--- title: "TRIAC light dimmer conduction angles and power" subject: "Advanced Electronics" chapter: 7 key: "semester_4/advanced_electronics:triac-light-dimmer" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/triac-light-dimmer site: BCT Engineering --- # TRIAC light dimmer conduction angles and power Advanced Electronics · Chapter 7: Introduction to Power Electronics - Asked in: 2070 Asar, 3+2+2 marks - Video: [AC Dimmer Circuit Using Phase Angle Control](https://www.youtube.com/watch?v=w70OqEhkgAw) — element14 presents - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/triac-light-dimmer - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-7 ## Question In the light dimmer circuit shown in figure below, find the minimum and maximum possible conduction angles. Draw the voltage waveform across the load for both cases. Also compute the average power delivered to the load at maximum conduction angle. Variable resistor R can be varied from 2 KΩ to 20 KΩ and break over voltage for diac is 12V. [Figure: 220 V / 50 Hz a.c. source; load RL = 15 Ω in series with the TRIAC across the source; variable resistor R in series with C = 1 μF across the source; a DIAC connects the R-C junction to the TRIAC gate.] ## Short answer With RC phase shift and DIAC breakover 12V, firing angle α ranges from 34.75° (R=2kΩ) to 95.16° (R=20kΩ), giving conduction angles of 145.25° and 84.84° respectively. At maximum conduction, RMS load voltage is 215.1V, average power ≈ 3.08kW for 220V rms and 15Ω load. ## Answer Given: 220 V (rms), 50 Hz supply; RL = 15 Ω; C = 1 μF; R = 2 kΩ to 20 kΩ; DIAC breakover VBO = 12 V. Assumptions: RC branch is across the supply, DIAC current loading on C is neglected, and the TRIAC drop is neglected. #### Firing angle and conduction angle The capacitor voltage lags the supply by φ = tan⁻¹(ωRC). The TRIAC fires when vC = VBO. ```text Xc = 1/(2πfC) = 1/(2π × 50 × 1×10⁻⁶) = 3183.1 Ω Vc = Vs·Xc/√(R² + Xc²) α = tan⁻¹(R/Xc) + sin⁻¹( VBO/(√2·Vc) ) Conduction angle = 180° − α ``` **R = 2 kΩ (maximum conduction):** ```text Z = √(2000² + 3183.1²) = 3759.3 Ω Vc = 220 × 3183.1/3759.3 = 186.28 V, peak 263.44 V φ = tan⁻¹(2000/3183.1) = 32.14° sin⁻¹(12/263.44) = 2.61° α_min = 34.75° → conduction = 145.25° ``` **R = 20 kΩ (minimum conduction):** ```text Z = √(20000² + 3183.1²) = 20251.7 Ω Vc = 220 × 3183.1/20251.7 = 34.58 V, peak 48.90 V φ = tan⁻¹(20000/3183.1) = 80.96° sin⁻¹(12/48.90) = 14.20° α_max = 95.16° → conduction = 84.84° ``` #### Load voltage waveforms ```text Max conduction (α = 34.75°) vL | /‾\ | | \ --+---+----+----------+----> ωt 0 35° 180° | / | | (same in −ve half, +‾‾‾ from 214.75° to 360°) Min conduction (α = 95.16°) vL | |\ | | \ --+------+--+-------+---> ωt 0 95° 180° | / | | (from 275.16° to 360°) +‾ ``` In each half cycle the load voltage is zero until α and then follows the supply sine wave up to the zero crossing. #### Average power at maximum conduction (α = 34.75°) ```text VL,rms = Vs·√[ (π − α)/π + sin2α/(2π) ] = 220 × √[ 0.8069 + 0.1491 ] = 220 × 0.9778 = 215.11 V P = VL,rms²/RL = 215.11²/15 P = 3084.7 W ``` **Answer: conduction angle ranges from 84.84° (R = 20 kΩ) to 145.25° (R = 2 kΩ); power at maximum conduction ≈ 3.08 kW** (full-wave power would be 220²/15 = 3226.7 W). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the conduction angle related to firing angle? A. 180° − α B. α C. α − 180° D. 360° − α Answer: A. Conduction occurs from α to 180° in each half cycle. (Key point: Conduction angle) 2. Maximum conduction occurs at which R? A. R = 20 kΩ, giving 145.25° B. R = 2 kΩ, giving conduction 145.25° C. R = 20 kΩ, giving 84.84° D. R = 2 kΩ, giving 84.84° Answer: B. Small R gives small lag and early firing at 34.75°. (Key point: Maximum conduction) 3. What does the load voltage look like in each half cycle? A. Follows sine until α then zero B. Constant at VBO C. Zero until α, then follows the sine wave to the zero crossing D. Square wave Answer: C. The TRIAC is OFF before firing. (Key point: Waveform) 4. How is the power at maximum conduction found? A. P = Vs²/RL = 3226.7 W B. P = Vdc·Idc C. P = VBO²/RL D. P = VL,rms²/RL with VL,rms = Vs·sqrt[(π−α)/π + sin2α/(2π)], about 3.08 kW Answer: D. RMS reflects the chopped sine wave. 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