What the Star-Delta Starter Calculator does
This calculator sizes the three contactors and the overload relay for a star-delta starter, and shows the starting current and torque compared with a direct-on-line start. The headline figure is the overload relay setting, because getting it wrong is the classic star-delta mistake and it leaves the motor with no overload protection at all.
The relay sits in the line to the delta contactor, in series with the windings, so it sees the winding current - the line current divided by root-three, or 57.7% of it. Setting it to the nameplate full-load current means it will not operate until the motor is drawing about 173% of rated current, by which time the windings are cooking.
How to use it
- Enter the motor's full-load line current - the nameplate figure for the supply voltage, in delta running.
- Enter the supply voltage. The motor's windings must be rated for delta operation at that voltage, or star-delta starting is simply not available: a motor marked 400 V star / 690 V delta cannot be star-delta started on 400 V.
- Enter the locked-rotor multiple from the data sheet, or leave it at the conventional 6.
- Read the three contactor currents. The main and delta contactors each carry 58% of the line current; the star contactor carries 33%.
- Set the overload relay to the winding current shown, not to the nameplate line current.
Reading the results
The main and delta contactors both sit in series with the windings in delta running, so both carry the winding current - 57.7% of the line current.
The star contactor only conducts during the star period, when the line current is a third of the direct-on-line figure, so it is conventionally sized at 33% of full-load current.
In star, both the starting current and the starting torque are exactly a third of their direct-on-line values. Every reduced-voltage method trades torque for current, and star-delta trades them one for one.
The changeover is where star-delta is least elegant. An open transition disconnects the motor briefly and reconnects it in delta while it is still turning, which can produce a current surge larger than the direct-on-line inrush.
Worked example: a 30 kW pump motor drawing 54.14 A at 400 V
The winding current in delta running is 54.14 / root-three = 31.26 A. Both the main and the delta contactor carry that, so both are chosen in AC-3 at 31.26 A or the next size up - typically a 32 A frame rather than the 55 A frame the line current would suggest, which is where much of the cost saving in a star-delta starter comes from.
The star contactor carries 54.14 / 3 = 18.05 A during the star period only, so it can be smaller again.
The overload relay goes in the line to the delta contactor and must be set to 31.26 A. Set to the nameplate 54.14 A it would only operate at 54.14 x root-three = 93.8 A of line current - about 173% of full load, which no motor survives for long.
With a locked-rotor multiple of 6, a direct-on-line start would draw 325 A. In star that falls to 108 A, and the winding current during star is 108 / root-three = 62.5 A. The torque available is a third of direct on line, which for a centrifugal pump is more than enough - its torque demand at standstill is close to zero and rises with the square of speed.
A loaded screw compressor is the counter-example. Its breakaway torque can exceed a third of direct-on-line torque, so it would sit near standstill drawing 108 A indefinitely until the overload tripped - and if the timer changed over anyway, the delta reconnection at near-zero speed would draw the full 325 A after all.
Formulas and scoring rules
- Winding current in delta
I_winding = I_line / root-three = 0.577 x I_line- Main and delta contactor rating
0.577 x full-load line currentBoth sit in series with the windings in delta running.- Star contactor rating
I_line / 3 = 0.33 x full-load line currentIt only carries current during the star period.- Overload relay setting
0.577 x full-load line currentBecause the relay is in series with the windings, not in the incoming line.- Star starting current
1/3 of the direct-on-line starting current- Star starting torque
1/3 of the direct-on-line torqueTorque follows the square of the winding voltage: (1/root-three)^2 = 1/3.- Why both are exactly 1/3
winding voltage x 1/root-three, so winding current x 1/root-three, and line current a further x 1/root-three
The overload relay position, in detail
A star-delta starter breaks the motor's three windings out to six terminals. In delta running, the main contactor feeds one end of each winding from the supply, and the delta contactor joins the other ends to form the delta. The overload relay is placed between the main contactor and the winding ends - in series with the windings - so the current through it is the winding current.
Some starter designs instead place the relay in the incoming line before the main contactor, where it does see the full line current and is set to the nameplate value. Both arrangements exist, which is precisely why the mistake is so common: the correct setting depends on where the relay physically sits, and you have to look at the wiring rather than assume. This page gives the setting for the usual arrangement, in the delta line.
Open transition and why it can surprise you
At changeover the star contactor opens, there is a short dead time, and the delta contactor closes. During the dead time the motor is disconnected but still turning, and it generates a decaying back-EMF whose phase drifts away from the supply. If the delta contactor closes when the two are out of phase, the effective voltage across the windings is briefly higher than the supply and the current surge can exceed the direct-on-line inrush - the opposite of what the starter was fitted to achieve.
The practical mitigations are a short dead time, typically 30 to 100 milliseconds, and changing over only when the motor is close to full speed so the back-EMF has not had time to drift far. Closed-transition starters add a resistor bank to keep the motor energised through the transition and avoid the problem entirely, at additional cost.
Limitations: what the result does not prove
- It sizes by current. Contactors are selected by utilisation category as well - AC-3 for normal motor duty, AC-4 for jogging and plugging, which demands much larger devices for the same motor.
- It does not check whether the load can actually accelerate on a third of the torque. That needs the load's torque-speed curve and the combined inertia.
- The changeover time given is guidance, not a calculated value. The correct moment depends on the load's inertia and is best set by watching the starting current fall and flatten.
- Short-circuit protection, coordination type (type 1 or type 2) and the tested combination of breaker, contactors and relay all come from the manufacturer's coordination tables.
- Nothing here is a compliance certificate. A qualified engineer must verify the design against the applicable standard and the local regulations.
Privacy: where your data goes
Everything you paste, type or drop is processed in this browser tab. It is not uploaded, logged, stored or sent to analytics. Session recording and tag-manager scripts are switched off on this page.
Standards and sources
- IEC 60947-2 - Low-voltage switchgear: circuit-breakers - checked 19 Sep 2026
- IEC 60034-12 - Starting performance of cage induction motors - checked 19 Sep 2026
- IEC 60947-4-1 - Contactors and motor-starters: electromechanical
Frequently asked questions
What should the overload relay be set to on a star-delta starter?
To the winding current, which is the motor's full-load line current divided by root-three - 57.7% of it. For a 54 A motor that is 31.2 A. This is for the usual arrangement with the relay in the delta line; if your starter has the relay in the incoming line instead, it sees the full line current and is set to the nameplate value.
Why is the star-delta contactor only 58% of the motor current?
Because in delta running the main and delta contactors sit in series with individual windings rather than in the supply line, and the winding current is the line current divided by root-three. That is why a star-delta starter uses three smaller contactors rather than one large one, and much of its cost advantage comes from it.
Does star-delta reduce starting current to a third?
Yes, exactly a third of the direct-on-line value - and the starting torque falls to a third as well. The two reductions are the same because torque follows the square of the winding voltage while current follows it linearly, and the winding voltage in star is 1/root-three of the line voltage.
What loads suit star-delta starting?
Loads whose torque demand is very low at standstill and rises with speed: centrifugal pumps, fans, blowers and compressors started unloaded. Conveyors, positive-displacement pumps, loaded compressors and anything with high breakaway friction generally cannot accelerate on a third of the torque.
Can any motor be star-delta started?
Only one whose windings are rated for delta operation at the supply voltage and which brings all six winding ends out to the terminal box. A motor marked 400 V star / 690 V delta is designed to run in star on 400 V, so there is no delta configuration available at that voltage and star-delta starting is impossible.
How long should the star period be?
Long enough for the motor to reach roughly 80 to 90% of full speed - typically 5 to 15 seconds on a fan or pump and longer on a high-inertia load. The reliable method is to watch the starting current: it falls and then flattens as the motor runs up, and that flattening is the moment to change over.
Why is my motor drawing a huge current at changeover?
Open transition. During the dead time the motor is disconnected but still spinning and generating a back-EMF that drifts out of phase with the supply; reconnecting out of phase produces a surge that can exceed the direct-on-line inrush. Shorten the dead time, change over closer to full speed, or use a closed-transition starter.
Is a soft starter better than star-delta?
For most new installations, yes. A soft starter lets you choose the current limit rather than accepting a fixed third, ramps smoothly instead of stepping, has no open-transition surge, and often includes overload protection and diagnostics. Star-delta survives because it is cheap, simple, repairable with parts from any wholesaler, and already installed in enormous numbers.
Last reviewed by the A2Z.Tools team against the sources listed above.
