Power supply for 16 2-phase multiple stepper motors

Hello to everyone,

I’d like to drive 16 stepper motor that have a specs as shown below. What should be specs of the power supply to handle this operation ? (V, amp, watt)
Thank’s to all for your interests

That motor needs 4.5A * 1.9V = 8.55 watts per coil - say 20 watts to be on the safe side.

16 of them will require 320 watts.

If you are using a 24v power supply then it would need to be able to produce 13 amps minimum. Really it should have at least 20 amps to ensure the power supply is not operating at its limit.

...R Stepper Motor Basics Simple Stepper Code

Thank you so much for your interest Robin,

When i wrote to the company that selling this motor for power supply selection, they told me this motor can be driven with CVD245 and for the power supply as stated in this document (https://www.orientalmotor.com/products/pdfs/opmanuals/HM-60128-8E.pdf#page=10) CVD-245 needs power supply current 3.9A or more. In a conclusion, i was told i need 3.9A * 16 = 64A power supply and maybe i should use multiple power supply. Your calculations seem very reasonable to me but i am a little confused with company's techs offers. What is your opinion about it ?

CVD245 https://catalog.orientalmotor.com/item/cvd-dc-input-2-phase-bipolar-drivers/cvd-2-phase-bipolar-stepper-motor-drivers/cvd245br-k

I’d suggest measure the consumption of one motor and its driver with a bench supply, then make your
decision based on data from the actual pattern of usage you are wanting. Add a healthy safety margin
for the supply (ie if you calculate 25A are needed, a 30A supply is a poor choice compared to a 40A
supply).

brunopacheco: Your calculations seem very reasonable to me but i am a little confused with company's techs offers. What is your opinion about it ?

No comment.

I am not a professional engineer.

Feel free to discuss my computations with the company.

...R

i was told i need 3.9A * 16 = 64A power supply

At what voltage?

MarkT: I'd suggest measure the consumption of one motor and its driver with a bench supply, then make your decision based on data from the actual pattern of usage you are wanting. Add a healthy safety margin for the supply (ie if you calculate 25A are needed, a 30A supply is a poor choice compared to a 40A supply).

Let's assume that consumption of one motor is measured as 3A. In that case, for 16 motors do i need a 3A x 16 = 48A Power Supply ? Or as the Robin's power computations , do i need to calculate the total power consumption of motors and divide it to the voltage value of the power supply ?

Robin2: No comment.

I am not a professional engineer.

Feel free to discuss my computations with the company.

...R

I just wonder is it possible to supply motors which draw total 50A, with a power supply that has 20A current value ?

JCA34F: At what voltage?

It is stated as +24 VDC±10% in the document https://www.orientalmotor.com/products/pdfs/opmanuals/HM-60128-8E.pdf#page=10)

brunopacheco: I just wonder is it possible to supply motors which draw total 50A, with a power supply that has 20A current value ?

The combination of a stepper motor and stepper driver acts like a buck converter changing a low current at a high voltage into a higher current at a lower voltage.

...R

It all makes sense now thanks to all

Since the motor is rated at 4.5A, it will actually be using 3.2A per winding at startup (the current is 2-phase quadrature, and at startup the driver chooses 45 electrical degrees, so that both windings take 1/sqrt(2) times the nominal current.

The driver, being a power converter, takes (much) less current from the supply.

People are often confused by specs for stepper current, as its a quadrature amplitude, ie the sum of the squares of the two winding currents equals the square of the nominal current setting.

Think of one winding taking I sin(a) and the other I cos(a), where I is the nominal current, and a is the number of electrical degrees (which cycles 360 for every 4 full steps).

Datasheets for chips like the DRV8825 typically give tables explaining all this.

Since the sum of the squares of the currents is constant, the motor dissipates constant power as winding heating depends on I-squared.

Yeah i got the point thanks for your help markT

Guys i told i got the point but there is smt puzzling me. Why in the drivers datasheet it says i need 3.9A or more for power supply current capacity for the one driver ? You can easily see it from below. https://www.orientalmotor.com/products/pdfs/opmanuals/HM-60128-8E.pdf#page=10)

brunopacheco: Why in the drivers datasheet it says i need 3.9A or more for power supply current capacity for the one driver ?

You need to ask the authors of the datasheet.

...R

Maybe 4.5A is the current required to produce the claimed holding torque (110Nm) or the current the motor can tolerate continuously without igniting. ?

JCA34F: Maybe 4.5A is the current required to produce the claimed holding torque (110Nm) or the current the motor can tolerate continuously without igniting. ?

Yes rated current is the limit at which the motor begins to overheat and it is provided by driver to motor. My curiousity is why 3.9A should be required from psu to provide 4.5A as stated in drivers datasheet. It should requires much less current since it works as a power converter reducing voltage while increasing the current as robin and marks pointed out

brunopacheco: My curiousity is why 3.9A should be required from psu to provide 4.5A as stated in drivers datasheet. It should requires much less current since it works as a power converter reducing voltage while increasing the current as robin and marks pointed out

Speculating about this is a complete waste of time. You need to ask the authors of the datasheet.

There are two possibilities {A} they authors are correct and will explain their reasoning or {B} they authors have made a mistake. Nobody except the authors can clarify that.

...R