# rewinding AC induction motor to work with lower supply voltage

I have small single phase AC induction motor. It runs on mains voltage (230 V) and I'd like to modify it to run from 12 V transformer. Motor has two stator coils.

How to calculate new number of turns for the coils?

My guess is that stator must give same magnetic flux before and after rewinding. Therefore number of ampturns must stay constant.

Current with no load is 0.09 A and with blocked shaft it is 0.1 A. Apparent power is about 230 * 0.1 = 23 VA. To get the same power but with 12 V current must be about 23 / 12 = 1.9 A. Solving for new number of turns I get: old number of turns * 0.1 / 1.9. Does it sound reasonable?

As for wire thickness I guess that I have to keep DC resistance constant and scale up the wire size.

• If the current varies by 10% between stall and no-load this is a VERY inefficient motor - what is it, a small fan? That might significantly change the answer. – Brian Drummond Aug 25 '16 at 18:43
• yes, it is a motor from broken small fan – dmz Aug 26 '16 at 7:26

For your problem, if you are trying to convert a 230 V motor to a 12 V motor you would need to divide the number of turns by $230/12 = 19.2$ and multiply the cross-sectional area of each turn by 19.2. If your motor has 95 turns and 25 AWG wire, then your 12 V winding would have 5 turns and 12 AWG or 13 AWG wire.
If there is a capacitor for this motor, it will change inversely proportional to the voltage ratio squared. So in your case, the capacitor will have to be $19.2^2 = 368$ times bigger.