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I am running a shapeoko2 from inventables and bought a 300 Watt quiet cut DC spindle to run on it. The operating voltage range is 24V to 48V. My power source is adjustable to 48V but can only deliver 8.3 amps. I have currently dialed down the power source to 24V to power the stepper motors via the smoothie board.

I assume then my 300 Watt motor won't be able to run at full efficiency since 300W/24V = 12.5 Amps. What happens when my motor can't pull that current from it's power source. Does that mean my spindle can only produce 199.2 Watts of cutting power if I am running the power supply at 24V, and the stepper motors aren't running? If I am running the spindle off of the same power supply while the motors are running will I see the cutting power vary when the motors are pulling current?

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  • \$\begingroup\$ Not directly related to your question, but if you are operating a small CNC why dont you just find an appropriate PSU? With these specs wont be hard nor expensive to find one. \$\endgroup\$ – Wesley Lee Feb 4 '17 at 21:55
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The actual current that your spindle motor draws depends on how heavily loaded it is. If you load it down to where the power supply reaches its current limit, what happens next is determined by the design of the power supply.

If it simply goes into a constant current mode, then the spindle speed will drop. If it has a "foldback" mode or other protection mechanism, everything will pretty much just stop.

If you are powering other motors from the same supply, then when they are drawing current, the peak current that the spindle can draw will be reduced correspondingly.

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    \$\begingroup\$ In other words, use a slightly smaller diameter cutter to reduce the load and you'll probably be OK. \$\endgroup\$ – Brian Drummond Aug 31 '16 at 22:14
  • \$\begingroup\$ Could not repeat this often enough: Electrical current is proportional to mechanical torque. And with a non-overloaded DC motor speed is nearly proportional to voltage. \$\endgroup\$ – Janka Apr 12 '17 at 20:43

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