IN THE EE SYLLABUS

Line-current harmonics, power factor and distortion factor

A diode or thyristor rectifier draws a non-sinusoidal line current, so the supply sees harmonics, and reactive power when the converter is phase-controlled, even when the load current is pure DC. This topic gives the harmonic orders of single-phase and three-phase converters, their magnitudes and phases, and the displacement factor, distortion factor and power factor of fully controlled, half-controlled and uncontrolled (diode) bridges.

6 previous-year questions · 6 solved · 6 practice questions

Setting and conventions

Take a rectifier fed from an ideal, balanced, sinusoidal AC source. The load current $I_{d}$ is ripple-free (a very large load inductance) and conduction is continuous. The source inductance is zero unless an overlap angle $μ$ is mentioned. The firing angle $α$ is measured from the natural commutation point, the instant a diode in the same position would start to conduct. $V_{m}$ is the peak of the supply phase voltage. $V_{s}$ is the rms supply voltage of a single-phase converter. $V_{LL}$ is the rms line-to-line voltage of a three-phase supply. $I_{s}$ is the rms line (supply) current. A diode (uncontrolled) rectifier behaves like the matching fully controlled converter with $α = 0$ .

Under these assumptions each thyristor (or diode) carries $I_{d}$ for a fixed part of the cycle, so the line current is a rectangular (square-type) wave. The DC side is smooth, and the AC side carries the harmonics.

AVERAGE OUTPUT VOLTAGE, SINGLE-PHASE FULLY CONTROLLED BRIDGE

V_{d} = \frac{2 2}{π} V_{s} cos α = 0.9 V_{s} cos α

Continuous conduction, zero source inductance.

AVERAGE OUTPUT VOLTAGE, THREE-PHASE FULLY CONTROLLED BRIDGE

V_{d} = \frac{3 2}{π} V_{LL} cos α = 1.35 V_{LL} cos α

Continuous conduction, zero source inductance. At $α = 0$ this is the diode bridge value $V_{d 0} = 1.35 V_{LL}$ .

RMS LINE CURRENT, THREE-PHASE BRIDGE

I_{s} = \frac{2}{3} I_{d} = 0.816 I_{d}

In each cycle a line carries $+ I_{d}$ for $12 0^{\circ}$ , nothing for $6 0^{\circ}$ , $- I_{d}$ for $12 0^{\circ}$ , and nothing for $6 0^{\circ}$ . With an ideal ripple-free $I_{d}$ this does not depend on $α$ .

Fourier series of the line current

The quasi-square line current of a six-pulse bridge has no even harmonics, because of half-wave symmetry. It also has no triplen harmonics (3rd, 9th, 15th, ...). The $12 0^{\circ}$ conduction blocks make each harmonic amplitude proportional to $sin (nπ /3)$ , which is zero for every multiple of 3. This agrees with the fact that a three-wire system gives balanced triplen (zero-sequence) currents no path. The only orders present are $n = 6 k \pm 1$ ( $k = 1, 2, 3, \dots$ ), that is $5, 7, 11, 13, 17, 19, \dots$

In general, a $p$ -pulse converter with ripple-free $I_{d}$ has characteristic line-current harmonics of order $n = k p \pm 1$ and characteristic DC-side voltage harmonics of order $k p$ . A twelve-pulse converter is two six-pulse bridges fed through star-star and star-delta transformers, with their DC outputs in series. Its supply current contains orders $11, 13, 23, 25, \dots$ , because the 5th and 7th cancel.

LINE CURRENT OF PHASE A (SIX-PULSE BRIDGE)

i_{a} (t) = \frac{2 3}{π} I_{d} [cos (ω t - α) - \frac{1}{5} cos 5 (ω t - α) + \frac{1}{7} cos 7 (ω t - α) - \frac{1}{11} cos 11 (ω t - α) + \frac{1}{13} cos 13 (ω t - α) - \dots]

Reference $v_{an} = V_{m} cos ω t$ . At $α = 0$ the positive $12 0^{\circ}$ block is centred on the positive peak of $v_{an}$ . For $α > 0$ the whole waveform is the $α = 0$ waveform delayed by $α / ω$ . The peak of the $n$ -th term is $\frac{4 I _{d}}{nπ} sin \frac{nπ}{3}$ . Zero overlap, ripple-free $I_{d}$ .

RMS FUNDAMENTAL LINE CURRENT, THREE-PHASE BRIDGE

I_{s 1} = \frac{6}{π} I_{d} = 0.78 I_{d}

The peak of the fundamental is $\frac{2 3}{π} I_{d}$ . Divide by $2$ for the rms value.

RMS OF THE N-TH HARMONIC, THREE-PHASE BRIDGE

I_{s n} = \frac{I _{s 1}}{n}, n = 6 k \pm 1

Six-pulse bridge, zero overlap, ripple-free $I_{d}$ . Relative to the fundamental the magnitude falls as $1/ n$ , and it does not depend on $α$ .

Phase. Delaying the waveform by $α / ω$ shifts the fundamental by $α$ and the $n$ -th harmonic by $n α$ , each measured in its own cycle, since the argument is $n (ω t - α)$ . So the fundamental current lags the phase voltage by $α$ . In a fully controlled bridge the harmonic magnitudes stay the same, and only their phases move with $α$ .

Order n	Rms relative to fundamental	Rms for $I_{d} = 1$ pu
1	1	0.780
5	0.200	0.156
7	0.143	0.111
11	0.091	0.071
13	0.077	0.060

LINE CURRENT, SINGLE-PHASE FULLY CONTROLLED BRIDGE

i_{s} (t) = \frac{4 I _{d}}{π} [sin (ω t - α) + \frac{1}{3} sin 3 (ω t - α) + \frac{1}{5} sin 5 (ω t - α) + \dots], I_{s n} = \frac{2 2 I _{d}}{nπ}

Reference $v_{s} = V_{m} sin ω t$ . The current is a square wave of $\pm I_{d}$ delayed by $α$ , so every odd order $n$ is present. Ripple-free $I_{d}$ , zero overlap. The magnitudes do not depend on $α$ .

LINE CURRENT, SINGLE-PHASE HALF-CONTROLLED BRIDGE

I_{s n} = \frac{2 2 I _{d}}{nπ} cos \frac{n α}{2} (n odd), I_{s} = I_{d} \frac{π - α}{π}

Half-controlled bridge (semi-converter), or a full bridge with a freewheeling diode, with ripple-free $I_{d}$ . The line current is $+ I_{d}$ from $α$ to $π$ , zero while the load freewheels, then $- I_{d}$ from $π + α$ to $2 π$ . Here the harmonic magnitudes do depend on $α$ , and the fundamental lags the supply voltage by $α /2$ .

Distortion factor, displacement factor and power factor

With a sinusoidal supply voltage, only the fundamental current carries average power. Harmonic currents add to the rms current without delivering average power, so they lower the power factor.

DISPLACEMENT FACTOR

DPF = cos ϕ_{1}

$ϕ_{1}$ is the phase angle between the supply voltage and the fundamental line current. With zero overlap, $ϕ_{1} = α$ for a fully controlled bridge and $ϕ_{1} = α /2$ for a single-phase half-controlled bridge.

DISTORTION FACTOR

DF = \frac{I _{s 1}}{I _{s}}

The ratio of the fundamental rms current to the total rms line current. DF $\leq 1$ .

EXAM TIP

Names vary. Some texts write DF for the displacement factor, call $I_{s 1} / I_{s}$ the current distortion factor, and call the THD the harmonic factor. Use the definition the question gives.

INPUT POWER FACTOR

PF = \frac{P}{S} = \frac{I _{s 1}}{I _{s}} cos ϕ_{1} = DF \times DPF

Sinusoidal supply voltage. For the three-phase case $P = 3 V_{LL} I_{s 1} cos ϕ_{1}$ and $S = 3 V_{LL} I_{s}$ . For the single-phase case $P = V_{s} I_{s 1} cos ϕ_{1}$ and $S = V_{s} I_{s}$ .

TOTAL HARMONIC DISTORTION OF THE LINE CURRENT

THD = \frac{I _{s}^{2} - I _{s 1}^{2}}{I _{s 1}} = \frac{1}{DF ^{2}} - 1

Harmonics are measured relative to the fundamental. Equivalently, $DF = 1/ 1 + THD^{2}$ .

SIX-PULSE BRIDGE, IDEAL INDUCTIVE LOAD

DF = \frac{3}{π} = 0.955, PF = \frac{3}{π} cos α, THD = \frac{π ^{2}}{9} - 1 \approx 31%

Fully controlled bridge, ripple-free $I_{d}$ , zero overlap. DF and THD do not change with $α$ . Only DPF does. A diode bridge ( $α = 0$ ) has PF $= 0.955$ .

SINGLE-PHASE HALF-CONTROLLED BRIDGE, IDEAL INDUCTIVE LOAD

DPF = cos \frac{α}{2}, PF = \frac{2 2}{π} \frac{cos ^{2} ( α /2 )}{( π - α ) / π} = \frac{2 ( 1 + cos α )}{π ( π - α )}

Ripple-free $I_{d}$ , zero overlap. At $α = 0$ this gives PF $= 0.900$ , the same as the full bridge. For $α > 0$ its PF is higher than the full bridge's $0.9 cos α$ .

EXAM TIP

For the three-phase bridge the 0.955 factor comes entirely from distortion, and the $cos α$ factor from displacement. Do not multiply by $cos α$ twice, and do not apply $3/ π$ to the DC-side power.

Effect of firing angle and overlap; reactive power

Increasing $α$ moves the whole current wave later, so the fundamental lags the voltage by more. With continuous conduction, a fully controlled converter draws lagging fundamental reactive power for every $0 < α < 18 0^{\circ}$ , in rectifier mode ( $α < 9 0^{\circ}$ ) and in inverter mode ( $α > 9 0^{\circ}$ ) alike. Active power $P = V_{d} I_{d}$ changes sign at $α = 9 0^{\circ}$ . $Q_{1} \propto sin α$ stays positive: it rises to its maximum at $α = 9 0^{\circ}$ and falls again towards $18 0^{\circ}$ .

POWER AND REACTIVE POWER, THREE-PHASE FULLY CONTROLLED BRIDGE

P = V_{d} I_{d} = V_{d 0} I_{d} cos α, Q_{1} = 3 V_{LL} I_{s 1} sin α = V_{d 0} I_{d} sin α

Zero overlap, lossless converter. $Q_{1}$ is the fundamental reactive power. With $V_{d 0} = \frac{3 2}{π} V_{LL}$ , the fundamental apparent power is $P^{2} + Q_{1}^{2} = V_{d 0} I_{d}$ .

Overlap. With a source inductance $L_{s}$ per phase, the current edges no longer jump. They ramp over the overlap angle $μ$ . This lowers the output voltage and delays the effective centre of the current wave, so the displacement angle is larger than $α$ . The sloping edges also generally reduce the magnitudes of the higher-order harmonics. The standard results are below.

OUTPUT VOLTAGE WITH OVERLAP, THREE-PHASE FULLY CONTROLLED BRIDGE

V_{d} = V_{d 0} cos α - \frac{3 ω L _{s}}{π} I_{d} = V_{d 0} \frac{cos α + cos ( α + μ )}{2}, cos α - cos (α + μ) = \frac{2 ω L _{s} I _{d}}{V _{LL}}

$L_{s}$ is the source inductance per phase, with ripple-free $I_{d}$ . For the single-phase fully controlled bridge the matching result is $V_{d} = 0.9 V_{s} cos α - \frac{2 ω L _{s}}{π} I_{d}$ .

DISPLACEMENT FACTOR WITH OVERLAP (APPROXIMATE)

cos ϕ_{1} \approx \frac{cos α + cos ( α + μ )}{2} = \frac{V _{d}}{V _{d 0}}

The ratio $V_{d} / V_{d 0}$ is approximately the displacement factor, since $P = V_{d} I_{d} \approx V_{d 0} I_{d} cos ϕ_{1}$ when $I_{s 1}$ is taken as its zero-overlap value. Use it only when a question involves overlap.

Quantity	Single-phase fully controlled bridge	Three-phase fully controlled bridge
Line current wave	Square, $\pm I_{d}$	Quasi-square, $12 0^{\circ}$ blocks
Harmonic orders	$3, 5, 7, \dots$ (all odd)	$6 k \pm 1$ only
$I_{s 1}$ (rms)	$\frac{2 2}{π} I_{d} = 0.90 I_{d}$	$\frac{6}{π} I_{d} = 0.78 I_{d}$
$I_{s n}$	$I_{s 1} / n$	$I_{s 1} / n$
Total rms $I_{s}$	$I_{d}$	$2/3 I_{d}$
DF	$\frac{2 2}{π} = 0.900$	$\frac{3}{π} = 0.955$
THD	$\approx 48%$	$\approx 31%$
PF	$0.900 cos α$	$0.955 cos α$
$P$	$0.9 V_{s} I_{d} cos α$	$V_{d 0} I_{d} cos α$
$Q_{1}$	$0.9 V_{s} I_{d} sin α$	$V_{d 0} I_{d} sin α$

Worked examples

WORKED EXAMPLE

A three-phase fully controlled bridge is fed from a 400 V (line-to-line, rms), 50 Hz supply with negligible source inductance. The load current is a constant 20 A and $α = 3 0^{\circ}$ . Find the rms fundamental line current, the 5th and 7th harmonic currents, the rms line current, the input power factor, and the real and reactive power.

Solution

$I_{s 1} = \frac{6}{π} \times 20 = 0.7797 \times 20 = 15.59$ A.
$I_{s 5} = 15.59/5 = 3.12$ A and $I_{s 7} = 15.59/7 = 2.23$ A.
$I_{s} = 2/3 \times 20 = 16.33$ A.
$DF = 3/ π = 0.955$ and $DPF = cos 3 0^{\circ} = 0.866$ , so $PF = 0.955 \times 0.866 = 0.827$ .
$V_{d 0} = 1.3505 \times 400 = 540.2$ V. $P = V_{d 0} I_{d} cos α = 540.2 \times 20 \times 0.866 = 9356$ W.
$Q_{1} = V_{d 0} I_{d} sin α = 540.2 \times 20 \times 0.5 = 5402$ var.
Check: $S = 3 \times 400 \times 16.33 = 11314$ VA and $P / S = 0.827$ .

Answer: I_s1 = 15.59 A, I_5 = 3.12 A, I_7 = 2.23 A, I_s = 16.33 A, PF = 0.827 (lagging), P ≈ 9.36 kW, Q1 ≈ 5.40 kvar.

WORKED EXAMPLE

A single-phase fully controlled bridge supplies a ripple-free load current of 10 A with $α = 6 0^{\circ}$ and zero source inductance. Find the 3rd harmonic current, the total harmonic distortion, and the input power factor.

Solution

The line current is a square wave of amplitude $10$ A, so $I_{s} = 10$ A.
$I_{s 1} = \frac{2 2}{π} \times 10 = 9.003$ A, and $I_{s 3} = 9.003/3 = 3.00$ A.
$DF = I_{s 1} / I_{s} = 0.9003$ .
$THD = 1/ DF^{2} - 1 = π^{2} /8 - 1 = 0.2337 = 0.483$ .
$PF = DF cos α = 0.9003 \times 0.5 = 0.450$ .

Answer: I_3 = 3.00 A, THD = 48.3 %, PF = 0.450 (lagging).

Traps and quick checks

EXAM TIP

Orders present: a single-phase bridge gives all odd harmonics, and a three-phase six-pulse bridge gives only $6 k \pm 1$ . With a balanced supply, triplen harmonics do not appear in the line current of the three-wire three-phase bridge.

EXAM TIP

In a fully controlled bridge with ideal ripple-free $I_{d}$ and no overlap, the harmonic magnitudes do not depend on $α$ . Only their phases change, along with the fundamental's displacement. So DF and THD are fixed and PF falls as $cos α$ .

EXAM TIP

In a half-controlled bridge the conduction width $π - α$ changes with $α$ , so the harmonic magnitudes, DF and THD all change with $α$ , and DPF $= cos (α /2)$ , not $cos α$ .

EXAM TIP

Check which factor a question asks for. DF is $I_{s 1} / I_{s}$ , DPF is $cos ϕ_{1}$ , and PF is their product. 'Power factor of the AC supply' means the full PF, including distortion.

EXAM TIP

A quick consistency test for the three-phase bridge: $PF = P / S = V_{d} I_{d} / (3 V_{LL} I_{s})$ . When overlap is neglected, it must equal $0.955 cos α$ .

PREVIOUS-YEAR QUESTIONS

6 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 2025 · GATE EE 2024 · GATE EE 2022 · GATE EE 2020 · GATE EE 2018

GATE 2025 EE Q.27

GATE 2025 · EEQ.27 · 1 mark · MCQ

The input voltage $v (t)$ and current $i (t)$ of a converter are given by, $v (t) = 300 sin (ω t)$ V, $i (t) = 10 sin (ω t - \frac{π}{6}) + 2 sin (3 ω t + \frac{π}{6}) + sin (5 ω t + \frac{π}{2})$ A, where $ω = 2 π \times 50$ rad/s. The input power factor of the converter is closest to

(A)
0.845
(B)
0.867
(C)
0.887
(D)
1.0

Answer and solution

GATE 2025 EE Q.27: answer and solution

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GATE 2024 EE Q.62

GATE 2024 · EEQ.62 · 2 marks · NAT

A single-phase half-controlled bridge converter supplies an inductive load with ripple free load current. The triggering angle of the converter is 60°. The ratio of the rms value of the fundamental component of the input current to the rms value of the total input current of the bridge is __________ (rounded off to 3 decimal places).

Answer and solution

GATE 2024 EE Q.62: answer and solution

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GATE 2022 EE Q.33

GATE 2022 · EEQ.33 · 1 mark · NAT

The voltage at the input of an AC-DC rectifier is given by $v (t) = 2302 sin ω t$ where $ω = 2 π \times 50$ rad/s. The input current drawn by the rectifier is given by $i (t) = 10 sin (ω t - \frac{π}{3}) + 4 sin (3 ω t - \frac{π}{6}) + 3 sin (5 ω t - \frac{π}{3})$ . The input power factor, (rounded off to two decimal places), is, ________ lag.

Answer and solution

GATE 2022 EE Q.33: answer and solution

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GATE 2022 EE Q.59

GATE 2022 · EEQ.59 · 2 marks · NAT

For the ideal AC-DC rectifier circuit shown in the figure below, the load current magnitude is $I_{d c} = 15$ A and is ripple free. The thyristors are fired with a delay angle of $4 5^{\circ}$ . The amplitude of the fundamental component of the source current, in amperes, is ______. (round off to two decimal places)

GATE 2022 EE question 59: Single-phase bridge fed from an AC source. — Figure from the GATE 2022 EE paper. It is part of the question.

Answer and solution

GATE 2022 EE Q.59: answer and solution

GATE 2020 EE Q.11

GATE 2020 · EEQ.11 · 1 mark · MCQ

A single-phase, full-bridge diode rectifier fed from a 230 V, 50 Hz sinusoidal source supplies a series combination of finite resistance, R, and a very large inductance, L. The two most dominant frequency components in the source current are:

(A)
50 Hz, 0 Hz
(B)
50 Hz, 100 Hz
(C)
50 Hz, 150 Hz
(D)
150 Hz, 250 Hz

Answer and solution

GATE 2020 EE Q.11: answer and solution

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GATE 2018 EE Q.23

GATE 2018 · EEQ.23 · 1 mark · NAT

The waveform of the current drawn by a semi-converter from a sinusoidal AC voltage source is shown in the figure. If $I_{0} = 20$ A, the rms value of fundamental component of the current is ______ A (up to 2 decimal places).

GATE 2018 EE question 23: Horizontal axis ω t, vertical axis 'voltage and current'. — Figure from the GATE 2018 EE paper. It is part of the question.

Answer and solution

GATE 2018 EE Q.23: answer and solution

PRACTICE

6 questions to test yourself

PRACTICE 11 mark · MCQ

A three-phase fully controlled bridge rectifier feeds a highly inductive load (ripple-free $I_{d}$ ) from a balanced supply. The source inductance is negligible. Which set lists the lowest four harmonic orders present in the line current?

(A)
3, 5, 7, 9
(B)
5, 7, 11, 13
(C)
2, 4, 6, 8
(D)
3, 6, 9, 12

Answer and solution

Answer: (B)

Half-wave symmetry rules out even harmonics.
The $12 0^{\circ}$ blocks make the amplitude proportional to $sin (nπ /3)$ , which is zero for multiples of 3, so there are no triplen harmonics.
The orders present are $6 k \pm 1$ : 5, 7, 11, 13, ...

PRACTICE 21 mark · NAT

A three-phase diode bridge rectifier supplies a ripple-free load current of 30 A. Source inductance is negligible. Find the rms value of the 11th harmonic of the line current, in amperes. Give your answer correct to two decimal places.

Answer and solution

Answer: $2.13$ (accepted range $\pm 0.02$ )

$I_{s 1} = \frac{6}{π} \times 30 = 0.7797 \times 30 = 23.39$ A.
$I_{s 11} = I_{s 1} /11 = 23.39/11 = 2.13$ A.

PRACTICE 32 marks · NAT

A three-phase fully controlled bridge converter operates with a firing angle of $4 0^{\circ}$ and a ripple-free load current. Source inductance is negligible and the supply is sinusoidal. Find the power factor at the AC input. Give your answer correct to three decimal places.

Answer and solution

Answer: $0.731$ (accepted range $\pm 0.005$ )

$DF = 3/ π = 0.9549$ .
$DPF = cos 4 0^{\circ} = 0.7660$ .
$PF = 0.9549 \times 0.7660 = 0.7315 \approx 0.731$ .

PRACTICE 42 marks · NAT

A single-phase fully controlled bridge is supplied from a 230 V rms, 50 Hz source with negligible source inductance. The load current is a constant 10 A and the firing angle is $4 5^{\circ}$ . Find the reactive power drawn at the fundamental frequency, in var. Give your answer to the nearest whole number.

Answer and solution

Answer: $1464$ (accepted range $\pm 5$ )

The line current is a square wave of amplitude 10 A, so $I_{s 1} = \frac{2 2}{π} \times 10 = 9.003$ A.
The fundamental lags the voltage by $ϕ_{1} = α = 4 5^{\circ}$ .
$Q_{1} = V_{s} I_{s 1} sin ϕ_{1} = 230 \times 9.003 \times 0.7071 = 1464$ var.

PRACTICE 52 marks · MSQ

For a three-phase fully controlled bridge with a balanced supply, a ripple-free load current and negligible source inductance, which of the following statements are correct?

(A)
The line current contains no 3rd or 9th harmonic.
(B)
The fundamental line current lags the phase voltage by the firing angle $α$ .
(C)
The distortion factor decreases as $α$ increases.
(D)
The total harmonic distortion of the line current is about 31% for every $α$ .

Answer and solution

Answer: (A), (B), (D)

A is true. The $12 0^{\circ}$ blocks give zero amplitude for every multiple of 3.
B is true. The current wave is the $α = 0$ wave delayed by $α$ , so $ϕ_{1} = α$ .
C is false. DF $= 3/ π$ for every $α$ , because the harmonic magnitudes do not change.
D is true. THD $= π^{2} /9 - 1 \approx 0.311$ for all $α$ .

PRACTICE 62 marks · NAT

A single-phase half-controlled bridge (semi-converter) supplies a ripple-free load current from a sinusoidal source with negligible source inductance. The firing angle is $9 0^{\circ}$ . Find the input power factor. Give your answer correct to three decimal places.

Answer and solution

Answer: $0.637$ (accepted range $\pm 0.005$ )

The line current is $\pm I_{d}$ for $π - α = 9 0^{\circ}$ in each half-cycle, so $I_{s} = I_{d} (π - α) / π = I_{d} / 2 = 0.7071 I_{d}$ .
$I_{s 1} = \frac{2 2}{π} I_{d} cos \frac{α}{2} = 0.9003 \times 0.7071 I_{d} = 0.6366 I_{d}$ .
$DF = 0.6366/0.7071 = 0.9003$ , and $DPF = cos (α /2) = cos 4 5^{\circ} = 0.7071$ .
$PF = 0.9003 \times 0.7071 = 0.6366 = 2/ π \approx 0.637$ .

OFFICIAL GATE 2027 SYLLABUS WORDING

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GATE EE

Magnitude and Phase of line current harmonics for uncontrolled and thyristor based converters; Power factor and Distortion Factor of AC to DC converters

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