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# Motors power supply - star-delta connection

kozi966 18450 5
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• kozi966
Moderator of Electricians group
Power supply to motors - the problem of star-delta switching

Problem : When can I connect a star-delta switch?

The engine, as an electrical machine capable of converting electrical energy into mechanical energy, requires an appropriate power supply.
The nameplate should describe all the necessary information, but as it turns out, not everything is so clear.
Each motor has in its construction the basic elements that enable it to work - windings.
The winding of the motor (coil), as an element usually made of copper wire, has its operating parameters, and to put it simply, the voltage for which it was built.
For example as below:

You can often find such records on engines:
230 (D) / 400 (Y)
400 (D) / 690 (Y)
For example:

It just means that the proper connection of the motor is required.
The motor that is to work in a typical Polish 230 / 400V network, for such designation 230/400 D / Y, can only work in the star (Y) configuration.
Why?
Because its coils are designed to work with 230V. Working in a delta configuration, it will give the coils a voltage of 400V.

Going back to our motor windings (coils), we mark their ends as follows:

We can use them to connect two systems:

Why then do we have to be careful to what network (or voltage) we connect the above systems?
Answer in the figures below.

Power supply for the standard system ("home" , has two voltages - phase (230V) and phase-to-phase (400V). Therefore, when connecting the 230 (D) / 400 (Y) motor in a triangle (D) configuration, we supply the coil with 400V, where only 230V is allowed. In the star (Y) system, the motor is connected to 400V, but the coils are connected to each other in such a way that only 230V will appear on a single coil.
Therefore, such an engine can't work with the GWIAZDA-TRÓJKĄT switch in this network.
Such an engine can't also work in a triangle pattern in this network.
It can only work in the connection system: Star.

The situation is similar in this case.
The power supply for the industrial system has two voltages - phase (400V) and phase-to-phase (690V). Therefore, when connecting the 230 (D) / 400 (Y) motor in a triangle (D) configuration, we apply the 690V coil, where only 230V is allowed.
Such a motor cannot work in a 400 / 690V industrial network.
In this system, the correct motor for connection may be the motor 400 (D) / 690 (Y), connected in a star configuration.

So what does star-delta switching involve?
It bindsit is, however, that the motor coil will work first in the star, which means that the voltage will be "given" one root of three lower than the rated voltage. This is done to start the motor "with a lower current" - to reduce the starting current. The delta switching (after starting) already supplies the rated voltage to the motor and allows for full power.

Thus, the motor suitable for operation in a star-delta switch system, for a 230 / 400V network system will be a motor with the following parameters:
400 (D) / 690 (Y)

What is the current and voltage situation?
Oh yes:

As you can see, in the case of star connection, the current and voltage are less by a root of three than the rated values for which the device is built.
This will not cause the engine to overheat or burn the coils, it can only cause the "power shortage" effect and, consequently, the engine will not start.
• qrnick
Level 10
Overall a very useful analysis, especially for beginners.
There was, however, a mistake in the last picture. I do not know what the subscripts of current I in the formula shown on the right should mean (whether mf means the machine phase or the phase-to-phase current), but for sure in a three-phase receiver connected in a triangle, the currents inside the triangle's circuit are root z 3 times lower than the current wired (the one flowing from the source to the receiver).

It is also worth noting that for the same supply voltage, the line current supplying the receiver for a system of the same impedance connected in a delta is 3 times greater than for a system of the same impedance values connected in a star, and in the case of the current flowing in individual phases, it is the root of 3 times larger in a triangular system in relation to the stellar system. Perhaps this was what the original drawing was about, but due to the fact that the markings were left in the layout on the left, some distinction would have been useful.

• Aleksander_01
Level 41
qrnick wrote:

It is also worth noting that for the same supply voltage, the line current supplying the receiver for a system of the same impedance connected in a delta is 3 times greater than for a system of the same impedance values connected in a star, and in the case of the current flowing in individual phases, it is the root of 3 times larger in a triangular system in relation to the stellar system. Perhaps this was what the original drawing was about, but due to the fact that the markings were left in the layout on the left, some distinction would have been useful.

What is this gibberish, the line current is never 3 times more, always by ?3.
Maybe the sentence is poorly worded and it was about power.
• qrnick
Level 10
Maybe instead of offending interlocutors, it is worth counting something.
I am talking about the line current, not flowing through the motor phase with unchanged power supply and unchanged impedance of a single motor phase for different receiver phase connection systems.

If a few simple calculations are too much effort, I refer you to "Collection of tasks on the basics of electric drive" Chodnikiewicz K., Moszczyński L., Oficyna Wydawnicza Politechniki Warszawskiej, Warsaw 2014 p. 127.

• Aleksander_01
Level 41
We misunderstood each other. If we treat the motor as a "black box" where the power supply is always 400 V (wire) and we only change the connection arrangement inside the "box", the current will decrease 3 times as much as possible.
No calculations are needed here, it is known that when changing ? to ?, the power changes three times, and since the power supply of the box is of the same value, the current must change three times.

I "figured it out" too much.
Sorry and not angry, after all, the one who does nothing is wrong.
• qrnick
Level 10
That's it. Calmly.
The power supply network we have at our disposal has a constant RMS voltage (whether wire or phase) and when we do not have any additional devices (transformer, converter, etc.), it is important information for the motor user to know what consequences the use of these two types of winding connections will have. . As a rule, we choose the engine to the existing network, not the other way around. That is why I mentioned it and suggested to distinguish between the markings in the diagram. It is also worth explaining these kind of truisms here, since beginners are looking at it. The fact that something "is known" is worth knowing why and rather it should result from the ability of some basic conversion and not, for example, forging information into a sheet of metal, hence I quoted a fragment of the book with the rest quite useful.