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I have just bought online what I thought was a synchro as described in this page, but I am trying to get my head around how to use it since it does not correspond to any of the types I currently know: it has 7 wires.

It's not a resolver transmitter, because it does not have 6 wires (or 4 if all returns are tied together), and it's not a synchro as defined above because it has more than 5 wires (4 if the wires are tied together). I thought it could be redundant as well but this doesn't check out either. I hope it's two-speeds somehow.

My objective is to do do a proof of concept of a resolver (transmitter)-to-digital chain. I have a Scott-T transformer. What do you reckon is this component exactly, and what do the wires do? I have not found any datasheet online, it's too old.

Resolver

The nameplate reads "Microtecnica Autosyn resolver AY 528 A 45 A 2, manufactured under Bendix licence. rotor 2 phase 26V stator 2 phase 11.8V". Wires on a rear view:

  • At 0° (vertical): black 1, yellow 1, red 1, blue
  • At 45° CW: red 2
  • At 180°: black 2
  • At -45° CW: yellow 2

Edit: - Black 2 to yellow 2: 202Ohm - Black 2 to red 2: 200Ohm - Yellow 2 to red 2: 400Ohm - Yellow 1 to blue 1: 40 Ohm (not 400) - Black 1 to red 1: 40Ohm - Everything else > 1Mohm

It looks like one side has two phases back-to-back (centerpoint), and another has two isolated phases. Nonetheless this does not correspond to anything I know. Perhaps the primary (rotor, then) is compatible with 2 different excitation voltages (and the higher resistance would mean less current draw + lower output impedance)? I'll test this theory this week-end if no one answers.

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  • \$\begingroup\$ Is one of the wires a ground? What do you measure when you check resistances between wires? \$\endgroup\$ Commented Jun 5, 2020 at 12:52
  • \$\begingroup\$ Thanks for the comment, Spehro Pefhany. I updated my post. \$\endgroup\$ Commented Jun 5, 2020 at 13:33

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I believe this is a "differential resolver" since it has two phase input and two phase output. That could have 8 wires but since there's a common Black2 only 7 are required.

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  • \$\begingroup\$ Good idea, that means I should short circuit one half of the primary and use this short circuit as my return for the other end, to get the angle of the differential or 90° from it - right (sin(0°)=0, cos(0°)=sin(carrier_pulsation))? \$\endgroup\$ Commented Jun 5, 2020 at 15:12
  • \$\begingroup\$ If you only power one side you should get Asin(theta) and Acos(theta) out of the two output windings. Not sure whether it's better to short the unused coil or leave it open. ` \$\endgroup\$ Commented Jun 5, 2020 at 17:49

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