---
title: "SCR two-transistor analogy; TRIAC firing angles"
subject: "Advanced Electronics"
chapter: 7
key: "semester_4/advanced_electronics:scr-analogy-triac-firing-25v"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/scr-analogy-triac-firing-25v
site: BCT Engineering
---

# SCR two-transistor analogy; TRIAC firing angles

Advanced Electronics · Chapter 7: Introduction to Power Electronics

- Asked in: 2074 Chaitra, 3+5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/scr-analogy-triac-firing-25v
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-7

## Question

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.]

## 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).

## Answer

#### 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.

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

```text
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:

```text
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):**

```text
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):**

```text
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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What makes the SCR latch in the two-transistor analogy?
   A. α1 + α2 approaching 0
   B. α1 + α2 approaching 1
   C. α1 = 0
   D. IG = ICBO
   Answer: B. The denominator 1 − (α1 + α2) tends to zero, so IA becomes very large. (Key point: Gain condition)

2. What is the lag of vC behind the supply in the RC network?
   A. φ = tan⁻¹(Xc/R)
   B. φ = sin⁻¹(R/Z)
   C. φ = tan⁻¹(R/Xc)
   D. φ = 0
   Answer: C. The capacitor voltage lags the supply by tan⁻¹(R/Xc). (Key point: Capacitor phase lag)

3. Which formula gives the firing angle?
   A. α = φ − sin⁻¹(VBO/(√2·Vc))
   B. α = sin⁻¹(VBO/Vs)
   C. α = tan⁻¹(VBO/Vc)
   D. α = φ + sin⁻¹(VBO/(√2·Vc))
   Answer: D. The TRIAC fires when √2·Vc·sin(α − φ) = VBO. (Key point: Firing angle formula)

4. Which resistance gives the minimum firing angle (about 25.4°)?
   A. R = 2 kΩ, the smallest
   B. R = 20 kΩ, the largest
   C. R = 0
   D. It is independent of R
   Answer: A. Smaller R means less phase lag and larger Vc, so earlier firing; larger R gives about 92.6°. (Key point: Minimum angle)

## Related in Advanced Electronics

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- [DIAC drawback; TRIAC four-quadrant operation](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Adiac-drawback-triac-quadrants) — Advanced Electronics, chapter 7

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