Power Electronics
GATE EE syllabus for power electronics, with thyristor, MOSFET and IGBT characteristics, controlled rectifiers, buck, boost and buck-boost converters, inverters and sinusoidal PWM, harmonics, power factor and distortion factor, with solved examples.
Syllabus
Static V-I characteristics and gating circuits for thyristor, MOSFET and IGBT; DC-DC conversion (buck, boost, buck-boost); single and three-phase uncontrolled rectifiers; voltage and current commutated thyristor converters; bidirectional AC-DC voltage source converters; line current harmonics; power factor and distortion factor of AC-DC converters; single and three-phase voltage and current source inverters; sinusoidal PWM.
Key concepts and formulas
Devices. The SCR turns on with a gate pulse when forward biased and turns off only when its current falls below the holding current (latching current > holding current). MOSFETs are voltage-controlled and fast (low-voltage, high-frequency use); IGBTs combine MOS gate drive with bipolar conduction (medium power and frequency).
Controlled rectifiers (supply peak Vm, firing angle α, continuous conduction unless stated):
| Converter | Average output voltage |
|---|---|
| Single-phase half-wave, R load | (Vm/2π)(1 + cos α) |
| Single-phase full converter (highly inductive load) | (2Vm/π) cos α |
| Single-phase semi-converter | (Vm/π)(1 + cos α) |
| Three-phase full converter | (3√3Vm/π) cos α = (3Vm,L/π) cos α ≈ 1.35VL,rms cos α |
| Three-phase diode bridge | 1.35VL,rms |
A full converter inverts (negative average voltage, power flows back to the AC side) for α > 90°.
DC-DC converters (duty ratio D, continuous conduction, ideal):
- Buck: Vo = DVs
- Boost: Vo = Vs/(1 − D)
- Buck-boost: Vo = −DVs/(1 − D) (polarity reversed)
- Buck inductor ripple ΔI = VsD(1 − D)/(Lf).
Inverters. Single-phase full-bridge square wave: the fundamental RMS is 0.9Vdc (peak 4Vdc/π), and only odd harmonics appear. Sinusoidal PWM: modulation index ma = Vcontrol/Vcarrier; for ma ≤ 1 the fundamental amplitude is maVdc (full bridge) or maVdc/2 (per half bridge leg).
Harmonics and power factor. Total harmonic distortion THD = √(Irms² − I₁²)/I₁. Power factor = distortion factor × displacement factor = (I₁/Irms) × cos φ₁. A square-wave line current (as with a single-phase full converter on a highly inductive load) has I₁/Irms = 2√2/π ≈ 0.9, and its displacement angle equals α.
Worked example
A single-phase full converter on a 230 V, 50 Hz supply feeds a highly inductive load at α = 30°: find Vdc and the input power factor. Find the outputs of buck, boost and buck-boost converters with Vs = 24 V and D = 0.4, and the Vdc of a three-phase full converter on 415 V at α = 45°.
import math
Vm = 230 * math.sqrt(2)
a = math.radians(30)
print(f"full converter Vdc = {2 * Vm / math.pi * math.cos(a):.2f} V")
print(f"input power factor = {2 * math.sqrt(2) / math.pi * math.cos(a):.4f}")
Vs, D = 24, 0.4
print(f"buck = {D * Vs:.2f} V, boost = {Vs / (1 - D):.2f} V, buck-boost = {-D * Vs / (1 - D):.2f} V")
VL = 415
print(f"3-phase full converter Vdc = {3 * math.sqrt(2) * VL / math.pi * math.cos(math.radians(45)):.2f} V")
full converter Vdc = 179.33 V input power factor = 0.7797 buck = 9.60 V, boost = 40.00 V, buck-boost = -16.00 V 3-phase full converter Vdc = 396.30 V
Practice questions
A boost converter has an input of 12 V and a duty ratio of 0.75. What is its output voltage (ideal, continuous conduction)?
Show answer and explanation
Answer: 48Vo = Vs/(1 − D) = 12/0.25 = 48 V.
A single-phase full converter with a highly inductive load is fed from a 200 V peak supply. What is the average output voltage at α = 60°? Give the answer to two decimal places.
Show answer and explanation
Answer: 63.66Vdc = (2Vm/π) cos α = (400/π) × 0.5 ≈ 63.66 V.
A buck converter steps 48 V down to 12 V. What is its duty ratio?
Show answer and explanation
Answer: 0.25Vo = DVs, so D = 12/48 = 0.25.
A current waveform has a fundamental RMS of 9 A and a total RMS of 10 A, and its displacement factor is 0.8. What is the power factor?
Show answer and explanation
Answer: 0.72PF = (I₁/Irms) cos φ₁ = 0.9 × 0.8 = 0.72.
For which range of firing angle does a single-phase full converter (continuous conduction) operate as an inverter?
Show answer and explanation
Answer: BVdc ∝ cos α becomes negative for α > 90°, so with a suitable DC source the power flows from the DC side to the AC side.