Can a Motor be Converted to a Dynamo and How to Do It?

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A motor can be converted to a dynamo, particularly if it's a permanent magnet motor, but the type of motor matters. Brushed motors are suitable for conversion, while brushless motors typically do not contain permanent magnets and may function differently. The conversion process involves turning the motor and wiring it to create an output current, but adjustments may be necessary, such as advancing the commutator timing to enhance efficiency. Understanding the specific motor type and its operation is crucial for successful conversion. Overall, the feasibility of converting a motor to a dynamo depends on the motor's design and characteristics.
B_Mulcahy
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Hey,

I understand that essentially a dynamo is the inverse of a motor however I was wondering if I could have a little more detail on the subject, how exactly can you convert a motor to a dynamo? Is it as simple as turning the motor and wiring it up to an output to create the current or do you need to wire the motor differently?

Thanks
 
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Depending upon the motor, yes. Permanent magnet motors such as would be used in a power window in a vehicle would work this way.
 
Averagesupernova said:
Depending upon the motor, yes. Permanent magnet motors such as would be used in a power window in a vehicle would work this way.

I assume then as long as the magnet is permanent the type of motor does not matter? (I.e brushed or brushless or in-runner vs out-runner motor)

Thanks
 
B_Mulcahy said:
I assume then as long as the magnet is permanent the type of motor does not matter? (I.e brushed or brushless or in-runner vs out-runner motor)

Thanks
No. It would need to be a brushed type motor. I don't think brushless motors even have magnets. I have no experience with them other than AC induction motors and I suspect most DC brushless motors are in fact induction motors with a variable frequency drive incorporated. A little googling should turn up many results on basic motor operation.
 
You will probably have to advance the commutator timing in generator mode to prevent wear and inefficiency.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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