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Uncontrolled and controlled rectifiers, AC-DC converters

How diodes and thyristors turn AC into DC: average and rms output, ripple, the effect of the firing angle, source inductance and overlap, inversion, thyristor commutation, and the converters built from them (dual converters, AC voltage controllers, PWM voltage source converters and DC drives). Every result is given with the conduction assumption it needs, because the same circuit gives a different output under a different assumption.

12 previous-year questions · 11 solved · 7 practice questions

Single-phase uncontrolled rectifiers

Conventions: $V_{m}$ is the peak of the supply voltage, $V_{s} = V_{m} / 2$ is its rms value, the diodes are ideal and the load is resistive unless stated. Average output is $V_{d c}$ and rms output is $V_{r m s}$ . Form factor is $F F = V_{r m s} / V_{d c}$ . Ripple factor is $R F = V_{a c, r m s} / V_{d c}$ , where $V_{a c, r m s}$ is the rms of the output with its average removed.

RIPPLE FACTOR FROM FORM FACTOR

R F = (\frac{V _{r m s}}{V _{d c}})^{2} - 1 = F F^{2} - 1

Holds for any output waveform; $V_{r m s}$ is the total rms including the DC part.

HALF-WAVE RECTIFIER, R LOAD

V_{d c} = \frac{V _{m}}{π}, V_{r m s} = \frac{V _{m}}{2}, F F = \frac{π}{2} = 1.57, R F = 1.21, P I V = V_{m}

One diode conducts for 180 degrees of each cycle.

FULL-WAVE RECTIFIER (CENTRE-TAP OR BRIDGE), R LOAD

V_{d c} = \frac{2 V _{m}}{π}, V_{r m s} = \frac{V _{m}}{2}, F F = \frac{π}{2 2} = 1.11, R F = 0.483

For the centre-tap type $V_{m}$ is the peak of each half of the secondary. PIV is $V_{m}$ for the bridge and $2 V_{m}$ for the centre-tap.

Quantity	Half-wave	Full-wave bridge	Full-wave centre-tap
$V_{d c}$ (R load)	$V_{m} / π$	$2 V_{m} / π$	$2 V_{m} / π$
$V_{r m s}$ (R load)	$V_{m} /2$	$V_{m} / 2$	$V_{m} / 2$
Ripple factor	1.21	0.483	0.483
Diodes	1	4	2
PIV per diode	$V_{m}$	$V_{m}$	$2 V_{m}$
Output ripple frequency	$f$	$2 f$	$2 f$

HALF-WAVE RECTIFIER, RL LOAD (NO FREEWHEELING DIODE)

V_{d c} = \frac{V _{m}}{2 π} (1 - cos β), sin (β - ϕ) + sin ϕ e^{- β / t a n ϕ} = 0

$ϕ = tan^{- 1} (ω L / R)$ . The current is $i = \frac{V _{m}}{Z} [sin (ω t - ϕ) + sin ϕ e^{- ω t / t a n ϕ}]$ with $Z = R^{2} + (ω L)^{2}$ . The current continues after the supply goes negative, up to the extinction angle $β > π$ where $i = 0$ , so the output voltage has a negative portion from $π$ to $β$ .

A freewheeling diode across the load removes the negative portion. $V_{d c}$ returns to $V_{m} / π$ and the load current becomes smoother. With a very large load inductance, the bridge output current is constant at $I_{d c}$ . The source current is then a square wave of height $\pm I_{d c}$ , with rms value $I_{d c}$ and fundamental rms value $\frac{2 2}{π} I_{d c} \approx 0.9 I_{d c}$ , so the input power factor is $0.9$ .

HALF-WAVE DIODE CHARGING A BATTERY $E$ THROUGH $R$ (NO INDUCTANCE)

θ_{1} = sin^{- 1} \frac{E}{V _{m}}, I_{d c} = \frac{1}{2 π R} [2 V_{m} cos θ_{1} - E (π - 2 θ_{1})]

The diode conducts only while the instantaneous supply voltage exceeds $E$ , from $θ_{1}$ to $π - θ_{1}$ . In a full-wave bridge the same interval repeats in every half cycle, so the average current doubles.

EXAM TIP

With an R + E load and no inductance, current flows only while $v_{s} > E$ . Find $θ_{1} = sin^{- 1} (E / V_{m})$ first; the conduction angle is $π - 2 θ_{1}$ in each conducting half cycle.

Three-phase uncontrolled rectifiers

Convention: $V_{m}$ is the peak of the phase voltage and $V_{LL, m} = 3 V_{m}$ is the peak of the line voltage. $V_{LL}$ without a subscript $m$ is the rms line voltage. The load current is continuous and ripple-free ( $I_{d c}$ ) unless stated.

THREE-PHASE HALF-WAVE (3-PULSE) DIODE RECTIFIER

V_{d c} = \frac{3 3}{2 π} V_{m} = 0.827 V_{m}, P I V = 3 V_{m}

Each diode conducts for 120 degrees; its average current is $I_{d c} /3$ and its rms current is $I_{d c} / 3$ . The output ripple frequency is $3 f$ .

THREE-PHASE BRIDGE (6-PULSE) DIODE RECTIFIER

V_{d c} = \frac{3 V _{LL, m}}{π} = \frac{3 3}{π} V_{m} = 1.35 V_{LL}

The output is made of six line-voltage segments per cycle, so the ripple frequency is $6 f$ . PIV $= V_{LL, m}$ .

BRIDGE CURRENTS WITH CONSTANT $I_{DC}$

I_{D, a v g} = \frac{I _{d c}}{3}, I_{D, r m s} = \frac{I _{d c}}{3}, I_{s, r m s} = \frac{2}{3} I_{d c}, I_{s 1, r m s} = \frac{6}{π} I_{d c}, P F = \frac{3}{π} = 0.955

Each diode conducts 120 degrees. The line current has one positive and one negative 120-degree block per cycle, and its fundamental is $0.78 I_{d c}$ .

At any instant the top diode connected to the phase with the highest voltage and the bottom diode connected to the phase with the lowest voltage conduct.
The output equals the largest line-to-line voltage at that instant.
Each line-voltage segment lasts 60 degrees, and the conducting diode pair changes every 60 degrees.

EXAM TIP

For the bridge, $V_{d c} = 1.35 V_{LL}$ where $V_{LL}$ is the rms line voltage. For the half-wave circuit use the phase voltage: $V_{d c} = 1.17 V_{p h}$ with $V_{p h}$ in rms.

Single-phase phase-controlled rectifiers

Convention: the firing angle $α$ is measured from the natural commutation point, which is the instant a diode in the same position would start conducting. For a single-phase supply this is the zero crossing of the supply voltage. Continuous conduction is assumed unless stated.

HALF-WAVE CONTROLLED RECTIFIER, R LOAD

V_{d c} = \frac{V _{m}}{2 π} (1 + cos α)

Applies to an R load and also to an RL load with a freewheeling diode. Range $0 \leq α \leq π$ .

CONTROLLED RECTIFIERS WITH R LOAD: RMS OUTPUT

V_{r m s} = \frac{V _{m}}{2} \frac{1}{π} (π - α + \frac{sin 2 α}{2}) (half-wave), V_{r m s} = \frac{V _{m}}{2} \frac{1}{π} (π - α + \frac{sin 2 α}{2}) (bridge)

Pure R load, so the output follows the supply from $α$ to $π$ in each conducting half cycle. The bridge result holds for both the fully controlled and the half-controlled bridge.

FULLY CONTROLLED BRIDGE, CONTINUOUS CONDUCTION (LARGE L)

V_{d c} = \frac{2 V _{m}}{π} cos α

Positive for $α < 9 0^{\circ}$ (rectification) and negative for $α > 9 0^{\circ}$ (inversion, which needs an active load such as a reversed battery or a regenerating machine). With an RLE load and continuous current, $I_{d c} = (V_{d c} - E) / R$ .

HALF-CONTROLLED BRIDGE, OR FULLY CONTROLLED BRIDGE WITH R LOAD OR FREEWHEELING DIODE

V_{d c} = \frac{V _{m}}{π} (1 + cos α)

The output cannot go negative, so only the rectifying quadrant is available.

FULLY CONTROLLED BRIDGE WITH LARGE L: SOURCE CURRENT

I_{s, r m s} = I_{d c}, I_{s 1, r m s} = \frac{2 2}{π} I_{d c} = 0.9 I_{d c}, D P F = cos α, P F = \frac{2 2}{π} cos α

Square-wave source current; each thyristor conducts for 180 degrees, with average current $I_{d c} /2$ and rms current $I_{d c} / 2$ . The source current has odd harmonics $n = 3, 5, 7, \dots$ with $I_{n} = I_{s 1} / n$ , and THD $= 48.3%$ . Input powers: $P = \frac{2 V _{m}}{π} I_{d c} cos α = V_{d c} I_{d c}$ and $Q = \frac{2 V _{m}}{π} I_{d c} sin α$ .

HALF-CONTROLLED BRIDGE WITH LARGE L: SOURCE CURRENT

I_{s, r m s} = I_{d c} \frac{π - α}{π}, I_{s 1, r m s} = \frac{2 2}{π} I_{d c} cos \frac{α}{2}, D P F = cos \frac{α}{2}, P F = \frac{2 ( 1 + cos α )}{π ( π - α )}

The source current is $+ I_{d c}$ from $α$ to $π$ , $- I_{d c}$ from $π + α$ to $2 π$ , and zero while the load freewheels. Its displacement factor $cos (α /2)$ is better than the fully controlled bridge's $cos α$ .

Circuit	Quadrant(s)	Output-voltage range
Half-wave controlled (R load, or RL with a freewheeling diode)	1 only	$0$ to $V_{m} / π$
Half-controlled bridge (any load)	1 only	$0$ to $2 V_{m} / π$
Fully controlled bridge, large L	1 and 4 (two-quadrant converter)	$+ 2 V_{m} / π$ to $- 2 V_{m} / π$

WORKED EXAMPLE

A single-phase fully controlled bridge is fed from a 230 V, 50 Hz supply. The load current is constant at 10 A and the firing angle is 60 degrees. Source inductance is negligible. Find the average output voltage, the output power and the input power factor.

Solution

$V_{m} = 2302 = 325.3$ V.
$V_{d c} = \frac{2 V _{m}}{π} cos 6 0^{\circ} = \frac{2 ( 325.3 )}{π} (0.5) = 103.5$ V.
$P = V_{d c} I_{d c} = 103.5 \times 10 = 1035$ W.
$P F = \frac{2 2}{π} cos α = 0.9003 \times 0.5 = 0.45$ .

Answer: $V_{d c} \approx 103.5$ V, $P \approx 1035$ W, $P F \approx 0.45$ lagging.

EXAM TIP

For an R load the bridge output is $\frac{V _{m}}{π} (1 + cos α)$ , not $\frac{2 V _{m}}{π} cos α$ . The cosine law holds only when the load current is continuous, which in practice needs enough load inductance.

Three-phase controlled rectifiers

Convention: $V_{m}$ is the peak phase voltage and $V_{LL, m} = 3 V_{m}$ is the peak line voltage. In a three-phase circuit the natural commutation point is where two phase voltages cross, 30 degrees after the zero crossing of the phase voltage. The firing angle $α$ is measured from there. Each thyristor in a bridge conducts for 120 degrees when conduction is continuous.

THREE-PHASE FULLY CONTROLLED BRIDGE, CONTINUOUS CONDUCTION

V_{d c} = \frac{3 V _{LL, m}}{π} cos α = \frac{3 3}{π} V_{m} cos α = 1.35 V_{LL} cos α

Negative for $α > 9 0^{\circ}$ , which is inversion. The load current stays continuous for all $α$ when $L$ is large enough.

THREE-PHASE HALF-CONTROLLED BRIDGE

V_{d c} = \frac{3 V _{LL, m}}{2 π} (1 + cos α)

The output never goes negative, so $0 \leq α \leq 18 0^{\circ}$ gives rectification only.

THREE-PHASE HALF-WAVE CONTROLLED CONVERTER

V_{d c} = \frac{3 3}{2 π} V_{m} cos α

Continuous conduction. Each thyristor conducts 120 degrees.

THREE-PHASE HALF-WAVE, R LOAD, DISCONTINUOUS ($\ALPHA>30^\CIRC$)

V_{d c} = \frac{3 V _{m}}{2 π} [1 + cos (α + 3 0^{\circ})]

Valid for $3 0^{\circ} \leq α \leq 15 0^{\circ}$ ; the output reaches zero at $α = 15 0^{\circ}$ . For $α \leq 3 0^{\circ}$ with an R load the current stays continuous and the cosine law holds.

FULLY CONTROLLED BRIDGE, R LOAD, DISCONTINUOUS ($\ALPHA>60^\CIRC$)

V_{d c} = \frac{3 V _{LL, m}}{π} [1 + cos (α + 6 0^{\circ})]

Valid for $6 0^{\circ} \leq α \leq 12 0^{\circ}$ ; the output reaches zero at $α = 12 0^{\circ}$ . For $α \leq 6 0^{\circ}$ with an R load the output stays positive and the cosine law $\frac{3 V _{LL, m}}{π} cos α$ still holds.

The bridge firing order is $T_{1} \to T_{2} \to \dots \to T_{6}$ in steps of 60 degrees, with each thyristor fired 120 degrees after the previous one in its own half of the bridge.

THREE-PHASE FULLY CONTROLLED BRIDGE WITH LARGE L: SOURCE CURRENT

I_{s, r m s} = \frac{2}{3} I_{d c}, I_{s 1, r m s} = \frac{6}{π} I_{d c}, D P F = cos α, P F = \frac{3}{π} cos α

The line current has one positive and one negative 120-degree block per cycle. Its harmonics are of order $n = 6 k \pm 1$ ( $5, 7, 11, 13, \dots$ ) with $I_{n} = I_{s 1} / n$ , and THD $= 31.1%$ . Input powers: $P = \frac{3 V _{LL, m}}{π} I_{d c} cos α$ and $Q = \frac{3 V _{LL, m}}{π} I_{d c} sin α$ .

EXAM TIP

When $α$ is just above 60 degrees, check whether the problem has an RL load with large $L$ (use $cos α$ ) or a pure R load (use the $1 + cos (α + 6 0^{\circ})$ form).

Source inductance, overlap and inversion

With source inductance $L_{s}$ the current cannot change instantly, so the outgoing and incoming devices conduct together during commutation, for an overlap angle $μ$ . During overlap the output voltage is the average of the outgoing and incoming voltages. In a single-phase bridge this average is zero, because all four thyristors conduct and short the supply. As a result $V_{d c}$ drops. Here $I_{d c}$ is constant and $ω$ is the supply angular frequency.

SINGLE-PHASE FULLY CONTROLLED BRIDGE WITH $L_S$

V_{d c} = \frac{2 V _{m}}{π} cos α - \frac{2 ω L _{s}}{π} I_{d c}, cos α - cos (α + μ) = \frac{2 ω L _{s} I _{d c}}{V _{m}}

Large L, continuous conduction. The drop equals $4 f L_{s} I_{d c}$ and does not depend on $α$ .

THREE-PHASE FULLY CONTROLLED BRIDGE WITH $L_S$

V_{d c} = \frac{3 V _{LL, m}}{π} cos α - \frac{3 ω L _{s}}{π} I_{d c}, cos α - cos (α + μ) = \frac{2 ω L _{s} I _{d c}}{V _{LL, m}}

Constant $I_{d c}$ ; $L_{s}$ is per phase. The drop equals $6 f L_{s} I_{d c}$ .

THREE-PHASE HALF-WAVE CONVERTER WITH $L_S$

Δ V = \frac{3 ω L _{s}}{2 π} I_{d c}, cos α - cos (α + μ) = \frac{2 ω L _{s} I _{d c}}{3 V _{m}}

Three commutations per cycle, each driven by the line voltage between the two phases involved.

Inversion: for $9 0^{\circ} < α < 18 0^{\circ}$ the average converter output voltage is negative while the current stays positive, so power flows from the DC side to the AC side. This needs a DC source in the load that drives the current against the negative converter voltage, such as a DC machine regenerating. The extinction angle is $γ = 18 0^{\circ} - α - μ$ , and the outgoing thyristor is reverse-biased for $γ / ω$ . If $γ / ω$ is less than the thyristor turn-off time $t_{q}$ , the outgoing thyristor is forward-biased again before it has recovered. It then resumes conduction and commutation fails. The firing angle must therefore satisfy $α \leq 18 0^{\circ} - μ - ω t_{q}$ .

WORKED EXAMPLE

A three-phase fully controlled bridge is supplied from a 400 V (rms line), 50 Hz source with $ω L_{s} = 0.2 Ω$ per phase. The load current is a constant 50 A and $α = 3 0^{\circ}$ . Find the average output voltage and the overlap angle.

Solution

$V_{LL, m} = 4002 = 565.7$ V.
Ideal output: $\frac{3 ( 565.7 )}{π} cos 3 0^{\circ} = 467.8$ V.
Drop: $\frac{3 ω L _{s} I _{d c}}{π} = \frac{3 ( 0.2 ) ( 50 )}{π} = 9.55$ V.
$V_{d c} = 467.8 - 9.55 = 458.3$ V.
$cos α - cos (α + μ) = \frac{2 ( 0.2 ) ( 50 )}{565.7} = 0.0354$ , so $cos (α + μ) = 0.8660 - 0.0354 = 0.8307$ .
$α + μ = 33. 8^{\circ}$ , so $μ \approx 3. 8^{\circ}$ .

Answer: $V_{d c} \approx 458.3$ V and $μ \approx 3. 8^{\circ}$ .

EXAM TIP

The overlap formula uses $2 ω L_{s} I_{d c}$ with the peak of the voltage driving the commutation: $V_{m}$ for a single-phase bridge and $V_{LL, m}$ for a three-phase bridge.

Thyristor commutation: line, voltage and current commutation

A thyristor turns off when its current falls below the holding current and it is then kept reverse-biased for longer than its turn-off time $t_{q}$ . The time for which the circuit reverse-biases it is the circuit turn-off time $t_{c}$ , and the requirement is $t_{c} > t_{q}$ . In a phase-controlled converter the AC supply does this: when the next thyristor is fired, the supply voltage reverse-biases the outgoing one. This is natural, or line, commutation, and its circuit turn-off time is $t_{c} = γ / ω$ . Where no alternating voltage is available, for example in a thyristor chopper or inverter, a forced-commutation circuit turns the thyristor off.

VOLTAGE (IMPULSE) COMMUTATION

t_{c} = \frac{C V _{C}}{I _{0}}

An auxiliary thyristor switches a capacitor $C$ , charged to $V_{C}$ , across the conducting thyristor, reverse-biasing it at once. With the load current $I_{0}$ constant during commutation, the capacitor voltage changes linearly from $- V_{C}$ to zero in time $t_{c}$ .

CURRENT (RESONANT-PULSE) COMMUTATION

I_{c p} = V_{C} \frac{C}{L} > I_{0}, t_{c} = L C [π - 2 sin^{- 1} \frac{I _{0}}{I _{c p}}]

An LC circuit drives a resonant current pulse of peak $I_{c p}$ through the thyristor, opposite to its load current $I_{0}$ , assumed constant during commutation. The thyristor current reaches zero when the pulse reaches $I_{0}$ . A diode across the thyristor then carries the excess, and its forward drop reverse-biases the thyristor until the pulse falls back to $I_{0}$ . Commutation needs $I_{c p} > I_{0}$ .

Dual converters, AC voltage controllers, voltage source converters and DC drives

A dual converter is two fully controlled converters connected back-to-back across one load. One carries positive current and the other carries negative current, so the drive can work in all four quadrants. The two firing angles satisfy $α_{1} + α_{2} = 18 0^{\circ}$ . Their average output voltages are then equal at the load terminals, with one converter rectifying and the other inverting. Their instantaneous voltages still differ. In the circulating-current mode both converters conduct all the time, with a reactor between them to limit the circulating current that this difference drives. In the non-circulating mode only the converter that carries the load current is gated.

AC VOLTAGE CONTROLLER, R LOAD (SINGLE-PHASE, PHASE CONTROL)

V_{o, r m s} = V_{s} \frac{1}{π} (π - α + \frac{sin 2 α}{2})

Two antiparallel thyristors (or a triac), fired symmetrically. $V_{s}$ is the rms supply voltage and $α$ is measured from the voltage zero crossing. Control range $0 \leq α \leq 18 0^{\circ}$ . The output has no DC component. Input power factor $= V_{o, r m s} / V_{s}$ .

With an RL load the load angle is $ϕ = tan^{- 1} (ω L / R)$ , and phase control works for $ϕ \leq α \leq 18 0^{\circ}$ . For $α < ϕ$ with a sustained gate signal or a pulse train, the load current is a continuous sinusoid lagging by $ϕ$ . The full supply voltage then appears across the load, so firing control is lost. With a single short gate pulse and $α < ϕ$ , only one thyristor conducts, and the load current then has a DC component.

BIDIRECTIONAL AC-DC VOLTAGE SOURCE CONVERTER (PWM RECTIFIER)

P = \frac{V _{s} V _{c}}{X} sin δ, Q_{s} = \frac{V _{s}^{2} - V _{s} V _{c} cos δ}{X}

Per phase, with a lossless line reactance $X = ω L$ . $V_{s}$ is the rms source voltage, $V_{c}$ is the rms fundamental converter voltage, and $δ$ is the angle by which $V_{s}$ leads $V_{c}$ . $Q_{s}$ is the reactive power sent by the source. For a three-phase converter, multiply both by 3 when phase rms values are used. $δ > 0$ gives power flow from AC to DC (rectification) and $δ < 0$ gives power flow from DC to AC (regeneration).

A voltage source converter uses fully controlled switches with antiparallel diodes and a DC-link capacitor. It can draw a nearly sinusoidal current at unity power factor, and it lets power flow in either direction. The DC-link voltage must exceed the peak supply voltage: $V_{d c} > V_{m}$ for a single-phase converter and $V_{d c} > V_{LL, m}$ for a three-phase one. Below that, the antiparallel diodes conduct as an uncontrolled diode bridge, and the current can no longer be controlled. The converter therefore works as a boost rectifier. In contrast, a thyristor converter draws a current whose fundamental lags by $α$ , and it can invert only when an opposing DC source drives the current.

RECTIFIER-FED SEPARATELY EXCITED DC MOTOR

V_{d c} = E_{b} + I_{a} R_{a}, E_{b} = K ϕ ω_{m}

Steady state with continuous armature current. Speed is raised by increasing $V_{d c}$ (reducing $α$ ).

Drive converter	Quadrants	Notes
Single-phase half-controlled bridge	1	Motoring only, no regeneration
Single-phase fully controlled bridge	1 and 4	Regeneration (as inverter) when $α > 9 0^{\circ}$ with a reversed $E_{b}$
Dual converter	1, 2, 3 and 4	Both directions of speed and torque

EXAM TIP

A fully controlled converter gives $V_{d c}$ of either sign but $I_{a} > 0$ only. $V_{d c} < 0$ with $I_{a} > 0$ means power flows from the machine to the supply. This needs $E_{b}$ reversed, for example by an overhauling load driving the motor backwards, and it is regenerative braking.

PREVIOUS-YEAR QUESTIONS

12 questions from the EE papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE EE 2026 · GATE EE 2025 · GATE EE 2024 · GATE EE 2023 · GATE EE 2022 · GATE EE 2020 · GATE EE 2018

GATE 2026 EE Q.34

GATE 2026 · EEQ.34 · 1 mark · NAT

Consider the circuit shown. Assume that the diode D is ideal. The supply voltage v_s = 325 sin(2π50t) V, L = 500 µH, and R = 10 Ω. The peak diode current (in amperes) is ______________ (Round off to one decimal place)

GATE 2026 EE question 34: A single series loop. — Figure from the GATE 2026 EE paper. It is part of the question.

Answer and solution

GATE 2026 EE Q.34: answer and solution

GATE 2026 EE Q.61

GATE 2026 · EEQ.61 · 2 marks · NAT

Consider the circuit shown. Assume that the diode (D) is ideal. Given $v_{s} = 100 sin (2 π 50 t)$ V, $V_{d c} = 50$ V, and $R = 10 Ω$ , the average value of the current through the diode is ____ A (Round off to two decimal places).

GATE 2026 EE question 61: A single series loop. — Figure from the GATE 2026 EE paper. It is part of the question.

Answer and solution

GATE 2026 EE Q.61: answer and solution

GATE 2025 EE Q.47

GATE 2025 · EEQ.47 · 2 marks · MCQ

An ideal sinusoidal voltage source $v (t) = 2302 sin (2 π \times 50 t)$ V feeds an ideal inductor $L$ through an ideal SCR with firing angle $α = 0^{\circ}$ . If $L = 100$ mH, then the peak of the inductor current, in ampere, is closest to

GATE 2025 EE question 47: Single loop. — Figure from the GATE 2025 EE paper. It is part of the question.

(A)
20.71
(B)
0
(C)
10.35
(D)
7.32

Answer and solution

GATE 2025 EE Q.47: answer and solution

GATE 2025 EE Q.48

GATE 2025 · EEQ.48 · 2 marks · MCQ

In the following circuit, the average voltage $V_{o} = 400 (1 + \frac{c o s α}{3})$ V, where $α$ is the firing angle. If the power dissipated in the resistor is 64 W, then the closest value of $α$ in degrees is

GATE 2025 EE question 48: A three-phase star-connected AC source (three sinusoidal sources joined at a common star point) feeds a box containing a… — Figure from the GATE 2025 EE paper. It is part of the question.

(A)
35.9
(B)
46.4
(C)
41.4
(D)
0

Answer and solution

GATE 2025 EE Q.48: answer and solution

GATE 2024 EE Q.28

GATE 2024 · EEQ.28 · 1 mark · MCQ

A single-phase triac based AC voltage controller feeds a series RL load. The input AC supply is 230 V, 50 Hz. The values of R and L are 10 Ω and 18.37 mH, respectively. The minimum triggering angle of the triac to obtain controllable output voltage is

(A)
15°
(B)
30°
(C)
45°
(D)
60°

Answer and solution

GATE 2024 EE Q.28: answer and solution

See the question in the paper

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GATE 2024 EE Q.28: answer and solution

GATE 2024 EE Q.65

GATE 2024 · EEQ.65 · 2 marks · NAT

A single-phase full-controlled thyristor converter bridge is used for regenerative braking of a separately excited DC motor with the following specifications: rated armature voltage 210 V; rated armature current 10 A; rated speed 1200 rpm; armature resistance 1 Ω; input to the converter bridge 240 V at 50 Hz. The armature of the DC motor is fed from the full-controlled bridge and the field current is kept constant. Assume that the motor is running at 600 rpm and the armature terminals of the motor are suitably reversed for regenerative braking. If the armature current of the motor is to be maintained at the rated value, the triggering angle of the converter bridge in degrees should be ____________ (rounded off to 2 decimal places).

Answer and solution

GATE 2024 EE Q.65: answer and solution

See the question in the paper

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GATE 2024 EE Q.65: answer and solution

GATE 2023 EE Q.58

GATE 2023 · EEQ.58 · 2 marks · NAT

The single phase rectifier consisting of three thyristors $T_{1}$ , $T_{2}$ , $T_{3}$ and a diode $D_{1}$ feed power to a 10 A constant current load. $T_{1}$ and $T_{3}$ are fired at $α = 6 0^{\circ}$ and $T_{2}$ is fired at $α = 24 0^{\circ}$ . The reference for $α$ is the positive zero crossing of $V_{in}$ . The average voltage $V_{O}$ across the load in volts is _____ (Round off to 2 decimal places).

GATE 2023 EE question 58: Bridge with two legs between a top rail and a bottom rail. — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

GATE 2023 EE Q.58: answer and solution

GATE 2022 EE Q.32

GATE 2022 · EEQ.32 · 1 mark · NAT

A single-phase full-bridge diode rectifier feeds a resistive load of 50 Ω from a 200 V, 50 Hz single phase AC supply. If the diodes are ideal, then the active power, in watts, drawn by the load is ______. (round off to nearest integer)

Answer and solution

GATE 2022 EE Q.32: answer and solution

See the question in the paper

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GATE 2022 EE Q.32: answer and solution

GATE 2020 EE Q.12

GATE 2020 · EEQ.12 · 1 mark · MCQ

Thyristor $T_{1}$ is triggered at an angle $α$ (in degree), and $T_{2}$ at angle $18 0^{\circ} + α$ , in each cycle of the sinusoidal input voltage. Assume both thyristors to be ideal. To control the load power over the range 0 to 2 kW, the minimum range of variation in $α$ is:

GATE 2020 EE question 12: Single-phase AC voltage controller. — Figure from the GATE 2020 EE paper. It is part of the question.

(A)
0° to 60°
(B)
0° to 120°
(C)
60° to 120°
(D)
60° to 180°

Official answer

Official answer: marks to all candidatesSource: Official answer key, GATE 2020 EE (IIT Delhi)

GATE 2020 EE Q.12: official answer

GATE 2020 EE Q.51

GATE 2020 · EEQ.51 · 2 marks · NAT

A single-phase, full-bridge, fully controlled thyristor rectifier feeds a load comprising a 10 Ω resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz sinusoidal source through cables which have negligible internal resistance and a total inductance of 2.28 mH. If the thyristors are triggered at an angle α = 45°, the commutation overlap angle in degree (rounded off to 2 decimal places) is ______.

Answer and solution

GATE 2020 EE Q.51: answer and solution

See the question in the paper

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GATE 2020 EE Q.51: answer and solution

GATE 2018 EE Q.3

GATE 2018 · EEQ.3 · 1 mark · MCQ

A single phase fully controlled rectifier is supplying a load with an anti-parallel diode as shown in the figure. All switches and diodes are ideal. Which one of the following is true for instantaneous load voltage and current?

GATE 2018 EE question 3: A single-phase thyristor bridge fed by an AC source (sine symbol) connected between the two mid-points of the two bridge… — Figure from the GATE 2018 EE paper. It is part of the question.

(A)
v_o ≥ 0 & i_o < 0
(B)
v_o < 0 & i_o < 0
(C)
v_o ≥ 0 & i_o ≥ 0
(D)
v_o < 0 & i_o ≥ 0

Answer and solution

GATE 2018 EE Q.3: answer and solution

GATE 2018 EE Q.49

GATE 2018 · EEQ.49 · 2 marks · NAT

A phase controlled single phase rectifier, supplied by an AC source, feeds power to an R-L-E load as shown in the figure. The rectifier output voltage has an average value given by $V_{o} = \frac{V _{m}}{2 π} (3 + cos α)$ , where $V_{m} = 80 π$ volts and $α$ is the firing angle. If the power delivered to the lossless battery is 1600 W, $α$ in degree is ________ (up to 2 decimal places).

GATE 2018 EE question 49: Source V_m sin (ω t) on the left feeds a rectifier block (a box containing a single thyristor symbol with a gate terminal). — Figure from the GATE 2018 EE paper. It is part of the question.

Answer and solution

GATE 2018 EE Q.49: answer and solution

PRACTICE

7 questions to test yourself

PRACTICE 11 mark · NAT

A single-phase full-wave bridge diode rectifier is fed from a 230 V rms, 50 Hz supply. The load is purely resistive and the diodes are ideal. Find the average output voltage in volts, correct to one decimal place.

Answer and solution

Answer: $207.1$ (accepted range $\pm 0.5$ )

$V_{m} = 2302 = 325.27$ V.
$V_{d c} = \frac{2 V _{m}}{π} = \frac{2 ( 325.27 )}{π} = 207.1$ V.

PRACTICE 22 marks · MCQ

A three-phase fully controlled thyristor bridge is fed from a 400 V rms line-to-line supply. The load contains a DC source that keeps the load current large and continuous. Source inductance is negligible. The firing angle is 120 degrees. The average output voltage is closest to which of the following?

(A)
+270 V
(B)
-270 V
(C)
-540 V
(D)
0 V

Answer and solution

Answer: (B)

$V_{LL, m} = 4002 = 565.7$ V.
$V_{d c} = \frac{3 V _{LL, m}}{π} cos α = 540.2 \times cos 12 0^{\circ} = - 270.1$ V.
The converter is in the inverting mode, so the answer is $- 270$ V.

PRACTICE 32 marks · MSQ

A single-phase fully controlled thyristor bridge supplies a highly inductive load with constant current, with $α = 4 5^{\circ}$ and negligible source inductance. Which of the following statements are correct?

(A)
The average output voltage is positive and equals $\frac{2 V _{m}}{π} cos 4 5^{\circ}$ .
(B)
The rms value of the source current equals the load current.
(C)
The displacement factor of the fundamental source current is $cos 4 5^{\circ}$ .
(D)
Each thyristor conducts for 90 degrees in a cycle.

Answer and solution

Answer: (A), (B), (C)

A is true: $α < 9 0^{\circ}$ , so $V_{d c} = \frac{2 V _{m}}{π} cos α > 0$ .
B is true: the source current is a square wave of height $I_{d c}$ , so its rms value is $I_{d c}$ .
C is true: the fundamental lags the supply by $α$ .
D is false: with continuous conduction each thyristor conducts for 180 degrees.

PRACTICE 42 marks · NAT

A single-phase AC voltage controller using two antiparallel thyristors supplies a 10 ohm resistor from a 230 V rms, 50 Hz source. The firing angle of each thyristor is 90 degrees. Find the power delivered to the resistor in watts.

Answer and solution

Answer: $2645$ (accepted range $\pm 5$ )

$V_{o, r m s} = V_{s} \frac{1}{π} (π - α + \frac{sin 2 α}{2})$ .
At $α = 9 0^{\circ}$ : $sin 18 0^{\circ} = 0$ , so $V_{o, r m s} = V_{s} \frac{1}{2} = 162.6$ V.
$P = \frac{V _{o, r m s}^{2}}{R} = \frac{23 0 ^{2} /2}{10} = 2645$ W.

PRACTICE 52 marks · NAT

A single-phase fully controlled bridge is fed from a 230 V rms, 50 Hz supply through a source inductance of 2 mH. The load current is constant at 20 A and the firing angle is 30 degrees. Find the average output voltage in volts, correct to one decimal place.

Answer and solution

Answer: $171.3$ (accepted range $\pm 0.5$ )

$V_{m} = 2302 = 325.27$ V and $ω L_{s} = 314.16 \times 0.002 = 0.6283 Ω$ .
Ideal output: $\frac{2 V _{m}}{π} cos 3 0^{\circ} = 207.07 \times 0.8660 = 179.33$ V.
Drop: $\frac{2 ω L _{s} I _{d c}}{π} = \frac{2 ( 0.6283 ) ( 20 )}{π} = 8.00$ V.
$V_{d c} = 179.33 - 8.00 = 171.3$ V.

PRACTICE 62 marks · NAT

A single-phase half-controlled bridge supplies a highly inductive load with constant, ripple-free current. Source inductance is negligible and the firing angle is 60 degrees. Find the input power factor, correct to two decimal places.

Answer and solution

Answer: $0.83$ (accepted range $\pm 0.01$ )

Source current: $+ I_{d c}$ from $α$ to $π$ and $- I_{d c}$ from $π + α$ to $2 π$ , so $I_{s, r m s} = I_{d c} (π - α) / π = I_{d c} 2/3 = 0.8165 I_{d c}$ .
$I_{s 1, r m s} = \frac{2 2}{π} I_{d c} cos 3 0^{\circ} = 0.9003 \times 0.8660 I_{d c} = 0.7797 I_{d c}$ .
Displacement factor $= cos (α /2) = cos 3 0^{\circ} = 0.8660$ .
$P F = \frac{I _{s 1, r m s}}{I _{s, r m s}} cos \frac{α}{2} = \frac{0.7797}{0.8165} \times 0.8660 = 0.9549 \times 0.8660 = 0.827$ .
Check: $\frac{2 ( 1 + cos 6 0 ^{\circ} )}{π ( 2 π /3 )} = \frac{2.1213}{2.5651} = 0.827$ .

PRACTICE 72 marks · NAT

A thyristor carrying a constant load current of 100 A is turned off by a resonant-pulse (current) commutation circuit with $L = 10 μ$ H and $C = 40 μ$ F, the capacitor being initially charged to 200 V. A diode across the thyristor carries the resonant current in excess of the load current. Find the circuit turn-off time in microseconds, correct to one decimal place.

Answer and solution

Answer: $52.7$ (accepted range $\pm 0.5$ )

$I_{c p} = V_{C} C / L = 200 40/10 = 400$ A, which exceeds $I_{0} = 100$ A, so commutation succeeds.
$L C = 10 \times 1 0^{- 6} \times 40 \times 1 0^{- 6} = 20 μ$ s.
$sin^{- 1} (I_{0} / I_{c p}) = sin^{- 1} (0.25) = 0.2527$ rad.
$t_{c} = L C [π - 2 (0.2527)] = 20 \times 2.6362 = 52.7 μ$ s.

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE EE

Single and three-phase configuration of uncontrolled rectifiers; Voltage and Current commutated Thyristor based converters; Bidirectional ac to dc voltage source converters

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rectifiers

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