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#1
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I been looking for a way to fined out what my step mote vdc is. When I go to buy the power supply and drives, I see this and need to make sure that I get the right size to power and drive everything. The motors are Keling KL23H286-20-8 I will run them as bipolar 2 phase. The info on it is Rated Voltage 4.17 V current/phase 2.8 A Resistance 1.5 O Inductance 6.8 mH. Now with this info can someone show me how to come up with a way of finding what the vdc is. I see in the info that it is Rated for 4.17 Volts.... I believe this Vac. |
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#2
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| 4.17V is the DC voltage. For best performance, run them at 32 times the square root of the inductance, or about 84V, if your drives can handle that.
__________________ Gerry Mach3 2010 Screenset http://home.comcast.net/~cncwoodworker/2010.html (Note: The opinions expressed in this post are my own and are not necessarily those of CNCzone and its management) |
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#4
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| Voltage will be the same for all motors. Adding each additional motor will require more current from the power supply (amps). You only need a driver that can handle 2.8A per motor. Typically you would get a supply that has around 70% of the total current for all your motors since they're not likely to all be running at full power at the same time, so if you have 4 of these motors, that would be 4x2.8=11.2. 11.2*.7=7.84A. Your power supply needs to be able to provide this current or higher. Taking a typical example, a Gecko G203V will handle 80V and 7A. So you would probably be more interested in the G251 or the G540 which can handle 3.5A. However they can only handle up to a 50V power supply. So your motors will not run quite as fast, but this probably will be okay depending on the specifics of your machine. Here's an FAQ with some more info: http://www.cnczone.com/forums/showthread.php?t=52090 |
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