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I am more in mechanics then in electrical engineering so sorry if this question sounds silly for you.

I have a little CNC machine with NEMA 17 stepper motors but their power is not enough for me, so I decided to change them for NEMA 23 stepper motors. I am not sure if my current adapter of 12V and 5A is enough to run 3x NEMA 23 motors of 3A each?

How could I calculate that?

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    \$\begingroup\$ The exact answer depends on the drivers and the way they are controlled, but, as a role-of-thumb, steppers draw current even when stalled (not rotating). 3 motors @ 3A each, results in 9A. You said that you are replacing the NEMA17 motors because they are not powerful enough; but, with the same 5A supply, where do you expect that the new motors will get the energy that they need to deliver the higher power that you want? Replace this supply too, period. Get a 10A one, or higher. \$\endgroup\$
    – mguima
    Sep 10 at 13:04
  • \$\begingroup\$ Thanks for your help. I wanted also to deliver the driver type, but I didn't know how to check it, because it is just a blue board with a lot of white letters, but nowhere is the type written. \$\endgroup\$
    – elano7
    Sep 10 at 13:55
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    \$\begingroup\$ The most common drivers are: DRV8825, A4988, and the TMC2xxx ones. Can you see the letters in the IC that is in the driver piece? But, anyways, I doubt that even the most perfect driver would allow using this 5A supply. Good luck! \$\endgroup\$
    – mguima
    Sep 10 at 22:48
  • \$\begingroup\$ @mguima great! You were right. After a bit of researching I found my driver type. It is A4988 \$\endgroup\$
    – elano7
    Sep 11 at 14:19
  • \$\begingroup\$ @elano7 Since you've now said the driver is an A4988 then you will not be able to set the phase current above 2A peak. In all probability your RMS current is actually going to be limited to about 1.2A RMS max by the module design. \$\endgroup\$ Sep 11 at 14:41
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This question highlights the need to understand the problem space and shows that leaving out information makes providing accurate answers very difficult.

To summarize the question with most of the necessary information: I have a little CNC machine with NEMA 17 stepper motors and a 12V/5A power supply but their power is insufficient. I want to change to NEMA 23 stepper motors rated at 3A RMS and continue to use my current controller (using A4988 drivers) and power supply. Will this work?

Short answer: Yes

Long answer:

  • The drivers you have (A4988) are limited to 2A PEAK per phase. In reality for typical designs you will not get more than about 1.2 - 1.3A PEAK, set by the small potentiometer on the module.
  • Given the limitation in #1, you will get slightly improved torque because you moved from a NEMA 17 magnetic stack to a NEMA 23 magnetic stack. You would also get a gain in torque if you moved to a NEMA 17 stepper with a longer magnetic stack. If you compare steppers here you will see that a longer stack can get you more torque for the same frame size.
  • Changing to larger NEMA 23 steppers will give you a boost in torque, and the motors will operate at less than their rated current, but the gains you achieve will be marginal. I'd suggest you simply get longer (larger magnetic stack NEMA 17 steppers).
  • Given the above, the power supply you have will be adequate as it's sized for the A4988 driver and you won't get to move to 3A. If you really want to increase the power to the stepper motor you will need to change both the power supply and the controllers driving the steppers. Something like a TB6600 driver will work up to 4A, and you'd need a power supply rated at 300W or more and a voltage of 30V or above.

Defining the real problem: Your question is still incomplete of course. What is the reason that you think you have insufficient power (torque?) in the stepper motors you have??

  • Do you lose steps when running a job?

  • Do you have poor surface finish?

A common problem for small CNC (subtractive not 3D printer) machines is that the unit is run with too high a microstep count. As you increase the microstep count the torque to move each step reduces markedly. This results in poor surface finish and poor repeatability for jobs. Keeping the same stepper motors and putting a belt reduction (3 or 4:1) and reducing to Half step (MS1,2,3 on the A4988) drive may help you achieve better results with less re-engineering effort.

It is incumbent on you to accurately define your problem/question so you can get worthwhile answers.

Note: This answer on PEAK vs RMS current may also help you understand some of the complexity in defining power usage. The rules for power usage change when comparing a CNC (subtractive) machine to a 3D printer. The worst case power for CNC is a spline (all three axis move at once). For a 3D printer the Z axis typically moves on its own (layer height setting) and only the X,Y axis move in tandem.

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  • \$\begingroup\$ Thank you for your time and all the help. This is a really detailed and useful answer. You asked why the power is not sufficient. It is not easy to write it in one comment so I will just say that the cnc machine was supposed to work with a 15w laser or a milling machine. I need it for some milling processes but the possible feed rate now is pretty small and the motors just stop if i try with bigger feed rates. This is a shortened explanation about the problem and purpose. \$\endgroup\$
    – elano7
    Sep 11 at 20:24
  • \$\begingroup\$ @elano7 feed rate is only one part of the equation. Feed rate and depth of cut are interdependent. You are also limited by the spindle motor. These small machine are really nothing like a true mill so you have to understand the relationship between depth of cut, feed rate along with the material being cut. This might help you: shopbottools.com/ShopBotDocs/files/FeedsandSpeeds.pdf \$\endgroup\$ Sep 11 at 20:53
  • \$\begingroup\$ thanks. The task is to cut the material in one pass, so the depth is constant. You and the other guys that answered and commented forwarded me miles and miles ahead. \$\endgroup\$
    – elano7
    Sep 12 at 20:10
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You can calculate that 3 motors requiring 3 A each require 3x3 = 9 amps. Even two motors would require 2x3 = 6 amps.

Your power supply is therefore only adequate to run one motor, making no other assumptions.

Sometimes stepper motor drivers are set up to use a lower holding current than their stepping current, so would use less when stationary. So the next step is to estimate, or simulate the operation of the machine, to see whether more than one motor is required to move at the same time.

If I was investing in replacing all three motors, I would replace the power supply as well, so that I could use all three motors at the same time.

Of course, being retired, I'm way behind the times. As Jack points out in comments, the drivers are probably switchers rather than the old fashioned linear types. With linear drivers, supply current is motor current. With switchers, supply power is motor power, and, as the motor voltage is usually a fraction of the supply voltage, supply current can be a fraction of the motor current. So, find out what your drivers are, you may be pleasantly surprised.

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    \$\begingroup\$ Do they run only from 12V tho? If they can run from 5V then one could get more current with buck converter. If it's possible, of course \$\endgroup\$
    – Ilya
    Sep 10 at 13:16
  • \$\begingroup\$ Thanks. I was not sure how to properly calculate the resulting amps. I thought maybe there are some coefficients that make the resulting value more or less than 9A. \$\endgroup\$
    – elano7
    Sep 10 at 13:54
  • \$\begingroup\$ @elano7 The motors will need 9 A, but the drivers themselves will require a bit. That should be specified in the driver documentation, or it may be so trivial compared to the motor current that they don't mention it. \$\endgroup\$
    – Neil_UK
    Sep 10 at 15:00
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    \$\begingroup\$ This answer is probably incorrect. The stepper motors are rated at 3A each, BUT the driver is most likely a switching driver. At 3A the motors probably have only 3-4V across them, so will draw about 9W each. A 12V 5A power supply can deliver 60W. While the current set will determine PEAK current drawn from the supply, the majority of this can be supplied from a capacitor. I'd suggest that all the OP needs to do is ensure sufficient capacitance on the power supply output (perhaps 470-1000uf). \$\endgroup\$ Sep 10 at 16:08
  • \$\begingroup\$ @Jack interesting conclusion, thanks. I've arranged some motors and adapters for testing. After trying them out in the next few weeks I will write the results here. \$\endgroup\$
    – elano7
    Sep 11 at 13:43

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