Phase-sequence indicator uses few passive components

In a three-phase ac system, a power source with three wires delivers ac potentials of equal frequency and amplitudes with respect to a zero-potential wire, each shifted in phase by 120° from one wire to the next. Two possibilities exist for establishing a phase sequence. In the first, voltage on the second wire shifts by 120° relative to the first, and, in the second, a –120° shift occurs with respect to the first wire. Phase order determines the direction of rotation of three-phase ac motors and affects other equipment that requires the correct phase sequence: a positive 120° shift. You can use a few low-cost passive components to build a phase-sequence indicator.

Figure 1 shows a conceptual circuit that can detect both phase sequences. For certain component values, the following conditions apply: The voltages across R₁ and C₂ are equal—that is, their magnitudes and phases are the same—only when V_S2 occurs exactly 120° ahead of V_S1, which indicates the correct phase sequence. In this case, the voltage between points A and B is zero. Conversely, the voltages across C₂ and R₃ are equal only when V_S2 is ahead of V_S3 by 120°, which corresponds to a reversed sequence.

Referring to the phasor diagram in Figure 2, when the voltages across R₁ and C₂ are equal, V_C1=–V_R2, V_C1+V_R1=V_S1, and V_C2+V_R2=V_S2. The following equations satisfy these conditions: |V_R1| = |V_C2| =(½)| V_S2|=(½)|V_S1|, and |V_C1| = |V_R2| =cos(30°)|V_S1| =cos(30°)|V_S2|. You calculate the component values by solving the following equations: |X_C1| = tan(60°)×R₁= and R₂=tan(60°)×|X_C2|, where X_C=–j[1/(2π×f×C)], and f represents the frequency of the V_S voltages.

Also, to ensure detection of a reversed phase sequence, C₁=C₃, and R₁=R₃; that is, the components in the third branch are identical to those in the first branch. The phase-sequence-detection circuit in Figure 3 eliminates the requirement for an accessible ground wire by adding resistors R₄ and R₅ that connect in parallel with the first and third branches. Eliminating the ground-wire requirement also dictates a ratio between |X_C1+R₁| and |X_C2+R₂|. For no current to flow to ground from Node G, the sum of currents in the branches must equal zero, and, if you disconnect Node G from ground, its potential with respect to ground is also zero.

A vector diagram of the currents shows that adding two currents, each with magnitudes equal to I₃ and the same phases as V_S1 and V_S3, produces a summed current with the same magnitude and phase as I₃; therefore, the total current at Node G is zero: I₁+I₂+I₃+I₁'+I₃'=I₁+I₂+2×I₃=0. To make the sum of the currents equal zero, R₄=R₅=|R₁+X_C1| = |R₁–j[1/(2π×f×C₁)]|. The two LEDs in Figure 3 indicate correct or reversed-phase sequence. When LED₂ lights and LED₁ remains dark, the voltage between nodes A and B is 0V, which corresponds to a correct phase sequence. A reversed-phase sequence lights LED₁ while LED₂ remains dark. The diodes connected in parallel with the LEDs protect against exceeding the LEDs' reverse-breakdown voltages, and resistors R₆ and R₇ limit forward currents through the LEDs. For greater sensitivity, you can replace the LEDs with high-input-impedance ac-detector circuits.

The circuit's final version includes indicators that show whether all three phases carry voltage. In the circuit in Figure 3, a phase that carries 0V lights both LEDs. Depending on your application, you can connect voltage-detection circuits comprising LEDs and protection diodes in series with current-limiting resistors between V_S1, V_S2, and V_S3 and Node G. You can also use low-wattage neon lamps with appropriate series-current-limiting resistors.

When selecting components, ensure that their values conform to the following proportions. For an arbitrarily chosen value for C₁, R₁=R₂=R₃=1/(2π×f×C₁×tan(60°)), C₁=C₃, C₂=3C₁, and R₄=R₅=2×R₁. When you select a value for C₁, the currents through the detection circuitry should be significantly lower than the currents through the branches, which excludes arbitrarily low values for C₁.