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TRIAC firing angles (lamp circuit)

Short answer

With Vs=230V, C=0.5µF, Xc=6366Ω, firing occurs when capacitor voltage reaches DIAC VBO=20 V. Minimum firing angle = 42.6° (R=5kΩ), maximum = 76.1° (R=15kΩ).

In the Triac firing circuit, input voltage is 230 V, 50 Hz, Diac breakdown voltage is 20 V, capacitance C = 0.5 μF, R can be varied from 5 KΩ to 15 KΩ. Find the minimum and maximum firing angle of Triac. [Figure: AC source in series with a lamp and the TRIAC; variable resistor R in series with capacitor C is connected across the TRIAC; a DIAC connects the R-C junction to the TRIAC gate.]

Given: Vs = 230 V (rms), 50 Hz, DIAC breakover VBO = 20 V, C = 0.5 μF, R = 5 kΩ to 15 kΩ.

Assumption: before the TRIAC fires it is OFF, so the RC network across the TRIAC sees almost the full supply (lamp resistance is small compared with R and Xc). DIAC current loading on C is neglected.

Method

The capacitor voltage lags the supply by φ = tan⁻¹(ωRC). The TRIAC fires when vC reaches the DIAC breakover voltage.

Xc = 1/(2πfC) = 1/(2π × 50 × 0.5×10⁻⁶) = 6366.2 Ω
Vc = Vs·Xc/√(R² + Xc²)        (rms)
φ  = tan⁻¹(R/Xc)
α  = φ + sin⁻¹( VBO/(√2·Vc) )

Minimum firing angle (R = 5 kΩ)

Z  = √(5000² + 6366.2²) = 8095.0 Ω
Vc = 230 × 6366.2/8095.0 = 180.88 V
Vc(peak) = √2 × 180.88 = 255.80 V
φ  = tan⁻¹(5000/6366.2) = 38.15°
sin⁻¹(20/255.80) = 4.48°
α_min = 38.15° + 4.48° = 42.63°

Maximum firing angle (R = 15 kΩ)

Z  = √(15000² + 6366.2²) = 16295.0 Ω
Vc = 230 × 6366.2/16295.0 = 89.86 V
Vc(peak) = √2 × 89.86 = 127.08 V
φ  = tan⁻¹(15000/6366.2) = 67.00°
sin⁻¹(20/127.08) = 9.06°
α_max = 67.00° + 9.06° = 76.06°

Answer: minimum firing angle ≈ 42.6°, maximum firing angle ≈ 76.1°

So the conduction angle in each half cycle ranges from about 137° (brightest lamp) to about 104° (dimmest lamp).

✓ Answer checked by Claude Opus 5.5 (high effort)

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