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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.

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  • \$\begingroup\$ 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. \$\endgroup\$ – Brian Drummond Aug 25 '16 at 18:43
  • \$\begingroup\$ yes, it is a motor from broken small fan \$\endgroup\$ – dmz Aug 26 '16 at 7:26
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The number of turns should be proportional to the voltage. Your calculation gives you that. The power dissipated in the resistance of the wire probably needs to be held constant. To do that I squared R must remain the same. You could also calculate the wire size based on current density. If there is a capacitor in series with one winding, you need to determine how to deal with that.

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To change voltages on a motor but keep the same performance, the number of turns and the wire cross-sectional area have to change proportionally to the ratio of the 2 voltages. If you want to increase voltage, the number of turns increases and the cross-sectional area of the wire decreases. If you want to decrease voltage, the number of turns decreases and cross-sectional area of the wire increases.

One hint to help with this is that in order to double the cross-sectional area of standard magnet wire, you need to decrease the wire a gauge by 3. So, for example, to double the cross-sectional area of 25 AWG wire, you need 22 AWG wire.

Also note that doubling the cross-sectional area of the wire decreases the resistance by half, so if you reduce the number of turns by half and double the cross-sectional area, your total resistance will be 1/4th the original resistance.

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.

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