# Building a three phase motor -- need help figuring out how many poles

## Summary:

Building a three phase motor need help figuring out how many poles.

## Main Question or Discussion Point

Hello. I am building a three phase motor for a science project of mine. I am trying to match my motor with an inverter so i need to count how many poles I have in this motor. The problem is that I have read about this and some people say that this.

https://electronics.stackexchange.c...e-induction-motors-with-poles-not-equal-to-3x

There is a thread it shows two diagrams one is the top one that shows a 2 pole three phase motor. There is another below it that shows a 4 pole three phase motor. This explanation should be straight forward but i also see motors with this coil arrangement. How do you calculate the poles of a three phase motor with 3 coils per phase? Thanks for any help.

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Baluncore
2019 Award
How do you calculate the poles of a three phase motor with 3 coils per phase?
Poles = number of coils per phase, multiplied by two.
The number of poles and supply frequency decides the synchronous speed.

• russ_watters and johnboyman
Poles = number of coils per phase, multiplied by two.
The number of poles and supply frequency decides the synchronous speed.
Thanks that's very easy for me to understand it soaked in asap. No one else could say it like that.

• anorlunda
Two threads on the same subject merged
Hello. I am working on a science project of mine which is a three phase motor. I have a 220 volt 1.5 kw inverter and i want to build a motor for it.

I have been experimenting on building rectangular loop inductors and everthing is going well. I have been doing a lot of research and I am part way there and partially lost. I know that the output of the inverter the 3 wires that hook up to my motor may have an amperage of 6.8 when the phase is active. 1500 watts / 220 = 6.8 amps.

An inductor has a tendency to oppose a change in the electric current flowing through it. I know that in a 3 phase motor if the coils are too small it will short out the inverter so i need my coils in my 3 phase motor to match the output of the inverter. My motor has three coils per phase. I am trying to find a formula that will help me match inverter with motor.

Can i add up the inductance of three coils and find out how much current will flow through them per second. This is as far as my mind has gone. Does anyone have any experience with this kind of thing? thanks.

berkeman
Mentor
Hello. I am working on a science project of mine which is a three phase motor. I have a 220 volt 1.5 kw inverter and i want to build a motor for it.

I have been experimenting on building rectangular loop inductors and everthing is going well. I have been doing a lot of research and I am part way there and partially lost. I know that the output of the inverter the 3 wires that hook up to my motor may have an amperage of 6.8 when the phase is active. 1500 watts / 220 = 6.8 amps.

An inductor has a tendency to oppose a change in the electric current flowing through it. I know that in a 3 phase motor if the coils are too small it will short out the inverter so i need my coils in my 3 phase motor to match the output of the inverter. My motor has three coils per phase. I am trying to find a formula that will help me match inverter with motor.

Can i add up the inductance of three coils and find out how much current will flow through them per second. This is as far as my mind has gone. Does anyone have any experience with this kind of thing? thanks.
What safety precautions do you have between the Inverter and the motor? How much experience do you have so far in working with AC Mains projects?

Before any assembly or connection work, discharge the VFD. Verify that the VFD is discharged completely.
Do not touch the terminals because the capacitors may still be charged. I have moderate experience with AC Mains projects. I am very careful always. Thanks

Three Phase Power is Vline-line * I * Sqrt(3) = so really about 4 A - but that is hardly enough difference to worry about for building a motor from scratch, while higher winding resistance (smaller wire or more turns) will provide some fault tolerance, it also hurts the efficiency.

You will start the motor at a low voltage - Speed is proportional to the voltage.

• johnboyman
Baluncore
2019 Award
You will start the motor at a low voltage - Speed is proportional to the voltage.
Unless it is a synchronous or induction motor, when;
Synchronous RPM = 120 * Frequency in Hz / Number of poles.
Notice that the synchronous RPM is independent of the number of phases, or the voltage.
The number of poles is always even because coils make magnetic poles in N-S pairs.

• johnboyman
Three Phase Power is Vline-line * I * Sqrt(3) = so really about 4 A - but that is hardly enough difference to worry about for building a motor from scratch, while higher winding resistance (smaller wire or more turns) will provide some fault tolerance, it also hurts the efficiency.

You will start the motor at a low voltage - Speed is proportional to the voltage.
So your talking about about 4 amps for my inverter? Sounds good. I have an induction motor it is 0 - 60 hz, when i change the frequency doesn't that change the speed. You said speed is proportional to the voltage are you talking about the inverter? thanks

Unless it is a synchronous or induction motor, when;
Synchronous RPM = 120 * Frequency in Hz / Number of poles.
Notice that the synchronous RPM is independent of the number of phases, or the voltage.
The number of poles is always even because coils make magnetic poles in N-S pairs.
Thanks.

This is all very valuable to me. I have one more thing that i am trying to finish. I have a motor that i made to test out on. I wanted to figure out what size inverter I need for it. I need match the amperage of the motor with the output power per phase witch is around 4 amps for my inverter like was mentioned earlier. I have a motor with 9 rectangular coils arranged in a y formation so far. I was thinking about using an ammeter to measure the amps used by one coil then multiplying it by 3 because i have 3 coils per phase. I am not the best physicist so my math isn't sharp enough. Does anyone know what i am getting at? Hope someone has experience thanks.

Baluncore
2019 Award
• johnboyman
You will need to vary the Output frequency in addition to modulating the voltage, to match the speed, but I was referring to needing to control the current when you turn on, it is one thing to make an inverter, another to make a motor drive.

Yes - about 4A RMS, per phase for a 1.5KW motor at 220VAC Phase to Phase

• johnboyman
I am not making an inverter. I want to know how to match the coils in the motor with the inverter. I am guessing if i have 9 coils total in my motor then it should be 3 coils running per phase. If the inverter uses 4 amps per phase how do i match the motors coils with the inverter amperage? Can i use am ammeter and find out how much current is running through one coil alone, then times that by three. If i can match that number with the inverters output phase power will that work?

I have seen various formulas that come close to what i am saying but i am not the best at math. Thanks

Tom.G
If your motor efficiency is anywhere near reasonable, the no-load running current will be quite low, determined mostly by bearing friction and moving the air around in the motor.

At start-up the current will be rather high, known as the 'locked rotor current.' Before the rotor starts turning, the current is limited by the winding resistance and inductance. VFD's are designed to handle the initial current surge.

The other characteristic to pay attention to is the Volts-per-Hertz. You stated 220V and 60Hz I believe. That works out to 3.67V per Hertz. If you decrease the frequency from the 60Hz to slow down the motor, you need to also decrease the supply voltage to the motor by the same ratio. For instance to run at 3/4 speed you would need 45Hz and 165V. Note that your VFD may automatically do this, but it may have to be programmed to do so.

The main reason for this is:
While running, there is a voltage induced in the coils from the spinning rotor. This voltage opposes the applied voltage, giving a lower effective voltage on the windings. With the rotor turning slower, there is less induced voltage in the windings to oppose the supply voltage; therefore the windings will draw more current; just as in the start-up condition.

Cheers,
Tom

• johnboyman
Ok thanks. I found a formula I= 1/L∫Vdt, . It is an Inductor current formula. So i can find out how much time has past for the frequency i was using and plug that into T. Then find out the voltage for that frequency and plug it in for V, then I can figure out the inductance for my coil and plug in L then isolate I. Do that for 3 coils and add them all together. Should be able to figure out that maximum current that my motor can take. Does that sound about right. What do you think about that. Thanks for the advice.

Tom.G
Yes, you could do that -- and that would only give you a poor approximation of the locked rotor current.

Is that what you are really after? ... the start-up current? If so, why? Much easier to measure it; or measure the inductance of the completed and assembled motor and compute the current based on that.

• johnboyman
Tom.G
I= E/XL

Where:
I= current in Amps
E= voltage in Volts
XL= inductive reactance= 2πFL
π= 3.14
F= frequency in Hertz
L= inductance in Henries

If the coils have a high DC resistance (more than 15% of XL), you can get a better approximation of current by replaceing XL in the first formula with [√(R2+XL2)]

Note that the inductance measurement is to be done with the motor completed and assembled. It may change slightly with rotational position of the rotor; for a conservative estimate use lowest reading.

Cheers,
Tom

• johnboyman
What is R in
[√(R2+XL2)]

Baluncore
2019 Award
What is R in
[√(R2+XL2)]
Resistance.
√ ( R2 + XL2 )
When you have a resistor in series with the reactance of an inductor, the total impedance is the root of the sum of the squares, because reactance is at right angles to resistance, and the impedance is the length of the hypotenuse of that right angle triangle.

• johnboyman
Resistance.
√ ( R2 + XL2 )
When you have a resistor in series with the reactance of an inductor, the total impedance is the root of the sum of the squares, because reactance is at right angles to resistance, and the impedance is the length of the hypotenuse of that right angle triangle.
So resistance of the inductor? I can get that value with and ohmmeter?

Tom.G