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I have a motor from a treadmill and I would like to know what size wire to ues to get the most amp/volts out of it? Answered

What I have is a 17.2 amps ,7099 rpm ,25 c , calss b 2 1/2 hp @ 130 VDC CW rotation. With a drill motor attached I can get 21 v out of it ,today it went into the air a top a 21ft pole and we had a storm come through I had hooked a extention cord to it so I cold get to it on the ground and I was not even getting 5 volts out of it in the height of the wind. It is a 50ft orange12 or 14 gage wire cord . Its hooked up to a blocking dioed that a brother of mine helped me wire via the internetand my load was a 12v battery with 5.95 v charge in it and now it has none. What do I need to do to make it right?



9 years ago

Hi, The main thing that is holding you back is the motor itself...it has a bad voltage/rpm ratio.....you want the two numbers to be as close to each other as possible....I have a treadmill motor 90VDC at 3500rpm...I can hand crank this thing and get 6v out of it...I would suggest gearing down the motor (big gear on blades, small gear on motor) and try to maximize the surface area of your blades as much as possible....Also, you don't need a full wave bridge rectifier like you have in your schematic. This could be loosing power....this is used to convert AC to DC but since you already have a DC motor all that is needed is one diode to regulate the direction of the current...good luck


10 years ago

Your circuit diagram is hard to read - but you are basically wiring the motor up to the battery in the correct polarity for rotation, but with the diode preventing the battery from preventing the motor being powered.

At 7100rpm you're expecting 118 rotations per second - the wind isn't going to put that out.
Getting 5 volts out of a 130 volt motor tells me you're getting 5/130*7100 = 270 rpm.

Since the motor is not putting out near its maximum - you should be okay using that wire.

To get your rpm's up you could use a steeper inclined blade - remember a large resistance on the line (like charging a battery) will act as a brake on the system, slowing it down - reducing the available voltage.

There are blade design programs available (and web wizards) that let you figure out what wattage a given design should put out at varying wind speed - and you can design the load (charger) around that.


Answer 10 years ago

correction* 'large resistance on the line' should read 'large load on the line'