Phasor diagram
Learn what each arrow represents, how a fault, a reversed CT or a Dyn11 transformer reads, and use the Advanced mode with your real measurements: symmetrical components, powers, impedance per loop and load-check diagnosis.
IEEE C37.118.1§ 4.1phasor definition, RMS valueIEC 60050-131IEV 131-11-26, phasorIEC 60076-1clock numberIEC 60909-0§ 5.3.2symmetrical components
Results
—
—
— °
—
—
— %
Charts
How it was calculated and standards
Load-check diagnosis
Verified rules for commissioning. It never replaces the engineer's judgement nor gives setting values.
Phasors, sequences and powers
| Phasor | Polar | Rectangular |
|---|
Powers and impedances: the three voltages and the three currents of side 1 are needed.
| Phase | P | Q | S | PF |
|---|
Angle of each phasor from another
Row from column: positive, the row leads.
| Loop | Z | R + jX |
|---|
Relay did not operate as expected? We review its settings and correct them remotely.
What it is for
A phasor diagram sums up in one picture how the voltages and currents of a three-phase circuit are: whether they are balanced, in which order they turn, whether energy goes in or out and what kind of fault there is. It is the tool used to review a commissioning and to analyse a trip.
- Essential mode: type the voltages and currents of the three phases (magnitude and angle) and see the diagram, the sine waves, the rotation, P, Q, the power factor, the unbalance and the load-check diagnosis.
- Advanced mode: your measurements in primary or secondary, one or two sides (87T, 87L) or pre-fault and fault; symmetrical components, powers, apparent impedance per loop, angles between phasors and the diagnosis of the load and polarity check. Exports to PNG, SVG, PDF and CSV.
- Learn to read phasor diagrams: a separate option, with the button in the input panel. Guided lessons from what a phasor is, the rectangular form, the j operator and adding and subtracting phasors to how faults, a reversed CT, a Dyn11 transformer or an open delta look, plus challenges to check what you learned. Each lesson has its own link.
How to read it
Each rule of the lessons and of the diagnosis, with its source.
IEEE C37.118.1§ 4.1phasor definitionIEC 60050-131IEV 131-11-26
What a phasor is. The wave x(t) = Xm·cos(ωt + φ) is represented by the complex number X = (Xm/√2)∠φ: the length is the RMS value and the angle, the initial phase. The arrow turns counter-clockwise at ω; its projection on the real axis, multiplied by √2, is the instantaneous value. That is why a phasor diagram is the snapshot of the instant ωt = 0.
IEC 60050-131IEV 131-11-26IEEE C37.118.1§ 4.1RMS value
Rectangular form and the j operator. The same phasor is written |X|∠θ (polar) or x + jy (rectangular), with x = |X|·cos θ, y = |X|·sin θ, |X| = √(x² + y²) and θ = atan2(y, x); atan2 looks at the signs of x and y, so it never confuses opposite phasors. Multiplying by j = 1∠90° turns the phasor 90° without changing its length (j² = −1), and by −j turns it −90°. The length is the RMS value (IEEE C37.118.1); the peak of the wave is √2 times larger (Blackburn, Symmetrical Components for Power Systems Engineering).
IEC 60909-0§ 5.3.2operator a
The a operator. a = 1∠120° = −1/2 + j√3/2 and a² = 1∠240°: a³ = 1 and 1 + a + a² = 0. IEC 60909-0 writes the symmetrical components with it: in ABC, VB = a²·VA and VC = a·VA.
IEC 60050-131IEV 131-11-26
Sum and difference. Two phasors of the same frequency are added part by part in rectangular form, which is the parallelogram rule; the difference X1 − X2 is the sum with −X2, the other diagonal. The instantaneous value of the sum is the sum of the instantaneous values. Phasors of different frequencies cannot be added this way, because the angle between them changes at every instant. This gives the line-to-line voltage VAB = VA − VB and the residual current 3I0 = IA + IB + IC (Blackburn, Protective Relaying, «Phasors and Polarity»).
IEC 60909-0§ 5.3.2operator a
Phase rotation. In ABC, VB = a²·VA (lags by 120°) and VC = a·VA, with a = 1∠120°. In ACB, B and C swap. The line voltage VAB = VA − VB is √3 times the phase voltage and leads by 30° in ABC (lags by 30° in ACB): Glover, Sarma and Overbye, Power System Analysis and Design, ch. 2.
IEC 60050-131IEV 131-11-39IEC 62053-23four quadrantsIEC 60375sign conventions
Current versus voltage. Complex power is S = V·I*. With the load convention (current arrow entering the object), a positive φ = ∠V − ∠I is an inductive load (Q > 0) and a negative one is capacitive. Quadrant I: absorbs P and Q; II: delivers P and absorbs Q; III: delivers P and Q; IV: absorbs P and delivers Q. The generator convention reverses the arrow and, with it, the signs (IEEE 1459 for the definitions of P, Q, S and PF).
IEC 60076-1clock numberIEEE C57.12.00angular displacement
The transformer clock. The HV phase-A voltage goes at 12 o'clock and each hour is 30°: the number says how much the LV lags (Dyn1: 30°; Dyn11: 330°, that is, it leads by 30°). IEEE C57.12.00 sets that in Y-Δ and Δ-Y the HV leads the LV by 30°. After compensating the group, the currents of both sides of an 87T must be equal and at 180° (Costello, SEL).
SELRelay-Assisted Commissioning
- Load check: with balanced load you detect a CT connected to the wrong tap, crossed phases and reversed polarity by comparing the angle of each phase with a reference (Labuschagne and Fischer, «Relay-Assisted Commissioning», SEL, WPRC 2005).
- Reversed CT: that current appears turned 180° (IB at 60° instead of −120°) and the residual equals twice the phase current; in the test by Peterson, Scharlach and Baltazar (IEEE PPIC 2009), reversing IA gave 3I2 ≈ 212 A and IG ≈ 213 A.
- Crossed phases: the currents turn the other way and almost all the load shows as I2; the SEL check declares crossed phases when I1 is below 10 % of I2.
- Before handover: verify with the line in service the magnitude and angle of voltages and currents (NERC, Lessons Learned LL20200702), using the relay metering, its events or synchrophasors (Zimmerman, SEL, 2008).
Faults. Single line to ground: I0 = I1 = I2 and 3I0 appears. Phase-phase: I2 = −I1, no I0. Two phases to ground: the healthy phase carries no fault current and there is I0. Three-phase: positive sequence only. Open phase without neutral: I2 equal to I1 (Blackburn, Symmetrical Components for Power Systems Engineering, and Protective Relaying, chapter «Phasors and Polarity»).
Open delta and broken delta. Two VTs A-B and B-C give the three line-to-line voltages (VCA = −(VAB + VBC)) but not the phase voltages or V0. Three VTs with the secondary in delta open at one corner give 3V0 for the 59N (Basler, note PC-59N01).
Motor starting. Locked-rotor current close to 6 times rated with a power factor of 0.15 to 0.20 (IEEE 399, default values of motor-starting studies): a large, strongly lagging and balanced current.
How it is calculated
The total power factor uses the vector apparent power |ΣS| (IEEE 1459).
IEC 60909-0§ 5.3.2symmetrical components
X0 = (Xa + Xb + Xc)/3 · X1 = (Xa + a·Xb + a²·Xc)/3 · X2 = (Xa + a²·Xb + a·Xc)/3The same functions as the site's symmetrical components calculator. Voltage unbalance u2 = V2/V1 (IEC 61000-4-30; 2 % compatibility level in IEC 61000-2-2) and, for reference, the NEMA MG 1 definition: maximum deviation from the average divided by the average.
Six loops: AG, BG, CG, AB, BC and CA (Kockott, Line Distance Protection Fundamentals, ABB, 2015; Dase et al., SEL; Ziegler, Numerical Distance Protection, 3.2). Primary to secondary: Zsec = Zpri·CTR/VTR.
Faults in the lessons. Sequence networks with the source behind the relay and the fault at the end of the line section, no pre-fault load and Z2 = Z1: single line to ground I0 = I1 = I2 = E/(2Z1 + Z0 + 3Rf); phase-phase I1 = −I2 = E/(2Z1 + Rf); two phases to ground I1 = E/(Z1 + Z2‖(Z0 + 3Rf)); three-phase I1 = E/(Z1 + Rf). The tests check that every faulted loop measures the line impedance.
87T. Each side is taken to primary amperes with its CT ratio, multiplied by its rated voltage (common MVA base) and the vector-group compensation matrix of the 87T compensation tool is applied. Differential = |I1 + I2| compensated, in % of the larger one.
Diagnosis thresholds
- Balanced set: I2/I1 and I0/I1 below 10 % (this tool's convention, equal to the SEL crossed-phase criterion).
- One phase angle out of line: the three voltage-current angles spread over more than 10° (this tool's convention).
- Differential after compensation above 10 % of the through current, or a leftover shift above 20° (this tool's convention for the load check; the relay threshold is set by its settings, not by this tool).
- Lost voltage: a phase below 50 % of the largest; open current: below 10 % (this tool's convention). Voltage unbalance: 2 % (IEC 61000-2-2).
Limits of the calculation
- It works with fundamental-frequency phasors taken at the same instant and with the same reference. In an oscillography with DC offset or harmonics, use the filtered phasors of the relay or of the COMTRADE viewer.
- The diagnosis recognises wiring patterns with reasonably balanced load; with very unbalanced or near-zero load the rules lose strength and it says so.
- The faults in the lessons are ideal (no pre-fault load, simple radial network): they are for learning to read them, not a short-circuit study.
- It never gives setting values: the diagnosis thresholds are for review, not for protection.
Frequently asked questions
How do you read a phasor diagram?
Each arrow is a voltage or a current: its length is the RMS value and its angle, the shift from a reference (usually VA at 0°). Angles grow counter-clockwise and the arrow that is ahead in that rotation leads. You read it by comparing: three equal voltages at 120°, each current close to its voltage and the angle between them according to the type of load.
How are two phasors added?
Convert them to rectangular form, add the real parts and the imaginary parts separately and take the result back to polar form: magnitude √(x² + y²) and angle atan2(y, x). That is the parallelogram rule. For example, 80∠0° + 60∠90° = 80 + j60 = 100∠36.87°, not 140. It only holds for phasors of the same frequency.
Why does the angle between voltage and current show the direction of power?
With the load convention, complex power is S = V·I*. If the current is within 90° of its voltage, active power enters the measured object; if it is more than 90° away, it leaves. The sign of the reactive power says whether the current lags (inductive) or leads (capacitive).
How do you detect a CT with reversed polarity in the load check?
The current of that phase appears turned 180° from where it should be, close to the opposite voltage, and with balanced load the residual current equals twice the phase current. Turning that current by 180° makes the set balanced again: that is how this tool identifies it.
We set and test your relays
If your load check shows a reversed polarity, crossed phases or a differential that does not close, we review it with you. We set and verify your protections remotely, and on site we do the secondary injection and the load check.