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SCR two-transistor analogy; TRIAC firing angles

Short answer

Two-transistor analogy models SCR as interconnected PNP and NPN transistors; latching occurs when α1+α2=1. For the TRIAC circuit, capacitor voltage phase lag causes firing when vC=DIAC VBO. Minimum α=25.4° (R=2kΩ), maximum α=92.6° (R=20kΩ).

Explain the two transistor analogy of SCR with necessary mathematical expression. In the TRIAC firing circuit the parameters are input voltage (Vs) = 230V, 50Hz, DIAC breakdown voltage is 25V, C = 0.6μF, R can be varied from 2000Ω to 20000Ω. Find the minimum and maximum firing angle. [Figure: source Vs in series with load RL and the TRIAC; variable resistor R in series with capacitor C connected across the supply; a DIAC connects the R-C junction to the TRIAC gate.]

Two-transistor analogy of SCR

The PNPN layers of an SCR are split into a PNP transistor Q1 (P1-N1-P2) and an NPN transistor Q2 (N1-P2-N2). The collector of each transistor drives the base of the other, so they form a regenerative loop.

      A
      |
     Q1 (PNP)
      |   \
      |    +-- IB1 = IC2
     IC1   |
      |    |
  G --+-- Q2 (NPN)
           |
           K

IC1 = α1·IA + ICBO1
IC2 = α2·IK + ICBO2,   IK = IA + IG
IA  = IC1 + IC2
IA  = (α2·IG + ICBO1 + ICBO2) / (1 − (α1 + α2))

When gate current raises α1 + α2 towards 1, IA rises sharply (limited only by the load) and the SCR latches ON.

TRIAC firing angles

Given: Vs = 230 V (rms), 50 Hz, VBO(DIAC) = 25 V, C = 0.6 μF, R = 2 kΩ to 20 kΩ.

The capacitor voltage lags the supply. The TRIAC fires when vC reaches the DIAC breakover voltage:

Xc = 1/(2πfC) = 1/(2π × 50 × 0.6×10⁻⁶) = 5305.2 Ω
Vc = Vs·Xc/√(R² + Xc²)   (rms)
φ  = tan⁻¹(R/Xc)         (lag of vC behind vs)
vC = √2·Vc·sin(ωt − φ)
Fires when √2·Vc·sin(α − φ) = VBO
α  = φ + sin⁻¹( VBO / (√2·Vc) )

Minimum firing angle (R = 2 kΩ):

Z  = √(2000² + 5305.2²) = 5669.6 Ω
Vc = 230 × 5305.2/5669.6 = 215.21 V,  peak 304.36 V
φ  = tan⁻¹(2000/5305.2) = 20.66°
sin⁻¹(25/304.36) = 4.71°
α_min = 20.66° + 4.71° = 25.37°

Maximum firing angle (R = 20 kΩ):

Z  = √(20000² + 5305.2²) = 20691.7 Ω
Vc = 230 × 5305.2/20691.7 = 58.97 V,  peak 83.40 V
φ  = tan⁻¹(20000/5305.2) = 75.14°
sin⁻¹(25/83.40) = 17.44°
α_max = 75.14° + 17.44° = 92.59°

Answer: minimum firing angle ≈ 25.4°, maximum firing angle ≈ 92.6° (DIAC/TRIAC gate loading on the RC network neglected).

✓ Answer checked by Claude Opus 5.5 (high effort)

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