I’d like to use a BLDC motor & control on a drill press. I’ll explain what I think I can do… and you can tell me where I’ve gone wrong.

It appears 1/2 HP is quite enough for most drill press operations. Thus a 400 watt motor & controller should do the job. The DC power supply would need to supply the motor’s voltage (12,24,48, etc) at, minimum - 400 watts. I'm also thinking a 3-phase motor with Hall should be easy to setup and run

The desired speed range is 100 to 3000 rpm.

If I used a 400 watt BLDC with a max speed of 12,000 rpm, and used 4:1 timing pulleys to get the 3000 top end - would I get the expected torque increase? I’m mostly concerned about the low-speed torque and I understand BLDC’s are good at sustaining this.

It all looks to simple - I just get the the 400w motor, power supply, controller and pulleys, and I’d be good to go - basically. Essentially I’m tinker-toying a system together... So now I need a reality check to see what is REALLY needed!

  • \$\begingroup\$ Yes, a 4:1 pulley will give you approximately 4x increased torque (minus losses.) \$\endgroup\$ – Jon Watte Jul 8 '17 at 3:25
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    \$\begingroup\$ I don't see why you would use a 12,000 RPM motor for a drill press. If you use a motor with about the same speed as your current motor (probably either a little under 1800 or 3600 rpm if you are in the US), then you can re-use the existing belt and pulley system on the drill-press. A 12,000 RPM motor with 400 W mechanical output will only put out 400W at 12,000 RPM. It will more or less linearly de-rate to 0 W at zero RPM. \$\endgroup\$ – mkeith Jul 8 '17 at 5:05
  • \$\begingroup\$ Note that if the motor mechanical output is 400W, the input electrical power will be higher. Maybe plan for 600 W. Likewise, if the input power is 400W, the mechanical output power will be less than 400W. \$\endgroup\$ – mkeith Jul 8 '17 at 5:06
  • \$\begingroup\$ The current arrangement provides that speed range with a mechanical solution: A Reeves drive with 2 pulleys that expand and contact to proved a variable ratio. However, it is so noisy and difficult to adjust I'd like to replace it. The ease of use of the BLDC controls on the lathe & mill have spoiled me! Plus I'm just a hobbyist and like to tinker... BLDC would NOT be a cost-effective solution but I believe quite elegant! \$\endgroup\$ – Harrzack Jul 8 '17 at 9:56
  • \$\begingroup\$ Here is a link to the system I'm considering: \$\endgroup\$ – Harrzack Jul 8 '17 at 10:15

There is a complex set of tradeoffs at work here that you may not understand. I will try to explain. I may make some generalizations that are not 100% true, but I want to focus on the big picture.

The output power (mechanical power) of a motor is torque * speed. If you use N-m for torque and radians/second for speed, then the units will be Watts. This is very convenient.

Consider your motor rated at 400W at 12,000 RPM. 12,000 RPM is 1257 rad/sec. So the output torque at that speed is 400/1257 = 0.318 N-m. This is likely the maximum sustained torque that the motor is capable of. Sure, you can generate more torque for a short time, but if you attempt to do it for an extended period, the motor windings will overheat. So if you run your 12,000 RPM motor at 1000 RPM, you will still have to live with 0.318 N-m. 0.318 N-m * 1000 rpm * 2pi / 60 = 33 Watts. That is not much mechanical power available at 1000 RPM.

So, as soon as you move down from the rated speed, the available power drops off linearly to zero power at zero speed. This is why you don't want to run a motor far below its rated speed when you are trying to use the motor to do work.

Now consider the case for a motor rated at 400W at 3000 rpm. By simple ratios, that motor will produce 4x more torque (12,000/3000). A 2000 rpm motor would produce 6x more torque than the 12,000 rpm motor.

I am not trying to suggest that you shouldn't use speed control on your motor. I am just pointing out that the multi-speed pulley systems maintain full motor power output over a wide range of speeds, but speed control of the motor will inherently give up power. So, if you opt to have a spindle motor, you should select the slowest motor that will do the job, which means being realistic about maximum required speed.

There is one more thing. It is often possible to drive a motor faster than its rated speed at reduced torque. So if you need to drill a small diameter hole in a soft material like wood or aluminum, you may be able to overspeed your motor considerably, as long as you also reduce current. VFD's and BLDC drivers may have the ability to do this. So this is another argument in favor of getting a lower speed motor rather than a 12,000 rpm motor. It is somewhat common to run induction motors up to 2x rated speed. For example, the typical 4-pole induction motor rated for a bit under 1800 RPM can be run all the way up to 3600 RPM and still produce full rated power (but not full rated torque).

Hope this helps.

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    \$\begingroup\$ Thanks! Very good info indeed. Will digest and look for lower speed motors. The hi speed one I found was a convient package, and the 12000 rpm was a bit startling! The DP I have now (with Reeves variable speed) offers a speed range of 500 to 3000 - I'm taking as my base range. So if I can find a moto with a max speed closer to 3000 and limit/eliminate the amount of "pulley" reduction ahead I should get a workable system. \$\endgroup\$ – Harrzack Jul 8 '17 at 19:56
  • \$\begingroup\$ I would say the more you work with wood and aluminum, the less crucial low-speed torque is. The more you work with steel, especially if you want to drill large holes in steel, the more important it is to harness full power at low speed. A lot of the machine tool forum members seem to like using 3-phase 1800 RPM induction motors with VFD speed control. These can give rated power from 1800 to 3600, and half rated power at 900 RPM. But your spindle motor and drive is very affordable! It would be a great deal for a CNC router for sure. \$\endgroup\$ – mkeith Jul 8 '17 at 20:05

yes and No.

Yes you can flatten the torque curve but not the power curve.

Since RPM at no load is voltage control and Torque is controlled by V/DCR at 0 RPM and then reduces with rising RPM due to back EMF.

Thus lowering the voltage to reduce RPM increases available torque and by using full bridge low impedance PWM control the average DC voltage can be controlled and maintain a constant driving impedance.

In the low speeds higher motor currents when more torque but lower voltage increases conduction losses significantly .

Perhaps you can improve cooling by some means at low RPM such as a water cooled engine.

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Do you really need constant power from 100 to 3000 RPM?

2 sets of 4 pulleys is the most common method with a fan belt.


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    \$\begingroup\$ A BLDC with appropriate control can also give you constant torque, because torque is directly proportional to current through windings. The reason you need VFD for AC induction motors is because of the way the induction needs to be out of phase; a good BLDC controller could achieve the same thing for three-phase BLDC. \$\endgroup\$ – Jon Watte Jul 8 '17 at 3:25
  • \$\begingroup\$ I guess I miss-spoke when I may have said CONSTANT power. I have a sieg lathe & mill that both use BLDC motors and have GOOD torque at low speeds. \$\endgroup\$ – Harrzack Jul 8 '17 at 3:35
  • \$\begingroup\$ ...I used a VFD on a bowl lathe I designed some years ago and was quite dismayed at the little torque that was available at low speeds. I ended up getting a 2-speed gearbox to get a decent low end. And ai was using a 3-phase AC motor \$\endgroup\$ – Harrzack Jul 8 '17 at 3:41
  • \$\begingroup\$ (Edit isn't working on the iPad) - back in those days there were only very expensive Brushed D.C. motors available, and the VFD (new kid on the block) was a less costly alternative. I'd just like to reproduce what Sieg has done in powering their machines. \$\endgroup\$ – Harrzack Jul 8 '17 at 3:44
  • \$\begingroup\$ VFDs are better now than they used to be. Or so I have been told. You can program them to allow higher torque at low RPM, and they have sensorless vector control, etc. \$\endgroup\$ – mkeith Jul 8 '17 at 5:09

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