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#1
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stepper motor of course Can motor drivers only be paired with specific motors? For example, I noticed that most of the kits for the NEMA34 1200 oz-in motors are often paired with the KL-9082 drivers, but it seems like the KL-8060 might also work with the NEMA34 1200 oz-in motors as well.... at least from a current control capability. I think I have enough sence to know that I cannot run a 4030 with a 1200 oz-in motor simpily because the 4030 cannot support the current requirements for the bigger motor....but are current capabilities the only thing you look at when you match a driver to a motor? What does the 9082 provide for the 1200 oz-in motors that the 8060 does not? |
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#2
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| Because the 9082 is designed to deliver 8.2 amps, it should run cooler, and possibly be more durable than the 8060, which would be running at it's max capabilities.
__________________ 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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#3
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| I have also noticed that both of the drivers have a voltage range from 24 to 80V. My current plan is to use an extra 36 volt 9 amp keling power supply I have lying around to power a single newer + larger motor yet to be determined. Does it mater what voltage you choose for your motor as long as you have enough amps at the power supply and its voltage is within the range of the driver? additionally I have noticed the the amperage dip switches on the 9082 do not include a combination of switch positions that provide exactly 6 amps or 3 amps which are the only two wiring methods shown on the 1200 oz-in motor that produce the rated power or 1200 oz-in. if I used the 9082 and wired the 1200 oz-in motor for 6 amps, would I set the 9082 to 5.5 or 6.4 amps? The 8060 has a dip switch combination that would allow it to achieve exactly 6 amps which makes the 8060 look like a better choice for me......bu again,I am a newbie so I may be completely wrong |
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#4
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| Speed is proportional to voltage, and torque is proportional to current. Running at 5.5 amps will give you 91% of the rated holding torque. It's a bad idea to run motors at more than their rated current. For maximum performance, the voltage should be 32 times the square root of the motors rated inductance. For the one with an inductance of 6, that would be about 78 volts. Running it on 36Volts would make your potential top speed half of what it would be at 78V. But it'll probably run a bit cooler.
__________________ 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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#5
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| just to double check my units???? Inductance is given in mili-henry's. I take the 2root of that and multiply by 32 and that gives me the voltage required to achieve max performance? The solution is in volts....not mili-volts or kilo-volts... Sorry if this sounds stupid, My electronics experience is non existant outside of a simple seires or parallel lightbulb circuit and good'ole ohm's law |
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