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Discuss Single phase induction motor schematic in the Electrical Forum area at ElectriciansForums.net

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Hello, I've found an old motor originally designed for operating jealousies.
Could you help me read and better understand the following schematic?
As far as I understand, it is a single phase induction motor, there is 47 uF capacitor connected to it in order to start.
There are 3 contacts marked 1, 2, 3 + PE. There is also a switch button.
What would be the correct way to wire 1, 2, 3 contacts? What is the purpose of the switch button? (change of rotation direction?)
I measured resistance between these:
12 - 121.7 ohm
23 - 242 ohhm (both windings?)
13 - 120.5 ohm

Thank you

20190914_180406.jpg
 
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Lucien Nunes

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The motor is a permanent split capacitor type (it is not a starting capacitor, it is in circuit all the time the motor is running, and is probably 4.5μF not 45.)

The circuit for these is standard. Ignoring the limit switches etc and concentrating only on the motor, the neutral connection goes to the junction between the windings, terminal 6, the capacitor connects between the two other ends as shown (at terminals 4 & 5). Connecting line to terminal 4 will run the motor one way, terminal 5 will run it the other way. You will note that in each case, one winding is energised directly from the supply, the other via the capacitor. The direction of rotation depends on which winding receives a leading phase shift by being connected to the capacitor.

If you can only get to terminals 1,2,3 and not 4,5,6, you can connect neutral to terminal 1, line to 2 or 3 for forwards or reverse. However there will be a limit switch connected in one direction, the manual switch in the other, and what looks like a thermal cutout overall. If it is indeed a thermal cutout, you should make the neutral connection to terminal 1 (not 6) anyway.

Note that the motor is short-time rated only (KB= Kurzbelastung) 4 minutes maximum followed by a cooling period typically of 15 mins. This is on account of the compact size of the motor not having enough surface area to dissipate the heat produced.
 
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The motor is a permanent split capacitor type (it is not a starting capacitor, it is in circuit all the time the motor is running, and is probably 4.5μF not 45.)

The circuit for these is standard. Ignoring the limit switches etc and concentrating only on the motor, the neutral connection goes to the junction between the windings, terminal 6, the capacitor connects between the two other ends as shown (at terminals 4 & 5). Connecting line to terminal 4 will run the motor one way, terminal 5 will run it the other way. You will note that in each case, one winding is energised directly from the supply, the other via the capacitor. The direction of rotation depends on which winding receives a leading phase shift by being connected to the capacitor.

If you can only get to terminals 1,2,3 and not 4,5,6, you can connect neutral to terminal 1, line to 2 or 3 for forwards or reverse. However there will be a limit switch connected in one direction, the manual switch in the other, and what looks like a thermal cutout overall. If it is indeed a thermal cutout, you should make the neutral connection to terminal 1 (not 6) anyway.

Note that the motor is short-time rated only (KB= Kurzbelastung) 4 minutes maximum followed by a cooling period typically of 15 mins. This is on account of the compact size of the motor not having enough surface area to dissipate the heat produced.
Thank you for explaining the principal and clarification. It works exactly the way you described. You helped me a lot, have a good day!
 
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