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Rewinding PM motor to change voltage 1
- Thread starter Bigjohn01
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I think I understand why the rating will be cut in haif, which I do'nt want to do.By halving the wire size, the resistance of the original turns will be doubled. Doubling the length will make the resistance 4 times the original,which results in the 24v current being i/4 the value at 12v. My understanding the power is dependent on "ampere-turns". 1/4 the current x twice the number of turns = 1/4 x 2 = 1/2 the number of ampere-turns. What determines the rpm of the motor? The present windings occupy less than half the space in the armature. Seems like maybe a combination of wire size and number of turns to keep the ampere-turns the same is necessary??? I may never understand it!!
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Hi Bigjohn,
If you double the number of turns, your resistance will go up by a factor of four. You will not halve the wire diameter. You will halve the cross sectional area. So the diameter would go up by about the square root of two. For instance, if the wire diameter is 1mm it would be 1.414mm for double the turns. The voltage constant (KE)is what limits the speed of your motor at no load (mV/rpm). You can estimate this by calculating:
mV/rpm=1000*(VT-INL*RT)/SNL
Where:
VT is applied terminal voltage
INL is no load current
RT is armature resistance
SNL is no load speed
When you double the number of turns the KE doubles so you reach your terminal voltage at a lower speed. You also double your torque constant (KT). This means that you have twice the torque per amp. If you go from a 12 volt 5500rpm motor to a 24 volt 5500 rpm motor, you will not halve the maximum output power (PMAX).
PMAX=(RT/4) * (VT/RT-INL)^2
Try this website for a pretty useful commentary.
If you double the number of turns, your resistance will go up by a factor of four. You will not halve the wire diameter. You will halve the cross sectional area. So the diameter would go up by about the square root of two. For instance, if the wire diameter is 1mm it would be 1.414mm for double the turns. The voltage constant (KE)is what limits the speed of your motor at no load (mV/rpm). You can estimate this by calculating:
mV/rpm=1000*(VT-INL*RT)/SNL
Where:
VT is applied terminal voltage
INL is no load current
RT is armature resistance
SNL is no load speed
When you double the number of turns the KE doubles so you reach your terminal voltage at a lower speed. You also double your torque constant (KT). This means that you have twice the torque per amp. If you go from a 12 volt 5500rpm motor to a 24 volt 5500 rpm motor, you will not halve the maximum output power (PMAX).
PMAX=(RT/4) * (VT/RT-INL)^2
Try this website for a pretty useful commentary.
Sorry, I made a mistake:
"So the diameter would go up by about the square root of two. For instance, if the wire diameter is 1mm it would be 1.414mm for double the turns."
The diameter would of course be reduced. In this case if the wire diameter were 1.414mm it would be 1mm.
Monday half and double trouble.
"So the diameter would go up by about the square root of two. For instance, if the wire diameter is 1mm it would be 1.414mm for double the turns."
The diameter would of course be reduced. In this case if the wire diameter were 1.414mm it would be 1mm.
Monday half and double trouble.
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- #10
edison123:
I think you have it backwards. Bigjohn is requesting that his no-load speed at 24V be cut in half. This means that he wants to cut his back-EMF constant Ke in half. The torque constant Kt of a motor must always vary with Ke, so this must be cut in half as well.
The torque generated is proportional to the armature magnetic flux, and this flux is equal to the magnetomotive force (MMF) divided by the reluctance of the magnetic circuit. The reluctance does not vary with winding (barring magnetic saturation), so flux and torque are proportional to the MMF.
The MMF generated by the coils is just n*I, the number of turns multiplied by the current. This means that Kt is proportional to the turns count n, and Bigjohn should cut his turns count in half.
If he does this with the same diameter wire, reducing the size of his winding, he cuts his resistance in half. In this case his mechanical time constant Tm [=(J*R)/(Kt*Ke)] is doubled (assuming no significant J change).
If he does this with wire of double the cross-sectional area, keeping is overall winding size the same, he reduces his resistance by a factor of 4, so his mechanical time constant stays the same.
Curt Wilson
Delta Tau Data Systems
I think you have it backwards. Bigjohn is requesting that his no-load speed at 24V be cut in half. This means that he wants to cut his back-EMF constant Ke in half. The torque constant Kt of a motor must always vary with Ke, so this must be cut in half as well.
The torque generated is proportional to the armature magnetic flux, and this flux is equal to the magnetomotive force (MMF) divided by the reluctance of the magnetic circuit. The reluctance does not vary with winding (barring magnetic saturation), so flux and torque are proportional to the MMF.
The MMF generated by the coils is just n*I, the number of turns multiplied by the current. This means that Kt is proportional to the turns count n, and Bigjohn should cut his turns count in half.
If he does this with the same diameter wire, reducing the size of his winding, he cuts his resistance in half. In this case his mechanical time constant Tm [=(J*R)/(Kt*Ke)] is doubled (assuming no significant J change).
If he does this with wire of double the cross-sectional area, keeping is overall winding size the same, he reduces his resistance by a factor of 4, so his mechanical time constant stays the same.
Curt Wilson
Delta Tau Data Systems
Arrgh....
Never mind! (The perils of trying to post while doing other things...) Of course, for the back EMF voltage to reach the supply voltage at half the speed, the back EMF constant must double, and therefore so must the torque constant, and therefore so must the turns count.
With the same diameter wire (assuming there is room, as Bigjohn implies), the mechanical time constant is cut in half.
With wire of half the cross-sectional area to keep the same "fill", the mechanical time constant stays the same.
Sorry about that!
Curt Wilson
Delta Tau Data Systems
Never mind! (The perils of trying to post while doing other things...) Of course, for the back EMF voltage to reach the supply voltage at half the speed, the back EMF constant must double, and therefore so must the torque constant, and therefore so must the turns count.
With the same diameter wire (assuming there is room, as Bigjohn implies), the mechanical time constant is cut in half.
With wire of half the cross-sectional area to keep the same "fill", the mechanical time constant stays the same.
Sorry about that!
Curt Wilson
Delta Tau Data Systems
If this is just for starting model engines. My experience with this from 20 years ago if the electric motor has enough torque to get the engine spinning even 2000 RPM the free running speed of the motor does not matter that much. If the engine is in good order it should only take a fraction of a second so even if the starter motor is pulling 300 or 400 percent of its rated current its not a big deal, duty cycle is almost zero unless you have a lot of people lined up all ready to start up at the same time.
I would just try it. If the motor does not have enough torque find a different motor, one of those 90 volt tread mill motors you find at surplus places may even work, I know they would if you jumped up to 48 volts.
Unless you just want to do the rewind for the fun of it, but the wire is going to cost as much as a surplus motor.
I would just try it. If the motor does not have enough torque find a different motor, one of those 90 volt tread mill motors you find at surplus places may even work, I know they would if you jumped up to 48 volts.
Unless you just want to do the rewind for the fun of it, but the wire is going to cost as much as a surplus motor.
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- #15
I am looking for a motor to power a 6/0 Penn fishing reel. Years ago, some fishermen in my city used a geardrive from WW2 aircraft to build some successful units. Needless to say, they're getting scarce and the motors are no longer available. They are 24-27v, field wound, shunt wired and very compact. Most of the motor failures were due to the shaft gear teeth wear or stripping. Locally, the 24v units outperform the factory made 12v units so everyone wants 24v motors. I have a solution to the gear problem but I feel 24v is the way to go.
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- #16
Found the answer to my situation. Acquired a 24v motor presently used by a manufacturer of electric drives for reels. Also found a 12v starter motor used on 6.5hp Briggs lawnmower engines. The two armatures are the same diameter and the 12v laminations are almost half again longer than the 24v. Both have 10 slots. I rewound the 12v armature with the same size wire and number of turns as the 24v. Haven't had time to compare performance of the two, but expect the rewind job to fill my needs. The starter motors are easy to find at repair shops. Sems a lot of people balk at the price of a replacement battery. Thanks everyone for your input.
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