I am working on a DIY Rotary Phase converter with pony motor. This will use single phase 240v and run capacitors to feed 2 legs of a 3 phase motor, which will generate the 3rd leg.

It uses a 1/2 HP 120v motor (pony motor, starter motor) to start a 5 HP 3 phase motor. I would like auto-start and cutout, so the idler and starter are never energized separately and the starter cuts out after the idler spins up.

The approximate design is below. A 120v switch turns on the starter motor and a 2 pole relay to power L1 and L2 of the idler. The idler has run capacitors between L1-L3 and L2-L3. When the idler starts, L3 will energize the cutout relay (240v NC), which turns off the starter motor. L1, L2, and L3 would be used to drive a 3 phase load.

I used ground symbol for neutral in the schematic.

My primary concern is with the cutout relay wired in parallel with the load on L3. Will this cause problems? If so is there an easy modification I can make to the circuit? Will the cut out relay energize prematurely due to the run capacitors?


simulate this circuit – Schematic created using CircuitLab

Information regarding rotary phase converters: http://www.metalwebnews.com/howto/ph-conv/ph-conv.html http://www.paragoncode.com/shop/rotary_converter/

Final simplified design:


simulate this circuit

The circuit diagram tool doesn't have an option for aux contactors or 3 pole relays, so RLY AUX indicates a 3rd pole or an AUX connection on the contactor

  • \$\begingroup\$ Why are there lamps in series with the start motor and the cutout relay? Are you using a lamp to drop the voltage because the cutout relay has a 120 volt coil? Is the voltage across L1 and L2 240 volts from two 120 volt hot lines? Please revise the question to make it more clear. \$\endgroup\$ – Charles Cowie Jan 15 '17 at 19:44
  • \$\begingroup\$ If you are trying to implement a scheme that you found on the internet, a link to the original description might help. \$\endgroup\$ – Charles Cowie Jan 15 '17 at 20:07
  • \$\begingroup\$ Lamps were superfluous. L1-L2 is 240v. The cutout relay would be a 240v NC relay. The phase generation part isn't really where I have questions, I'm trying to see if powering a relay off of 1 leg of the 3 phase side will cause problems, or energize so fast that the starter won't spin up. \$\endgroup\$ – user15741 Jan 15 '17 at 20:48

The RLY1 needs to be delayed until the start or "pony" motor has brought the idler motor up to speed. Delaying the energization of the start motor until the idler is up to speed is what avoids the "monstrous" inrush current. There will be no voltage on L3 until RLY1 is energized, so RLY2 may not de-energize the start motor soon enough. You will need to either use a manual starting sequence as shown in the paragon code link or use a time delay relay.

Probably something like this:

enter image description here

  • \$\begingroup\$ Will L3 be instantly energized to ~240v when L1 and L2 are, or will the idler need to get to speed first? Are you wanting start motor cut out as soon as RLY3 is on to prevent overspeed on start motor(due to pulley reduction)? I had planned for the start motor to just replace a very large start capacitor by providing a starting nudge (like the rope method), not spinning up to speed. But your in design RLY2 and RLY3 could be the NO and NC connections of one relay, so a 1 pole switch, delay timer, and a DPDT relay do it all. \$\endgroup\$ – user15741 Jan 16 '17 at 17:21
  • \$\begingroup\$ I was assuming that the start motor was directly coupled to the idler and that the 240 volt circuit breaker would trip if the 5 hp was connected at standstill. The start motor should cut out as soon as RLY3 is on to prevent one motor from driving the other while energized. Yes relays could be combined. The ones I am familiar with that have multiple contact options could be fairly expensive. \$\endgroup\$ – Charles Cowie Jan 16 '17 at 17:44

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