How can I make my 380/380 volts motor run on 220 volts?

Question

I bought a small and old combi woodworking machine "Emco Start". The machine runs on 380V. The problem is that I only have 220 at home. I did some research and i found out that it is possible to get it run on 220 volts, by adding a capacitor.

My problem is that I am not sure how to wire it to the motor, and I am not sure what is the needed value of the capacitor. The machine has a switch, so it runs with two speeds.

The label on the motor shows :

• Volts : 380/380
• KW : 0,37/0,51
• U/min : 1430/2860

The wiring taken from the schematics is attached here.

U, V, U => from the motor to the switch (speed 1) WV, MV, VU => from the motor to the switch (speed 2)

Answer 1

The connection of a capacitor to a three-phase motor for single-phase operation is called a Steinmetz connection. If you search "Steinmetz connection" you will quite a bit of information about that.

If the motor has only six leads or terminals available for external connections, it can only be operated at 380 volts at either of the two speeds indicated. For low speed, U4, V4 and W4 are connected together and three-phase power is connected to U2, V2 and W2. For high-speed operation, there is no connection to U2, T2 and W2, and power is connected to Uw, T4 and W4. The mechanical power rating is the same for both speeds, so the torque available at the high speed is half of the low speed torque. You could use a variable frequency drive (VFD) with 380 volt output for either of those connections.

If each end of each winding is available independent external connection, 12-leads or terminals, the windings could possibly be connected in a parallel delta configuration. That should be suitable for 220 volt, 3-phase power. I believe that would still be the 4-pole, low-speed configuration. You could use a VFD with 220 volt output for that connection.

You should have no problem finding a VFD with 220 volt, single phase input and 220 volt, three-phase output. You might be able to find a VFD with a built-in voltage boost circuit to give 380 volt, three-phase output with 220 volt, single-phase input. Otherwise you would need an input transformer for the VFD and a 380 volt VFD that accepts single-phase input.

I don't know what all the options are with a Steinmetz connection.

Unless the existing motor has a special shaft or a gear mounted directly to it. The best option may be to buy a different motor and perhaps a VFD for speed control.

See diagram below:

For U2, V2 and W2, two motor coils are connected together inside the motor or in the motor terminal box. If you can break that connection, you can reconnect the coils as shown by the red lines. I am pretty sure that will allow the motor to run at the high speed on 220 volts. For single phase, connect a capacitor from one of the power lines to the point where the missing phase would be connected. That allows the motor to run on single-phase, but the torque capability is greatly reduced. This is the Steinmetz connection. You should be able to find capacitor values and other information by searching "Steinmetz connection."

Answer 2

"The label on the motor shows : Volts : 380/380 KW : 0,37/0,51 U/min : 1430/2860"

A little confusion here, so let's put this in columns:

• Volts: 380/380
• KW: 0,37/0,51
• U/min: 1430/2860

In plain English:

• It's a 380 volt two-speed motor.
• At 1430 RPM, it draws 0.37 kW at full load
• At 2860 RPM, it draws 0.51 kW at full load

So now that we understand it's not a 380 kW motor, we can better solve this. :-)

If you have single phase 220 volts, there are phase converters (Google is your friend) to convert to 220 three phase. Then , you could use either:

• A 220 to 380v 3 phase .5 kva step-up transformer
• Three single-phase transformers wired as a 3 phase.

You may find the latter arrangement to be less expensive than a 3 phase transformer.

Answer 3

A simple answer: 3 phase motors have 3 power connections (one can ignore U2, V2, W2). Connect mains L&N to 2 of those, and a capacitor from the 3rd to L or N. Your motor now runs.

The extra motor tappings & switches complicate the picture to add features.

Answer 4

I don't think the diagram is correct. The phase rotation looks to be opposite for speed 2. I suspect that 2 of the 3: U2 V2 & W2 should be shown connected & labelled the other way round. I can't think of any better way to do it than how Charles Cowie did it, separating the joints U2, V2 & W2 & paralleling the windings. You will only have low speed unless you reverse all the P1,2 coils. Motor o/p power is 0.37kW, so low speed current is 370÷380÷√3÷efficiency÷power factor. Assuming efficiency is about 80% & PF≈.6, Current≈370÷380÷√3÷.8÷.6≈1.17A. On single phase, that is 370÷220÷.8÷.6 ≈ 3.5A Coil voltage on low speed is √3÷2x220 = .866x220V, so, on 220V, the load component of input current is .866 x normal for the same power, but the magnetisation current is increased by 15.5%, which brings it back up a little bit. About 3.17A. Phase currents will be 3.17÷√3=1.83A. Xc=220÷current≈120.2Ω. So C1≈26.48uF, C2≈36.5uF-C1≈10uF. Try something around that value or a bit less, and, if you can measure the current while keeping the load constant, you can adjust the value so the current is equal to the other phases. I did some analysis of how the motor worked in the diagram below. I also drew a 220V conversion diagram, but you would need an extra 3 wires from the motor and a cam switch with 16 (8 n/o & 8 n/c), AND, all the coils wired correctly. The motor will be phased if you wire one backwards. 😀

I found a perfect switch on ebay: https://www.ebay.com.au/itm/Changeover-Switch-3-Position-Rotary-Selector-Cam-Switch-32-Terminal-Latching-10A-/233283092114?_trksid=p2385738.m4383.l4275.c10 I made a diagram on how to connect it.