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TRIAC light dimmer conduction angles and power

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.

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.]

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.

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):

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):

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

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°)

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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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