Knowledge / Electrical and Electronics Engineering / AC voltage controllers and cycloconverters

AC voltage controllers and cycloconverters

Electrical and Electronics Engineering (B.E./B.Tech) · Power electronics and drives · free study notes for exams, placements and interviews

In short

An AC voltage controller takes a fixed-voltage, fixed-frequency AC supply and delivers a variable RMS voltage at the same frequency, using anti-parallel thyristors or a TRIAC fired at a chosen angle in each half-cycle. A cycloconverter goes further: by selecting segments of the input waveform through two groups of thyristors, it synthesises a variable-voltage, variable-frequency AC output directly, without any DC link. Both are single-stage, line-commutated AC-to-AC converters and both are heavy harmonic polluters, which is why the modern alternative is a PWM AC chopper or an inverter-fed drive. The topic is examined every year in GATE EE (firing-angle and RMS voltage numericals), in university Power Electronics papers (derivations of V_rms, extinction angle, cycloconverter output frequency), and is used in Indian industry in fan regulators, lamp dimmers, induction heating, resistance welding, soft starters (L&T, Schneider, Siemens, Danfoss) and large low-speed cement/mine mill drives.

Explained

Classification of AC-to-AC converters

Single-phase full-wave AC voltage controller

Integral cycle (on-off) control and burst firing

Three-phase AC voltage controllers

Cycloconverter: principle and types

Harmonics, power factor and Indian standards

Key points

Formulas and examples

Single-phase full-wave controller, R load: V_o(rms) = V_s √[ (1/π)(π − α + (sin 2α)/2) ]
where V_s = RMS supply voltage (V), α = firing angle (rad). Power P = V_o(rms)²/R (W).
Average and RMS SCR current (R load): I_T(avg) = (√2 V_s /2πR)(1 + cos α); I_T(rms) = I_o(rms)/√2, with I_o(rms) = V_o(rms)/R (A).
Single-phase half-wave (unidirectional) controller, R load: V_o(rms) = V_s √[ (1/2π)(2π − α + (sin 2α)/2) ] — minimum obtainable is V_s/√2, not zero.
RL load current (while conducting): i(ωt) = (√2 V_s/Z)[ sin(ωt − φ) − sin(α − φ) e^{(α − ωt)/tan φ} ], Z = √(R² + (ωL)²) Ω, φ = tan⁻¹(ωL/R).
Extinction angle β from the transcendental equation sin(β − φ) = sin(α − φ) e^{(α − β)/tan φ} — solved by iteration.
Integral cycle control: V_o(rms) = V_s √k, k = n/(n+m); P = k V_s²/R; input pf = √k.
Three-phase controller, star R load, 0 ≤ α ≤ 60°: V_ph(rms) = √3 V_ph √[ (1/π)(π/3 − α/2 + (sin 2α)/4) ] type expressions apply mode-wise; per-phase RMS is most safely obtained from the mode-wise integral given in the syllabus text.
Input power factor, phase control with R load: pf = V_o(rms)/V_s (since input RMS current = output RMS current and the source is sinusoidal).
Cycloconverter output frequency (one output cycle per N input cycles of the group): f_o = f_s / N, N integer ≥ 2 for simple circuits; general control gives f_o ≤ f_s/3 for low distortion.
Cycloconverter output fundamental (p-pulse group, cosine control): V_o1(peak) = (p/π) V_m sin(π/p) × cos α_min, so V_o(rms) = V_do × r / √2 with modulation ratio r = V_o(peak)/V_do, 0 ≤ r ≤ 1.
Firing angle modulation (cosine-crossing): α(t) = cos⁻¹[ r sin(ω_o t) ], where ω_o = 2π f_o (rad/s).
Induction motor with stator voltage control: T ∝ V², starting current I_st ∝ V; soft-start current I = x·I_LRC, starting torque = x² × DOL torque, with x = applied voltage fraction.

Example 1 — Single-phase controller with R load. A 230 V, 50 Hz supply feeds a 10 Ω heater through a TRIAC fired at α = 60°.

Example 2 — Half power point and comparison with integral cycle control. Find α for 50% power in Example 1, then do the same by burst firing.

Example 3 — RL load and the φ limit. A 230 V, 50 Hz controller feeds R = 8 Ω, L = 20 mH.

Example 4 — Soft starter sizing. A 30 kW, 415 V, 50 Hz induction motor has full-load current 55 A, locked-rotor current 6.5 × FLC and DOL starting torque 1.8 × full-load torque. A soft starter is set to 50% voltage.

Example 5 — Cycloconverter output frequency and voltage. A three-phase, 6-pulse cycloconverter is fed from a 415 V, 50 Hz supply (phase voltage 239.6 V) and drives a mill motor at 10 Hz.

FeatureAC voltage controllerCycloconverterAC chopper (PWM)
Output frequencySame as inputVariable, ≤ f_s/3Same as input
Output voltageVariable RMSVariable RMSVariable RMS
Devices2 SCRs / 1 TRIAC per phase18–36 SCRs2 IGBTs + diodes per phase
CommutationNatural (line)Natural (line)Forced (self)
HarmonicsLow-order, severeLow-order + interharmonicsHigh-frequency only
Input pfPoor, falls with αPoor at low outputNear unity displacement
Typical ratingW to MW (heating)MW (mill, mine, ship)kW
Typical useDimmer, heater, soft startLow-speed high-torque driveFan/light dimming, voltage regulator
Control methodV_o(rms)Harmonic signatureBest load
Phase controlV_s√[(1/π)(π−α+sin2α/2)]3rd, 5th, 7th of 50 HzLamps, motors, small heaters
Integral cycleV_s√kSub-harmonics, flickerFurnaces, ovens (long time constant)
PWM choppingV_s × √(duty) approx.kHz-range, easily filteredLighting, fans, AC regulators

Where it is used

Interview questions

Common mistakes

Related topics

Power semiconductor devices · Rectifiers and converters · Inverters · Electric drives · Electric traction · Power quality and harmonics

Open in the Knowledge reader →Find jobs →