-1
\$\begingroup\$

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.

enter image description here

enter image description here

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

\$\endgroup\$
10
  • 3
    \$\begingroup\$ That's a three phase motor. How many 220V phases do you have at home? \$\endgroup\$
    – Jack B
    Commented Jan 13, 2017 at 13:08
  • 1
    \$\begingroup\$ "380 kW" and "small" are contradictory IMHO. Resolve the anomaly then buy a 220 V motor if you value your time. \$\endgroup\$
    – Andy aka
    Commented Jan 13, 2017 at 13:11
  • 2
    \$\begingroup\$ Look for "Variable Frequency Drive" with a competent supplier. Some (but not all) may be able to do what you need - providing 3 phase from single phase input. Note - some may only provide 3 phase 208V so don't just assume any VFD will work. \$\endgroup\$
    – user16324
    Commented Jan 13, 2017 at 13:18
  • \$\begingroup\$ @Andy : that's "Volts : 380" and "Kw: 0.37" so pretty small. \$\endgroup\$
    – user16324
    Commented Jan 13, 2017 at 13:19
  • 1
    \$\begingroup\$ @BrianDrummond I'll take your word for it LOL \$\endgroup\$
    – Andy aka
    Commented Jan 13, 2017 at 13:26

4 Answers 4

5
\$\begingroup\$

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."

enter image description here

\$\endgroup\$
8
  • \$\begingroup\$ Hi and thank you for a detailed explanation. I will be using the multi tool machine for hobby purposes only. So i wont really want to invest to much money, and i can see a variable frequency drive can be a bit expensive. I updated my post with the picture of the wirings from the motors to the switch \$\endgroup\$
    – mounim
    Commented Jan 19, 2017 at 20:07
  • \$\begingroup\$ I don't think the switch wiring detail adds anything to my assessment of the problem. Unless you can re-connect the motor for 220 volt operation, you will need a transformer to step the voltage up to 380 volts and capacitor rated for 380 volts. The reconnection must be done with wires that are not now coming out of the motor. I will try to post a sketch. \$\endgroup\$
    – user80875
    Commented Jan 19, 2017 at 23:34
  • \$\begingroup\$ Thanks, i understand ... one more banal question really. So cannot just use the end wires as shown on the picture to make the conversion ? i should open the whole motor. Because it is embeded inside the machine, it is quite difficult to get it out, and to get to the casing it self. it is mach easier to just manipulate the end wirings. \$\endgroup\$
    – mounim
    Commented Jan 20, 2017 at 11:59
  • \$\begingroup\$ You can not use the six wires from the motor to the switch. Most motors have a box or a small cover on the side or end. There the internal motor wire are connected to longer wires going to the switch. If that is opened, you may find two wires spliced to each wire going to the switch. If that is the case, you have what you need. Continued - \$\endgroup\$
    – user80875
    Commented Jan 20, 2017 at 12:15
  • \$\begingroup\$ If you change those connections you must be very careful not to lose track of which wire is connected to each external wire. If you find only six wires, then the motor must be completely opened up. Finding the internal connections, disconnecting them and connecting them to 12 external wires is likely to be very difficult. The wires are solid rather than stranded. They are somewhat fragile. There is very little length to spare. They are soldered or welded. Special high-temperature insulating tape is used to insulate the connections. I have practically no experience with that. \$\endgroup\$
    – user80875
    Commented Jan 20, 2017 at 12:22
2
\$\begingroup\$

"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.

\$\endgroup\$
5
  • \$\begingroup\$ The motor does not "draw" any kW. It could be said to draw X current under some stated load conditions. According to international standards (IEC and NEMA), power ratings listed on motor labels are the rated mechanical output. Input power is not listed, only the rated current and power factor. \$\endgroup\$
    – user80875
    Commented Jan 19, 2017 at 22:04
  • \$\begingroup\$ @Charles Cowie - I assumed that, but I edited it with your clarification. Thanks. \$\endgroup\$ Commented Jan 19, 2017 at 22:20
  • \$\begingroup\$ Technically, it does not draw 0.37 kW at full load, it produces .37 kW of mechanical power at full load and "draws" 0.37 kW plus losses. \$\endgroup\$
    – user80875
    Commented Jan 19, 2017 at 22:42
  • \$\begingroup\$ thanks guys .. very advanced discussion for me. transformers seems like an expensive solution. i guess i would just keep it with the capacitor solution. \$\endgroup\$
    – mounim
    Commented Jan 20, 2017 at 0:03
  • \$\begingroup\$ @Mike .. when talking about "220 to 380v 3 phase .5 kva step-up transformer" .. is this model suitable ? galco.com/buy/ABB/ACS143-1K6-3-U \$\endgroup\$
    – mounim
    Commented Jan 20, 2017 at 12:32
1
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\begingroup\$ Thanks for the answer, but i am really not sure what capacitor should i use, should i use one according to the 0,37 or for the 0,51 Kw, and is a single capacitor is enough or do i need a starting capacitor as well ?. i am also confused about the wiring at the switch how to connect my capacitor to that . i just updated my question with a photo pf the switch \$\endgroup\$
    – mounim
    Commented Jan 19, 2017 at 20:08
  • \$\begingroup\$ @mounim C1 is according to 0.37kW, and C2 was to make up the difference to 0.51kW, but I modified my diagram. C2 is now according to 0.51kW (for speed 2). The upper formulas are to calculate the value using actual current. The lower ones are so you can use the 380V current rating without multiplying by √3. \$\endgroup\$ Commented May 27, 2021 at 6:52
1
\$\begingroup\$

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.

enter image description here

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ Please provide a link or citation for the graphics you copied into your answer. We need to give credit to the original creator. \$\endgroup\$ Commented Oct 29, 2020 at 13:35
  • \$\begingroup\$ @Elliot Alderson Source of diagram: Part extract from circuit diagram. The rest, my own work. \$\endgroup\$ Commented Oct 29, 2020 at 23:58
  • \$\begingroup\$ Seeing as the coils are all overlapping: To check the order of the poles, just connect a torch battery across the coil and probe around the inside with a magnet or a compass, you can check that the poles of all 3 phases are in the same order NS, NS, NS, or SN, SN, SN. \$\endgroup\$ Commented Nov 3, 2020 at 0:54
  • \$\begingroup\$ @winny I see in the report, it says anonymous user? I edited it myself when I did some research into what cam switches were available, and came up with a better way to prevent C2 from damaging the contacts when discharged, or connected to C1 with power on. \$\endgroup\$ Commented Nov 12, 2020 at 1:38

Not the answer you're looking for? Browse other questions tagged or ask your own question.