find the average output voltage for a highly inductive (continuous current) load.
Answer: V_dc = (2 V_m / π) cos α, where V_m = 230√2 = 325.3 V. V_dc = (2 × 325.3 / π) cos 60° = (650.6 / 3.1416) × 0.5 = 207.0 × 0.5 = 103.5 V.
For a fully controlled single-phase bridge with continuous current, V_dc = (2 V_m/π) cos α. At α = 0, this gives the maximum output (2 V_m/π = 207 V). At α = 60°, cos 60° = 0.5, so V_dc = 103.5 V. The formula changes for a half-controlled bridge: V_dc = (V_m/π)(1 + cos α). Always confirm whether the bridge is fully controlled (4 thyristors) or half-controlled (2 thyristors + 2 diodes).
A boost converter operates at f = 50 kHz with V_in = 12 V, V_out = 48 V, and output current I_out = 1 A. Find the duty cycle D, input current I_in, and the minimum inductance to maintain CCM given ΔI_L_max = 20% of I_in.
Answer: D = 1 - V_in/V_out = 1 - 12/48 = 0.75. By power conservation: I_in = I_out/(1-D) = 1/0.25 = 4 A. ΔI_L = 0.20 × 4 = 0.8 A. L_min = V_in × D / (f × ΔI_L) = 12 × 0.75 / (50000 × 0.8) = 9 / 40000 = 225 μH.
For an ideal boost converter, power in = power out: V_in I_in = V_out I_out => I_in = 4 A. Inductor current ripple: ΔI_L = V_in D / (L f). Solving for L: L = V_in D / (f ΔI_L) = 12 × 0.75 / (5×10^4 × 0.8) = 225 μH. This is the minimum L for CCM at this operating point. Below this inductance, the converter enters discontinuous conduction mode (DCM) where the conversion ratio formula changes.
A three-phase uncontrolled diode bridge rectifier is supplied by a 3-phase 415 V (line-to-line) 50 Hz source. Calculate the average DC output voltage and the ripple frequency.
Answer: V_line = 415 V. V_m_line = 415√2 = 586.9 V. For a 3-phase bridge: V_dc = (3√3 / π) V_m_phase = (3/π) V_m_line. V_dc = (3 × 586.9) / π = 1760.7 / 3.1416 = 560.4 V. Ripple frequency = 6 × supply frequency = 6 × 50 = 300 Hz.
The three-phase bridge has 6 pulses per cycle (p = 6), so ripple frequency = p × f_supply = 300 Hz. The formula V_dc = (3/π) V_m_line uses the peak line-to-line voltage. Equivalently, V_dc = (3√3/π) V_m_phase where V_m_phase = V_line/√3 × √2 = 415/√3 × √2. The 6-pulse output has much lower ripple than a single-phase rectifier.
Advanced: GATE-style Questions
Advanced
A buck-boost converter has V_in = 20 V, D = 0.6, L = 100 μH, C = 100 μF, R = 10 Ω, f = 100 kHz. Find V_out, inductor peak-to-peak ripple current, and output voltage ripple.
Answer: V_out = -D V_in/(1-D) = -(0.6 × 20)/(0.4) = -30 V (magnitude 30 V). ΔI_L = V_in D/(L f) = 20 × 0.6/(100e-6 × 100e3) = 12/10 = 1.2 A. ΔV_out = D/(R C f) × V_out_magnitude = 0.6/(10 × 100e-6 × 100e3) × 30 = 0.6/100 × 30 = 0.18 V.
The trap is to ignore the negative sign and state V_out = +30 V. The buck-boost inverts the polarity. ΔI_L: during on-time, voltage across L = V_in = 20 V, time = D/f = 6 μs. ΔI_L = V_in D/(Lf) = 20 × 0.6/(100e-6 × 100e3) = 1.2 A. ΔV_out: during off-time, the capacitor supplies load current. ΔV_out = I_out × D / (C f) = (30/10) × 0.6 / (100e-6 × 100e3) = 3 × 0.6/10 = 0.18 V.
A single-phase semiconverter (half-controlled bridge) feeds an RL load (R = 5 Ω, L = large). The supply is 100 V rms at 50 Hz. At α = 90°, find the average load current and the power delivered to the load.
Answer: V_dc = (V_m/π)(1 + cos α) = (100√2/π)(1 + cos 90°) = (141.4/3.1416)(1 + 0) = 45.0 V. I_dc = V_dc / R = 45.0 / 5 = 9.0 A. P = I_dc^2 × R = 81 × 5 = 405 W.
The trap is to use the fully controlled bridge formula (2 V_m/π) cos α, which gives (2 × 141.4/π) × 0 = 0. The half-controlled bridge formula is V_dc = (V_m/π)(1 + cos α). For a large inductance (highly inductive) load, current is smooth and continuous, so I_dc ≈ V_dc / R. Power to the resistive part only: P = I_dc^2 R (no average power in L for DC steady state). At α = 90°, the freewheeling diode in the half-controlled bridge is active, maintaining current continuity.

Lab exercises

  • Build and measure a full-wave bridge rectifier with capacitor filter on a breadboard, plot the output waveform and measure ripple voltage: lab-bridge-rectifier-filter
  • Simulate a buck converter in LTspice, vary D from 0.2 to 0.8 in steps and record V_out, then verify against M = D: lab-buck-converter-ltspice
  • Observe thyristor firing angle control on a single-phase half-wave circuit using a DIAC-TRIAC trainer, measure V_dc vs α: lab-thyristor-firing-angle

Revision checklist

  • Can you write the average output voltage formulas for single-phase half-wave, full-wave, and fully controlled bridge rectifiers without notes?
  • Can you derive the volt-second balance equation for a buck converter to arrive at M = D?
  • Can you calculate the minimum capacitance to keep ripple below 1% for a given load and switching frequency?
  • Can you identify the continuous and discontinuous conduction boundary condition for a boost converter?
  • Can you state the effect of increasing firing angle α on the power factor of a controlled rectifier?
  • Can you sketch the output voltage waveform of a 3-phase bridge rectifier and mark the 6 conduction intervals?