Controlled Rectifiers

A phase-controlled rectifier varies its DC output by delaying the SCR firing angle α; for a single-phase full converter V_dc = (2V_m/π)cosα, which even goes negative (inversion) for α > 90°.

Key formulas & points

Skim these first — then read the full notes below.

  • Firing angle α controls average output voltage
  • Discontinuous conduction with RL load at large α
  • Inversion when α > 90° with active DC source

Topic details

Introduction

Controlled rectifiers convert AC to adjustable DC by firing SCRs at an angle α after the natural commutation point. Increasing α reduces the average output voltage. With a highly inductive load the current is continuous and the standard cosα formulas apply.

Scope in B.Tech and GATE syllabus

A full converter (four SCRs in single phase) can operate in inverter mode (α > 90°) if the DC side has an active EMF, returning power to the AC line — the basis of regenerative DC drives.

Key relations & formulas

Formulas (Indian textbook notation)

  • Singlephasehalfwave:Vdc=Vm(2π)(1+cosα)Single-phase half-wave: V_{dc} = \frac{V_{m}}{(2\pi)}(1 + cos \alpha)

Formulas (Indian textbook notation)

  • Singlephasefullconverter:Vdc=(2Vmπ)cosαSingle-phase full converter: V_{dc} = (2 \frac{V_{m}}{\pi}) cos \alpha

Formulas (Indian textbook notation)

  • Threephase6pulse:Vdc=(3VmLπ)cosαThree-phase 6-pulse: V_{dc} = (3 \frac{V_{mL}}{\pi}) cos \alpha

Notation and sign conventions

Relation 1 —
Singlephasehalfwave:Vdc=Vm/Single-phase half-wave: V_{dc} = V_{m}/

Formulas (Indian textbook notation)

  • Singlephasehalfwave:Vdc=Vm(2π)(1+cosα)Single-phase half-wave: V_{dc} = \frac{V_{m}}{(2\pi)}(1 + cos \alpha)
Write this relation with symbols exactly as in Power Electronics — PS Bimbhra before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
Singlephasefullconverter:Vdc=Single-phase full converter: V_{dc} =

Formulas (Indian textbook notation)

  • Singlephasefullconverter:Vdc=(2Vmπ)cosαSingle-phase full converter: V_{dc} = (2 \frac{V_{m}}{\pi}) cos \alpha
Write this relation with symbols exactly as in Power Electronics — PS Bimbhra before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
Threephase6pulse:Vdc=Three-phase 6-pulse: V_{dc} =

Formulas (Indian textbook notation)

  • Threephase6pulse:Vdc=(3VmLπ)cosαThree-phase 6-pulse: V_{dc} = (3 \frac{V_{mL}}{\pi}) cos \alpha
Write this relation with symbols exactly as in Power Electronics — PS Bimbhra before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Fundamentals and definitions

For continuous conduction with an RL load, the single-phase full-converter output is V_dc = (2V_m/π)cosα, where V_m is the peak of the AC supply. At α = 0 it equals the uncontrolled diode-bridge value; at α = 90° the average is zero.

Governing relations in practice

Input power factor degrades as α increases because the current lags the voltage by α (displacement) and is non-sinusoidal (distortion). This is why controlled rectifiers draw reactive power and harmonics.

Design and analysis considerations

A freewheeling diode across the load in a semiconverter improves power factor by preventing the output from going negative and allowing load current to circulate.

Assumptions and validity limits

State assumptions explicitly before using any relation for controlled rectifiers — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.
Wrong assumptions invalidate the entire solution even when the formula is correct. In Power Electronics viva and GATE descriptive questions, listing valid assumptions often earns separate marks.

Step-by-step problem approach

1. Read the question and list given data with SI units (common in Power Electronics papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to controlled rectifiers.
4. Use equation 1:
Singlephasehalfwave:Vdc=Vm/Single-phase half-wave: V_{dc} = V_{m}/
.
5. Use equation 2:
Singlephasefullconverter:Vdc=Single-phase full converter: V_{dc} =
.
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.

Applications & exam relevance

Controlled Rectifiers appears in drives, UPS, and grid interfaces. In Indian electrical curricula this topic is tested because it connects theory to controlled power conversion.
GATE and semester exams often combine controlled rectifiers with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use controlled rectifiers?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

• Using peak line-to-line voltage where phase peak is required (or vice versa) in three-phase converters
• Applying the continuous-conduction formula when conduction is actually discontinuous
• Forgetting the full converter gives 2V_m/π (twice the half-controlled average)
• Ignoring the input displacement factor cosα when computing power factor

Quick revision checklist

Before attempting controlled rectifiers problems, confirm you can:
1. Firing angle α controls average output voltage
2. Discontinuous conduction with RL load at large α
3. Inversion when α > 90° with active DC source
Revise the solved examples in Power Electronics — PS Bimbhra and one previous-year GATE or university paper for this unit.

Worked examples

Try the problem first — open the solution when you are ready to check.

Output voltage of a full converter

Problem

A single-phase full converter is fed from 230 V (rms), 50 Hz and fires at α = 60° into a highly inductive load. Find the average DC output voltage.

Solution

Peak V_m = √2 × 230 = 325.3 V.
V_dc = (2V_m/π) cosα = (2 × 325.3/π) × cos60°.
2×325.3/π = 207.1 V; cos60° = 0.5.
V_dc = 207.1 × 0.5 = 103.5 V.

Conceptual check — Controlled Rectifiers

Problem

In a Power Electronics semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of controlled rectifiers." What should a complete answer include?

Exams & GATE

PS Bimbhra Ch. 4 — output voltage vs firing angle plots.

📖 Standard books (India)

  • Power ElectronicsPS Bimbhra

    Read: Syllabus unit

    Rectifiers, choppers, inverters — Indian standard