Controlled rectifiers and choppers

Phase-controlled SCR rectifiers (single- and three-phase, resistive and continuous-current loads, inversion) and buck, boost and buck-boost choppers, with converter, chopper-ripple and regeneration examples.

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Why it matters

Controlled rectifiers turn the fixed AC mains into an adjustable DC voltage, and choppers turn a fixed DC voltage into an adjustable one. Together they drive DC motors in rolling mills, cranes, electric traction and battery vehicles, charge batteries and feed DC links in servo drives. A mechatronics engineer who sizes a drive or reads its datasheet must know how the firing angle or duty cycle sets the output voltage, and what that does to current ripple and supply power factor.

Key ideas

Phase control with an SCR. A thyristor (SCR) blocks forward voltage until a gate pulse arrives, then conducts until its current falls below the holding current. In a rectifier the gate pulse is delayed by the firing angle α, measured from the instant the device would have started conducting as a diode (the zero crossing for a single-phase circuit). Increasing α removes more of each half-cycle, so the average DC voltage falls. The SCR turns off naturally when the AC supply reverses its current (line or natural commutation), so no extra turn-off circuit is needed in rectifiers.

The load decides the formula. The average output voltage depends on whether the load current flows continuously:

  • Resistive load: the output follows the supply from α to π and is zero after the current reaches zero. Output voltage can never be negative.
  • Highly inductive load (continuous current): in a fully controlled converter the SCRs keep conducting past the voltage zero, so the output includes negative portions. The average becomes proportional to cos α.
  • A freewheeling diode across the load, or a semi-converter (half SCRs, half diodes), clamps the output at zero instead of letting it go negative. The formula is then the same as for a resistive load.

Converter types.

  • Single-phase half-wave controlled rectifier: one SCR; simple but with a large ripple and a DC component in the supply transformer.
  • Single-phase full converter (four-SCR bridge or two-SCR midpoint): two-quadrant operation, positive current with positive or negative average voltage.
  • Single-phase semi-converter: two SCRs and two diodes; one quadrant only, better input power factor.
  • Three-phase full converter (six SCRs): six-pulse output, ripple frequency 6f (300 Hz on a 50 Hz supply), used for drives above a few kilowatts.

Inversion. In a fully controlled converter with continuous current, α > 90° makes the average output voltage negative while the current is still positive, so power flows from the DC side back to the AC supply. This needs a DC source on the load side (a DC motor running as a generator during regenerative braking, or an HVDC line). In practice α is limited to about 150°–160° to leave time for the outgoing SCR to turn off (commutation overlap and turn-off margin).

Supply power factor. A phase-controlled rectifier draws current whose fundamental lags the voltage by about α (full converter, continuous current). Large α therefore means a poor displacement power factor, plus harmonic distortion. This is a real cost of phase control, not a benefit.

Choppers (DC-DC converters). A chopper switches a DC source on and off at a fixed frequency f (switching period T = 1/f). The duty cycle D = T_on/T sets the average output.

  • Step-down (buck): switch in series with the load, freewheeling diode across the load. V_o = D·V_s, always below V_s.
  • Step-up (boost): inductor in series with the source, switch across it, diode to the output capacitor. When the switch opens, the inductor voltage adds to the source, so V_o = V_s/(1 − D).
  • Buck-boost: output polarity reversed, magnitude D·V_s/(1 − D), above or below V_s.
  • Output control by varying T_on at constant f (pulse-width modulation, the usual method) or by varying f at constant T_on (frequency modulation, harder to filter).
  • Modern choppers use MOSFETs (low voltage, high frequency) or IGBTs (hundreds of volts, a few to tens of kHz). Older SCR choppers needed forced-commutation circuits because DC never passes through zero.

Quadrants. A first-quadrant (class A) chopper motors a DC machine; a second-quadrant (class B) chopper returns braking energy to the source; two-quadrant and four-quadrant (H-bridge) choppers allow reversal and regeneration — the basis of a DC servo amplifier.

Ripple. The load inductance smooths the current. Ripple falls as switching frequency or inductance rises, and in a buck chopper it is largest at D = 0.5.

Formulas

V_dc = (V_m / 2π)·(1 + cos α) — single-phase half-wave controlled rectifier, resistive load. V_dc = (V_m / π)·(1 + cos α) — single-phase full-wave (bridge or midpoint) with resistive load, and single-phase semi-converter or full converter with freewheeling diode. V_dc = (2·V_m / π)·cos α — single-phase full converter, continuous load current. V_dc = (3·√2·V_L / π)·cos α ≈ 1.35·V_L·cos α — three-phase full converter, continuous current.

  • V_dc: average output voltage (V); V_m: peak of the AC supply voltage, V_m = √2·V_rms (V); V_L: RMS line-to-line voltage (V); α: firing angle (rad or degrees, use the same in cos).

D = T_on / T = T_on·f V_o = D·V_s (buck) V_o = V_s / (1 − D) (boost) |V_o| = D·V_s / (1 − D) (buck-boost)

  • V_s: source voltage (V); V_o: average output voltage (V); T_on: on-time (s); T: switching period (s); f: switching frequency (Hz). Boost and buck-boost formulas assume continuous inductor current and ideal components.

ΔI = V_s·D·(1 − D) / (L·f) — peak-to-peak current ripple in a buck chopper feeding an R-L load (or the inductor of a buck converter), valid when L/R ≫ T.

  • L: inductance (H); maximum ripple V_s / (4·L·f) at D = 0.5.

Worked examples

Example 1 (standard: single-phase full converter). A single-phase fully controlled bridge is fed from 230 V, 50 Hz and supplies a load of R = 10 Ω with a large series inductance, so the current is continuous and ripple-free. Firing angle α = 45°. Find the average output voltage and current.

  1. V_m = √2·V_rms = 1.4142 × 230 = 325.3 V.
  2. V_dc = (2·V_m/π)·cos α = (2 × 325.3 / 3.1416) × 0.7071.
  3. V_dc = 207.1 × 0.7071 = 146.4 V.
  4. The inductor has no average voltage, so I_dc = V_dc/R = 146.4/10 = 14.64 A.

V_dc = 146.4 V, I_dc = 14.6 A. (With a resistive load only, the same bridge would give (V_m/π)(1 + cos 45°) = 176.7 V.)

Example 2 (GATE level: buck chopper ripple). A step-down chopper with V_s = 200 V switches at 10 kHz with D = 0.6 into a load of R = 10 Ω in series with L = 5 mH. Find the average output voltage, average current and the maximum and minimum load current.

  1. V_o = D·V_s = 0.6 × 200 = 120 V.
  2. I_avg = V_o/R = 120/10 = 12 A.
  3. Check the approximation: L/R = 0.5 ms ≫ T = 0.1 ms, so the ripple formula applies.
  4. ΔI = V_s·D·(1 − D)/(L·f) = 200 × 0.6 × 0.4 / (5×10⁻³ × 10⁴) = 48/50 = 0.96 A.
  5. I_max = I_avg + ΔI/2 = 12.48 A, I_min = I_avg − ΔI/2 = 11.52 A.

V_o = 120 V, I_avg = 12 A, I_max ≈ 12.48 A, I_min ≈ 11.52 A.

Example 3 (GATE level: three-phase converter, rectifying and inverting). A three-phase full converter on a 400 V, 50 Hz supply drives a separately excited DC motor with continuous current. Find the average DC voltage at α = 30° and at α = 120°, and say which way power flows.

  1. V_dc = (3·√2·V_L/π)·cos α = 1.3505 × 400 × cos α = 540.2·cos α.
  2. α = 30°: V_dc = 540.2 × 0.8660 = 467.8 V — rectifying, power from AC to motor.
  3. α = 120°: V_dc = 540.2 × (−0.5) = −270.1 V. Current direction cannot reverse through the SCRs, so power flows from the DC side to the AC supply. This works only if the motor's back EMF has been reversed (by reversing the field) so that it acts as a source — regenerative braking.

α = 30°: 467.8 V (rectifier). α = 120°: −270.1 V (inverter, power returned to the supply).

Common mistakes

  • Using (2V_m/π)cos α for a resistive load, or (V_m/π)(1 + cos α) for a continuous-current full converter. Read the load and circuit before choosing the formula.
  • Using the RMS supply voltage as V_m; V_m = √2·V_rms.
  • Using phase voltage instead of line voltage in the three-phase formula (1.35 multiplies V_L).
  • Claiming the output is zero at α = 90° for every load: that is true only for a full converter with continuous current.
  • Calling a controlled rectifier "better" for power factor: its displacement factor falls roughly as cos α.
  • Writing boost output as D·V_s or forgetting that the buck-boost output is inverted.
  • Confusing firing angle α with conduction angle (π − α for a resistive load).

For GATE ME

Expect short numericals: average output voltage of single-phase and three-phase converters for a given α and load type, the effect of a freewheeling diode, firing angle needed for a required DC voltage, output voltage and duty cycle of buck, boost and buck-boost choppers, and current ripple for a given L and f. Conceptual MCQs test inversion (α > 90°), quadrant of operation and why choppers need devices with gate turn-off. Practise picking the right formula from the load description.

Quick check

  1. A single-phase full converter with continuous current operates at α = 90°. What is the average output voltage?
  2. A buck chopper from 48 V must give 18 V. What duty cycle is needed?
  3. A boost converter from 24 V runs at D = 0.75. What is the ideal output voltage?
  4. What is the ripple frequency of the output of a three-phase full converter on a 50 Hz supply?
  5. Why can a single-phase semi-converter not regenerate?

Answers: 1. 0 V. 2. 0.375. 3. 96 V. 4. 300 Hz. 5. Its freewheeling diodes clamp the output at zero, so the average voltage can never be negative.

Interview questions

All Electrical Circuits and Electronics interview questions

Try answering each one aloud before you open it.

  1. 1.What is a controlled rectifier and how does it differ from an uncontrolled rectifier?Concept

    A controlled rectifier uses thyristors (SCRs) in place of some or all of the diodes, and delays each SCR's gate pulse by a firing angle α after the point where a diode would have started conducting. Increasing α removes part of each half-cycle, so the average DC output can be varied, for example V_dc = (2·V_m/π)·cos α for a single-phase full converter with continuous current. A diode rectifier has no such control: its output is fixed by the supply voltage. A fully controlled converter can also invert (α > 90°) and return power to the AC supply when the DC side has a source such as a motor.

  2. 2.Explain the working principle of a chopper circuit.Concept

    A chopper is a DC-DC converter that switches a fixed DC supply on and off at high frequency with a MOSFET, IGBT or (in older designs) a force-commutated SCR. The fraction of each period the switch is on is the duty cycle D = T_on/T, and the average output follows it: V_o = D·V_s for a step-down chopper. An inductance in the load and a freewheeling diode keep the current nearly continuous while the switch is off. Varying D, usually by PWM at a constant frequency, gives smooth, efficient control of DC motor speed or battery charging.

  3. 3.Why are thyristors commonly used in controlled rectifiers?Application

    A thyristor can be turned on at any chosen instant by a gate pulse, which gives phase control of the output, and on an AC supply it turns off by itself when its current falls below the holding current as the supply reverses (natural or line commutation). So no turn-off circuit is needed, unlike in a chopper. Thyristors are also available with very high voltage and current ratings, low conduction loss and a high surge rating, and they are cheap and rugged, which suits large drives and HVDC converters.

  4. 4.What happens if the firing angle of a thyristor in a controlled rectifier is set to 90 degrees?Application

    It depends on the circuit and load. In a fully controlled converter with a highly inductive load (continuous current), V_dc = (2·V_m/π)·cos α, so at α = 90° the positive and negative parts of the output waveform are equal and the average output voltage is zero. With a resistive load, or with a freewheeling diode, the output cannot go negative, so at 90° the average is (V_m/π)·(1 + cos 90°) = V_m/π for a full-wave circuit, half the α = 0 value. Beyond 90° a fully controlled converter with continuous current gives a negative average voltage and works as an inverter.

  5. 5.How does a step-down chopper differ from a step-up chopper?Concept

    In a step-down (buck) chopper the switch is in series between the source and the load and a freewheeling diode carries the load current when it is off, so the average output is V_o = D·V_s, always below the source. In a step-up (boost) chopper an inductor is in series with the source and the switch is across it: with the switch on the inductor stores energy, and when it opens the inductor voltage adds to the source and drives current through a diode into the output capacitor. The ideal output is V_o = V_s/(1 − D), always above the source. A boost stage is also the basis of regenerative braking from a lower motor voltage into a higher DC supply.

  6. 6.In what applications would you use a chopper circuit?Application

    Chopper circuits are used in applications where variable DC voltage is required. Common applications include speed control of DC motors, regenerative braking systems, and power supplies for variable voltage devices. They are also used in renewable energy systems, such as solar power converters, to match the output voltage to the load requirements.

  7. 7.Calculate the average output voltage of a single-phase fully controlled bridge rectifier with a resistive load, a firing angle of 30 degrees and an AC input of 230 V RMS.Numerical

    With a resistive load the current stops at each voltage zero, so V_dc = (V_m/π)·(1 + cos α). The peak voltage is V_m = √2 × 230 = 325.3 V. Then V_dc = (325.3/π) × (1 + 0.866) = 103.5 × 1.866 ≈ 193.2 V. For the same bridge with continuous current (large inductance) the formula would instead be (2·V_m/π)·cos α, giving 179.3 V.

  8. 8.What are the advantages and drawbacks of a controlled rectifier compared with an uncontrolled rectifier?Application

    The main advantage is an adjustable DC output from the same supply, which allows speed control of DC motors, controlled battery charging and soft starting, and a fully controlled converter can also invert to regenerate braking energy into the supply. The drawbacks are that the supply-side displacement power factor falls roughly as cos α, so it is poor at large firing angles, and the input current is rich in harmonics. It also needs gate-firing circuits synchronised to the supply, which a diode rectifier does not.

  9. 9.Explain the role of freewheeling diodes in chopper circuits.Concept

    When the chopper switch turns off, the load inductance tries to keep its current flowing; without a path it would produce a large L·di/dt voltage spike that could destroy the switch. The freewheeling diode across the load provides that path, so the current decays gradually through the load and diode until the switch turns on again. This keeps the load current continuous with small ripple, clamps the output voltage at about zero during the off-time and returns the stored inductor energy to the load instead of dissipating it in the switch.

  10. 10.A chopper circuit has an input voltage of 100 V and a duty cycle of 0.4. Calculate the average output voltage.Numerical

    The average output voltage V_out of a chopper circuit is given by the formula: V_out = D * V_in, where D is the duty cycle and V_in is the input voltage. For a duty cycle of 0.4 and an input voltage of 100 V, V_out = 0.4 * 100 V = 40 V.

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