Type to search, in English or Thai

  1. Home
  2. English lessons
  3. AC Motor Control

// Chapter 21 · AC Motor Control · electrician → student → engineer

AC Motor Control with Magnetic Contactors and VFDs

Three-phase induction motors use more than half of industrial electricity. This course goes from how the motor turns, through reading the nameplate, connecting star or delta, choosing contactors and overloads and reading drawings, to designing a star–delta starter and setting up a VFD. Every example uses the real 230/400 V 50 Hz supply of Thailand, Laos, Cambodia, Vietnam, Myanmar, Malaysia and Singapore. There are formulas with citable IDs and worked examples.

By Ajarn ModUpdated
In 30 seconds
  • An induction motor turns slightly slower than its field: ns=120f/pn_s = 120f/p = 1500 rpm (4-pole, 50 Hz), actual ≈ 1470 rpm
  • Direct-on-line starting draws 6–8 × rated current, so voltage dip and the starting method matter
  • On 400 V: a 400/690 V motor runs in Δ and can be star–delta started · a 230/400 V motor runs in Y only
  • Star–delta cuts current and torque to 1/3 · overload at 0.58 In · needs interlocks and a dead time
  • A VFD varies f and V together (V/f): soft start, full torque and pump/fan energy savings

Learning path by level

Level 1 · electrician · vocational

Wire it right and safely

Nameplates, Y/Δ terminals, panel devices, self-holding circuits, reversing, star–delta

Connections → Devices → DOL → Star–delta

Level 2 · technician · supervisor

Select devices and troubleshoot

AC-3 contactor size, overload setting, cable size, voltage drop, VFDs and fault tables

Drawings → VFD setup → Design · faults

Level 3 · engineering student · engineer

Understand it with mathematics

Equivalent circuit, slip, torque T∝V2T \propto V^2, run-up dynamics J dω/dt=Tm−TLJ\,d\omega/dt = T_m - T_L, V/f and the affinity laws

Induction motor → Starting methods compared → VFD principles

Try it: start a 7.5 kW motor four ways

Pick a tab, press START and watch the plot: blue is line current, green is speed and orange is torque. Start direct-on-line first, then switch to star–delta and notice the starting current falling to about 1/3 and the current peak at the change to delta. Then raise the load until star can no longer accelerate it, and open the full simulator with worksheets.

Key equations used throughout

H1 Synchronous speed & slip

ns=120fps=ns−nnsn_s = \frac{120 f}{p} \qquad s = \frac{n_s - n}{n_s}

H2 Motor rated current

In=Pout3 VL η cos⁡φI_n = \frac{P_{out}}{\sqrt{3}\, V_L\, \eta\, \cos\varphi}

H3 Star–delta reduction

IL,Y=13IL,ΔTY=13TΔIr,OL=In3≈0.58 InI_{L,Y} = \tfrac{1}{3} I_{L,\Delta} \qquad T_Y = \tfrac{1}{3} T_\Delta \qquad I_{r,OL} = \frac{I_n}{\sqrt{3}} \approx 0.58\, I_n

H4 V/f and affinity laws

Vf=40050=8 V/HzP2P1=(n2n1)3\frac{V}{f} = \frac{400}{50} = 8\ \text{V/Hz} \qquad \frac{P_2}{P_1} = \left(\frac{n_2}{n_1}\right)^3

The supply used in this course: 230/400 V 50 Hz

All seven countries use 50 Hz three-phase four-wire systems with line voltages of 380–415 V, following IEC 60038 (230/400 V). Motor-control gear rated 380–415 V 50 Hz is therefore usable across the region. What differs is the declared site voltage, the earthing system and the utility's connection rules — details in 230/400 V 50 Hz supply.

CountryPhase / line voltage found on siteFrequencyNotes for motor work
Thailand230 / 400 V (formerly 220/380 V)50 HzMEA/PEA · TN-C-S · EIT wiring code
Laos230 / 400 V (220/380 V also found)50 HzGrid interconnected with Thailand, IEC equipment
Cambodia230 / 400 V (220/380 V also found)50 HzMeasure the real voltage before setting a VFD
Vietnam220 / 380 V50 Hz400 V motors work; 5 % undervoltage raises current slightly
Myanmar230 / 400 V50 HzVoltage varies widely: use voltage and phase relays
Malaysia230 / 400 V (formerly 240/415 V)50 HzMuch older 415 V-rated equipment still in service
Singapore230 / 400 V50 HzInstallation to SS 638 (based on IEC 60364)

“The same standard” means the same voltage level and frequency so equipment is interchangeable — not that every site has the same voltage. Measure the line voltage before setting overloads and drives.

How to choose a starting method

MethodStarting currentStarting torqueSpeed controlGood for
Direct-on-line6–8 InHighest (1.5–2.5 Tn)NoSmall motors, stiff supply, heavy starts
Star–delta≈ 2–2.7 In1/3 of DOLNoPumps, fans, compressors starting unloaded
Soft starter2–4 In (set)Falls as (I/Is)²NoPumps needing a soft stop (water hammer), conveyors
VFD≤ 1–1.5 InUp to 1.5 TnYesEnergy-saving pumps and fans, conveyors, process machines

Every page in this course

Part 3

Starter & Control Circuits

DOL, reversing, a detailed star-delta starter, soft starters and field-proven applied circuits

21.6 Direct-on-line starter: power circuit, self-holding control circuit, two-location control, jogging and step-by-step device selection Direct-on-line starter Self-holding (latching) contact, why STOP is a normally-closed button in series, multi-location control, jog, pilot lights, and selecting the MPCB, contactor and cable for a 7.5 kW 400 V motor Open →Formulas L1, L2, … 21.7 Three-phase motor reversing circuits: swapping two phases, electrical, mechanical and push-button interlocks, and automatic reversing with limit switches Three-phase motor reversing circuits Why swapping L1 and L3 reverses rotation, three layers of interlock against a phase-to-phase short, reversing via STOP vs direct, roller doors, hoists, conveyors and limit switches Open →Formulas R1, R2, … 21.8 The star–delta starter in detail: power and control circuits, 0.58·In contactor sizing, overload setting, timer and changeover time The star–delta starter in detail Current and torque reduced to 1/3, contactors KM1 KM2 KM3, overload at 0.58·In, timer setting, Y–Δ dead time, transition current peaks, phase order on the delta side, closed-transition starters and six-wire cabling Open →Formulas Y1, Y2, … 21.9 Star–Delta Analysis: Timer Setting, Star or Delta Operation, Currents, Energy and Numerical Research Star–Delta Analysis Timer from the mechanical time constant τm = J·ωs/Tn, dead time and residual voltage, the lagging KM3 order, closed transition, I²t, rotor heat, voltage dip, running in star at light load, case studies, past, present and future, and a dynamic simulation lab Open →Formulas E1, E2, … 21.10 Soft starters, autotransformer starters and a comparison of motor starting methods: current, torque, voltage dip and cost Soft starters, autotransformer starters and a comparison of motor starting methods Thyristor phase-angle control, voltage ramp, current limit, torque control, bypass contactor, soft stop against water hammer, autotransformer starting and a starting-method selection table Open →Formulas T1, T2, … 21.11 Applied motor-control circuits from real plants: sequence starting, float-switch pumps, duty/standby alternation, pressure-switch compressors, two-speed motors and phase-failure relays Applied motor-control circuits from real plants Timed sequence starting, conveyor chains, float or electrode pump control, duty/standby alternation, Hand-Off-Auto, compressors, Dahlander two-speed motors and phase-loss/phase-reversal relays Open →Formulas A1, A2, …

Search every formula in one place

Prefixes: H overview · M motor · S supply · C connections · D devices · G drawings · L DOL · R reversing · Y star–delta · T soft starters · A applied circuits · V VFD · P VFD applications · K design

  1. H1Synchronous speed & slip
    ns=120fps=ns−nnsn_s = \frac{120 f}{p} \qquad s = \frac{n_s - n}{n_s}
  2. H2Motor rated current
    In=Pout3 VL η cos⁡φI_n = \frac{P_{out}}{\sqrt{3}\, V_L\, \eta\, \cos\varphi}
  3. H3Star–delta reduction
    IL,Y=13IL,ΔTY=13TΔIr,OL=In3≈0.58 InI_{L,Y} = \tfrac{1}{3} I_{L,\Delta} \qquad T_Y = \tfrac{1}{3} T_\Delta \qquad I_{r,OL} = \frac{I_n}{\sqrt{3}} \approx 0.58\, I_n
  4. H4V/f and affinity laws
    Vf=40050=8 V/HzP2P1=(n2n1)3\frac{V}{f} = \frac{400}{50} = 8\ \text{V/Hz} \qquad \frac{P_2}{P_1} = \left(\frac{n_2}{n_1}\right)^3
  5. M1Synchronous speed
    ns=120 fp  rpm(p=number of poles)n_s = \frac{120\, f}{p}\ \ \text{rpm} \qquad (p = \text{number of poles})
  6. M2Slip
    s=ns−nnsn=ns(1−s)s = \frac{n_s - n}{n_s} \qquad n = n_s(1 - s)
  7. M3Rotor frequency
    fr=s ff_r = s\, f
  8. M4Torque from power and speed
    T=Pω=60 P2πn    ⇒    T [N⋅m]=9550 P [kW]n [rpm]T = \frac{P}{\omega} = \frac{60\,P}{2\pi n} \;\;\Rightarrow\;\; T\,[\text{N}{\cdot}\text{m}] = \frac{9550\, P\,[\text{kW}]}{n\,[\text{rpm}]}
  9. M5Power split in the rotor
    Pag=3I2′2R2′sPcu2=s PagPmech=(1−s) PagT=PagωsP_{ag} = 3 I_2'^2 \frac{R_2'}{s} \qquad P_{cu2} = s\,P_{ag} \qquad P_{mech} = (1 - s)\,P_{ag} \qquad T = \frac{P_{ag}}{\omega_s}
  10. M6Rated current
    In=Pout3 VL η cos⁡φPin=PoutηI_n = \frac{P_{out}}{\sqrt{3}\,V_L\,\eta\,\cos\varphi} \qquad P_{in} = \frac{P_{out}}{\eta}
  11. M7Starting current
    Ist=(IaIn)InIaIn≈6–8I_{st} = \left(\frac{I_a}{I_n}\right) I_n \qquad \frac{I_a}{I_n} \approx 6\text{–}8
  12. M8Torque versus voltage
    T(s)∝V2T2T1=(V2V1)2I(s)∝VT(s) \propto V^2 \qquad \frac{T_2}{T_1} = \left(\frac{V_2}{V_1}\right)^2 \qquad I(s) \propto V
  13. S1Line and phase voltage
    VL=3 Vph230×3=398≈400 VVpeak=2 VrmsV_L = \sqrt{3}\,V_{ph} \qquad 230 \times \sqrt3 = 398 \approx 400\ \text{V} \qquad V_{peak} = \sqrt2\, V_{rms}
  14. S2Three-phase power
    P=3 VLILcos⁡φS=3 VLILP = \sqrt{3}\,V_L I_L \cos\varphi \qquad S = \sqrt{3}\,V_L I_L
  15. S3Reversing rotation
    (L1,L2,L3)→(L3,L2,L1)  ⇒  ns→−ns(L1, L2, L3) \to (L3, L2, L1) \;\Rightarrow\; n_s \to -n_s
  16. S4Voltage unbalance (NEMA working definition)
    %VU=max⁡∣VL−VˉL∣VˉL×100VˉL=V12+V23+V313\%VU = \frac{\max |V_{L} - \bar V_L|}{\bar V_L} \times 100 \qquad \bar V_L = \frac{V_{12} + V_{23} + V_{31}}{3}
  17. S5Current unbalance estimate
    %Iunb≈IstIn %VU≈(6–10) %VU\%I_{unb} \approx \frac{I_{st}}{I_n}\,\%VU \approx (6\text{–}10)\,\%VU
  18. S660 Hz motor on 50 Hz
    Φ∝VfV50=V60⋅5060n50=56n60P50≈56P60\Phi \propto \frac{V}{f} \qquad V_{50} = V_{60}\cdot\frac{50}{60} \qquad n_{50} = \frac{5}{6} n_{60} \qquad P_{50} \approx \frac{5}{6} P_{60}
  19. C1Star (Y)
    Vph=VL3IL=IphV_{ph} = \frac{V_L}{\sqrt{3}} \qquad I_L = I_{ph}
  20. C2Delta (Δ)
    Vph=VLIL=3 IphV_{ph} = V_L \qquad I_L = \sqrt{3}\, I_{ph}
  21. C3Power in either connection
    P=3 Vph Iphcos⁡φ=3 VL ILcos⁡φP = 3\,V_{ph}\,I_{ph}\cos\varphi = \sqrt{3}\,V_L\,I_L\cos\varphi
  22. C4Same windings, same supply: Y vs Δ
    Vph,YVph,Δ=13  ⇒  IL,YIL,Δ=13TYTΔ=13\frac{V_{ph,Y}}{V_{ph,\Delta}} = \frac{1}{\sqrt3} \;\Rightarrow\; \frac{I_{L,Y}}{I_{L,\Delta}} = \frac{1}{3} \qquad \frac{T_Y}{T_\Delta} = \frac{1}{3}
  23. C5Direction of rotation
    L1 ⁣→ ⁣U1, L2 ⁣→ ⁣V1, L3 ⁣→ ⁣W1  ⇒  CW (viewed from shaft end)swap 2 lines⇒CCWL1\!\to\!U1,\ L2\!\to\!V1,\ L3\!\to\!W1 \;\Rightarrow\; \text{CW (viewed from shaft end)} \qquad \text{swap 2 lines} \Rightarrow \text{CCW}
  24. D1Coil operating range
    0.85 Uc≤Upick-up≤1.1 UcUdrop-out≈(0.2–0.75) Uc0.85\,U_c \le U_{pick\text{-}up} \le 1.1\,U_c \qquad U_{drop\text{-}out} \approx (0.2\text{–}0.75)\,U_c
  25. D2Contactor selection
    Ie(AC-3, 400 V)≥Infor AC-4 duty derate to about 0.3–0.5 Ie(AC-3)I_e(\text{AC-3},\,400\ \text{V}) \ge I_n \qquad \text{for AC-4 duty derate to about } 0.3\text{–}0.5\,I_e(\text{AC-3})
  26. D3Overload setting
    Ir=In  (DOL, reversing, soft starter)Ir=In3≈0.58 In  (star–delta, in the phase-current path)I_r = I_n \;(\text{DOL, reversing, soft starter}) \qquad I_r = \frac{I_n}{\sqrt3} \approx 0.58\,I_n \;(\text{star–delta, in the phase-current path})
  27. D4Trip class check
    ttrip(7.2 Ir)≤class [s]tstart<0.8×ttrip(Ist/Ir)t_{trip}(7.2\,I_r) \le \text{class [s]} \qquad t_{start} < 0.8 \times t_{trip}(I_{st}/I_r)
  28. D5Breaking capacity
    Icu≥Ik,max at the point of installationIk≈Str3 VL uk  (at the transformer terminals)I_{cu} \ge I_{k,max}\ \text{at the point of installation} \qquad I_{k} \approx \frac{S_{tr}}{\sqrt3\,V_L\,u_k} \;(\text{at the transformer terminals})
  29. D6No nuisance trip at start
    Im>(1.5–2) Ist≈(10–13) In(allowing for the asymmetric first half-cycle inrush)I_{m} > (1.5\text{–}2)\,I_{st} \approx (10\text{–}13)\,I_n \qquad (\text{allowing for the asymmetric first half-cycle inrush})
  30. D7Control transformer sizing
    Str≥∑Shold+Sinrush, maxcheck voltage dip at pick-up≤5–10 %S_{tr} \ge \sum S_{hold} + S_{inrush,\,max} \qquad \text{check voltage dip at pick-up} \le 5\text{–}10\,\%
  31. G1Series = AND
    K=A⋅BK = A \cdot B
  32. G2Parallel = OR
    K=A+BK = A + B
  33. G3NC contact = NOT
    K=A‾K = \overline{A}
  34. G4Self-holding (seal-in)
    K1=S0‾⋅F2‾⋅(S1+K1)K1 = \overline{S0}\cdot\overline{F2}\cdot(S1 + K1)
  35. L1DOL starting current
    Ist=kst Inkst=Ia/In≈6–8I_{st} = k_{st}\,I_n \qquad k_{st} = I_a/I_n \approx 6\text{–}8
  36. L2Voltage dip at the transformer
    ΔU%≈SstSsc+Sst×100Sst=3 VL IstSsc=Struk\Delta U\% \approx \frac{S_{st}}{S_{sc} + S_{st}}\times 100 \qquad S_{st} = \sqrt3\,V_L\,I_{st} \qquad S_{sc} = \frac{S_{tr}}{u_k}
  37. L3Cable voltage drop
    ΔV=3 I L (Rcos⁡φ+Xsin⁡φ)cos⁡φstart≈0.3\Delta V = \sqrt3\, I\, L\,(R\cos\varphi + X\sin\varphi) \qquad \cos\varphi_{start} \approx 0.3
  38. L4Seal-in logic
    K1new=F2‾⋅S0‾⋅(S1+K1old)K1_{\text{new}} = \overline{F2}\cdot\overline{S0}\cdot(S1 + K1_{\text{old}})
  39. R1Phase swap
    K1: (L1,L2,L3)→(U1,V1,W1)K2: (L3,L2,L1)→(U1,V1,W1)\text{K1: } (L1, L2, L3)\to(U1, V1, W1) \qquad \text{K2: } (L3, L2, L1)\to(U1, V1, W1)
  40. R2Fault if both close
    Ik2=VL1L32Zsrc=32Ik3≈0.87 Ik3I_{k2} = \frac{V_{L1L3}}{2 Z_{src}} = \frac{\sqrt3}{2} I_{k3} \approx 0.87\,I_{k3}
  41. R3Interlocked logic
    K1=S0‾ F2‾ (S1+K1) S2‾ K2‾K2=S0‾ F2‾ (S2+K2) S1‾ K1‾K1 = \overline{S0}\,\overline{F2}\,(S1 + K1)\,\overline{S2}\,\overline{K2} \qquad K2 = \overline{S0}\,\overline{F2}\,(S2 + K2)\,\overline{S1}\,\overline{K1}
  42. R4Slip during plugging
    s=−ns−n−ns=1+nns≈2(n≈ns)s = \frac{-n_s - n}{-n_s} = 1 + \frac{n}{n_s} \approx 2 \quad (n \approx n_s)
  43. R5Rotor heat per operation (load with no torque)
    Estart=12Jωs2Eplug (ωs→0)=32Jωs2Ereverse (ωs→−ωs)=4⋅12Jωs2E_{start} = \tfrac12 J\omega_s^2 \qquad E_{plug\,(\omega_s \to 0)} = \tfrac32 J\omega_s^2 \qquad E_{reverse\,(\omega_s \to -\omega_s)} = 4\cdot\tfrac12 J\omega_s^2
  44. Y1Starting current in star
    IL,Y=IL,Δ3=kst In3I_{L,Y} = \frac{I_{L,\Delta}}{3} = \frac{k_{st}\,I_n}{3}
  45. Y2Starting torque in star
    TY(n)=TΔ(n)3Tst,Y=Ta/Tn3 Tn≈0.7–0.8 TnT_Y(n) = \frac{T_\Delta(n)}{3} \qquad T_{st,Y} = \frac{T_a/T_n}{3}\,T_n \approx 0.7\text{–}0.8\,T_n
  46. Y3Main and delta contactors
    Ie(KM1)=Ie(KM3)≥In3≈0.58 In(AC-3)I_e(\text{KM1}) = I_e(\text{KM3}) \ge \frac{I_n}{\sqrt3} \approx 0.58\,I_n \quad (\text{AC-3})
  47. Y4Star contactor
    Ie(KM2)≥In3≈0.33 In(AC-3)I_e(\text{KM2}) \ge \frac{I_n}{3} \approx 0.33\,I_n \quad (\text{AC-3})
  48. Y5Overload setting (in the phase-current path)
    Ir=In3≈0.58 InI_r = \frac{I_n}{\sqrt3} \approx 0.58\,I_n
  49. Y6Changeover dead time
    tdead≈40–100 ms  >  tarc,KM2+tclose,KM3 margint_{dead} \approx 40\text{–}100\ \text{ms} \;>\; t_{arc,KM2} + t_{close,KM3}\ \text{margin}
  50. Y7Acceleration time
    Jdωdt=Tm−TL  ⇒  tacc≈J ΔωTm−TL‾J\frac{d\omega}{dt} = T_m - T_L \;\Rightarrow\; t_{acc} \approx \frac{J\,\Delta\omega}{\overline{T_m - T_L}}
  51. Y8Condition before changing to delta
    TY(n)>TL(n)  ∀ n<nchnch≈(0.85–0.9) nnT_Y(n) > T_L(n)\ \ \forall\, n < n_{ch} \qquad n_{ch} \approx (0.85\text{–}0.9)\,n_n
  52. Y9Cable sizing for six leads
    Iz≥In3 per conductorΔV from the phase current and the actual lengthI_z \ge \frac{I_n}{\sqrt3} \ \text{per conductor} \qquad \Delta V \text{ from the phase current and the actual length}
  53. E1Winding voltage in Y and Δ
    Vw,Y=VL3Vw,Δ=VLIL,Δ=3 Iw,ΔV_{w,Y} = \frac{V_L}{\sqrt3} \qquad V_{w,\Delta} = V_L \qquad I_{L,\Delta} = \sqrt3\,I_{w,\Delta}
  54. E2Line-current ratio at the same speed
    IL,YIL,Δ=VL/(3 Zw)3 VL/Zw=13Zw=winding impedance at that slip\frac{I_{L,Y}}{I_{L,\Delta}} = \frac{V_L/(\sqrt3\,Z_w)}{\sqrt3\,V_L/Z_w} = \frac13 \qquad Z_w = \text{winding impedance at that slip}
  55. E3Torque ratio (torque ∝ V²)
    TY(s)TΔ(s)=(Vw,YVw,Δ)2=13Tk,YTk=13,  sk unchanged\frac{T_Y(s)}{T_\Delta(s)} = \left(\frac{V_{w,Y}}{V_{w,\Delta}}\right)^{2} = \frac13 \qquad \frac{T_{k,Y}}{T_k} = \frac13,\ \ s_k \text{ unchanged}
  56. E4Input power and starting kVA
    Pin=3 VLILcos⁡φSst=3 VL IstSst,Y=13 Sst,ΔP_{in} = \sqrt3\,V_L I_L\cos\varphi \qquad S_{st} = \sqrt3\,V_L\,I_{st} \qquad S_{st,Y} = \tfrac13\,S_{st,\Delta}
  57. E5Kloss approximation
    TTk=2s/sk+sk/sTk,Y=Tk3\frac{T}{T_k} = \frac{2}{s/s_k + s_k/s} \qquad T_{k,Y} = \frac{T_k}{3}
  58. E6Slip in star near rated speed
    T≈3Vw2ωsR2′ s  ⇒  sY≈3 sΔ (same load torque)T \approx \frac{3V_w^2}{\omega_s R_2'}\,s \;\Rightarrow\; s_Y \approx 3\,s_\Delta \ \text{(same load torque)}
  59. E7Star equilibrium and feasibility
    TY(nY)=TL(nY),  d(TY−TL)dn∣nY<0TY(0)>TL(0),  nY≥0.9 nNT_Y(n_Y) = T_L(n_Y),\ \ \frac{d(T_Y - T_L)}{dn}\Big|_{n_Y} < 0 \qquad T_Y(0) > T_L(0),\ \ n_Y \ge 0.9\,n_N
  60. E8Run-up time in star
    Jdωdt=TY(ω)−TL(ω)  ⇒  t(ω1)=J∫0ω1dωTY(ω)−TL(ω)J\frac{d\omega}{dt} = T_Y(\omega) - T_L(\omega) \;\Rightarrow\; t(\omega_1) = J\int_0^{\omega_1}\frac{d\omega}{T_Y(\omega) - T_L(\omega)}
  61. E9Mechanical time constant
    τm=J ωsTnωs=2πns60Tn=9550 PnnN\tau_m = \frac{J\,\omega_s}{T_n} \qquad \omega_s = \frac{2\pi n_s}{60} \qquad T_n = \frac{9550\,P_n}{n_N}
  62. E10Timer setting rule
    tset=max⁡ ⁣(1.2 t98(Vn), t95(0.9Vn))=c τmt_{set} = \max\!\big(1.2\,t_{98}(V_n),\ t_{95}(0.9V_n)\big) = c\,\tau_m
  63. E11Average-torque estimate
    t≈J ΔωTY−TL‾t \approx \frac{J\,\Delta\omega}{\overline{T_Y - T_L}}
  64. E12No-load run-up with Kloss (closed form)
    t=JωsTk[1−se24sk+sk2ln⁡1se]tY=3 tΔt = \frac{J\omega_s}{T_k}\left[\frac{1-s_e^2}{4s_k} + \frac{s_k}{2}\ln\frac1{s_e}\right] \qquad t_Y = 3\,t_\Delta
  65. E13Residual voltage after opening
    e‾(t)≈e‾0 e−t/T0′ e j∫0t(p ωr−ωs) dτT0′=LrRr\underline e(t) \approx \underline e_0\,e^{-t/T_0'}\,e^{\,j\int_0^t (p\,\omega_r - \omega_s)\,d\tau} \qquad T_0' = \frac{L_r}{R_r}
  66. E14Phase drift during the dead time
    δ(td)≈sch ωs td+p TL2J td2(electrical rad)\delta(t_d) \approx s_{ch}\,\omega_s\,t_d + \frac{p\,T_L}{2J}\,t_d^{2} \quad \text{(electrical rad)}
  67. E15Transient driving voltage
    ∣ΔV‾∣=∣V‾Δ−e‾∣=VΔ2+e2−2VΔecos⁡θθ=±30∘−φ0−δ|\Delta\underline V| = |\underline V_\Delta - \underline e| = \sqrt{V_\Delta^2 + e^2 - 2V_\Delta e\cos\theta} \qquad \theta = \pm30^\circ - \varphi_0 - \delta
  68. E16Changeover peak estimate
    Ipk≈ka ∣ΔV‾∣VΔ ILRka≈1.35–1.45I_{pk} \approx k_a\,\frac{|\Delta\underline V|}{V_\Delta}\,I_{LR} \qquad k_a \approx 1.35\text{–}1.45
  69. E17Transition resistor
    Rt=r ZbZb=VLIn/3r≈0.1–0.3R_t = r\,Z_b \qquad Z_b = \frac{V_L}{I_n/\sqrt3} \qquad r \approx 0.1\text{–}0.3
  70. E18Rotor heat, no load
    Er=12Jωs2 (s12−s22)⇒Er(1→0)=12Jωs2 independent of voltageE_r = \tfrac12 J\omega_s^2\,(s_1^2 - s_2^2) \quad \Rightarrow \quad E_r(1\to0) = \tfrac12 J\omega_s^2 \ \text{independent of voltage}
  71. E19Rotor heat with load
    Er=Jωs2∫s2s1s Tm(s)Tm(s)−TL(s) dsE_r = J\omega_s^2\int_{s_2}^{s_1}\frac{s\,T_m(s)}{T_m(s) - T_L(s)}\,ds
  72. E20I²t and overload thermal budget
    Q=∫irms2 dtB=∫(I/Ir)2 dt7.22 tclassQ = \int i_{rms}^2\,dt \qquad B = \frac{\int (I/I_r)^2\,dt}{7.2^2\,t_{class}}
  73. E21Voltage dip at the PCC
    ΔV%≈SmSsc+Sm×100Ssc=STuk  or  SGxd′′Sm=3 VLI\Delta V\% \approx \frac{S_m}{S_{sc} + S_m}\times100 \qquad S_{sc} = \frac{S_T}{u_k}\ \ \text{or}\ \ \frac{S_G}{x_d''} \qquad S_m = \sqrt3\,V_L I
  74. E22Losses in Y versus Δ
    Ploss=PCu1+PCu2+PFe+PfwPFe,Y≈13PFe,ΔsY≈3 sΔP_{loss} = P_{Cu1} + P_{Cu2} + P_{Fe} + P_{fw} \qquad P_{Fe,Y} \approx \tfrac13 P_{Fe,\Delta} \qquad s_Y \approx 3\,s_\Delta
  75. E23Saving and break-even
    ΔP=Pin,Δ−Pin,YΔP(xbe)=0Eyear=ΔP⋅hcost=Eyear×tariff\Delta P = P_{in,\Delta} - P_{in,Y} \qquad \Delta P(x_{be}) = 0 \qquad E_{year} = \Delta P\cdot h \qquad \text{cost} = E_{year}\times\text{tariff}
  76. E24Torque margin in star
    Tk,YTL=Tk3 TL≥2.5  ⇒  TL≤Tk7.5≈0.4 Tn (Tk=3Tn)\frac{T_{k,Y}}{T_L} = \frac{T_k}{3\,T_L} \ge 2.5 \;\Rightarrow\; T_L \le \frac{T_k}{7.5} \approx 0.4\,T_n \ (T_k = 3T_n)
  77. T1Reduced-voltage starting
    VVn=k  ⇒  Ist=k Ist,DOLTst=k2 Tst,DOL\frac{V}{V_n} = k \;\Rightarrow\; I_{st} = k\,I_{st,DOL} \qquad T_{st} = k^2\,T_{st,DOL}
  78. T2Torque at a current limit
    Tst=Tst,DOL(IlimIst,DOL)2T_{st} = T_{st,DOL}\left(\frac{I_{lim}}{I_{st,DOL}}\right)^2
  79. T3RMS voltage vs firing angle (resistive approximation)
    VrmsV=1−απ+sin⁡2α2π\frac{V_{rms}}{V} = \sqrt{1 - \frac{\alpha}{\pi} + \frac{\sin 2\alpha}{2\pi}}
  80. aa soft starters & autotransformers i_{motor} = a\,i_{st,dol} qquad i_{line} = a^2\,i_{st,dol} qquad t_{st} = a^2\,t_{st,dol} ">T4Autotransformer starter at tap aa
    Imotor=a Ist,DOLIline=a2 Ist,DOLTst=a2 Tst,DOLI_{motor} = a\,I_{st,DOL} \qquad I_{line} = a^2\,I_{st,DOL} \qquad T_{st} = a^2\,T_{st,DOL}
  81. T5Supply dip for any method
    ΔU%≈3 VL Iline,stStr/uk+3 VL Iline,st×100\Delta U\% \approx \frac{\sqrt3\,V_L\,I_{line,st}}{S_{tr}/u_k + \sqrt3\,V_L\,I_{line,st}}\times 100
  82. A1Sequence interlock
    K2=K1⋅TK3T⋅S0‾K1=S0‾ (S1+K1)K2 = K1 \cdot T_{K3T} \cdot \overline{S0} \qquad K1 = \overline{S0}\,(S1 + K1)
  83. A2Maximum starts per hour
    tcycle=VQp−q+Vq  ≥  4VQp  (q=Qp/2)Nmax=Qp4Vt_{cycle} = \frac{V}{Q_p - q} + \frac{V}{q} \;\ge\; \frac{4V}{Q_p}\ \ (q = Q_p/2) \qquad N_{max} = \frac{Q_p}{4V}
  84. A3Equal running hours
    h1≈h2≈htotal2the standby pump must run at least every 1–2 weeksh_1 \approx h_2 \approx \frac{h_{total}}{2} \qquad \text{the standby pump must run at least every 1–2 weeks}
  85. A4Receiver size for compressor cycling
    toff=Vtank Δpp0 quseton=Vtank Δpp0 (Qc−quse)t_{off} = \frac{V_{tank}\,\Delta p}{p_0\,q_{use}} \qquad t_{on} = \frac{V_{tank}\,\Delta p}{p_0\,(Q_c - q_{use})}
  86. A5Pole-changing speeds
    ns,low=120f2pns,high=120fp=2 ns,lown_{s,low} = \frac{120f}{2p} \qquad n_{s,high} = \frac{120f}{p} = 2\,n_{s,low}
  87. V1DC bus voltage
    Vdc≈1.35 VL=540 V (loaded)Vdc,0=2 VL=566 V (no load)V_{dc} \approx 1.35\,V_L = 540\ \text{V (loaded)} \qquad V_{dc,0} = \sqrt2\,V_L = 566\ \text{V (no load)}
  88. V2Maximum output voltage
    VLL,rms=0.612 m Vdc (SPWM,m≤1)VLL,max≈0.707 Vdc (SVPWM)V_{LL,rms} = 0.612\,m\,V_{dc}\ (\text{SPWM}, m \le 1) \qquad V_{LL,max} \approx 0.707\,V_{dc}\ (\text{SVPWM})
  89. V3Constant V/f
    V(f)=Vnffn+Vboost(1−ffn)Vnfn=40050=8 V/HzV(f) = V_n\frac{f}{f_n} + V_{boost}\left(1 - \frac{f}{f_n}\right) \qquad \frac{V_n}{f_n} = \frac{400}{50} = 8\ \text{V/Hz}
  90. V4Speed with a VFD
    n=120fp−ΔnΔn≈sn ns,n  (constant at rated torque)n = \frac{120 f}{p} - \Delta n \qquad \Delta n \approx s_n\,n_{s,n}\ \ (\text{constant at rated torque})
  91. V5Field weakening (f > f_n)
    Tcont≈Tn fnfTmax≈Tmax,n(fnf)2P≈PnT_{cont} \approx T_n\,\frac{f_n}{f} \qquad T_{max} \approx T_{max,n}\left(\frac{f_n}{f}\right)^2 \qquad P \approx P_n
  92. V6Braking resistor
    E=12J(ω12−ω22)Ppeak≈J ω1 ΔωtdecR≤Vbr2Ppeak, R≥RminE = \tfrac12 J(\omega_1^2 - \omega_2^2) \qquad P_{peak} \approx \frac{J\,\omega_1\,\Delta\omega}{t_{dec}} \qquad R \le \frac{V_{br}^2}{P_{peak}},\ R \ge R_{min}
  93. P1Minimum ramp time without tripping
    tacc,min≈J ωnTlim−TL‾Tlim≈1.5 Tn  (set by the drive current limit)t_{acc,min} \approx \frac{J\,\omega_n}{T_{lim} - \overline{T_L}} \qquad T_{lim} \approx 1.5\,T_n\ \ (\text{set by the drive current limit})
  94. P2Analog reference scaling
    f=fmin+(fmax−fmin) x−x0x1−x0x:0–10 V or 4–20 mAf = f_{min} + (f_{max} - f_{min})\,\frac{x - x_0}{x_1 - x_0} \qquad x: 0\text{–}10\ \text{V or } 4\text{–}20\ \text{mA}
  95. P3Affinity law — flow
    Q2Q1=n2n1\frac{Q_2}{Q_1} = \frac{n_2}{n_1}
  96. P4Affinity law — head
    H2H1=(n2n1)2\frac{H_2}{H_1} = \left(\frac{n_2}{n_1}\right)^2
  97. P5Affinity law — power
    P2P1=(n2n1)3\frac{P_2}{P_1} = \left(\frac{n_2}{n_1}\right)^3
  98. P6System with static head
    Hsys=Hst+kQ2nn1≈h0+(1−h0)q2(h0=Hst/H1, q=Q/Q1)H_{sys} = H_{st} + kQ^2 \qquad \frac{n}{n_1} \approx \sqrt{h_0 + (1 - h_0)q^2} \quad (h_0 = H_{st}/H_1,\ q = Q/Q_1)
  99. P7Energy saved per year
    ΔE=(Pthrottle−PVFD) hmoney=ΔE×tariff per kWh\Delta E = (P_{throttle} - P_{VFD})\,h \qquad \text{money} = \Delta E \times \text{tariff per kWh}
  100. P8Simple payback
    tpayback=installed VFD costsavings per yeart_{payback} = \frac{\text{installed VFD cost}}{\text{savings per year}}
  101. K1Shaft power of a pump or a drive
    Ppump=ρgQHηpP=T ω=T n9550 (kW)P_{pump} = \frac{\rho g Q H}{\eta_p} \qquad P = T\,\omega = \frac{T\,n}{9550}\ \text{(kW)}
  102. K2Motor size margin
    Pmotor≥(1.10–1.15) Pload,max  →  next standard sizeP_{motor} \ge (1.10\text{–}1.15)\,P_{load,max} \;\rightarrow\; \text{next standard size}
  103. K3Cable ampacity after derating
    Iz ktemp kgroup≥InIz≥Ir (overload protection setting)I_z\,k_{temp}\,k_{group} \ge I_n \qquad I_z \ge I_r\ (\text{overload protection setting})
  104. K4Running voltage drop
    ΔV%=3 In L (Rcos⁡φ+Xsin⁡φ)VL×100≤3–5 %R=ρA, ρCu,70∘C≈0.0225 Ω mm2/m\Delta V\% = \frac{\sqrt3\,I_n\,L\,(R\cos\varphi + X\sin\varphi)}{V_L}\times100 \le 3\text{–}5\,\% \qquad R = \frac{\rho}{A},\ \rho_{Cu,70^\circ C} \approx 0.0225\ \Omega\,\text{mm}^2/\text{m}
  105. K5Starting voltage drop
    ΔVst%≈ΔVtr%+3 Ist L (R⋅0.3+X⋅0.95)VL×100≤10–15 %\Delta V_{st}\% \approx \Delta V_{tr}\% + \frac{\sqrt3\,I_{st}\,L\,(R\cdot0.3 + X\cdot0.95)}{V_L}\times100 \le 10\text{–}15\,\%
  106. K6Earth fault at the cable end (TN system)
    If=U0Zs≥IaZs≈Zsrc+(1+m) ρLAph  (m=Aph/APE)I_{f} = \frac{U_0}{Z_s} \ge I_a \qquad Z_s \approx Z_{src} + (1 + m)\,\frac{\rho L}{A_{ph}}\ \ (m = A_{ph}/A_{PE})
  107. K7Enclosure temperature rise and fan airflow
    ΔT=Plossk Ae  (k≈5.5 W/m2K painted steel)V˙ [m3/h]≈3.1 Ploss [W]ΔTallow\Delta T = \frac{P_{loss}}{k\,A_e}\ \ (k \approx 5.5\ \text{W/m}^2\text{K painted steel}) \qquad \dot V\,[\text{m}^3/\text{h}] \approx \frac{3.1\,P_{loss}\,[\text{W}]}{\Delta T_{allow}}
  108. K8Insulation resistance acceptance
    Riso≥1 MΩ @ 500 V DC (IEC 60204-1)R40∘C≥5 MΩ windings<1 kV (IEEE 43)PI=R10 minR1 min≥2R_{iso} \ge 1\ \text{M}\Omega\ @\ 500\ \text{V DC (IEC 60204-1)} \qquad R_{40^\circ C} \ge 5\ \text{M}\Omega\ \text{windings} < 1\ \text{kV (IEEE 43)} \qquad PI = \frac{R_{10\,min}}{R_{1\,min}} \ge 2

Standards and references

  • IEC 60034-1 (motor ratings) · IEC 60034-8 (terminal markings and rotation) · IEC 60034-30-1 (IE efficiency classes)
  • IEC 60947-4-1 (contactors and starters, AC-3, Type 1/2 coordination) · IEC 60947-4-2 (soft starters) · IEC 60947-2 (circuit-breakers)
  • IEC 60204-1 (electrical equipment of machines) · IEC 60617 (symbols) · IEC 81346 (reference designations)
  • IEC 61800 (adjustable-speed drives) · IEC 60364 / Thai EIT wiring code · IEC 61439 (low-voltage assemblies)
  • S. J. Chapman, Electric Machinery Fundamentals · B. K. Bose, Modern Power Electronics and AC Drives · manufacturers' starter selection guides

Frequently asked questions

Which motors can be star–delta started?

Only motors whose delta voltage on the nameplate equals the supply line voltage. On a 400 V system that is a motor marked 400/690 V Δ/Y (usually 4 kW and above). A 230/400 V Δ/Y motor must run in star on 400 V and cannot be star–delta started.

Can the same equipment be used in Thailand and its neighbours?

In practice yes. Every country covered here uses 50 Hz three-phase four-wire systems with line voltages of 380–415 V (IEC 60038 nominal 230/400 V), so equipment rated 380–415 V 50 Hz fits all of them. Always measure the actual site voltage and check the local utility's rules.

What is the overload setting for a star–delta starter?

When the overload relay sits after the main contactor in the phase-current path (the most common arrangement) set it to 0.58 × the motor In, e.g. 8.5 A for a 14.6 A motor. If it sits in the line-current path before the split, set it to In.

When should I use a VFD instead of star–delta?

When you need variable speed, a soft start with high torque, energy savings on pumps and fans that often run below full flow (P ∝ n³), or when the starting voltage dip is a problem. If you only need to reduce the starting current of a light load at fixed speed, star–delta or a soft starter is cheaper.

Ajarn Mod · Teerawut Savangboon
Lecturer in Electrical Engineering, Dhonburi Rajabhat University, Samut Prakan · Profile